Random Six Character Strings Without Collisions











up vote
2
down vote

favorite












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?










share|improve this question


















  • 5




    replace RandomChoice with RandomSample?
    – kglr
    16 hours ago










  • @kglr: I will certainly look into that. Thanks for the pointer!
    – Moo
    16 hours ago






  • 2




    When a function is close to what you want, look at the See Also section of that function's documentation.
    – Bob Hanlon
    16 hours ago















up vote
2
down vote

favorite












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?










share|improve this question


















  • 5




    replace RandomChoice with RandomSample?
    – kglr
    16 hours ago










  • @kglr: I will certainly look into that. Thanks for the pointer!
    – Moo
    16 hours ago






  • 2




    When a function is close to what you want, look at the See Also section of that function's documentation.
    – Bob Hanlon
    16 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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?










share|improve this question













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?







list-manipulation random






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share|improve this question











share|improve this question




share|improve this question










asked 16 hours ago









Moo

6941515




6941515








  • 5




    replace RandomChoice with RandomSample?
    – kglr
    16 hours ago










  • @kglr: I will certainly look into that. Thanks for the pointer!
    – Moo
    16 hours ago






  • 2




    When a function is close to what you want, look at the See Also section of that function's documentation.
    – Bob Hanlon
    16 hours ago














  • 5




    replace RandomChoice with RandomSample?
    – kglr
    16 hours ago










  • @kglr: I will certainly look into that. Thanks for the pointer!
    – Moo
    16 hours ago






  • 2




    When a function is close to what you want, look at the See Also section of that function's documentation.
    – Bob Hanlon
    16 hours ago








5




5




replace RandomChoice with RandomSample?
– kglr
16 hours ago




replace RandomChoice with RandomSample?
– kglr
16 hours ago












@kglr: I will certainly look into that. Thanks for the pointer!
– Moo
16 hours ago




@kglr: I will certainly look into that. Thanks for the pointer!
– Moo
16 hours ago




2




2




When a function is close to what you want, look at the See Also section of that function's documentation.
– Bob Hanlon
16 hours ago




When a function is close to what you want, look at the See Also section of that function's documentation.
– Bob Hanlon
16 hours ago










5 Answers
5






active

oldest

votes

















up vote
6
down vote



accepted










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"}







share|improve this answer























  • Is there a way to generate some number of these, keep the state of the RNG and use it to reseed from that point moving forward so we can maintain uniqueness of all six-characters over time?
    – Moo
    14 hours ago










  • @Moo Not that I'm aware of. Do you know how many unique strings you want in total?
    – Carl Woll
    14 hours ago










  • If I am doing my math correctly, we have 6-character strings where strings can repeat, then there are $26$ letters plus $10$ digits in the pool of characters, giving us $36$ in all per character. This means we would have $36^6 = 2176782336$ combinations.
    – Moo
    14 hours ago






  • 1




    @Moo There are 36^6 total strings, but how many do you want to generate? There is no point in using a random approach if you want to generate all of them.
    – Carl Woll
    14 hours ago










  • CarlWoll: The issue is that I would generate something like $10000$ at a time, use them and then generate a new batch and over time, duplicates matter. Maybe there is a better way based on that, but I haven't thought about it. I am not sure I would ever use up the total number shown, but it can certainly be over several million of them over time.
    – Moo
    13 hours ago




















up vote
3
down vote













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.






share|improve this answer























  • How does does guarantee no collisions?
    – Moo
    15 hours ago


















up vote
3
down vote













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.






share|improve this answer

















  • 1




    This becomes much less silly if you use RandomSample[0 ;; 36^6 - 1, 1000] instead. This way, the humongous list range need not be created and 1000 words are created within 4 ms. See also Carl's answer.
    – Henrik Schumacher
    9 hours ago












  • Holy cow I had no idea that worked!
    – Pillsy
    7 hours ago


















up vote
2
down vote













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 100000 sample.. Of course, it might be larger than 5.






share|improve this answer






























    up vote
    0
    down vote













    Sorry that I'm not a Mathematica user, so just provide my thought on solving this in pseudo code:



    while less than 5 strings in the collection:
    generate a random string
    if new string already exists in the collection: do nothing
    else: add it to the collection


    No collisions guaranteed, but might take a while.






    share|improve this answer








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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      6
      down vote



      accepted










      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"}







      share|improve this answer























      • Is there a way to generate some number of these, keep the state of the RNG and use it to reseed from that point moving forward so we can maintain uniqueness of all six-characters over time?
        – Moo
        14 hours ago










      • @Moo Not that I'm aware of. Do you know how many unique strings you want in total?
        – Carl Woll
        14 hours ago










      • If I am doing my math correctly, we have 6-character strings where strings can repeat, then there are $26$ letters plus $10$ digits in the pool of characters, giving us $36$ in all per character. This means we would have $36^6 = 2176782336$ combinations.
        – Moo
        14 hours ago






      • 1




        @Moo There are 36^6 total strings, but how many do you want to generate? There is no point in using a random approach if you want to generate all of them.
        – Carl Woll
        14 hours ago










      • CarlWoll: The issue is that I would generate something like $10000$ at a time, use them and then generate a new batch and over time, duplicates matter. Maybe there is a better way based on that, but I haven't thought about it. I am not sure I would ever use up the total number shown, but it can certainly be over several million of them over time.
        – Moo
        13 hours ago

















      up vote
      6
      down vote



      accepted










      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"}







      share|improve this answer























      • Is there a way to generate some number of these, keep the state of the RNG and use it to reseed from that point moving forward so we can maintain uniqueness of all six-characters over time?
        – Moo
        14 hours ago










      • @Moo Not that I'm aware of. Do you know how many unique strings you want in total?
        – Carl Woll
        14 hours ago










      • If I am doing my math correctly, we have 6-character strings where strings can repeat, then there are $26$ letters plus $10$ digits in the pool of characters, giving us $36$ in all per character. This means we would have $36^6 = 2176782336$ combinations.
        – Moo
        14 hours ago






      • 1




        @Moo There are 36^6 total strings, but how many do you want to generate? There is no point in using a random approach if you want to generate all of them.
        – Carl Woll
        14 hours ago










      • CarlWoll: The issue is that I would generate something like $10000$ at a time, use them and then generate a new batch and over time, duplicates matter. Maybe there is a better way based on that, but I haven't thought about it. I am not sure I would ever use up the total number shown, but it can certainly be over several million of them over time.
        – Moo
        13 hours ago















      up vote
      6
      down vote



      accepted







      up vote
      6
      down vote



      accepted






      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"}







      share|improve this answer














      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"}








      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 14 hours ago

























      answered 14 hours ago









      Carl Woll

      66.5k385174




      66.5k385174












      • Is there a way to generate some number of these, keep the state of the RNG and use it to reseed from that point moving forward so we can maintain uniqueness of all six-characters over time?
        – Moo
        14 hours ago










      • @Moo Not that I'm aware of. Do you know how many unique strings you want in total?
        – Carl Woll
        14 hours ago










      • If I am doing my math correctly, we have 6-character strings where strings can repeat, then there are $26$ letters plus $10$ digits in the pool of characters, giving us $36$ in all per character. This means we would have $36^6 = 2176782336$ combinations.
        – Moo
        14 hours ago






      • 1




        @Moo There are 36^6 total strings, but how many do you want to generate? There is no point in using a random approach if you want to generate all of them.
        – Carl Woll
        14 hours ago










      • CarlWoll: The issue is that I would generate something like $10000$ at a time, use them and then generate a new batch and over time, duplicates matter. Maybe there is a better way based on that, but I haven't thought about it. I am not sure I would ever use up the total number shown, but it can certainly be over several million of them over time.
        – Moo
        13 hours ago




















      • Is there a way to generate some number of these, keep the state of the RNG and use it to reseed from that point moving forward so we can maintain uniqueness of all six-characters over time?
        – Moo
        14 hours ago










      • @Moo Not that I'm aware of. Do you know how many unique strings you want in total?
        – Carl Woll
        14 hours ago










      • If I am doing my math correctly, we have 6-character strings where strings can repeat, then there are $26$ letters plus $10$ digits in the pool of characters, giving us $36$ in all per character. This means we would have $36^6 = 2176782336$ combinations.
        – Moo
        14 hours ago






      • 1




        @Moo There are 36^6 total strings, but how many do you want to generate? There is no point in using a random approach if you want to generate all of them.
        – Carl Woll
        14 hours ago










      • CarlWoll: The issue is that I would generate something like $10000$ at a time, use them and then generate a new batch and over time, duplicates matter. Maybe there is a better way based on that, but I haven't thought about it. I am not sure I would ever use up the total number shown, but it can certainly be over several million of them over time.
        – Moo
        13 hours ago


















      Is there a way to generate some number of these, keep the state of the RNG and use it to reseed from that point moving forward so we can maintain uniqueness of all six-characters over time?
      – Moo
      14 hours ago




      Is there a way to generate some number of these, keep the state of the RNG and use it to reseed from that point moving forward so we can maintain uniqueness of all six-characters over time?
      – Moo
      14 hours ago












      @Moo Not that I'm aware of. Do you know how many unique strings you want in total?
      – Carl Woll
      14 hours ago




      @Moo Not that I'm aware of. Do you know how many unique strings you want in total?
      – Carl Woll
      14 hours ago












      If I am doing my math correctly, we have 6-character strings where strings can repeat, then there are $26$ letters plus $10$ digits in the pool of characters, giving us $36$ in all per character. This means we would have $36^6 = 2176782336$ combinations.
      – Moo
      14 hours ago




      If I am doing my math correctly, we have 6-character strings where strings can repeat, then there are $26$ letters plus $10$ digits in the pool of characters, giving us $36$ in all per character. This means we would have $36^6 = 2176782336$ combinations.
      – Moo
      14 hours ago




      1




      1




      @Moo There are 36^6 total strings, but how many do you want to generate? There is no point in using a random approach if you want to generate all of them.
      – Carl Woll
      14 hours ago




      @Moo There are 36^6 total strings, but how many do you want to generate? There is no point in using a random approach if you want to generate all of them.
      – Carl Woll
      14 hours ago












      CarlWoll: The issue is that I would generate something like $10000$ at a time, use them and then generate a new batch and over time, duplicates matter. Maybe there is a better way based on that, but I haven't thought about it. I am not sure I would ever use up the total number shown, but it can certainly be over several million of them over time.
      – Moo
      13 hours ago






      CarlWoll: The issue is that I would generate something like $10000$ at a time, use them and then generate a new batch and over time, duplicates matter. Maybe there is a better way based on that, but I haven't thought about it. I am not sure I would ever use up the total number shown, but it can certainly be over several million of them over time.
      – Moo
      13 hours ago












      up vote
      3
      down vote













      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.






      share|improve this answer























      • How does does guarantee no collisions?
        – Moo
        15 hours ago















      up vote
      3
      down vote













      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.






      share|improve this answer























      • How does does guarantee no collisions?
        – Moo
        15 hours ago













      up vote
      3
      down vote










      up vote
      3
      down vote









      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.






      share|improve this answer














      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.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 15 hours ago

























      answered 15 hours ago









      David G. Stork

      22.9k22051




      22.9k22051












      • How does does guarantee no collisions?
        – Moo
        15 hours ago


















      • How does does guarantee no collisions?
        – Moo
        15 hours ago
















      How does does guarantee no collisions?
      – Moo
      15 hours ago




      How does does guarantee no collisions?
      – Moo
      15 hours ago










      up vote
      3
      down vote













      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.






      share|improve this answer

















      • 1




        This becomes much less silly if you use RandomSample[0 ;; 36^6 - 1, 1000] instead. This way, the humongous list range need not be created and 1000 words are created within 4 ms. See also Carl's answer.
        – Henrik Schumacher
        9 hours ago












      • Holy cow I had no idea that worked!
        – Pillsy
        7 hours ago















      up vote
      3
      down vote













      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.






      share|improve this answer

















      • 1




        This becomes much less silly if you use RandomSample[0 ;; 36^6 - 1, 1000] instead. This way, the humongous list range need not be created and 1000 words are created within 4 ms. See also Carl's answer.
        – Henrik Schumacher
        9 hours ago












      • Holy cow I had no idea that worked!
        – Pillsy
        7 hours ago













      up vote
      3
      down vote










      up vote
      3
      down vote









      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.






      share|improve this answer












      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.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 13 hours ago









      Pillsy

      12.7k13179




      12.7k13179








      • 1




        This becomes much less silly if you use RandomSample[0 ;; 36^6 - 1, 1000] instead. This way, the humongous list range need not be created and 1000 words are created within 4 ms. See also Carl's answer.
        – Henrik Schumacher
        9 hours ago












      • Holy cow I had no idea that worked!
        – Pillsy
        7 hours ago














      • 1




        This becomes much less silly if you use RandomSample[0 ;; 36^6 - 1, 1000] instead. This way, the humongous list range need not be created and 1000 words are created within 4 ms. See also Carl's answer.
        – Henrik Schumacher
        9 hours ago












      • Holy cow I had no idea that worked!
        – Pillsy
        7 hours ago








      1




      1




      This becomes much less silly if you use RandomSample[0 ;; 36^6 - 1, 1000] instead. This way, the humongous list range need not be created and 1000 words are created within 4 ms. See also Carl's answer.
      – Henrik Schumacher
      9 hours ago






      This becomes much less silly if you use RandomSample[0 ;; 36^6 - 1, 1000] instead. This way, the humongous list range need not be created and 1000 words are created within 4 ms. See also Carl's answer.
      – Henrik Schumacher
      9 hours ago














      Holy cow I had no idea that worked!
      – Pillsy
      7 hours ago




      Holy cow I had no idea that worked!
      – Pillsy
      7 hours ago










      up vote
      2
      down vote













      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 100000 sample.. Of course, it might be larger than 5.






      share|improve this answer



























        up vote
        2
        down vote













        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 100000 sample.. Of course, it might be larger than 5.






        share|improve this answer

























          up vote
          2
          down vote










          up vote
          2
          down vote









          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 100000 sample.. Of course, it might be larger than 5.






          share|improve this answer














          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 100000 sample.. Of course, it might be larger than 5.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 12 hours ago

























          answered 13 hours ago









          Okkes Dulgerci

          3,7451716




          3,7451716






















              up vote
              0
              down vote













              Sorry that I'm not a Mathematica user, so just provide my thought on solving this in pseudo code:



              while less than 5 strings in the collection:
              generate a random string
              if new string already exists in the collection: do nothing
              else: add it to the collection


              No collisions guaranteed, but might take a while.






              share|improve this answer








              New contributor




              Kyle R. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






















                up vote
                0
                down vote













                Sorry that I'm not a Mathematica user, so just provide my thought on solving this in pseudo code:



                while less than 5 strings in the collection:
                generate a random string
                if new string already exists in the collection: do nothing
                else: add it to the collection


                No collisions guaranteed, but might take a while.






                share|improve this answer








                New contributor




                Kyle R. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.




















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Sorry that I'm not a Mathematica user, so just provide my thought on solving this in pseudo code:



                  while less than 5 strings in the collection:
                  generate a random string
                  if new string already exists in the collection: do nothing
                  else: add it to the collection


                  No collisions guaranteed, but might take a while.






                  share|improve this answer








                  New contributor




                  Kyle R. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  Sorry that I'm not a Mathematica user, so just provide my thought on solving this in pseudo code:



                  while less than 5 strings in the collection:
                  generate a random string
                  if new string already exists in the collection: do nothing
                  else: add it to the collection


                  No collisions guaranteed, but might take a while.







                  share|improve this answer








                  New contributor




                  Kyle R. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|improve this answer



                  share|improve this answer






                  New contributor




                  Kyle R. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered 6 hours ago









                  Kyle R.

                  1




                  1




                  New contributor




                  Kyle R. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  Kyle R. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  Kyle R. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






























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