Bdtyktl

I want to generate a random list of six characters that only contain characters from "0...9" and "A...Z".

For this I can define

ToCharacterCode[{"ABCDEFGHIJKLMONPQRSTUVWXYZ", "0123456789"}]

I can modify the answer here as

rndnew = FromCharacterCode@

 RandomChoice[Join @@ Range[{48, 65}, {57, 90}], #] &

I can now generate five of these strings by doing

rndnew[{5, 6}]

This results in something like

{"35UVUS", "F7WIJG", "PQSBHF", "PIHTSW", "R3MDM6"}

My question is how can I guarantee that these strings have no collisions (I know this is a very large space)? Is there a better way to code this using random over the range of the size of this set (like a linear congruential generator with that range) to make sure the strings are unique or is there a facility in Mathematica to do that?

Just changing RandomChoice to RandomSample doesn't help, since RandomSample just means that none of the characters are repeated, and clearly you want to allow repeated characters, since one of your strings is "35UVUS". One idea is to oversample and remove duplicates. For example:

SeedRandom[1]

Take[

    DeleteDuplicates[rndnew[{10, 6}]],

    UpTo[5]

]

{"K1263G", "SXVXHC", "UO2PBL", "EC1FTJ", "0TKLEH"}

Another possibility is to index the possible random strings, and then do a random sample from the possible indices. For your case, the number of possible strings is simply 36^6. To convert from an index to a string:

characters = Join[CharacterRange["A","Z"], CharacterRange["0","9"]];



fromIndex[index_, len_] := StringJoin[

    characters[[1+IntegerDigits[index, 36, len]]]

]

Then, a function that returns n samples of length len strings:

sample[n_, len_] := fromIndex[#, len]& /@ RandomSample[1 ;; 36^len, n]

Example:

SeedRandom[1]

sample[5, 6]

{"0621O7", "0WH6XW", "ODLV4Z", "KWSN6U", "AMOSFA"}

RandomSample[

 Union[CharacterRange["a", "z"], CharacterRange["0", "9"]], 6]

or

RandomSample[Flatten[CharacterRange @@@ {{"a", "z"}, {"0", "9"}}], 6]

or

RandomSample[Join @@ (CharacterRange @@@ {{"a", "z"}, {"0", "9"}}), 6]

This guarantees no collisions because RandomSample selects elements without replacement.

Inspired by some of the other answers and comments, here's a rather silly brute-force approach. We can RandomSample a range that we can use to enumerate all combinations.

range = Range[0, 36^6 - 1];

(* This takes a little while: 15 seconds on my laptop. *)



chars = Union[CharacterRange["a", "z"], CharacterRange["0", "9"]];



makeString[n_] := ToString@Part[chars, 1 + IntegerDigits[n, 36, 6]];

Then

makeString /@ RandomSample[range, 1000]

will always return a bunch of unique strings.

Following is what you want. Unfortunately it takes time to generate all possible set for n=6. But for n=5 it works. And it is guaranteed that there is no repeated element.

val = Tuples[Join @@ (CharacterRange @@@ {{"A", "Z"}, {"0", "9"}}), 5];



    RandomSample[val, 5]

Much smart way is

DeleteDuplicates[

 Table[RandomSample[

   Join @@ (CharacterRange @@@ {{"A", "Z"}, {"0", "9"}}), 6], 5]]

It is very unlikely to generate the same element.

n = 10000;

Length@DeleteDuplicates[

   Table[RandomSample[

     Join @@ (CharacterRange @@@ {{"A", "Z"}, {"0", "9"}}), 6], n]] ==

  n

True

But when you increase n

Table[n = 100000;

  n - Length@

    DeleteDuplicates[

     Table[RandomSample[

       Join @@ (CharacterRange @@@ {{"A", "Z"}, {"0", "9"}}), 6], n]],

   10] // Max

5

This means 5 elements are repeated when we generate 1000000 sample.. Of course, it might be larger than 5.