Am I not good enough for you?
$begingroup$
Background:
The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.
Challenge
Given a positive integer through any standard input format, distinguish between whether it is perfect or not.
A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.
Test Cases:
Imperfect:
1,12,13,18,20,1000,33550335
Perfect:
6,28,496,8128,33550336,8589869056
Rules
- Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
- Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
code-golf number decision-problem number-theory factoring
$endgroup$
add a comment |
$begingroup$
Background:
The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.
Challenge
Given a positive integer through any standard input format, distinguish between whether it is perfect or not.
A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.
Test Cases:
Imperfect:
1,12,13,18,20,1000,33550335
Perfect:
6,28,496,8128,33550336,8589869056
Rules
- Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
- Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
code-golf number decision-problem number-theory factoring
$endgroup$
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
1 hour ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect
/Perfect
to make it clearer
$endgroup$
– Jo King
1 hour ago
add a comment |
$begingroup$
Background:
The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.
Challenge
Given a positive integer through any standard input format, distinguish between whether it is perfect or not.
A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.
Test Cases:
Imperfect:
1,12,13,18,20,1000,33550335
Perfect:
6,28,496,8128,33550336,8589869056
Rules
- Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
- Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
code-golf number decision-problem number-theory factoring
$endgroup$
Background:
The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.
Challenge
Given a positive integer through any standard input format, distinguish between whether it is perfect or not.
A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.
Test Cases:
Imperfect:
1,12,13,18,20,1000,33550335
Perfect:
6,28,496,8128,33550336,8589869056
Rules
- Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
- Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
code-golf number decision-problem number-theory factoring
code-golf number decision-problem number-theory factoring
edited 3 mins ago
Jo King
asked 1 hour ago
Jo KingJo King
24.6k357126
24.6k357126
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
1 hour ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect
/Perfect
to make it clearer
$endgroup$
– Jo King
1 hour ago
add a comment |
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
1 hour ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect
/Perfect
to make it clearer
$endgroup$
– Jo King
1 hour ago
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
1 hour ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
1 hour ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect
/Perfect
to make it clearer$endgroup$
– Jo King
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect
/Perfect
to make it clearer$endgroup$
– Jo King
1 hour ago
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
Japt -!
, 4 bytes
¥â¬x
For some reason ¦
doesnt work on tio so I need to use the -!
flag and ¥
instead
Try it online!
$endgroup$
add a comment |
$begingroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE
for perfect numbers ans FALSE
for imperfect ones.
$endgroup$
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
55 mins ago
$begingroup$
@CTHall I misread the spec; they originally mapped0
(perfect) toFALSE
and nonzero toTRUE
but I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumeric
tological
, often in conjunction withwhich
or[
.
$endgroup$
– Giuseppe
34 mins ago
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
$endgroup$
add a comment |
$begingroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
add a comment |
$begingroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
13 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
8 mins ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
$endgroup$
add a comment |
$begingroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true.
or false.
if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
$endgroup$
add a comment |
Your Answer
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Japt -!
, 4 bytes
¥â¬x
For some reason ¦
doesnt work on tio so I need to use the -!
flag and ¥
instead
Try it online!
$endgroup$
add a comment |
$begingroup$
Japt -!
, 4 bytes
¥â¬x
For some reason ¦
doesnt work on tio so I need to use the -!
flag and ¥
instead
Try it online!
$endgroup$
add a comment |
$begingroup$
Japt -!
, 4 bytes
¥â¬x
For some reason ¦
doesnt work on tio so I need to use the -!
flag and ¥
instead
Try it online!
$endgroup$
Japt -!
, 4 bytes
¥â¬x
For some reason ¦
doesnt work on tio so I need to use the -!
flag and ¥
instead
Try it online!
edited 1 hour ago
answered 1 hour ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
5,60821670
add a comment |
add a comment |
$begingroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE
for perfect numbers ans FALSE
for imperfect ones.
$endgroup$
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
55 mins ago
$begingroup$
@CTHall I misread the spec; they originally mapped0
(perfect) toFALSE
and nonzero toTRUE
but I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumeric
tological
, often in conjunction withwhich
or[
.
$endgroup$
– Giuseppe
34 mins ago
add a comment |
$begingroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE
for perfect numbers ans FALSE
for imperfect ones.
$endgroup$
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
55 mins ago
$begingroup$
@CTHall I misread the spec; they originally mapped0
(perfect) toFALSE
and nonzero toTRUE
but I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumeric
tological
, often in conjunction withwhich
or[
.
$endgroup$
– Giuseppe
34 mins ago
add a comment |
$begingroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE
for perfect numbers ans FALSE
for imperfect ones.
$endgroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE
for perfect numbers ans FALSE
for imperfect ones.
answered 59 mins ago
GiuseppeGiuseppe
16.8k31052
16.8k31052
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
55 mins ago
$begingroup$
@CTHall I misread the spec; they originally mapped0
(perfect) toFALSE
and nonzero toTRUE
but I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumeric
tological
, often in conjunction withwhich
or[
.
$endgroup$
– Giuseppe
34 mins ago
add a comment |
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
55 mins ago
$begingroup$
@CTHall I misread the spec; they originally mapped0
(perfect) toFALSE
and nonzero toTRUE
but I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumeric
tological
, often in conjunction withwhich
or[
.
$endgroup$
– Giuseppe
34 mins ago
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
55 mins ago
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
55 mins ago
$begingroup$
@CTHall I misread the spec; they originally mapped
0
(perfect) to FALSE
and nonzero to TRUE
but I removed one of them to reverse the mapping. It's a useful golfing trick to cast from numeric
to logical
, often in conjunction with which
or [
.$endgroup$
– Giuseppe
34 mins ago
$begingroup$
@CTHall I misread the spec; they originally mapped
0
(perfect) to FALSE
and nonzero to TRUE
but I removed one of them to reverse the mapping. It's a useful golfing trick to cast from numeric
to logical
, often in conjunction with which
or [
.$endgroup$
– Giuseppe
34 mins ago
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
$endgroup$
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
$endgroup$
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
$endgroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
8,50932674
add a comment |
add a comment |
$begingroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
add a comment |
$begingroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
add a comment |
$begingroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
$endgroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
edited 1 hour ago
answered 1 hour ago
zeveezevee
22016
22016
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
add a comment |
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
add a comment |
$begingroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
13 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
8 mins ago
add a comment |
$begingroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
13 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
8 mins ago
add a comment |
$begingroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
$endgroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
edited 13 mins ago
answered 14 mins ago
Neil A.Neil A.
1,278120
1,278120
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
13 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
8 mins ago
add a comment |
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
13 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
8 mins ago
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
13 mins ago
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
13 mins ago
1
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
8 mins ago
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
8 mins ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
$endgroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
edited 13 mins ago
answered 18 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
1,588124
add a comment |
add a comment |
$begingroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true.
or false.
if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
$endgroup$
add a comment |
$begingroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true.
or false.
if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
$endgroup$
add a comment |
$begingroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true.
or false.
if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
$endgroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true.
or false.
if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
answered 11 mins ago
Unrelated StringUnrelated String
91118
91118
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$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
1 hour ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect
/Perfect
to make it clearer$endgroup$
– Jo King
1 hour ago