Is there a math expression equivalent to the conditional ternary operator?
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Is there a math equivalent of the ternary conditional operator as used in programming?
a = b + (c > 0 ? 1 : 2)
The above means that if $c$ is greater than $0$ then $a = b + 1$, otherwise $a = b + 2$.
notation computer-science
edited 1 hour ago
Wharf Rat
999
asked Mar 5 at 5:23
dataphiledataphile
350310
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show 4 more comments
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Is there a math equivalent of the ternary conditional operator as used in programming?
a = b + (c > 0 ? 1 : 2)
The above means that if $c$ is greater than $0$ then $a = b + 1$, otherwise $a = b + 2$.
notation computer-science
edited 1 hour ago
Wharf Rat
999
asked Mar 5 at 5:23
dataphiledataphile
350310
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5
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@Alex $a = b + 2 - u(c)$
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– eyeballfrog
Mar 5 at 5:45
16
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A better question here is: what is good notation in this case. There are many ways of writing this, many ways which one would almost never use (like the accepted answer).
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– Winther
2 days ago
5
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In C-derived languages at least, you have made a grievous error! Your expression parses asa = (b + c) > 0 ? 1 : 2. I always use parentheses in these cases, even when they are not strictly necessary.
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– TonyK
2 days ago
3
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Just to clarify, by “ternary operator” you really mean “conditional operator”? Or are interested in any kind of ternary operator, regardless of semantics?
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– Konrad Rudolph
2 days ago
1
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What @KonradRudolph said; you are showing the conditional operator, which is a ternary (takes three arguments; compare unary and binary) operator. I don't think the C language has any other ternary operator, and some form of the conditional operator is found in many languages because it's very convenient to have, but there's no rule that says a language couldn't include other ternary operators alongside (or without having) a conditional operator.
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– a CVn
2 days ago
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show 4 more comments
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Is there a math equivalent of the ternary conditional operator as used in programming?
a = b + (c > 0 ? 1 : 2)
The above means that if $c$ is greater than $0$ then $a = b + 1$, otherwise $a = b + 2$.
notation computer-science
edited 1 hour ago
Wharf Rat
999
asked Mar 5 at 5:23
dataphiledataphile
350310
$endgroup$
Is there a math equivalent of the ternary conditional operator as used in programming?
a = b + (c > 0 ? 1 : 2)
The above means that if $c$ is greater than $0$ then $a = b + 1$, otherwise $a = b + 2$.
notation computer-science
notation computer-science
edited 1 hour ago
Wharf Rat
999
asked Mar 5 at 5:23
dataphiledataphile
350310
edited 1 hour ago
Wharf Rat
999
asked Mar 5 at 5:23
dataphiledataphile
350310
edited 1 hour ago
Wharf Rat
999
edited 1 hour ago
Wharf Rat
999
edited 1 hour ago
Wharf Rat
999
999
asked Mar 5 at 5:23
dataphiledataphile
350310
asked Mar 5 at 5:23
dataphiledataphile
350310
asked Mar 5 at 5:23
dataphiledataphile
350310
350310
5
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@Alex $a = b + 2 - u(c)$
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– eyeballfrog
Mar 5 at 5:45
16
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A better question here is: what is good notation in this case. There are many ways of writing this, many ways which one would almost never use (like the accepted answer).
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– Winther
2 days ago
5
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In C-derived languages at least, you have made a grievous error! Your expression parses asa = (b + c) > 0 ? 1 : 2. I always use parentheses in these cases, even when they are not strictly necessary.
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– TonyK
2 days ago
3
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Just to clarify, by “ternary operator” you really mean “conditional operator”? Or are interested in any kind of ternary operator, regardless of semantics?
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– Konrad Rudolph
2 days ago
1
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What @KonradRudolph said; you are showing the conditional operator, which is a ternary (takes three arguments; compare unary and binary) operator. I don't think the C language has any other ternary operator, and some form of the conditional operator is found in many languages because it's very convenient to have, but there's no rule that says a language couldn't include other ternary operators alongside (or without having) a conditional operator.
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– a CVn
2 days ago
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show 4 more comments
5
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@Alex $a = b + 2 - u(c)$
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– eyeballfrog
Mar 5 at 5:45
16
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A better question here is: what is good notation in this case. There are many ways of writing this, many ways which one would almost never use (like the accepted answer).
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– Winther
2 days ago
5
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In C-derived languages at least, you have made a grievous error! Your expression parses asa = (b + c) > 0 ? 1 : 2. I always use parentheses in these cases, even when they are not strictly necessary.
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– TonyK
2 days ago
3
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Just to clarify, by “ternary operator” you really mean “conditional operator”? Or are interested in any kind of ternary operator, regardless of semantics?
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– Konrad Rudolph
2 days ago
1
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What @KonradRudolph said; you are showing the conditional operator, which is a ternary (takes three arguments; compare unary and binary) operator. I don't think the C language has any other ternary operator, and some form of the conditional operator is found in many languages because it's very convenient to have, but there's no rule that says a language couldn't include other ternary operators alongside (or without having) a conditional operator.
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– a CVn
2 days ago
5
5
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@Alex $a = b + 2 - u(c)$
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– eyeballfrog
Mar 5 at 5:45
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@Alex $a = b + 2 - u(c)$
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– eyeballfrog
Mar 5 at 5:45
16
16
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A better question here is: what is good notation in this case. There are many ways of writing this, many ways which one would almost never use (like the accepted answer).
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– Winther
2 days ago
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A better question here is: what is good notation in this case. There are many ways of writing this, many ways which one would almost never use (like the accepted answer).
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– Winther
2 days ago
5
5
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In C-derived languages at least, you have made a grievous error! Your expression parses as
a = (b + c) > 0 ? 1 : 2. I always use parentheses in these cases, even when they are not strictly necessary.$endgroup$
– TonyK
2 days ago
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In C-derived languages at least, you have made a grievous error! Your expression parses as
a = (b + c) > 0 ? 1 : 2. I always use parentheses in these cases, even when they are not strictly necessary.$endgroup$
– TonyK
2 days ago
3
3
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Just to clarify, by “ternary operator” you really mean “conditional operator”? Or are interested in any kind of ternary operator, regardless of semantics?
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– Konrad Rudolph
2 days ago
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Just to clarify, by “ternary operator” you really mean “conditional operator”? Or are interested in any kind of ternary operator, regardless of semantics?
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– Konrad Rudolph
2 days ago
1
1
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What @KonradRudolph said; you are showing the conditional operator, which is a ternary (takes three arguments; compare unary and binary) operator. I don't think the C language has any other ternary operator, and some form of the conditional operator is found in many languages because it's very convenient to have, but there's no rule that says a language couldn't include other ternary operators alongside (or without having) a conditional operator.
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– a CVn
2 days ago
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What @KonradRudolph said; you are showing the conditional operator, which is a ternary (takes three arguments; compare unary and binary) operator. I don't think the C language has any other ternary operator, and some form of the conditional operator is found in many languages because it's very convenient to have, but there's no rule that says a language couldn't include other ternary operators alongside (or without having) a conditional operator.
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– a CVn
2 days ago
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11 Answers
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oldest
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From physics, I'm used to seeing the Kronecker delta,$$
{delta}_{ij}
equiv
left{
begin{array}{lll}
1 &text{if} & i=j \
0 &text{else}
end{array}
right. _{,}
$$and I think people who work with it find the slightly generalized notation$$
{delta}_{left[text{condition}right]}
equiv
left{
begin{array}{lll}
1 &text{if} & left[text{condition}right] \
0 &text{else}
end{array}
right.
$$to be pretty natural to them.
So, I tend to use $delta_{left[text{condition}right]}$ for a lot of things. Just seems so simple and well-understood.
Transforms:
Basic Kronecker delta:
To write the basic Kronecker delta in terms of the generalized Kronecker delta, it's just$$
delta_{ij}
Rightarrow
delta_{i=j}
,.$$It's almost the same notation, and I think most folks can figure it out pretty easily without needing it explained.Conditional operator:
The "conditional operator" or "ternary operator" for the simple case of?1:0:$$
begin{array}{ccccc}
boxed{
begin{array}{l}
texttt{if}~left(texttt{condition}right) \
{ \
~~~~texttt{return 1;} \
} \
texttt{else} \
{ \
~~~~texttt{return 0;} \
}
end{array}
~} &
Rightarrow &
boxed{~
texttt{condition ? 1 : 0}
~} &
Rightarrow &
delta_{i=j}
end{array}
_{.}
$$Then if you want a non-zero value for thefalse-case, you'd just add another Kronecker delta, $delta_{operatorname{NOT}left(left[text{condition}right]right)} ,$ e.g. $delta_{i neq j} .$Indicator function:
@SiongThyeGoh's answer recommended using indicator function notation. I'd rewrite their example like$$
begin{array}{ccccc}
underbrace{a=b+1+mathbb{1}_{(-infty, 0]}(c)}
_{text{their example}}
&
Rightarrow &
underbrace{a=b+1+ delta_{c in left(-infty, 0right]}}
_{text{direct translation}} &
Rightarrow &
underbrace{a=b+1+ delta_{c , {small{leq}} , 0}}
_{text{cleaner form}}
end{array}
,.
$$Iverson bracket:
Iverson bracket notation, as suggested in @FredH's answer, is apparently the same thing; according to Wikipedia, it's meant as a generalization of the Kronecker delta, except they drop the $delta$ entirely, just putting the condition in square-brackets. In a context in readers expect it, Iverson bracket notation might be preferable if conditionals will be used a lot.
Note: "Conditional operator" rather than "ternary operator".
The conditional operator, condition ? trueValue : falseValue, has 3 arguments, making it an example of a ternary operator. By contrast, most other operators in programming tend to be unary operators (which have 1 argument) or binary operators (which have 2 arguments).
Since the conditional operator is fairly unique in being a ternary operator, it's often been called "the ternary operator", leading many to believe that that's its name. However, "conditional operator" is more specific and should generally be preferred.
answered 2 days ago
NatNat
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wow, this is beautiful. I need it for an infinite series (with some elements excluded i.e. multiplied by zero and others included, multiplied by 1) so this cleaner form is exactly what I need.
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– dataphile
2 days ago
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@dataphile Yeah, the Kronecker delta's great for specifying elements like that; it comes in handy if you're using Sigma/Pi/Einstein notations. Worth noting that it's basically the integrated Dirac delta; if you're working in a continuous domain, then the Dirac delta tends to be preferred, though it's largely the same thing.
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– Nat
2 days ago
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The Kronecker delta isn’t a ternary operator, it’s a plain old binary operator, since it has two operands.
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– Konrad Rudolph
2 days ago
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The $1$-if-condition-else-$0$ generalisation of the Kronecker delta is called the Iverson bracket. We can generalise to any two input values we like with a suitable function, of domain ${0,,1}$, wraping the Iverson bracket.
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– J.G.
2 days ago
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Tangentially, folks really need to get over the "ternary operator" thing. It's a common name, if a misnomer, in programming; it's a strawman to misattribute the general definition to the question statement when it's clearly not what they meant. Further, arity's arbitrary anyway; we may as well say that the addition operator, $`` + " ,$ is actually unary because it doesn't act on two addends but rather the tuple of them. Like in counting degrees of freedom or assigning entropy, just pick some sensible convention and understand it to be subjective.
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– Nat
yesterday
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The expression b + (c > 0 ? 1 : 2) is not a ternary operator; it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
edited yesterday
J. W. Tanner
2,8421217
answered Mar 5 at 5:40
Ross MillikanRoss Millikan
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Ternary operator in programming means a conditional statement of form "if A then x else y" and uually is written just as presented in the OP, i.e. A ? x : y. It would also literally be termary if the OP wouldn't have inserted the particular values of 1 and 2.
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– Džuris
2 days ago
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@Džuris Ternary means "of arity 3", like binary means "of arity 2" and unary "of arity 1". The specific operator is the conditional ternary operator.
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– frabala
2 days ago
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Once again, clarity beats compactness when it comes to mathematics.
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– Asaf Karagila♦
2 days ago
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Well-chosen compact notations can contribute to clarity rather than detract from it, though.
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– Henning Makholm
2 days ago
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@Džuris No it isn’t. The link you cite says, in its very first sentence, that “?:is a ternary operator” (not “the ternary operator”; emphasis mine). Some people refer to it as “the ternary operator” but this is strictly incorrect (and merits a correction e.g. on Stack Overflow).
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– Konrad Rudolph
2 days ago
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In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered Mar 5 at 6:21
FredHFredH
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aka en.wikipedia.org/wiki/Iverson_bracket
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– qwr
Mar 5 at 6:50
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I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
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– Rahul
2 days ago
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I personally find Iverson brackets less cluttered than indicator functions, so I definitely recommend this notation.
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– J. M. is not a mathematician
2 days ago
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Surely this is a unary operator, which works much like for example!in the C programming language: Input is an arbitrary expression, the result is either 0 or 1 depending on the expression.
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– pipe
2 days ago
2
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Interesting note: Iverson, in his original book, used parentheses (x>0) and not brackets [x>0]. Apparently, it was Knuth and Patashnik that converted it into brackets; and they called it the "Iverson bracket".
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– safkan
yesterday
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Using the indicator function notation:$$a=b+1+mathbb{1}_{(-infty, 0]}(c)$$
answered Mar 5 at 5:34
Siong Thye GohSiong Thye Goh
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Indicator is definitely the way to go, since the conditional can define an arbitrary set.
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– eyeballfrog
Mar 5 at 5:47
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Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
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– dataphile
Mar 5 at 5:52
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I often use - and have seen others use - $mathbb{1}$ ("blackboard bold") to denote the indicator function to distinguish it from the number $1$
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– Christopher
2 days ago
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The indicator function isn’t a ternary operator, it only has two operands.
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– Konrad Rudolph
2 days ago
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I don't understand what your indicator means. I've only ever seen indicators used as $1_X$, where $X$ is a true/false-valued expression, so I'd write your indicator as $1_{c<0}$.
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– David Richerby
19 hours ago
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In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered Mar 5 at 5:35
Paul ChildsPaul Childs
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One should realize that operators are just a fancy way of using functions.
So a ternary operator is a function of 3 variables that is notated in a different way. Is that useful? The answer is mostly not. Also realize that any mathematician is allowed to introduce any notation he feels is illustrative.
Let's review why we use binary operators at all like in a+b*c . Because parameters and results are of the same type, it makes sense to leave out parentheses and introduce complicated priority rules. Imagine that a b c are numbers and we have a normal + and a peculiar * that results in dragons. Now the expression doesn't make sense (assumming a high priority *), because there is no way to add numbers and dragons. Thusly most ternary operators results in a mess.
With a proper notation there are examples of ternary operations. For example, there is a special notation for "sum for i from a to b of expression". This takes two boundaries (numbers) and a function from a number of that type that results in another number. (Mathematician, read "element of an addition group" for number.)
The notation for integration is similarly ternary.
So in short ternary operators exist, and you can define your own. They are in general accompagnied with a special notation, or they are not helpful.
Now back to the special case you mention.
Because truth values are implied in math, an expression like "if a then b else c" makes sense if a represens a truth value like (7<12). The above expression is understood in every mathematical context. However in a context where truth values are not considered a set, (if .. then .. else ..) would not be considered an operator/function, but a textual explanation. A general accepted notation could be useful in math, but I'm not aware there is one. That is probably, because like in the above, informal notations are readily understood.
answered 2 days ago
Albert van der HorstAlbert van der Horst
20125
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I think mathematicians should not be afraid to use the Iverson bracket, including when teaching, as this is a generally very useful notation whose only slightly unusual feature is to introduce a logical expression in the middle of an algebraic one (but one already regularly finds conditions inside set-theoretic expressions, so it really is not a big deal). It may avoid a lot of clutter, notably many instances of clumsy expressions by cases with a big unmatched brace (which is usually only usable as the right hand side of a definition). Since brackets do have many other uses in mathematics, I personally prefer a typographically distinct representation of Iverson brackets, rendering your example as
$$def[#1]{[![{#1}]!]}
a= b + [c>0]1 + [cnot>0]2.
$$
This works best in additive context (though one can use Iverson brackets in the exponent for optional multiplicative factors). It is not really ideal for general two-way branches, as the condition must be repeated twice, one of them in negated form, but it happens that most of the time one needs $0$ as the value for one branch anyway.
As a more concise two-way branch, I can recall that Algol68 introduced the notation $b+(c>0mid 1mid 2)$ for the right-hand side of your equation; though this is a programming language and not mathematics, it was designed by mathematicians. They also had notation for multi-way branching: thus the solution to the recursion $a_{n+2}=a_{n+1}-a_n$ with initial values $a_0=0$, $a_1=1$ can be written
$$
a_n=(nbmod 6+1mid 0,1,1,0,-1,-1)
$$
(where the "${}+1$" is needed because in 1968 they still counted starting from $1$, which is a mistake), which is reasonably concise and readable, compared to other ways to express this result. Also consider, for month $m$ in year $y$, the number
$$
( m mid 31,(ybmod 4=0land ybmod 100neq0lor ybmod400=0mid 29mid 28)
,31,30,31,30,31,31,30,31,30,31).
$$
answered yesterday
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
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The accepted answer from Nat, suggesting the Kronecker Delta, is correct. However, it is also important to note that one of the highly upvoted answers here, which claims the C ternary operator x?y:z is not ternary is incorrect.
Mathematically, the expression x?y:z can be expressed as a function of three variables:
$$f(x,y,z)=begin {cases} y&xneq 0\
z & x=0 end {cases}$$
Note that in programming an expression such as $a<b$ could be used for $x$. If the expression is true, then $x=1$, otherwise $x=0$.
About nomenclature: computer programmers have used the phrase the ternary operator to mean exactly this since at least the 1970s. Of course, among mathematicians, it would simply be a ternary operator and we would qualify it by either stating a programming language, e.g., the C ternary operator, or by calling it the conditional operator.
answered 13 hours ago
hackerb9hackerb9
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There are many good answers that give notation for, “if this condition holds, then 1, else 0.” This corresponds to an even simpler expression in C;(x>1) is equivalent to (x>1 ? 1 : 0).
It’s worth noting that the ternary operator is more general than that. If the arguments are elements of a ring, you could express c ? a : b with (using Iverson-bracket notation) $(a-b) cdot [c] + b$, but not otherwise. (And compilers frequently use this trick, in a Boolean ring, to compile conditionals without needing to execute a branch instruction.) In a C program, evaluating the expressions $a$ or $b$ might have side-effects, such as deleting a file or printing a message to the screen. In a mathematical function, this isn’t something you would worry about, and a programming language where this is impossible is called functional.
Ross Millikan gave the most standard notation, a cases block. The closest equivalent in mathematical computer science is the if-then-else function of Lambda Calculus.
answered 2 days ago
DavislorDavislor
2,360815
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"in C;(x>1) is equivalent to (x>1 ? 1 : 0)" Not exactly. Rather, C considers any nonzero value as equivalent from a truthiness perspective. So there is no difference between an integer expression taking the value of -1, 1, 42 orINT_MAXwhen that expression is treated as a boolean rvalue. In C, the one special integer value, when treated as a boolean, is 0, representing false. That said, if someone actually used(x>1)as a non-boolean expression and I noticed it, I would likely at least briefly try to find some heavy physical object that could be applied at high speed to their keyboard.
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– a CVn
2 days ago
1
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@aCVn The C11 standard states, “Each of the operators<(less than),>(greater than),<=(less than or equal to), and>=(greater than or equal to) shall yield1if the specified relation is true and0if it is false.” Similarly for equality and inequality, “Each of the operators yields1if the specified relation is true and0if it is false.”
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– Davislor
2 days ago
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@aCVn What you wrote is true, but the relational operators in C will only return the values1or0.
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– Davislor
2 days ago
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Following solution is not defined for $c = 0$; however it uses very basic operations only, which might be useful as you probably look for an expression to implement in a program:
$$a = b + 1lambda + 2(1-lambda)$$
where
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
You need to make the problem discrete and make a choice from two values. So, given some value $c in mathbb{R}$ we need to calculate some value $lambda in {0,1}$ depnding on $c<0$ or $c>0$.
Knowing that
$$frac{|c|}{c} in {1,-1}$$
we can calculate the $lambda$ as follows:
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
Now that our $lambda in {0,1}$ we can do the "choice" between the two constants $d$ and $e$ as follows:
$$dlambda + e(1-lambda)$$
which equals $d$ for $lambda = 1$, and $e$ for $lambda = 0$.
answered yesterday
TimurTimur
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And you're seriously proposing that I should write all of that instead of "$a=b+1$ if $c>0$ and $a=b+2$, otherwise"? Um...
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– David Richerby
19 hours ago
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Fundamentally, the non-answer is the answer. Whatever notation you think you might want to use to express "$a=b+1$ if $c>0$ and $a=b+2$ otherwise" or
$$a=begin{cases}b+1 &text{if }c>0\b+2 &text{otherwise}end{cases}$$
is much harder to read than either of those two things.
answered 19 hours ago
David RicherbyDavid Richerby
2,18711324
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The beauty of formal maths notation is that it expresses identities that can be used to solve / simplify / invert equations without translation to and from a natural language. You are right in a sense, I could just invent a new notation and describe it alongside the equation, but I wouldn't want to reinvent the wheel and risk confusion.
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– dataphile
16 hours ago
1
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Most of any mathematics paper is natural language. There's a reason for that.
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– David Richerby
15 hours ago
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$begingroup$
From physics, I'm used to seeing the Kronecker delta,$$
{delta}_{ij}
equiv
left{
begin{array}{lll}
1 &text{if} & i=j \
0 &text{else}
end{array}
right. _{,}
$$and I think people who work with it find the slightly generalized notation$$
{delta}_{left[text{condition}right]}
equiv
left{
begin{array}{lll}
1 &text{if} & left[text{condition}right] \
0 &text{else}
end{array}
right.
$$to be pretty natural to them.
So, I tend to use $delta_{left[text{condition}right]}$ for a lot of things. Just seems so simple and well-understood.
Transforms:
Basic Kronecker delta:
To write the basic Kronecker delta in terms of the generalized Kronecker delta, it's just$$
delta_{ij}
Rightarrow
delta_{i=j}
,.$$It's almost the same notation, and I think most folks can figure it out pretty easily without needing it explained.Conditional operator:
The "conditional operator" or "ternary operator" for the simple case of?1:0:$$
begin{array}{ccccc}
boxed{
begin{array}{l}
texttt{if}~left(texttt{condition}right) \
{ \
~~~~texttt{return 1;} \
} \
texttt{else} \
{ \
~~~~texttt{return 0;} \
}
end{array}
~} &
Rightarrow &
boxed{~
texttt{condition ? 1 : 0}
~} &
Rightarrow &
delta_{i=j}
end{array}
_{.}
$$Then if you want a non-zero value for thefalse-case, you'd just add another Kronecker delta, $delta_{operatorname{NOT}left(left[text{condition}right]right)} ,$ e.g. $delta_{i neq j} .$Indicator function:
@SiongThyeGoh's answer recommended using indicator function notation. I'd rewrite their example like$$
begin{array}{ccccc}
underbrace{a=b+1+mathbb{1}_{(-infty, 0]}(c)}
_{text{their example}}
&
Rightarrow &
underbrace{a=b+1+ delta_{c in left(-infty, 0right]}}
_{text{direct translation}} &
Rightarrow &
underbrace{a=b+1+ delta_{c , {small{leq}} , 0}}
_{text{cleaner form}}
end{array}
,.
$$Iverson bracket:
Iverson bracket notation, as suggested in @FredH's answer, is apparently the same thing; according to Wikipedia, it's meant as a generalization of the Kronecker delta, except they drop the $delta$ entirely, just putting the condition in square-brackets. In a context in readers expect it, Iverson bracket notation might be preferable if conditionals will be used a lot.
Note: "Conditional operator" rather than "ternary operator".
The conditional operator, condition ? trueValue : falseValue, has 3 arguments, making it an example of a ternary operator. By contrast, most other operators in programming tend to be unary operators (which have 1 argument) or binary operators (which have 2 arguments).
Since the conditional operator is fairly unique in being a ternary operator, it's often been called "the ternary operator", leading many to believe that that's its name. However, "conditional operator" is more specific and should generally be preferred.
answered 2 days ago
NatNat
1,0681713
$endgroup$
1
$begingroup$
wow, this is beautiful. I need it for an infinite series (with some elements excluded i.e. multiplied by zero and others included, multiplied by 1) so this cleaner form is exactly what I need.
$endgroup$
– dataphile
2 days ago
$begingroup$
@dataphile Yeah, the Kronecker delta's great for specifying elements like that; it comes in handy if you're using Sigma/Pi/Einstein notations. Worth noting that it's basically the integrated Dirac delta; if you're working in a continuous domain, then the Dirac delta tends to be preferred, though it's largely the same thing.
$endgroup$
– Nat
2 days ago
2
$begingroup$
The Kronecker delta isn’t a ternary operator, it’s a plain old binary operator, since it has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
1
$begingroup$
The $1$-if-condition-else-$0$ generalisation of the Kronecker delta is called the Iverson bracket. We can generalise to any two input values we like with a suitable function, of domain ${0,,1}$, wraping the Iverson bracket.
$endgroup$
– J.G.
2 days ago
1
$begingroup$
Tangentially, folks really need to get over the "ternary operator" thing. It's a common name, if a misnomer, in programming; it's a strawman to misattribute the general definition to the question statement when it's clearly not what they meant. Further, arity's arbitrary anyway; we may as well say that the addition operator, $`` + " ,$ is actually unary because it doesn't act on two addends but rather the tuple of them. Like in counting degrees of freedom or assigning entropy, just pick some sensible convention and understand it to be subjective.
$endgroup$
– Nat
yesterday
|
show 10 more comments
$begingroup$
From physics, I'm used to seeing the Kronecker delta,$$
{delta}_{ij}
equiv
left{
begin{array}{lll}
1 &text{if} & i=j \
0 &text{else}
end{array}
right. _{,}
$$and I think people who work with it find the slightly generalized notation$$
{delta}_{left[text{condition}right]}
equiv
left{
begin{array}{lll}
1 &text{if} & left[text{condition}right] \
0 &text{else}
end{array}
right.
$$to be pretty natural to them.
So, I tend to use $delta_{left[text{condition}right]}$ for a lot of things. Just seems so simple and well-understood.
Transforms:
Basic Kronecker delta:
To write the basic Kronecker delta in terms of the generalized Kronecker delta, it's just$$
delta_{ij}
Rightarrow
delta_{i=j}
,.$$It's almost the same notation, and I think most folks can figure it out pretty easily without needing it explained.Conditional operator:
The "conditional operator" or "ternary operator" for the simple case of?1:0:$$
begin{array}{ccccc}
boxed{
begin{array}{l}
texttt{if}~left(texttt{condition}right) \
{ \
~~~~texttt{return 1;} \
} \
texttt{else} \
{ \
~~~~texttt{return 0;} \
}
end{array}
~} &
Rightarrow &
boxed{~
texttt{condition ? 1 : 0}
~} &
Rightarrow &
delta_{i=j}
end{array}
_{.}
$$Then if you want a non-zero value for thefalse-case, you'd just add another Kronecker delta, $delta_{operatorname{NOT}left(left[text{condition}right]right)} ,$ e.g. $delta_{i neq j} .$Indicator function:
@SiongThyeGoh's answer recommended using indicator function notation. I'd rewrite their example like$$
begin{array}{ccccc}
underbrace{a=b+1+mathbb{1}_{(-infty, 0]}(c)}
_{text{their example}}
&
Rightarrow &
underbrace{a=b+1+ delta_{c in left(-infty, 0right]}}
_{text{direct translation}} &
Rightarrow &
underbrace{a=b+1+ delta_{c , {small{leq}} , 0}}
_{text{cleaner form}}
end{array}
,.
$$Iverson bracket:
Iverson bracket notation, as suggested in @FredH's answer, is apparently the same thing; according to Wikipedia, it's meant as a generalization of the Kronecker delta, except they drop the $delta$ entirely, just putting the condition in square-brackets. In a context in readers expect it, Iverson bracket notation might be preferable if conditionals will be used a lot.
Note: "Conditional operator" rather than "ternary operator".
The conditional operator, condition ? trueValue : falseValue, has 3 arguments, making it an example of a ternary operator. By contrast, most other operators in programming tend to be unary operators (which have 1 argument) or binary operators (which have 2 arguments).
Since the conditional operator is fairly unique in being a ternary operator, it's often been called "the ternary operator", leading many to believe that that's its name. However, "conditional operator" is more specific and should generally be preferred.
answered 2 days ago
NatNat
1,0681713
$endgroup$
1
$begingroup$
wow, this is beautiful. I need it for an infinite series (with some elements excluded i.e. multiplied by zero and others included, multiplied by 1) so this cleaner form is exactly what I need.
$endgroup$
– dataphile
2 days ago
$begingroup$
@dataphile Yeah, the Kronecker delta's great for specifying elements like that; it comes in handy if you're using Sigma/Pi/Einstein notations. Worth noting that it's basically the integrated Dirac delta; if you're working in a continuous domain, then the Dirac delta tends to be preferred, though it's largely the same thing.
$endgroup$
– Nat
2 days ago
2
$begingroup$
The Kronecker delta isn’t a ternary operator, it’s a plain old binary operator, since it has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
1
$begingroup$
The $1$-if-condition-else-$0$ generalisation of the Kronecker delta is called the Iverson bracket. We can generalise to any two input values we like with a suitable function, of domain ${0,,1}$, wraping the Iverson bracket.
$endgroup$
– J.G.
2 days ago
1
$begingroup$
Tangentially, folks really need to get over the "ternary operator" thing. It's a common name, if a misnomer, in programming; it's a strawman to misattribute the general definition to the question statement when it's clearly not what they meant. Further, arity's arbitrary anyway; we may as well say that the addition operator, $`` + " ,$ is actually unary because it doesn't act on two addends but rather the tuple of them. Like in counting degrees of freedom or assigning entropy, just pick some sensible convention and understand it to be subjective.
$endgroup$
– Nat
yesterday
|
show 10 more comments
$begingroup$
From physics, I'm used to seeing the Kronecker delta,$$
{delta}_{ij}
equiv
left{
begin{array}{lll}
1 &text{if} & i=j \
0 &text{else}
end{array}
right. _{,}
$$and I think people who work with it find the slightly generalized notation$$
{delta}_{left[text{condition}right]}
equiv
left{
begin{array}{lll}
1 &text{if} & left[text{condition}right] \
0 &text{else}
end{array}
right.
$$to be pretty natural to them.
So, I tend to use $delta_{left[text{condition}right]}$ for a lot of things. Just seems so simple and well-understood.
Transforms:
Basic Kronecker delta:
To write the basic Kronecker delta in terms of the generalized Kronecker delta, it's just$$
delta_{ij}
Rightarrow
delta_{i=j}
,.$$It's almost the same notation, and I think most folks can figure it out pretty easily without needing it explained.Conditional operator:
The "conditional operator" or "ternary operator" for the simple case of?1:0:$$
begin{array}{ccccc}
boxed{
begin{array}{l}
texttt{if}~left(texttt{condition}right) \
{ \
~~~~texttt{return 1;} \
} \
texttt{else} \
{ \
~~~~texttt{return 0;} \
}
end{array}
~} &
Rightarrow &
boxed{~
texttt{condition ? 1 : 0}
~} &
Rightarrow &
delta_{i=j}
end{array}
_{.}
$$Then if you want a non-zero value for thefalse-case, you'd just add another Kronecker delta, $delta_{operatorname{NOT}left(left[text{condition}right]right)} ,$ e.g. $delta_{i neq j} .$Indicator function:
@SiongThyeGoh's answer recommended using indicator function notation. I'd rewrite their example like$$
begin{array}{ccccc}
underbrace{a=b+1+mathbb{1}_{(-infty, 0]}(c)}
_{text{their example}}
&
Rightarrow &
underbrace{a=b+1+ delta_{c in left(-infty, 0right]}}
_{text{direct translation}} &
Rightarrow &
underbrace{a=b+1+ delta_{c , {small{leq}} , 0}}
_{text{cleaner form}}
end{array}
,.
$$Iverson bracket:
Iverson bracket notation, as suggested in @FredH's answer, is apparently the same thing; according to Wikipedia, it's meant as a generalization of the Kronecker delta, except they drop the $delta$ entirely, just putting the condition in square-brackets. In a context in readers expect it, Iverson bracket notation might be preferable if conditionals will be used a lot.
Note: "Conditional operator" rather than "ternary operator".
The conditional operator, condition ? trueValue : falseValue, has 3 arguments, making it an example of a ternary operator. By contrast, most other operators in programming tend to be unary operators (which have 1 argument) or binary operators (which have 2 arguments).
Since the conditional operator is fairly unique in being a ternary operator, it's often been called "the ternary operator", leading many to believe that that's its name. However, "conditional operator" is more specific and should generally be preferred.
answered 2 days ago
NatNat
1,0681713
$endgroup$
From physics, I'm used to seeing the Kronecker delta,$$
{delta}_{ij}
equiv
left{
begin{array}{lll}
1 &text{if} & i=j \
0 &text{else}
end{array}
right. _{,}
$$and I think people who work with it find the slightly generalized notation$$
{delta}_{left[text{condition}right]}
equiv
left{
begin{array}{lll}
1 &text{if} & left[text{condition}right] \
0 &text{else}
end{array}
right.
$$to be pretty natural to them.
So, I tend to use $delta_{left[text{condition}right]}$ for a lot of things. Just seems so simple and well-understood.
Transforms:
Basic Kronecker delta:
To write the basic Kronecker delta in terms of the generalized Kronecker delta, it's just$$
delta_{ij}
Rightarrow
delta_{i=j}
,.$$It's almost the same notation, and I think most folks can figure it out pretty easily without needing it explained.Conditional operator:
The "conditional operator" or "ternary operator" for the simple case of?1:0:$$
begin{array}{ccccc}
boxed{
begin{array}{l}
texttt{if}~left(texttt{condition}right) \
{ \
~~~~texttt{return 1;} \
} \
texttt{else} \
{ \
~~~~texttt{return 0;} \
}
end{array}
~} &
Rightarrow &
boxed{~
texttt{condition ? 1 : 0}
~} &
Rightarrow &
delta_{i=j}
end{array}
_{.}
$$Then if you want a non-zero value for thefalse-case, you'd just add another Kronecker delta, $delta_{operatorname{NOT}left(left[text{condition}right]right)} ,$ e.g. $delta_{i neq j} .$Indicator function:
@SiongThyeGoh's answer recommended using indicator function notation. I'd rewrite their example like$$
begin{array}{ccccc}
underbrace{a=b+1+mathbb{1}_{(-infty, 0]}(c)}
_{text{their example}}
&
Rightarrow &
underbrace{a=b+1+ delta_{c in left(-infty, 0right]}}
_{text{direct translation}} &
Rightarrow &
underbrace{a=b+1+ delta_{c , {small{leq}} , 0}}
_{text{cleaner form}}
end{array}
,.
$$Iverson bracket:
Iverson bracket notation, as suggested in @FredH's answer, is apparently the same thing; according to Wikipedia, it's meant as a generalization of the Kronecker delta, except they drop the $delta$ entirely, just putting the condition in square-brackets. In a context in readers expect it, Iverson bracket notation might be preferable if conditionals will be used a lot.
Note: "Conditional operator" rather than "ternary operator".
The conditional operator, condition ? trueValue : falseValue, has 3 arguments, making it an example of a ternary operator. By contrast, most other operators in programming tend to be unary operators (which have 1 argument) or binary operators (which have 2 arguments).
Since the conditional operator is fairly unique in being a ternary operator, it's often been called "the ternary operator", leading many to believe that that's its name. However, "conditional operator" is more specific and should generally be preferred.
answered 2 days ago
NatNat
1,0681713
edited 2 days ago
answered 2 days ago
NatNat
1,0681713
answered 2 days ago
NatNat
1,0681713
answered 2 days ago
NatNat
1,0681713
1,0681713
1
$begingroup$
wow, this is beautiful. I need it for an infinite series (with some elements excluded i.e. multiplied by zero and others included, multiplied by 1) so this cleaner form is exactly what I need.
$endgroup$
– dataphile
2 days ago
$begingroup$
@dataphile Yeah, the Kronecker delta's great for specifying elements like that; it comes in handy if you're using Sigma/Pi/Einstein notations. Worth noting that it's basically the integrated Dirac delta; if you're working in a continuous domain, then the Dirac delta tends to be preferred, though it's largely the same thing.
$endgroup$
– Nat
2 days ago
2
$begingroup$
The Kronecker delta isn’t a ternary operator, it’s a plain old binary operator, since it has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
1
$begingroup$
The $1$-if-condition-else-$0$ generalisation of the Kronecker delta is called the Iverson bracket. We can generalise to any two input values we like with a suitable function, of domain ${0,,1}$, wraping the Iverson bracket.
$endgroup$
– J.G.
2 days ago
1
$begingroup$
Tangentially, folks really need to get over the "ternary operator" thing. It's a common name, if a misnomer, in programming; it's a strawman to misattribute the general definition to the question statement when it's clearly not what they meant. Further, arity's arbitrary anyway; we may as well say that the addition operator, $`` + " ,$ is actually unary because it doesn't act on two addends but rather the tuple of them. Like in counting degrees of freedom or assigning entropy, just pick some sensible convention and understand it to be subjective.
$endgroup$
– Nat
yesterday
|
show 10 more comments
1
$begingroup$
wow, this is beautiful. I need it for an infinite series (with some elements excluded i.e. multiplied by zero and others included, multiplied by 1) so this cleaner form is exactly what I need.
$endgroup$
– dataphile
2 days ago
$begingroup$
@dataphile Yeah, the Kronecker delta's great for specifying elements like that; it comes in handy if you're using Sigma/Pi/Einstein notations. Worth noting that it's basically the integrated Dirac delta; if you're working in a continuous domain, then the Dirac delta tends to be preferred, though it's largely the same thing.
$endgroup$
– Nat
2 days ago
2
$begingroup$
The Kronecker delta isn’t a ternary operator, it’s a plain old binary operator, since it has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
1
$begingroup$
The $1$-if-condition-else-$0$ generalisation of the Kronecker delta is called the Iverson bracket. We can generalise to any two input values we like with a suitable function, of domain ${0,,1}$, wraping the Iverson bracket.
$endgroup$
– J.G.
2 days ago
1
$begingroup$
Tangentially, folks really need to get over the "ternary operator" thing. It's a common name, if a misnomer, in programming; it's a strawman to misattribute the general definition to the question statement when it's clearly not what they meant. Further, arity's arbitrary anyway; we may as well say that the addition operator, $`` + " ,$ is actually unary because it doesn't act on two addends but rather the tuple of them. Like in counting degrees of freedom or assigning entropy, just pick some sensible convention and understand it to be subjective.
$endgroup$
– Nat
yesterday
1
1
$begingroup$
wow, this is beautiful. I need it for an infinite series (with some elements excluded i.e. multiplied by zero and others included, multiplied by 1) so this cleaner form is exactly what I need.
$endgroup$
– dataphile
2 days ago
$begingroup$
wow, this is beautiful. I need it for an infinite series (with some elements excluded i.e. multiplied by zero and others included, multiplied by 1) so this cleaner form is exactly what I need.
$endgroup$
– dataphile
2 days ago
$begingroup$
@dataphile Yeah, the Kronecker delta's great for specifying elements like that; it comes in handy if you're using Sigma/Pi/Einstein notations. Worth noting that it's basically the integrated Dirac delta; if you're working in a continuous domain, then the Dirac delta tends to be preferred, though it's largely the same thing.
$endgroup$
– Nat
2 days ago
$begingroup$
@dataphile Yeah, the Kronecker delta's great for specifying elements like that; it comes in handy if you're using Sigma/Pi/Einstein notations. Worth noting that it's basically the integrated Dirac delta; if you're working in a continuous domain, then the Dirac delta tends to be preferred, though it's largely the same thing.
$endgroup$
– Nat
2 days ago
2
2
$begingroup$
The Kronecker delta isn’t a ternary operator, it’s a plain old binary operator, since it has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
$begingroup$
The Kronecker delta isn’t a ternary operator, it’s a plain old binary operator, since it has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
1
1
$begingroup$
The $1$-if-condition-else-$0$ generalisation of the Kronecker delta is called the Iverson bracket. We can generalise to any two input values we like with a suitable function, of domain ${0,,1}$, wraping the Iverson bracket.
$endgroup$
– J.G.
2 days ago
$begingroup$
The $1$-if-condition-else-$0$ generalisation of the Kronecker delta is called the Iverson bracket. We can generalise to any two input values we like with a suitable function, of domain ${0,,1}$, wraping the Iverson bracket.
$endgroup$
– J.G.
2 days ago
1
1
$begingroup$
Tangentially, folks really need to get over the "ternary operator" thing. It's a common name, if a misnomer, in programming; it's a strawman to misattribute the general definition to the question statement when it's clearly not what they meant. Further, arity's arbitrary anyway; we may as well say that the addition operator, $`` + " ,$ is actually unary because it doesn't act on two addends but rather the tuple of them. Like in counting degrees of freedom or assigning entropy, just pick some sensible convention and understand it to be subjective.
$endgroup$
– Nat
yesterday
$begingroup$
Tangentially, folks really need to get over the "ternary operator" thing. It's a common name, if a misnomer, in programming; it's a strawman to misattribute the general definition to the question statement when it's clearly not what they meant. Further, arity's arbitrary anyway; we may as well say that the addition operator, $`` + " ,$ is actually unary because it doesn't act on two addends but rather the tuple of them. Like in counting degrees of freedom or assigning entropy, just pick some sensible convention and understand it to be subjective.
$endgroup$
– Nat
yesterday
|
show 10 more comments
$begingroup$
The expression b + (c > 0 ? 1 : 2) is not a ternary operator; it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
edited yesterday
J. W. Tanner
2,8421217
answered Mar 5 at 5:40
Ross MillikanRoss Millikan
299k24200374
$endgroup$
14
$begingroup$
Ternary operator in programming means a conditional statement of form "if A then x else y" and uually is written just as presented in the OP, i.e. A ? x : y. It would also literally be termary if the OP wouldn't have inserted the particular values of 1 and 2.
$endgroup$
– Džuris
2 days ago
19
$begingroup$
@Džuris Ternary means "of arity 3", like binary means "of arity 2" and unary "of arity 1". The specific operator is the conditional ternary operator.
$endgroup$
– frabala
2 days ago
21
$begingroup$
Once again, clarity beats compactness when it comes to mathematics.
$endgroup$
– Asaf Karagila♦
2 days ago
8
$begingroup$
Well-chosen compact notations can contribute to clarity rather than detract from it, though.
$endgroup$
– Henning Makholm
2 days ago
7
$begingroup$
@Džuris No it isn’t. The link you cite says, in its very first sentence, that “?:is a ternary operator” (not “the ternary operator”; emphasis mine). Some people refer to it as “the ternary operator” but this is strictly incorrect (and merits a correction e.g. on Stack Overflow).
$endgroup$
– Konrad Rudolph
2 days ago
|
show 10 more comments
$begingroup$
The expression b + (c > 0 ? 1 : 2) is not a ternary operator; it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
edited yesterday
J. W. Tanner
2,8421217
answered Mar 5 at 5:40
Ross MillikanRoss Millikan
299k24200374
$endgroup$
14
$begingroup$
Ternary operator in programming means a conditional statement of form "if A then x else y" and uually is written just as presented in the OP, i.e. A ? x : y. It would also literally be termary if the OP wouldn't have inserted the particular values of 1 and 2.
$endgroup$
– Džuris
2 days ago
19
$begingroup$
@Džuris Ternary means "of arity 3", like binary means "of arity 2" and unary "of arity 1". The specific operator is the conditional ternary operator.
$endgroup$
– frabala
2 days ago
21
$begingroup$
Once again, clarity beats compactness when it comes to mathematics.
$endgroup$
– Asaf Karagila♦
2 days ago
8
$begingroup$
Well-chosen compact notations can contribute to clarity rather than detract from it, though.
$endgroup$
– Henning Makholm
2 days ago
7
$begingroup$
@Džuris No it isn’t. The link you cite says, in its very first sentence, that “?:is a ternary operator” (not “the ternary operator”; emphasis mine). Some people refer to it as “the ternary operator” but this is strictly incorrect (and merits a correction e.g. on Stack Overflow).
$endgroup$
– Konrad Rudolph
2 days ago
|
show 10 more comments
$begingroup$
The expression b + (c > 0 ? 1 : 2) is not a ternary operator; it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
edited yesterday
J. W. Tanner
2,8421217
answered Mar 5 at 5:40
Ross MillikanRoss Millikan
299k24200374
$endgroup$
The expression b + (c > 0 ? 1 : 2) is not a ternary operator; it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
edited yesterday
J. W. Tanner
2,8421217
answered Mar 5 at 5:40
Ross MillikanRoss Millikan
299k24200374
edited yesterday
J. W. Tanner
2,8421217
edited yesterday
J. W. Tanner
2,8421217
edited yesterday
J. W. Tanner
2,8421217
2,8421217
answered Mar 5 at 5:40
Ross MillikanRoss Millikan
299k24200374
answered Mar 5 at 5:40
Ross MillikanRoss Millikan
299k24200374
answered Mar 5 at 5:40
Ross MillikanRoss Millikan
299k24200374
299k24200374
14
$begingroup$
Ternary operator in programming means a conditional statement of form "if A then x else y" and uually is written just as presented in the OP, i.e. A ? x : y. It would also literally be termary if the OP wouldn't have inserted the particular values of 1 and 2.
$endgroup$
– Džuris
2 days ago
19
$begingroup$
@Džuris Ternary means "of arity 3", like binary means "of arity 2" and unary "of arity 1". The specific operator is the conditional ternary operator.
$endgroup$
– frabala
2 days ago
21
$begingroup$
Once again, clarity beats compactness when it comes to mathematics.
$endgroup$
– Asaf Karagila♦
2 days ago
8
$begingroup$
Well-chosen compact notations can contribute to clarity rather than detract from it, though.
$endgroup$
– Henning Makholm
2 days ago
7
$begingroup$
@Džuris No it isn’t. The link you cite says, in its very first sentence, that “?:is a ternary operator” (not “the ternary operator”; emphasis mine). Some people refer to it as “the ternary operator” but this is strictly incorrect (and merits a correction e.g. on Stack Overflow).
$endgroup$
– Konrad Rudolph
2 days ago
|
show 10 more comments
14
$begingroup$
Ternary operator in programming means a conditional statement of form "if A then x else y" and uually is written just as presented in the OP, i.e. A ? x : y. It would also literally be termary if the OP wouldn't have inserted the particular values of 1 and 2.
$endgroup$
– Džuris
2 days ago
19
$begingroup$
@Džuris Ternary means "of arity 3", like binary means "of arity 2" and unary "of arity 1". The specific operator is the conditional ternary operator.
$endgroup$
– frabala
2 days ago
21
$begingroup$
Once again, clarity beats compactness when it comes to mathematics.
$endgroup$
– Asaf Karagila♦
2 days ago
8
$begingroup$
Well-chosen compact notations can contribute to clarity rather than detract from it, though.
$endgroup$
– Henning Makholm
2 days ago
7
$begingroup$
@Džuris No it isn’t. The link you cite says, in its very first sentence, that “?:is a ternary operator” (not “the ternary operator”; emphasis mine). Some people refer to it as “the ternary operator” but this is strictly incorrect (and merits a correction e.g. on Stack Overflow).
$endgroup$
– Konrad Rudolph
2 days ago
14
14
$begingroup$
Ternary operator in programming means a conditional statement of form "if A then x else y" and uually is written just as presented in the OP, i.e. A ? x : y. It would also literally be termary if the OP wouldn't have inserted the particular values of 1 and 2.
$endgroup$
– Džuris
2 days ago
$begingroup$
Ternary operator in programming means a conditional statement of form "if A then x else y" and uually is written just as presented in the OP, i.e. A ? x : y. It would also literally be termary if the OP wouldn't have inserted the particular values of 1 and 2.
$endgroup$
– Džuris
2 days ago
19
19
$begingroup$
@Džuris Ternary means "of arity 3", like binary means "of arity 2" and unary "of arity 1". The specific operator is the conditional ternary operator.
$endgroup$
– frabala
2 days ago
$begingroup$
@Džuris Ternary means "of arity 3", like binary means "of arity 2" and unary "of arity 1". The specific operator is the conditional ternary operator.
$endgroup$
– frabala
2 days ago
21
21
$begingroup$
Once again, clarity beats compactness when it comes to mathematics.
$endgroup$
– Asaf Karagila♦
2 days ago
$begingroup$
Once again, clarity beats compactness when it comes to mathematics.
$endgroup$
– Asaf Karagila♦
2 days ago
8
8
$begingroup$
Well-chosen compact notations can contribute to clarity rather than detract from it, though.
$endgroup$
– Henning Makholm
2 days ago
$begingroup$
Well-chosen compact notations can contribute to clarity rather than detract from it, though.
$endgroup$
– Henning Makholm
2 days ago
7
7
$begingroup$
@Džuris No it isn’t. The link you cite says, in its very first sentence, that “
?: is a ternary operator” (not “the ternary operator”; emphasis mine). Some people refer to it as “the ternary operator” but this is strictly incorrect (and merits a correction e.g. on Stack Overflow).$endgroup$
– Konrad Rudolph
2 days ago
$begingroup$
@Džuris No it isn’t. The link you cite says, in its very first sentence, that “
?: is a ternary operator” (not “the ternary operator”; emphasis mine). Some people refer to it as “the ternary operator” but this is strictly incorrect (and merits a correction e.g. on Stack Overflow).$endgroup$
– Konrad Rudolph
2 days ago
|
show 10 more comments
$begingroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered Mar 5 at 6:21
FredHFredH
1,073612
$endgroup$
7
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
Mar 5 at 6:50
6
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
2 days ago
5
$begingroup$
I personally find Iverson brackets less cluttered than indicator functions, so I definitely recommend this notation.
$endgroup$
– J. M. is not a mathematician
2 days ago
4
$begingroup$
Surely this is a unary operator, which works much like for example!in the C programming language: Input is an arbitrary expression, the result is either 0 or 1 depending on the expression.
$endgroup$
– pipe
2 days ago
2
$begingroup$
Interesting note: Iverson, in his original book, used parentheses (x>0) and not brackets [x>0]. Apparently, it was Knuth and Patashnik that converted it into brackets; and they called it the "Iverson bracket".
$endgroup$
– safkan
yesterday
|
show 4 more comments
$begingroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered Mar 5 at 6:21
FredHFredH
1,073612
$endgroup$
7
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
Mar 5 at 6:50
6
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
2 days ago
5
$begingroup$
I personally find Iverson brackets less cluttered than indicator functions, so I definitely recommend this notation.
$endgroup$
– J. M. is not a mathematician
2 days ago
4
$begingroup$
Surely this is a unary operator, which works much like for example!in the C programming language: Input is an arbitrary expression, the result is either 0 or 1 depending on the expression.
$endgroup$
– pipe
2 days ago
2
$begingroup$
Interesting note: Iverson, in his original book, used parentheses (x>0) and not brackets [x>0]. Apparently, it was Knuth and Patashnik that converted it into brackets; and they called it the "Iverson bracket".
$endgroup$
– safkan
yesterday
|
show 4 more comments
$begingroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered Mar 5 at 6:21
FredHFredH
1,073612
$endgroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered Mar 5 at 6:21
FredHFredH
1,073612
edited 2 days ago
answered Mar 5 at 6:21
FredHFredH
1,073612
answered Mar 5 at 6:21
FredHFredH
1,073612
answered Mar 5 at 6:21
FredHFredH
1,073612
1,073612
7
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
Mar 5 at 6:50
6
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
2 days ago
5
$begingroup$
I personally find Iverson brackets less cluttered than indicator functions, so I definitely recommend this notation.
$endgroup$
– J. M. is not a mathematician
2 days ago
4
$begingroup$
Surely this is a unary operator, which works much like for example!in the C programming language: Input is an arbitrary expression, the result is either 0 or 1 depending on the expression.
$endgroup$
– pipe
2 days ago
2
$begingroup$
Interesting note: Iverson, in his original book, used parentheses (x>0) and not brackets [x>0]. Apparently, it was Knuth and Patashnik that converted it into brackets; and they called it the "Iverson bracket".
$endgroup$
– safkan
yesterday
|
show 4 more comments
7
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
Mar 5 at 6:50
6
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
2 days ago
5
$begingroup$
I personally find Iverson brackets less cluttered than indicator functions, so I definitely recommend this notation.
$endgroup$
– J. M. is not a mathematician
2 days ago
4
$begingroup$
Surely this is a unary operator, which works much like for example!in the C programming language: Input is an arbitrary expression, the result is either 0 or 1 depending on the expression.
$endgroup$
– pipe
2 days ago
2
$begingroup$
Interesting note: Iverson, in his original book, used parentheses (x>0) and not brackets [x>0]. Apparently, it was Knuth and Patashnik that converted it into brackets; and they called it the "Iverson bracket".
$endgroup$
– safkan
yesterday
7
7
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
Mar 5 at 6:50
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
Mar 5 at 6:50
6
6
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
2 days ago
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
2 days ago
5
5
$begingroup$
I personally find Iverson brackets less cluttered than indicator functions, so I definitely recommend this notation.
$endgroup$
– J. M. is not a mathematician
2 days ago
$begingroup$
I personally find Iverson brackets less cluttered than indicator functions, so I definitely recommend this notation.
$endgroup$
– J. M. is not a mathematician
2 days ago
4
4
$begingroup$
Surely this is a unary operator, which works much like for example
! in the C programming language: Input is an arbitrary expression, the result is either 0 or 1 depending on the expression.$endgroup$
– pipe
2 days ago
$begingroup$
Surely this is a unary operator, which works much like for example
! in the C programming language: Input is an arbitrary expression, the result is either 0 or 1 depending on the expression.$endgroup$
– pipe
2 days ago
2
2
$begingroup$
Interesting note: Iverson, in his original book, used parentheses (x>0) and not brackets [x>0]. Apparently, it was Knuth and Patashnik that converted it into brackets; and they called it the "Iverson bracket".
$endgroup$
– safkan
yesterday
$begingroup$
Interesting note: Iverson, in his original book, used parentheses (x>0) and not brackets [x>0]. Apparently, it was Knuth and Patashnik that converted it into brackets; and they called it the "Iverson bracket".
$endgroup$
– safkan
yesterday
|
show 4 more comments
$begingroup$
Using the indicator function notation:$$a=b+1+mathbb{1}_{(-infty, 0]}(c)$$
answered Mar 5 at 5:34
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
1
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
Mar 5 at 5:47
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
Mar 5 at 5:52
2
$begingroup$
I often use - and have seen others use - $mathbb{1}$ ("blackboard bold") to denote the indicator function to distinguish it from the number $1$
$endgroup$
– Christopher
2 days ago
1
$begingroup$
The indicator function isn’t a ternary operator, it only has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
$begingroup$
I don't understand what your indicator means. I've only ever seen indicators used as $1_X$, where $X$ is a true/false-valued expression, so I'd write your indicator as $1_{c<0}$.
$endgroup$
– David Richerby
19 hours ago
|
show 1 more comment
$begingroup$
Using the indicator function notation:$$a=b+1+mathbb{1}_{(-infty, 0]}(c)$$
answered Mar 5 at 5:34
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
1
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
Mar 5 at 5:47
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
Mar 5 at 5:52
2
$begingroup$
I often use - and have seen others use - $mathbb{1}$ ("blackboard bold") to denote the indicator function to distinguish it from the number $1$
$endgroup$
– Christopher
2 days ago
1
$begingroup$
The indicator function isn’t a ternary operator, it only has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
$begingroup$
I don't understand what your indicator means. I've only ever seen indicators used as $1_X$, where $X$ is a true/false-valued expression, so I'd write your indicator as $1_{c<0}$.
$endgroup$
– David Richerby
19 hours ago
|
show 1 more comment
$begingroup$
Using the indicator function notation:$$a=b+1+mathbb{1}_{(-infty, 0]}(c)$$
answered Mar 5 at 5:34
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
Using the indicator function notation:$$a=b+1+mathbb{1}_{(-infty, 0]}(c)$$
answered Mar 5 at 5:34
Siong Thye GohSiong Thye Goh
102k1467119
edited 2 days ago
answered Mar 5 at 5:34
Siong Thye GohSiong Thye Goh
102k1467119
answered Mar 5 at 5:34
Siong Thye GohSiong Thye Goh
102k1467119
answered Mar 5 at 5:34
Siong Thye GohSiong Thye Goh
102k1467119
102k1467119
1
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
Mar 5 at 5:47
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
Mar 5 at 5:52
2
$begingroup$
I often use - and have seen others use - $mathbb{1}$ ("blackboard bold") to denote the indicator function to distinguish it from the number $1$
$endgroup$
– Christopher
2 days ago
1
$begingroup$
The indicator function isn’t a ternary operator, it only has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
$begingroup$
I don't understand what your indicator means. I've only ever seen indicators used as $1_X$, where $X$ is a true/false-valued expression, so I'd write your indicator as $1_{c<0}$.
$endgroup$
– David Richerby
19 hours ago
|
show 1 more comment
1
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
Mar 5 at 5:47
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
Mar 5 at 5:52
2
$begingroup$
I often use - and have seen others use - $mathbb{1}$ ("blackboard bold") to denote the indicator function to distinguish it from the number $1$
$endgroup$
– Christopher
2 days ago
1
$begingroup$
The indicator function isn’t a ternary operator, it only has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
$begingroup$
I don't understand what your indicator means. I've only ever seen indicators used as $1_X$, where $X$ is a true/false-valued expression, so I'd write your indicator as $1_{c<0}$.
$endgroup$
– David Richerby
19 hours ago
1
1
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
Mar 5 at 5:47
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
Mar 5 at 5:47
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
Mar 5 at 5:52
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
Mar 5 at 5:52
2
2
$begingroup$
I often use - and have seen others use - $mathbb{1}$ ("blackboard bold") to denote the indicator function to distinguish it from the number $1$
$endgroup$
– Christopher
2 days ago
$begingroup$
I often use - and have seen others use - $mathbb{1}$ ("blackboard bold") to denote the indicator function to distinguish it from the number $1$
$endgroup$
– Christopher
2 days ago
1
1
$begingroup$
The indicator function isn’t a ternary operator, it only has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
$begingroup$
The indicator function isn’t a ternary operator, it only has two operands.
$endgroup$
– Konrad Rudolph
2 days ago
$begingroup$
I don't understand what your indicator means. I've only ever seen indicators used as $1_X$, where $X$ is a true/false-valued expression, so I'd write your indicator as $1_{c<0}$.
$endgroup$
– David Richerby
19 hours ago
$begingroup$
I don't understand what your indicator means. I've only ever seen indicators used as $1_X$, where $X$ is a true/false-valued expression, so I'd write your indicator as $1_{c<0}$.
$endgroup$
– David Richerby
19 hours ago
|
show 1 more comment
$begingroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered Mar 5 at 5:35
Paul ChildsPaul Childs
4328
$endgroup$
add a comment |
$begingroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered Mar 5 at 5:35
Paul ChildsPaul Childs
4328
$endgroup$
add a comment |
$begingroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered Mar 5 at 5:35
Paul ChildsPaul Childs
4328
$endgroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered Mar 5 at 5:35
Paul ChildsPaul Childs
4328
answered Mar 5 at 5:35
Paul ChildsPaul Childs
4328
answered Mar 5 at 5:35
Paul ChildsPaul Childs
4328
answered Mar 5 at 5:35
Paul ChildsPaul Childs
4328
4328
add a comment |
add a comment |
$begingroup$
One should realize that operators are just a fancy way of using functions.
So a ternary operator is a function of 3 variables that is notated in a different way. Is that useful? The answer is mostly not. Also realize that any mathematician is allowed to introduce any notation he feels is illustrative.
Let's review why we use binary operators at all like in a+b*c . Because parameters and results are of the same type, it makes sense to leave out parentheses and introduce complicated priority rules. Imagine that a b c are numbers and we have a normal + and a peculiar * that results in dragons. Now the expression doesn't make sense (assumming a high priority *), because there is no way to add numbers and dragons. Thusly most ternary operators results in a mess.
With a proper notation there are examples of ternary operations. For example, there is a special notation for "sum for i from a to b of expression". This takes two boundaries (numbers) and a function from a number of that type that results in another number. (Mathematician, read "element of an addition group" for number.)
The notation for integration is similarly ternary.
So in short ternary operators exist, and you can define your own. They are in general accompagnied with a special notation, or they are not helpful.
Now back to the special case you mention.
Because truth values are implied in math, an expression like "if a then b else c" makes sense if a represens a truth value like (7<12). The above expression is understood in every mathematical context. However in a context where truth values are not considered a set, (if .. then .. else ..) would not be considered an operator/function, but a textual explanation. A general accepted notation could be useful in math, but I'm not aware there is one. That is probably, because like in the above, informal notations are readily understood.
answered 2 days ago
Albert van der HorstAlbert van der Horst
20125
$endgroup$
add a comment |
$begingroup$
One should realize that operators are just a fancy way of using functions.
So a ternary operator is a function of 3 variables that is notated in a different way. Is that useful? The answer is mostly not. Also realize that any mathematician is allowed to introduce any notation he feels is illustrative.
Let's review why we use binary operators at all like in a+b*c . Because parameters and results are of the same type, it makes sense to leave out parentheses and introduce complicated priority rules. Imagine that a b c are numbers and we have a normal + and a peculiar * that results in dragons. Now the expression doesn't make sense (assumming a high priority *), because there is no way to add numbers and dragons. Thusly most ternary operators results in a mess.
With a proper notation there are examples of ternary operations. For example, there is a special notation for "sum for i from a to b of expression". This takes two boundaries (numbers) and a function from a number of that type that results in another number. (Mathematician, read "element of an addition group" for number.)
The notation for integration is similarly ternary.
So in short ternary operators exist, and you can define your own. They are in general accompagnied with a special notation, or they are not helpful.
Now back to the special case you mention.
Because truth values are implied in math, an expression like "if a then b else c" makes sense if a represens a truth value like (7<12). The above expression is understood in every mathematical context. However in a context where truth values are not considered a set, (if .. then .. else ..) would not be considered an operator/function, but a textual explanation. A general accepted notation could be useful in math, but I'm not aware there is one. That is probably, because like in the above, informal notations are readily understood.
answered 2 days ago
Albert van der HorstAlbert van der Horst
20125
$endgroup$
add a comment |
$begingroup$
One should realize that operators are just a fancy way of using functions.
So a ternary operator is a function of 3 variables that is notated in a different way. Is that useful? The answer is mostly not. Also realize that any mathematician is allowed to introduce any notation he feels is illustrative.
Let's review why we use binary operators at all like in a+b*c . Because parameters and results are of the same type, it makes sense to leave out parentheses and introduce complicated priority rules. Imagine that a b c are numbers and we have a normal + and a peculiar * that results in dragons. Now the expression doesn't make sense (assumming a high priority *), because there is no way to add numbers and dragons. Thusly most ternary operators results in a mess.
With a proper notation there are examples of ternary operations. For example, there is a special notation for "sum for i from a to b of expression". This takes two boundaries (numbers) and a function from a number of that type that results in another number. (Mathematician, read "element of an addition group" for number.)
The notation for integration is similarly ternary.
So in short ternary operators exist, and you can define your own. They are in general accompagnied with a special notation, or they are not helpful.
Now back to the special case you mention.
Because truth values are implied in math, an expression like "if a then b else c" makes sense if a represens a truth value like (7<12). The above expression is understood in every mathematical context. However in a context where truth values are not considered a set, (if .. then .. else ..) would not be considered an operator/function, but a textual explanation. A general accepted notation could be useful in math, but I'm not aware there is one. That is probably, because like in the above, informal notations are readily understood.
answered 2 days ago
Albert van der HorstAlbert van der Horst
20125
$endgroup$
One should realize that operators are just a fancy way of using functions.
So a ternary operator is a function of 3 variables that is notated in a different way. Is that useful? The answer is mostly not. Also realize that any mathematician is allowed to introduce any notation he feels is illustrative.
Let's review why we use binary operators at all like in a+b*c . Because parameters and results are of the same type, it makes sense to leave out parentheses and introduce complicated priority rules. Imagine that a b c are numbers and we have a normal + and a peculiar * that results in dragons. Now the expression doesn't make sense (assumming a high priority *), because there is no way to add numbers and dragons. Thusly most ternary operators results in a mess.
With a proper notation there are examples of ternary operations. For example, there is a special notation for "sum for i from a to b of expression". This takes two boundaries (numbers) and a function from a number of that type that results in another number. (Mathematician, read "element of an addition group" for number.)
The notation for integration is similarly ternary.
So in short ternary operators exist, and you can define your own. They are in general accompagnied with a special notation, or they are not helpful.
Now back to the special case you mention.
Because truth values are implied in math, an expression like "if a then b else c" makes sense if a represens a truth value like (7<12). The above expression is understood in every mathematical context. However in a context where truth values are not considered a set, (if .. then .. else ..) would not be considered an operator/function, but a textual explanation. A general accepted notation could be useful in math, but I'm not aware there is one. That is probably, because like in the above, informal notations are readily understood.
answered 2 days ago
Albert van der HorstAlbert van der Horst
20125
answered 2 days ago
Albert van der HorstAlbert van der Horst
20125
answered 2 days ago
Albert van der HorstAlbert van der Horst
20125
answered 2 days ago
Albert van der HorstAlbert van der Horst
20125
20125
add a comment |
add a comment |
$begingroup$
I think mathematicians should not be afraid to use the Iverson bracket, including when teaching, as this is a generally very useful notation whose only slightly unusual feature is to introduce a logical expression in the middle of an algebraic one (but one already regularly finds conditions inside set-theoretic expressions, so it really is not a big deal). It may avoid a lot of clutter, notably many instances of clumsy expressions by cases with a big unmatched brace (which is usually only usable as the right hand side of a definition). Since brackets do have many other uses in mathematics, I personally prefer a typographically distinct representation of Iverson brackets, rendering your example as
$$def[#1]{[![{#1}]!]}
a= b + [c>0]1 + [cnot>0]2.
$$
This works best in additive context (though one can use Iverson brackets in the exponent for optional multiplicative factors). It is not really ideal for general two-way branches, as the condition must be repeated twice, one of them in negated form, but it happens that most of the time one needs $0$ as the value for one branch anyway.
As a more concise two-way branch, I can recall that Algol68 introduced the notation $b+(c>0mid 1mid 2)$ for the right-hand side of your equation; though this is a programming language and not mathematics, it was designed by mathematicians. They also had notation for multi-way branching: thus the solution to the recursion $a_{n+2}=a_{n+1}-a_n$ with initial values $a_0=0$, $a_1=1$ can be written
$$
a_n=(nbmod 6+1mid 0,1,1,0,-1,-1)
$$
(where the "${}+1$" is needed because in 1968 they still counted starting from $1$, which is a mistake), which is reasonably concise and readable, compared to other ways to express this result. Also consider, for month $m$ in year $y$, the number
$$
( m mid 31,(ybmod 4=0land ybmod 100neq0lor ybmod400=0mid 29mid 28)
,31,30,31,30,31,31,30,31,30,31).
$$
answered yesterday
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
$endgroup$
add a comment |
$begingroup$
I think mathematicians should not be afraid to use the Iverson bracket, including when teaching, as this is a generally very useful notation whose only slightly unusual feature is to introduce a logical expression in the middle of an algebraic one (but one already regularly finds conditions inside set-theoretic expressions, so it really is not a big deal). It may avoid a lot of clutter, notably many instances of clumsy expressions by cases with a big unmatched brace (which is usually only usable as the right hand side of a definition). Since brackets do have many other uses in mathematics, I personally prefer a typographically distinct representation of Iverson brackets, rendering your example as
$$def[#1]{[![{#1}]!]}
a= b + [c>0]1 + [cnot>0]2.
$$
This works best in additive context (though one can use Iverson brackets in the exponent for optional multiplicative factors). It is not really ideal for general two-way branches, as the condition must be repeated twice, one of them in negated form, but it happens that most of the time one needs $0$ as the value for one branch anyway.
As a more concise two-way branch, I can recall that Algol68 introduced the notation $b+(c>0mid 1mid 2)$ for the right-hand side of your equation; though this is a programming language and not mathematics, it was designed by mathematicians. They also had notation for multi-way branching: thus the solution to the recursion $a_{n+2}=a_{n+1}-a_n$ with initial values $a_0=0$, $a_1=1$ can be written
$$
a_n=(nbmod 6+1mid 0,1,1,0,-1,-1)
$$
(where the "${}+1$" is needed because in 1968 they still counted starting from $1$, which is a mistake), which is reasonably concise and readable, compared to other ways to express this result. Also consider, for month $m$ in year $y$, the number
$$
( m mid 31,(ybmod 4=0land ybmod 100neq0lor ybmod400=0mid 29mid 28)
,31,30,31,30,31,31,30,31,30,31).
$$
answered yesterday
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
$endgroup$
add a comment |
$begingroup$
I think mathematicians should not be afraid to use the Iverson bracket, including when teaching, as this is a generally very useful notation whose only slightly unusual feature is to introduce a logical expression in the middle of an algebraic one (but one already regularly finds conditions inside set-theoretic expressions, so it really is not a big deal). It may avoid a lot of clutter, notably many instances of clumsy expressions by cases with a big unmatched brace (which is usually only usable as the right hand side of a definition). Since brackets do have many other uses in mathematics, I personally prefer a typographically distinct representation of Iverson brackets, rendering your example as
$$def[#1]{[![{#1}]!]}
a= b + [c>0]1 + [cnot>0]2.
$$
This works best in additive context (though one can use Iverson brackets in the exponent for optional multiplicative factors). It is not really ideal for general two-way branches, as the condition must be repeated twice, one of them in negated form, but it happens that most of the time one needs $0$ as the value for one branch anyway.
As a more concise two-way branch, I can recall that Algol68 introduced the notation $b+(c>0mid 1mid 2)$ for the right-hand side of your equation; though this is a programming language and not mathematics, it was designed by mathematicians. They also had notation for multi-way branching: thus the solution to the recursion $a_{n+2}=a_{n+1}-a_n$ with initial values $a_0=0$, $a_1=1$ can be written
$$
a_n=(nbmod 6+1mid 0,1,1,0,-1,-1)
$$
(where the "${}+1$" is needed because in 1968 they still counted starting from $1$, which is a mistake), which is reasonably concise and readable, compared to other ways to express this result. Also consider, for month $m$ in year $y$, the number
$$
( m mid 31,(ybmod 4=0land ybmod 100neq0lor ybmod400=0mid 29mid 28)
,31,30,31,30,31,31,30,31,30,31).
$$
answered yesterday
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
$endgroup$
I think mathematicians should not be afraid to use the Iverson bracket, including when teaching, as this is a generally very useful notation whose only slightly unusual feature is to introduce a logical expression in the middle of an algebraic one (but one already regularly finds conditions inside set-theoretic expressions, so it really is not a big deal). It may avoid a lot of clutter, notably many instances of clumsy expressions by cases with a big unmatched brace (which is usually only usable as the right hand side of a definition). Since brackets do have many other uses in mathematics, I personally prefer a typographically distinct representation of Iverson brackets, rendering your example as
$$def[#1]{[![{#1}]!]}
a= b + [c>0]1 + [cnot>0]2.
$$
This works best in additive context (though one can use Iverson brackets in the exponent for optional multiplicative factors). It is not really ideal for general two-way branches, as the condition must be repeated twice, one of them in negated form, but it happens that most of the time one needs $0$ as the value for one branch anyway.
As a more concise two-way branch, I can recall that Algol68 introduced the notation $b+(c>0mid 1mid 2)$ for the right-hand side of your equation; though this is a programming language and not mathematics, it was designed by mathematicians. They also had notation for multi-way branching: thus the solution to the recursion $a_{n+2}=a_{n+1}-a_n$ with initial values $a_0=0$, $a_1=1$ can be written
$$
a_n=(nbmod 6+1mid 0,1,1,0,-1,-1)
$$
(where the "${}+1$" is needed because in 1968 they still counted starting from $1$, which is a mistake), which is reasonably concise and readable, compared to other ways to express this result. Also consider, for month $m$ in year $y$, the number
$$
( m mid 31,(ybmod 4=0land ybmod 100neq0lor ybmod400=0mid 29mid 28)
,31,30,31,30,31,31,30,31,30,31).
$$
answered yesterday
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
answered yesterday
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
answered yesterday
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
answered yesterday
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
87.9k5111226
add a comment |
add a comment |
$begingroup$
The accepted answer from Nat, suggesting the Kronecker Delta, is correct. However, it is also important to note that one of the highly upvoted answers here, which claims the C ternary operator x?y:z is not ternary is incorrect.
Mathematically, the expression x?y:z can be expressed as a function of three variables:
$$f(x,y,z)=begin {cases} y&xneq 0\
z & x=0 end {cases}$$
Note that in programming an expression such as $a<b$ could be used for $x$. If the expression is true, then $x=1$, otherwise $x=0$.
About nomenclature: computer programmers have used the phrase the ternary operator to mean exactly this since at least the 1970s. Of course, among mathematicians, it would simply be a ternary operator and we would qualify it by either stating a programming language, e.g., the C ternary operator, or by calling it the conditional operator.
answered 13 hours ago
hackerb9hackerb9
1394
New contributor
hackerb9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
The accepted answer from Nat, suggesting the Kronecker Delta, is correct. However, it is also important to note that one of the highly upvoted answers here, which claims the C ternary operator x?y:z is not ternary is incorrect.
Mathematically, the expression x?y:z can be expressed as a function of three variables:
$$f(x,y,z)=begin {cases} y&xneq 0\
z & x=0 end {cases}$$
Note that in programming an expression such as $a<b$ could be used for $x$. If the expression is true, then $x=1$, otherwise $x=0$.
About nomenclature: computer programmers have used the phrase the ternary operator to mean exactly this since at least the 1970s. Of course, among mathematicians, it would simply be a ternary operator and we would qualify it by either stating a programming language, e.g., the C ternary operator, or by calling it the conditional operator.
answered 13 hours ago
hackerb9hackerb9
1394
New contributor
hackerb9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The accepted answer from Nat, suggesting the Kronecker Delta, is correct. However, it is also important to note that one of the highly upvoted answers here, which claims the C ternary operator x?y:z is not ternary is incorrect.
Mathematically, the expression x?y:z can be expressed as a function of three variables:
$$f(x,y,z)=begin {cases} y&xneq 0\
z & x=0 end {cases}$$
Note that in programming an expression such as $a<b$ could be used for $x$. If the expression is true, then $x=1$, otherwise $x=0$.
About nomenclature: computer programmers have used the phrase the ternary operator to mean exactly this since at least the 1970s. Of course, among mathematicians, it would simply be a ternary operator and we would qualify it by either stating a programming language, e.g., the C ternary operator, or by calling it the conditional operator.
answered 13 hours ago
hackerb9hackerb9
1394
New contributor
hackerb9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The accepted answer from Nat, suggesting the Kronecker Delta, is correct. However, it is also important to note that one of the highly upvoted answers here, which claims the C ternary operator x?y:z is not ternary is incorrect.
Mathematically, the expression x?y:z can be expressed as a function of three variables:
$$f(x,y,z)=begin {cases} y&xneq 0\
z & x=0 end {cases}$$
Note that in programming an expression such as $a<b$ could be used for $x$. If the expression is true, then $x=1$, otherwise $x=0$.
About nomenclature: computer programmers have used the phrase the ternary operator to mean exactly this since at least the 1970s. Of course, among mathematicians, it would simply be a ternary operator and we would qualify it by either stating a programming language, e.g., the C ternary operator, or by calling it the conditional operator.
answered 13 hours ago
hackerb9hackerb9
1394
New contributor
hackerb9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
answered 13 hours ago
hackerb9hackerb9
1394
New contributor
hackerb9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 13 hours ago
hackerb9hackerb9
1394
answered 13 hours ago
hackerb9hackerb9
1394
1394
New contributor
hackerb9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
hackerb9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
hackerb9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
There are many good answers that give notation for, “if this condition holds, then 1, else 0.” This corresponds to an even simpler expression in C;(x>1) is equivalent to (x>1 ? 1 : 0).
It’s worth noting that the ternary operator is more general than that. If the arguments are elements of a ring, you could express c ? a : b with (using Iverson-bracket notation) $(a-b) cdot [c] + b$, but not otherwise. (And compilers frequently use this trick, in a Boolean ring, to compile conditionals without needing to execute a branch instruction.) In a C program, evaluating the expressions $a$ or $b$ might have side-effects, such as deleting a file or printing a message to the screen. In a mathematical function, this isn’t something you would worry about, and a programming language where this is impossible is called functional.
Ross Millikan gave the most standard notation, a cases block. The closest equivalent in mathematical computer science is the if-then-else function of Lambda Calculus.
answered 2 days ago
DavislorDavislor
2,360815
$endgroup$
$begingroup$
"in C;(x>1) is equivalent to (x>1 ? 1 : 0)" Not exactly. Rather, C considers any nonzero value as equivalent from a truthiness perspective. So there is no difference between an integer expression taking the value of -1, 1, 42 orINT_MAXwhen that expression is treated as a boolean rvalue. In C, the one special integer value, when treated as a boolean, is 0, representing false. That said, if someone actually used(x>1)as a non-boolean expression and I noticed it, I would likely at least briefly try to find some heavy physical object that could be applied at high speed to their keyboard.
$endgroup$
– a CVn
2 days ago
1
$begingroup$
@aCVn The C11 standard states, “Each of the operators<(less than),>(greater than),<=(less than or equal to), and>=(greater than or equal to) shall yield1if the specified relation is true and0if it is false.” Similarly for equality and inequality, “Each of the operators yields1if the specified relation is true and0if it is false.”
$endgroup$
– Davislor
2 days ago
$begingroup$
@aCVn What you wrote is true, but the relational operators in C will only return the values1or0.
$endgroup$
– Davislor
2 days ago
add a comment |
$begingroup$
There are many good answers that give notation for, “if this condition holds, then 1, else 0.” This corresponds to an even simpler expression in C;(x>1) is equivalent to (x>1 ? 1 : 0).
It’s worth noting that the ternary operator is more general than that. If the arguments are elements of a ring, you could express c ? a : b with (using Iverson-bracket notation) $(a-b) cdot [c] + b$, but not otherwise. (And compilers frequently use this trick, in a Boolean ring, to compile conditionals without needing to execute a branch instruction.) In a C program, evaluating the expressions $a$ or $b$ might have side-effects, such as deleting a file or printing a message to the screen. In a mathematical function, this isn’t something you would worry about, and a programming language where this is impossible is called functional.
Ross Millikan gave the most standard notation, a cases block. The closest equivalent in mathematical computer science is the if-then-else function of Lambda Calculus.
answered 2 days ago
DavislorDavislor
2,360815
$endgroup$
$begingroup$
"in C;(x>1) is equivalent to (x>1 ? 1 : 0)" Not exactly. Rather, C considers any nonzero value as equivalent from a truthiness perspective. So there is no difference between an integer expression taking the value of -1, 1, 42 orINT_MAXwhen that expression is treated as a boolean rvalue. In C, the one special integer value, when treated as a boolean, is 0, representing false. That said, if someone actually used(x>1)as a non-boolean expression and I noticed it, I would likely at least briefly try to find some heavy physical object that could be applied at high speed to their keyboard.
$endgroup$
– a CVn
2 days ago
1
$begingroup$
@aCVn The C11 standard states, “Each of the operators<(less than),>(greater than),<=(less than or equal to), and>=(greater than or equal to) shall yield1if the specified relation is true and0if it is false.” Similarly for equality and inequality, “Each of the operators yields1if the specified relation is true and0if it is false.”
$endgroup$
– Davislor
2 days ago
$begingroup$
@aCVn What you wrote is true, but the relational operators in C will only return the values1or0.
$endgroup$
– Davislor
2 days ago
add a comment |
$begingroup$
There are many good answers that give notation for, “if this condition holds, then 1, else 0.” This corresponds to an even simpler expression in C;(x>1) is equivalent to (x>1 ? 1 : 0).
It’s worth noting that the ternary operator is more general than that. If the arguments are elements of a ring, you could express c ? a : b with (using Iverson-bracket notation) $(a-b) cdot [c] + b$, but not otherwise. (And compilers frequently use this trick, in a Boolean ring, to compile conditionals without needing to execute a branch instruction.) In a C program, evaluating the expressions $a$ or $b$ might have side-effects, such as deleting a file or printing a message to the screen. In a mathematical function, this isn’t something you would worry about, and a programming language where this is impossible is called functional.
Ross Millikan gave the most standard notation, a cases block. The closest equivalent in mathematical computer science is the if-then-else function of Lambda Calculus.
answered 2 days ago
DavislorDavislor
2,360815
$endgroup$
There are many good answers that give notation for, “if this condition holds, then 1, else 0.” This corresponds to an even simpler expression in C;(x>1) is equivalent to (x>1 ? 1 : 0).
It’s worth noting that the ternary operator is more general than that. If the arguments are elements of a ring, you could express c ? a : b with (using Iverson-bracket notation) $(a-b) cdot [c] + b$, but not otherwise. (And compilers frequently use this trick, in a Boolean ring, to compile conditionals without needing to execute a branch instruction.) In a C program, evaluating the expressions $a$ or $b$ might have side-effects, such as deleting a file or printing a message to the screen. In a mathematical function, this isn’t something you would worry about, and a programming language where this is impossible is called functional.
Ross Millikan gave the most standard notation, a cases block. The closest equivalent in mathematical computer science is the if-then-else function of Lambda Calculus.
answered 2 days ago
DavislorDavislor
2,360815
edited 2 days ago
answered 2 days ago
DavislorDavislor
2,360815
answered 2 days ago
DavislorDavislor
2,360815
answered 2 days ago
DavislorDavislor
2,360815
2,360815
$begingroup$
"in C;(x>1) is equivalent to (x>1 ? 1 : 0)" Not exactly. Rather, C considers any nonzero value as equivalent from a truthiness perspective. So there is no difference between an integer expression taking the value of -1, 1, 42 orINT_MAXwhen that expression is treated as a boolean rvalue. In C, the one special integer value, when treated as a boolean, is 0, representing false. That said, if someone actually used(x>1)as a non-boolean expression and I noticed it, I would likely at least briefly try to find some heavy physical object that could be applied at high speed to their keyboard.
$endgroup$
– a CVn
2 days ago
1
$begingroup$
@aCVn The C11 standard states, “Each of the operators<(less than),>(greater than),<=(less than or equal to), and>=(greater than or equal to) shall yield1if the specified relation is true and0if it is false.” Similarly for equality and inequality, “Each of the operators yields1if the specified relation is true and0if it is false.”
$endgroup$
– Davislor
2 days ago
$begingroup$
@aCVn What you wrote is true, but the relational operators in C will only return the values1or0.
$endgroup$
– Davislor
2 days ago
add a comment |
$begingroup$
"in C;(x>1) is equivalent to (x>1 ? 1 : 0)" Not exactly. Rather, C considers any nonzero value as equivalent from a truthiness perspective. So there is no difference between an integer expression taking the value of -1, 1, 42 orINT_MAXwhen that expression is treated as a boolean rvalue. In C, the one special integer value, when treated as a boolean, is 0, representing false. That said, if someone actually used(x>1)as a non-boolean expression and I noticed it, I would likely at least briefly try to find some heavy physical object that could be applied at high speed to their keyboard.
$endgroup$
– a CVn
2 days ago
1
$begingroup$
@aCVn The C11 standard states, “Each of the operators<(less than),>(greater than),<=(less than or equal to), and>=(greater than or equal to) shall yield1if the specified relation is true and0if it is false.” Similarly for equality and inequality, “Each of the operators yields1if the specified relation is true and0if it is false.”
$endgroup$
– Davislor
2 days ago
$begingroup$
@aCVn What you wrote is true, but the relational operators in C will only return the values1or0.
$endgroup$
– Davislor
2 days ago
$begingroup$
"in C;(x>1) is equivalent to (x>1 ? 1 : 0)" Not exactly. Rather, C considers any nonzero value as equivalent from a truthiness perspective. So there is no difference between an integer expression taking the value of -1, 1, 42 or
INT_MAX when that expression is treated as a boolean rvalue. In C, the one special integer value, when treated as a boolean, is 0, representing false. That said, if someone actually used (x>1) as a non-boolean expression and I noticed it, I would likely at least briefly try to find some heavy physical object that could be applied at high speed to their keyboard.$endgroup$
– a CVn
2 days ago
$begingroup$
"in C;(x>1) is equivalent to (x>1 ? 1 : 0)" Not exactly. Rather, C considers any nonzero value as equivalent from a truthiness perspective. So there is no difference between an integer expression taking the value of -1, 1, 42 or
INT_MAX when that expression is treated as a boolean rvalue. In C, the one special integer value, when treated as a boolean, is 0, representing false. That said, if someone actually used (x>1) as a non-boolean expression and I noticed it, I would likely at least briefly try to find some heavy physical object that could be applied at high speed to their keyboard.$endgroup$
– a CVn
2 days ago
1
1
$begingroup$
@aCVn The C11 standard states, “Each of the operators
< (less than), > (greater than), <= (less than or equal to), and >= (greater than or equal to) shall yield 1 if the specified relation is true and 0 if it is false.” Similarly for equality and inequality, “Each of the operators yields 1 if the specified relation is true and 0 if it is false.”$endgroup$
– Davislor
2 days ago
$begingroup$
@aCVn The C11 standard states, “Each of the operators
< (less than), > (greater than), <= (less than or equal to), and >= (greater than or equal to) shall yield 1 if the specified relation is true and 0 if it is false.” Similarly for equality and inequality, “Each of the operators yields 1 if the specified relation is true and 0 if it is false.”$endgroup$
– Davislor
2 days ago
$begingroup$
@aCVn What you wrote is true, but the relational operators in C will only return the values
1 or 0.$endgroup$
– Davislor
2 days ago
$begingroup$
@aCVn What you wrote is true, but the relational operators in C will only return the values
1 or 0.$endgroup$
– Davislor
2 days ago
add a comment |
$begingroup$
Following solution is not defined for $c = 0$; however it uses very basic operations only, which might be useful as you probably look for an expression to implement in a program:
$$a = b + 1lambda + 2(1-lambda)$$
where
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
You need to make the problem discrete and make a choice from two values. So, given some value $c in mathbb{R}$ we need to calculate some value $lambda in {0,1}$ depnding on $c<0$ or $c>0$.
Knowing that
$$frac{|c|}{c} in {1,-1}$$
we can calculate the $lambda$ as follows:
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
Now that our $lambda in {0,1}$ we can do the "choice" between the two constants $d$ and $e$ as follows:
$$dlambda + e(1-lambda)$$
which equals $d$ for $lambda = 1$, and $e$ for $lambda = 0$.
answered yesterday
TimurTimur
1193
New contributor
Timur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
And you're seriously proposing that I should write all of that instead of "$a=b+1$ if $c>0$ and $a=b+2$, otherwise"? Um...
$endgroup$
– David Richerby
19 hours ago
add a comment |
$begingroup$
Following solution is not defined for $c = 0$; however it uses very basic operations only, which might be useful as you probably look for an expression to implement in a program:
$$a = b + 1lambda + 2(1-lambda)$$
where
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
You need to make the problem discrete and make a choice from two values. So, given some value $c in mathbb{R}$ we need to calculate some value $lambda in {0,1}$ depnding on $c<0$ or $c>0$.
Knowing that
$$frac{|c|}{c} in {1,-1}$$
we can calculate the $lambda$ as follows:
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
Now that our $lambda in {0,1}$ we can do the "choice" between the two constants $d$ and $e$ as follows:
$$dlambda + e(1-lambda)$$
which equals $d$ for $lambda = 1$, and $e$ for $lambda = 0$.
answered yesterday
TimurTimur
1193
New contributor
Timur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
And you're seriously proposing that I should write all of that instead of "$a=b+1$ if $c>0$ and $a=b+2$, otherwise"? Um...
$endgroup$
– David Richerby
19 hours ago
add a comment |
$begingroup$
Following solution is not defined for $c = 0$; however it uses very basic operations only, which might be useful as you probably look for an expression to implement in a program:
$$a = b + 1lambda + 2(1-lambda)$$
where
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
You need to make the problem discrete and make a choice from two values. So, given some value $c in mathbb{R}$ we need to calculate some value $lambda in {0,1}$ depnding on $c<0$ or $c>0$.
Knowing that
$$frac{|c|}{c} in {1,-1}$$
we can calculate the $lambda$ as follows:
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
Now that our $lambda in {0,1}$ we can do the "choice" between the two constants $d$ and $e$ as follows:
$$dlambda + e(1-lambda)$$
which equals $d$ for $lambda = 1$, and $e$ for $lambda = 0$.
answered yesterday
TimurTimur
1193
New contributor
Timur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Following solution is not defined for $c = 0$; however it uses very basic operations only, which might be useful as you probably look for an expression to implement in a program:
$$a = b + 1lambda + 2(1-lambda)$$
where
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
You need to make the problem discrete and make a choice from two values. So, given some value $c in mathbb{R}$ we need to calculate some value $lambda in {0,1}$ depnding on $c<0$ or $c>0$.
Knowing that
$$frac{|c|}{c} in {1,-1}$$
we can calculate the $lambda$ as follows:
$$lambda = frac{ 1 + frac{|c|}{c} }{2}$$
Now that our $lambda in {0,1}$ we can do the "choice" between the two constants $d$ and $e$ as follows:
$$dlambda + e(1-lambda)$$
which equals $d$ for $lambda = 1$, and $e$ for $lambda = 0$.
answered yesterday
TimurTimur
1193
New contributor
Timur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
TimurTimur
1193
New contributor
Timur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
TimurTimur
1193
answered yesterday
TimurTimur
1193
1193
New contributor
Timur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Timur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Timur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
And you're seriously proposing that I should write all of that instead of "$a=b+1$ if $c>0$ and $a=b+2$, otherwise"? Um...
$endgroup$
– David Richerby
19 hours ago
add a comment |
$begingroup$
And you're seriously proposing that I should write all of that instead of "$a=b+1$ if $c>0$ and $a=b+2$, otherwise"? Um...
$endgroup$
– David Richerby
19 hours ago
$begingroup$
And you're seriously proposing that I should write all of that instead of "$a=b+1$ if $c>0$ and $a=b+2$, otherwise"? Um...
$endgroup$
– David Richerby
19 hours ago
$begingroup$
And you're seriously proposing that I should write all of that instead of "$a=b+1$ if $c>0$ and $a=b+2$, otherwise"? Um...
$endgroup$
– David Richerby
19 hours ago
add a comment |
$begingroup$
Fundamentally, the non-answer is the answer. Whatever notation you think you might want to use to express "$a=b+1$ if $c>0$ and $a=b+2$ otherwise" or
$$a=begin{cases}b+1 &text{if }c>0\b+2 &text{otherwise}end{cases}$$
is much harder to read than either of those two things.
answered 19 hours ago
David RicherbyDavid Richerby
2,18711324
$endgroup$
$begingroup$
The beauty of formal maths notation is that it expresses identities that can be used to solve / simplify / invert equations without translation to and from a natural language. You are right in a sense, I could just invent a new notation and describe it alongside the equation, but I wouldn't want to reinvent the wheel and risk confusion.
$endgroup$
– dataphile
16 hours ago
1
$begingroup$
Most of any mathematics paper is natural language. There's a reason for that.
$endgroup$
– David Richerby
15 hours ago
add a comment |
$begingroup$
Fundamentally, the non-answer is the answer. Whatever notation you think you might want to use to express "$a=b+1$ if $c>0$ and $a=b+2$ otherwise" or
$$a=begin{cases}b+1 &text{if }c>0\b+2 &text{otherwise}end{cases}$$
is much harder to read than either of those two things.
answered 19 hours ago
David RicherbyDavid Richerby
2,18711324
$endgroup$
$begingroup$
The beauty of formal maths notation is that it expresses identities that can be used to solve / simplify / invert equations without translation to and from a natural language. You are right in a sense, I could just invent a new notation and describe it alongside the equation, but I wouldn't want to reinvent the wheel and risk confusion.
$endgroup$
– dataphile
16 hours ago
1
$begingroup$
Most of any mathematics paper is natural language. There's a reason for that.
$endgroup$
– David Richerby
15 hours ago
add a comment |
$begingroup$
Fundamentally, the non-answer is the answer. Whatever notation you think you might want to use to express "$a=b+1$ if $c>0$ and $a=b+2$ otherwise" or
$$a=begin{cases}b+1 &text{if }c>0\b+2 &text{otherwise}end{cases}$$
is much harder to read than either of those two things.
answered 19 hours ago
David RicherbyDavid Richerby
2,18711324
$endgroup$
Fundamentally, the non-answer is the answer. Whatever notation you think you might want to use to express "$a=b+1$ if $c>0$ and $a=b+2$ otherwise" or
$$a=begin{cases}b+1 &text{if }c>0\b+2 &text{otherwise}end{cases}$$
is much harder to read than either of those two things.
answered 19 hours ago
David RicherbyDavid Richerby
2,18711324
answered 19 hours ago
David RicherbyDavid Richerby
2,18711324
answered 19 hours ago
David RicherbyDavid Richerby
2,18711324
answered 19 hours ago
David RicherbyDavid Richerby
2,18711324
2,18711324
$begingroup$
The beauty of formal maths notation is that it expresses identities that can be used to solve / simplify / invert equations without translation to and from a natural language. You are right in a sense, I could just invent a new notation and describe it alongside the equation, but I wouldn't want to reinvent the wheel and risk confusion.
$endgroup$
– dataphile
16 hours ago
1
$begingroup$
Most of any mathematics paper is natural language. There's a reason for that.
$endgroup$
– David Richerby
15 hours ago
add a comment |
$begingroup$
The beauty of formal maths notation is that it expresses identities that can be used to solve / simplify / invert equations without translation to and from a natural language. You are right in a sense, I could just invent a new notation and describe it alongside the equation, but I wouldn't want to reinvent the wheel and risk confusion.
$endgroup$
– dataphile
16 hours ago
1
$begingroup$
Most of any mathematics paper is natural language. There's a reason for that.
$endgroup$
– David Richerby
15 hours ago
$begingroup$
The beauty of formal maths notation is that it expresses identities that can be used to solve / simplify / invert equations without translation to and from a natural language. You are right in a sense, I could just invent a new notation and describe it alongside the equation, but I wouldn't want to reinvent the wheel and risk confusion.
$endgroup$
– dataphile
16 hours ago
$begingroup$
The beauty of formal maths notation is that it expresses identities that can be used to solve / simplify / invert equations without translation to and from a natural language. You are right in a sense, I could just invent a new notation and describe it alongside the equation, but I wouldn't want to reinvent the wheel and risk confusion.
$endgroup$
– dataphile
16 hours ago
1
1
$begingroup$
Most of any mathematics paper is natural language. There's a reason for that.
$endgroup$
– David Richerby
15 hours ago
$begingroup$
Most of any mathematics paper is natural language. There's a reason for that.
$endgroup$
– David Richerby
15 hours ago
add a comment |
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5
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
Mar 5 at 5:45
16
$begingroup$
A better question here is: what is good notation in this case. There are many ways of writing this, many ways which one would almost never use (like the accepted answer).
$endgroup$
– Winther
2 days ago
5
$begingroup$
In C-derived languages at least, you have made a grievous error! Your expression parses as
a = (b + c) > 0 ? 1 : 2. I always use parentheses in these cases, even when they are not strictly necessary.$endgroup$
– TonyK
2 days ago
3
$begingroup$
Just to clarify, by “ternary operator” you really mean “conditional operator”? Or are interested in any kind of ternary operator, regardless of semantics?
$endgroup$
– Konrad Rudolph
2 days ago
1
$begingroup$
What @KonradRudolph said; you are showing the conditional operator, which is a ternary (takes three arguments; compare unary and binary) operator. I don't think the C language has any other ternary operator, and some form of the conditional operator is found in many languages because it's very convenient to have, but there's no rule that says a language couldn't include other ternary operators alongside (or without having) a conditional operator.
$endgroup$
– a CVn
2 days ago