To string or not to string

Given an input string, output at random the unique combinations with repetition of the characters in the input string, from length 1 up to the length of the input string, with an equal chance of each one occurring.

Example: given the input abcd (or any combination thereof of the four characters a,b,c,d) , there is an equal chance of outputting:

a b c d aa ab ac ad ba bb bc bd ca cb cc cd da db dc dd aaa aab aac aad aba abb abc abd aca acb acc acd ada adb adc add baa bab bac bad bba bbb bbc bbd bca bcb bcc bcd bda bdb bdc bdd caa cab cac cad cba cbb cbc cbd cca ccb ccc ccd cda cdb cdc cdd daa dab dac dad dba dbb dbc dbd dca dcb dcc dcd dda ddb ddc ddd aaaa aaab aaac aaad aaba aabb aabc aabd aaca aacb aacc aacd aada aadb aadc aadd abaa abab abac abad abba abbb abbc abbd abca abcb abcc abcd abda abdb abdc abdd acaa acab acac acad acba acbb acbc acbd acca accb accc accd acda acdb acdc acdd adaa adab adac adad adba adbb adbc adbd adca adcb adcc adcd adda addb addc addd baaa baab baac baad baba babb babc babd baca bacb bacc bacd bada badb badc badd bbaa bbab bbac bbad bbba bbbb bbbc bbbd bbca bbcb bbcc bbcd bbda bbdb bbdc bbdd bcaa bcab bcac bcad bcba bcbb bcbc bcbd bcca bccb bccc bccd bcda bcdb bcdc bcdd bdaa bdab bdac bdad bdba bdbb bdbc bdbd bdca bdcb bdcc bdcd bdda bddb bddc bddd caaa caab caac caad caba cabb cabc cabd caca cacb cacc cacd cada cadb cadc cadd cbaa cbab cbac cbad cbba cbbb cbbc cbbd cbca cbcb cbcc cbcd cbda cbdb cbdc cbdd ccaa ccab ccac ccad ccba ccbb ccbc ccbd ccca cccb cccc cccd ccda ccdb ccdc ccdd cdaa cdab cdac cdad cdba cdbb cdbc cdbd cdca cdcb cdcc cdcd cdda cddb cddc cddd daaa daab daac daad daba dabb dabc dabd daca dacb dacc dacd dada dadb dadc dadd dbaa dbab dbac dbad dbba dbbb dbbc dbbd dbca dbcb dbcc dbcd dbda dbdb dbdc dbdd dcaa dcab dcac dcad dcba dcbb dcbc dcbd dcca dccb dccc dccd dcda dcdb dcdc dcdd ddaa ddab ddac ddad ddba ddbb ddbc ddbd ddca ddcb ddcc ddcd ddda dddb dddc dddd

Example: given the input efgh (or any combination thereof of the four characters e,f,g,h), there is an equal chance of outputting:

e f g h ee ef eg eh fe ff fg fh ge gf gg gh he hf hg hh eee eef eeg eeh efe eff efg efh ege egf egg egh ehe ehf ehg ehh fee fef feg feh ffe fff ffg ffh fge fgf fgg fgh fhe fhf fhg fhh gee gef geg geh gfe gff gfg gfh gge ggf ggg ggh ghe ghf ghg ghh hee hef heg heh hfe hff hfg hfh hge hgf hgg hgh hhe hhf hhg hhh eeee eeef eeeg eeeh eefe eeff eefg eefh eege eegf eegg eegh eehe eehf eehg eehh efee efef efeg efeh effe efff effg effh efge efgf efgg efgh efhe efhf efhg efhh egee egef egeg egeh egfe egff egfg egfh egge eggf eggg eggh eghe eghf eghg eghh ehee ehef eheg eheh ehfe ehff ehfg ehfh ehge ehgf ehgg ehgh ehhe ehhf ehhg ehhh feee feef feeg feeh fefe feff fefg fefh fege fegf fegg fegh fehe fehf fehg fehh ffee ffef ffeg ffeh fffe ffff fffg fffh ffge ffgf ffgg ffgh ffhe ffhf ffhg ffhh fgee fgef fgeg fgeh fgfe fgff fgfg fgfh fgge fggf fggg fggh fghe fghf fghg fghh fhee fhef fheg fheh fhfe fhff fhfg fhfh fhge fhgf fhgg fhgh fhhe fhhf fhhg fhhh geee geef geeg geeh gefe geff gefg gefh gege gegf gegg gegh gehe gehf gehg gehh gfee gfef gfeg gfeh gffe gfff gffg gffh gfge gfgf gfgg gfgh gfhe gfhf gfhg gfhh ggee ggef ggeg ggeh ggfe ggff ggfg ggfh ggge gggf gggg gggh gghe gghf gghg gghh ghee ghef gheg gheh ghfe ghff ghfg ghfh ghge ghgf ghgg ghgh ghhe ghhf ghhg ghhh heee heef heeg heeh hefe heff hefg hefh hege hegf hegg hegh hehe hehf hehg hehh hfee hfef hfeg hfeh hffe hfff hffg hffh hfge hfgf hfgg hfgh hfhe hfhf hfhg hfhh hgee hgef hgeg hgeh hgfe hgff hgfg hgfh hgge hggf hggg hggh hghe hghf hghg hghh hhee hhef hheg hheh hhfe hhff hhfg hhfh hhge hhgf hhgg hhgh hhhe hhhf hhhg hhhh

Example: given the input ijkl (or any combination thereof of the four characters i,j,k,l), there is an equal chance of outputting:

i j k l ii ij ik il ji jj jk jl ki kj kk kl li lj lk ll iii iij iik iil iji ijj ijk ijl iki ikj ikk ikl ili ilj ilk ill jii jij jik jil jji jjj jjk jjl jki jkj jkk jkl jli jlj jlk jll kii kij kik kil kji kjj kjk kjl kki kkj kkk kkl kli klj klk kll lii lij lik lil lji ljj ljk ljl lki lkj lkk lkl lli llj llk lll iiii iiij iiik iiil iiji iijj iijk iijl iiki iikj iikk iikl iili iilj iilk iill ijii ijij ijik ijil ijji ijjj ijjk ijjl ijki ijkj ijkk ijkl ijli ijlj ijlk ijll ikii ikij ikik ikil ikji ikjj ikjk ikjl ikki ikkj ikkk ikkl ikli iklj iklk ikll ilii ilij ilik ilil ilji iljj iljk iljl ilki ilkj ilkk ilkl illi illj illk illl jiii jiij jiik jiil jiji jijj jijk jijl jiki jikj jikk jikl jili jilj jilk jill jjii jjij jjik jjil jjji jjjj jjjk jjjl jjki jjkj jjkk jjkl jjli jjlj jjlk jjll jkii jkij jkik jkil jkji jkjj jkjk jkjl jkki jkkj jkkk jkkl jkli jklj jklk jkll jlii jlij jlik jlil jlji jljj jljk jljl jlki jlkj jlkk jlkl jlli jllj jllk jlll kiii kiij kiik kiil kiji kijj kijk kijl kiki kikj kikk kikl kili kilj kilk kill kjii kjij kjik kjil kjji kjjj kjjk kjjl kjki kjkj kjkk kjkl kjli kjlj kjlk kjll kkii kkij kkik kkil kkji kkjj kkjk kkjl kkki kkkj kkkk kkkl kkli kklj kklk kkll klii klij klik klil klji kljj kljk kljl klki klkj klkk klkl klli kllj kllk klll liii liij liik liil liji lijj lijk lijl liki likj likk likl lili lilj lilk lill ljii ljij ljik ljil ljji ljjj ljjk ljjl ljki ljkj ljkk ljkl ljli ljlj ljlk ljll lkii lkij lkik lkil lkji lkjj lkjk lkjl lkki lkkj lkkk lkkl lkli lklj lklk lkll llii llij llik llil llji lljj lljk lljl llki llkj llkk llkl llli lllj lllk llll

Every individual character within the string has equal probability as well.

For example, given an output of aaab, this does not mean that a has more probability of occuring, rather a and b have equal probability of occurence.

Another example, given an output of aaabeeeee, this does not mean that e has the highest probability of occuring, rather a and b and e all have equal probability of occurence.

The input is guranteed to have at least two distinct characters.

edited yesterday

asked yesterday

Flog Edoc

413

New contributor

$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– DJMcMayhem♦
yesterday

$begingroup$
@EmbodimentofIgnorance Thanks, fixed.
$endgroup$
– Flog Edoc
yesterday

$begingroup$
Is the input guaranteed to have at least two distinct characters?
$endgroup$
– Neil
yesterday

1

$begingroup$
@Flog Edoc: (2) There is a real difference between 'will contain at least two distinct characters' and 'all characters will be distinct'. Which do you intend?
$endgroup$
– Chas Brown
23 hours ago

1

$begingroup$
do you mean given an *input* of aaab? Can we have examples that aren't isomorphic to each other?
$endgroup$
– Jo King
22 hours ago

|
show 9 more comments

Example: given the input abcd (or any combination thereof of the four characters a,b,c,d) , there is an equal chance of outputting:

a b c d aa ab ac ad ba bb bc bd ca cb cc cd da db dc dd aaa aab aac aad aba abb abc abd aca acb acc acd ada adb adc add baa bab bac bad bba bbb bbc bbd bca bcb bcc bcd bda bdb bdc bdd caa cab cac cad cba cbb cbc cbd cca ccb ccc ccd cda cdb cdc cdd daa dab dac dad dba dbb dbc dbd dca dcb dcc dcd dda ddb ddc ddd aaaa aaab aaac aaad aaba aabb aabc aabd aaca aacb aacc aacd aada aadb aadc aadd abaa abab abac abad abba abbb abbc abbd abca abcb abcc abcd abda abdb abdc abdd acaa acab acac acad acba acbb acbc acbd acca accb accc accd acda acdb acdc acdd adaa adab adac adad adba adbb adbc adbd adca adcb adcc adcd adda addb addc addd baaa baab baac baad baba babb babc babd baca bacb bacc bacd bada badb badc badd bbaa bbab bbac bbad bbba bbbb bbbc bbbd bbca bbcb bbcc bbcd bbda bbdb bbdc bbdd bcaa bcab bcac bcad bcba bcbb bcbc bcbd bcca bccb bccc bccd bcda bcdb bcdc bcdd bdaa bdab bdac bdad bdba bdbb bdbc bdbd bdca bdcb bdcc bdcd bdda bddb bddc bddd caaa caab caac caad caba cabb cabc cabd caca cacb cacc cacd cada cadb cadc cadd cbaa cbab cbac cbad cbba cbbb cbbc cbbd cbca cbcb cbcc cbcd cbda cbdb cbdc cbdd ccaa ccab ccac ccad ccba ccbb ccbc ccbd ccca cccb cccc cccd ccda ccdb ccdc ccdd cdaa cdab cdac cdad cdba cdbb cdbc cdbd cdca cdcb cdcc cdcd cdda cddb cddc cddd daaa daab daac daad daba dabb dabc dabd daca dacb dacc dacd dada dadb dadc dadd dbaa dbab dbac dbad dbba dbbb dbbc dbbd dbca dbcb dbcc dbcd dbda dbdb dbdc dbdd dcaa dcab dcac dcad dcba dcbb dcbc dcbd dcca dccb dccc dccd dcda dcdb dcdc dcdd ddaa ddab ddac ddad ddba ddbb ddbc ddbd ddca ddcb ddcc ddcd ddda dddb dddc dddd

Example: given the input efgh (or any combination thereof of the four characters e,f,g,h), there is an equal chance of outputting:

e f g h ee ef eg eh fe ff fg fh ge gf gg gh he hf hg hh eee eef eeg eeh efe eff efg efh ege egf egg egh ehe ehf ehg ehh fee fef feg feh ffe fff ffg ffh fge fgf fgg fgh fhe fhf fhg fhh gee gef geg geh gfe gff gfg gfh gge ggf ggg ggh ghe ghf ghg ghh hee hef heg heh hfe hff hfg hfh hge hgf hgg hgh hhe hhf hhg hhh eeee eeef eeeg eeeh eefe eeff eefg eefh eege eegf eegg eegh eehe eehf eehg eehh efee efef efeg efeh effe efff effg effh efge efgf efgg efgh efhe efhf efhg efhh egee egef egeg egeh egfe egff egfg egfh egge eggf eggg eggh eghe eghf eghg eghh ehee ehef eheg eheh ehfe ehff ehfg ehfh ehge ehgf ehgg ehgh ehhe ehhf ehhg ehhh feee feef feeg feeh fefe feff fefg fefh fege fegf fegg fegh fehe fehf fehg fehh ffee ffef ffeg ffeh fffe ffff fffg fffh ffge ffgf ffgg ffgh ffhe ffhf ffhg ffhh fgee fgef fgeg fgeh fgfe fgff fgfg fgfh fgge fggf fggg fggh fghe fghf fghg fghh fhee fhef fheg fheh fhfe fhff fhfg fhfh fhge fhgf fhgg fhgh fhhe fhhf fhhg fhhh geee geef geeg geeh gefe geff gefg gefh gege gegf gegg gegh gehe gehf gehg gehh gfee gfef gfeg gfeh gffe gfff gffg gffh gfge gfgf gfgg gfgh gfhe gfhf gfhg gfhh ggee ggef ggeg ggeh ggfe ggff ggfg ggfh ggge gggf gggg gggh gghe gghf gghg gghh ghee ghef gheg gheh ghfe ghff ghfg ghfh ghge ghgf ghgg ghgh ghhe ghhf ghhg ghhh heee heef heeg heeh hefe heff hefg hefh hege hegf hegg hegh hehe hehf hehg hehh hfee hfef hfeg hfeh hffe hfff hffg hffh hfge hfgf hfgg hfgh hfhe hfhf hfhg hfhh hgee hgef hgeg hgeh hgfe hgff hgfg hgfh hgge hggf hggg hggh hghe hghf hghg hghh hhee hhef hheg hheh hhfe hhff hhfg hhfh hhge hhgf hhgg hhgh hhhe hhhf hhhg hhhh

Example: given the input ijkl (or any combination thereof of the four characters i,j,k,l), there is an equal chance of outputting:

i j k l ii ij ik il ji jj jk jl ki kj kk kl li lj lk ll iii iij iik iil iji ijj ijk ijl iki ikj ikk ikl ili ilj ilk ill jii jij jik jil jji jjj jjk jjl jki jkj jkk jkl jli jlj jlk jll kii kij kik kil kji kjj kjk kjl kki kkj kkk kkl kli klj klk kll lii lij lik lil lji ljj ljk ljl lki lkj lkk lkl lli llj llk lll iiii iiij iiik iiil iiji iijj iijk iijl iiki iikj iikk iikl iili iilj iilk iill ijii ijij ijik ijil ijji ijjj ijjk ijjl ijki ijkj ijkk ijkl ijli ijlj ijlk ijll ikii ikij ikik ikil ikji ikjj ikjk ikjl ikki ikkj ikkk ikkl ikli iklj iklk ikll ilii ilij ilik ilil ilji iljj iljk iljl ilki ilkj ilkk ilkl illi illj illk illl jiii jiij jiik jiil jiji jijj jijk jijl jiki jikj jikk jikl jili jilj jilk jill jjii jjij jjik jjil jjji jjjj jjjk jjjl jjki jjkj jjkk jjkl jjli jjlj jjlk jjll jkii jkij jkik jkil jkji jkjj jkjk jkjl jkki jkkj jkkk jkkl jkli jklj jklk jkll jlii jlij jlik jlil jlji jljj jljk jljl jlki jlkj jlkk jlkl jlli jllj jllk jlll kiii kiij kiik kiil kiji kijj kijk kijl kiki kikj kikk kikl kili kilj kilk kill kjii kjij kjik kjil kjji kjjj kjjk kjjl kjki kjkj kjkk kjkl kjli kjlj kjlk kjll kkii kkij kkik kkil kkji kkjj kkjk kkjl kkki kkkj kkkk kkkl kkli kklj kklk kkll klii klij klik klil klji kljj kljk kljl klki klkj klkk klkl klli kllj kllk klll liii liij liik liil liji lijj lijk lijl liki likj likk likl lili lilj lilk lill ljii ljij ljik ljil ljji ljjj ljjk ljjl ljki ljkj ljkk ljkl ljli ljlj ljlk ljll lkii lkij lkik lkil lkji lkjj lkjk lkjl lkki lkkj lkkk lkkl lkli lklj lklk lkll llii llij llik llil llji lljj lljk lljl llki llkj llkk llkl llli lllj lllk llll

Every individual character within the string has equal probability as well.

For example, given an output of aaab, this does not mean that a has more probability of occuring, rather a and b have equal probability of occurence.

Another example, given an output of aaabeeeee, this does not mean that e has the highest probability of occuring, rather a and b and e all have equal probability of occurence.

The input is guranteed to have at least two distinct characters.

edited yesterday

asked yesterday

Flog Edoc

413

New contributor

$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– DJMcMayhem♦
yesterday

$begingroup$
@EmbodimentofIgnorance Thanks, fixed.
$endgroup$
– Flog Edoc
yesterday

$begingroup$
Is the input guaranteed to have at least two distinct characters?
$endgroup$
– Neil
yesterday

1

$begingroup$
@Flog Edoc: (2) There is a real difference between 'will contain at least two distinct characters' and 'all characters will be distinct'. Which do you intend?
$endgroup$
– Chas Brown
23 hours ago

1

$begingroup$
do you mean given an *input* of aaab? Can we have examples that aren't isomorphic to each other?
$endgroup$
– Jo King
22 hours ago

|
show 9 more comments

Example: given the input abcd (or any combination thereof of the four characters a,b,c,d) , there is an equal chance of outputting:

a b c d aa ab ac ad ba bb bc bd ca cb cc cd da db dc dd aaa aab aac aad aba abb abc abd aca acb acc acd ada adb adc add baa bab bac bad bba bbb bbc bbd bca bcb bcc bcd bda bdb bdc bdd caa cab cac cad cba cbb cbc cbd cca ccb ccc ccd cda cdb cdc cdd daa dab dac dad dba dbb dbc dbd dca dcb dcc dcd dda ddb ddc ddd aaaa aaab aaac aaad aaba aabb aabc aabd aaca aacb aacc aacd aada aadb aadc aadd abaa abab abac abad abba abbb abbc abbd abca abcb abcc abcd abda abdb abdc abdd acaa acab acac acad acba acbb acbc acbd acca accb accc accd acda acdb acdc acdd adaa adab adac adad adba adbb adbc adbd adca adcb adcc adcd adda addb addc addd baaa baab baac baad baba babb babc babd baca bacb bacc bacd bada badb badc badd bbaa bbab bbac bbad bbba bbbb bbbc bbbd bbca bbcb bbcc bbcd bbda bbdb bbdc bbdd bcaa bcab bcac bcad bcba bcbb bcbc bcbd bcca bccb bccc bccd bcda bcdb bcdc bcdd bdaa bdab bdac bdad bdba bdbb bdbc bdbd bdca bdcb bdcc bdcd bdda bddb bddc bddd caaa caab caac caad caba cabb cabc cabd caca cacb cacc cacd cada cadb cadc cadd cbaa cbab cbac cbad cbba cbbb cbbc cbbd cbca cbcb cbcc cbcd cbda cbdb cbdc cbdd ccaa ccab ccac ccad ccba ccbb ccbc ccbd ccca cccb cccc cccd ccda ccdb ccdc ccdd cdaa cdab cdac cdad cdba cdbb cdbc cdbd cdca cdcb cdcc cdcd cdda cddb cddc cddd daaa daab daac daad daba dabb dabc dabd daca dacb dacc dacd dada dadb dadc dadd dbaa dbab dbac dbad dbba dbbb dbbc dbbd dbca dbcb dbcc dbcd dbda dbdb dbdc dbdd dcaa dcab dcac dcad dcba dcbb dcbc dcbd dcca dccb dccc dccd dcda dcdb dcdc dcdd ddaa ddab ddac ddad ddba ddbb ddbc ddbd ddca ddcb ddcc ddcd ddda dddb dddc dddd

Example: given the input efgh (or any combination thereof of the four characters e,f,g,h), there is an equal chance of outputting:

e f g h ee ef eg eh fe ff fg fh ge gf gg gh he hf hg hh eee eef eeg eeh efe eff efg efh ege egf egg egh ehe ehf ehg ehh fee fef feg feh ffe fff ffg ffh fge fgf fgg fgh fhe fhf fhg fhh gee gef geg geh gfe gff gfg gfh gge ggf ggg ggh ghe ghf ghg ghh hee hef heg heh hfe hff hfg hfh hge hgf hgg hgh hhe hhf hhg hhh eeee eeef eeeg eeeh eefe eeff eefg eefh eege eegf eegg eegh eehe eehf eehg eehh efee efef efeg efeh effe efff effg effh efge efgf efgg efgh efhe efhf efhg efhh egee egef egeg egeh egfe egff egfg egfh egge eggf eggg eggh eghe eghf eghg eghh ehee ehef eheg eheh ehfe ehff ehfg ehfh ehge ehgf ehgg ehgh ehhe ehhf ehhg ehhh feee feef feeg feeh fefe feff fefg fefh fege fegf fegg fegh fehe fehf fehg fehh ffee ffef ffeg ffeh fffe ffff fffg fffh ffge ffgf ffgg ffgh ffhe ffhf ffhg ffhh fgee fgef fgeg fgeh fgfe fgff fgfg fgfh fgge fggf fggg fggh fghe fghf fghg fghh fhee fhef fheg fheh fhfe fhff fhfg fhfh fhge fhgf fhgg fhgh fhhe fhhf fhhg fhhh geee geef geeg geeh gefe geff gefg gefh gege gegf gegg gegh gehe gehf gehg gehh gfee gfef gfeg gfeh gffe gfff gffg gffh gfge gfgf gfgg gfgh gfhe gfhf gfhg gfhh ggee ggef ggeg ggeh ggfe ggff ggfg ggfh ggge gggf gggg gggh gghe gghf gghg gghh ghee ghef gheg gheh ghfe ghff ghfg ghfh ghge ghgf ghgg ghgh ghhe ghhf ghhg ghhh heee heef heeg heeh hefe heff hefg hefh hege hegf hegg hegh hehe hehf hehg hehh hfee hfef hfeg hfeh hffe hfff hffg hffh hfge hfgf hfgg hfgh hfhe hfhf hfhg hfhh hgee hgef hgeg hgeh hgfe hgff hgfg hgfh hgge hggf hggg hggh hghe hghf hghg hghh hhee hhef hheg hheh hhfe hhff hhfg hhfh hhge hhgf hhgg hhgh hhhe hhhf hhhg hhhh

Example: given the input ijkl (or any combination thereof of the four characters i,j,k,l), there is an equal chance of outputting:

i j k l ii ij ik il ji jj jk jl ki kj kk kl li lj lk ll iii iij iik iil iji ijj ijk ijl iki ikj ikk ikl ili ilj ilk ill jii jij jik jil jji jjj jjk jjl jki jkj jkk jkl jli jlj jlk jll kii kij kik kil kji kjj kjk kjl kki kkj kkk kkl kli klj klk kll lii lij lik lil lji ljj ljk ljl lki lkj lkk lkl lli llj llk lll iiii iiij iiik iiil iiji iijj iijk iijl iiki iikj iikk iikl iili iilj iilk iill ijii ijij ijik ijil ijji ijjj ijjk ijjl ijki ijkj ijkk ijkl ijli ijlj ijlk ijll ikii ikij ikik ikil ikji ikjj ikjk ikjl ikki ikkj ikkk ikkl ikli iklj iklk ikll ilii ilij ilik ilil ilji iljj iljk iljl ilki ilkj ilkk ilkl illi illj illk illl jiii jiij jiik jiil jiji jijj jijk jijl jiki jikj jikk jikl jili jilj jilk jill jjii jjij jjik jjil jjji jjjj jjjk jjjl jjki jjkj jjkk jjkl jjli jjlj jjlk jjll jkii jkij jkik jkil jkji jkjj jkjk jkjl jkki jkkj jkkk jkkl jkli jklj jklk jkll jlii jlij jlik jlil jlji jljj jljk jljl jlki jlkj jlkk jlkl jlli jllj jllk jlll kiii kiij kiik kiil kiji kijj kijk kijl kiki kikj kikk kikl kili kilj kilk kill kjii kjij kjik kjil kjji kjjj kjjk kjjl kjki kjkj kjkk kjkl kjli kjlj kjlk kjll kkii kkij kkik kkil kkji kkjj kkjk kkjl kkki kkkj kkkk kkkl kkli kklj kklk kkll klii klij klik klil klji kljj kljk kljl klki klkj klkk klkl klli kllj kllk klll liii liij liik liil liji lijj lijk lijl liki likj likk likl lili lilj lilk lill ljii ljij ljik ljil ljji ljjj ljjk ljjl ljki ljkj ljkk ljkl ljli ljlj ljlk ljll lkii lkij lkik lkil lkji lkjj lkjk lkjl lkki lkkj lkkk lkkl lkli lklj lklk lkll llii llij llik llil llji lljj lljk lljl llki llkj llkk llkl llli lllj lllk llll

Every individual character within the string has equal probability as well.

For example, given an output of aaab, this does not mean that a has more probability of occuring, rather a and b have equal probability of occurence.

Another example, given an output of aaabeeeee, this does not mean that e has the highest probability of occuring, rather a and b and e all have equal probability of occurence.

The input is guranteed to have at least two distinct characters.

edited yesterday

asked yesterday

Flog Edoc

413

New contributor

Example: given the input abcd (or any combination thereof of the four characters a,b,c,d) , there is an equal chance of outputting:

a b c d aa ab ac ad ba bb bc bd ca cb cc cd da db dc dd aaa aab aac aad aba abb abc abd aca acb acc acd ada adb adc add baa bab bac bad bba bbb bbc bbd bca bcb bcc bcd bda bdb bdc bdd caa cab cac cad cba cbb cbc cbd cca ccb ccc ccd cda cdb cdc cdd daa dab dac dad dba dbb dbc dbd dca dcb dcc dcd dda ddb ddc ddd aaaa aaab aaac aaad aaba aabb aabc aabd aaca aacb aacc aacd aada aadb aadc aadd abaa abab abac abad abba abbb abbc abbd abca abcb abcc abcd abda abdb abdc abdd acaa acab acac acad acba acbb acbc acbd acca accb accc accd acda acdb acdc acdd adaa adab adac adad adba adbb adbc adbd adca adcb adcc adcd adda addb addc addd baaa baab baac baad baba babb babc babd baca bacb bacc bacd bada badb badc badd bbaa bbab bbac bbad bbba bbbb bbbc bbbd bbca bbcb bbcc bbcd bbda bbdb bbdc bbdd bcaa bcab bcac bcad bcba bcbb bcbc bcbd bcca bccb bccc bccd bcda bcdb bcdc bcdd bdaa bdab bdac bdad bdba bdbb bdbc bdbd bdca bdcb bdcc bdcd bdda bddb bddc bddd caaa caab caac caad caba cabb cabc cabd caca cacb cacc cacd cada cadb cadc cadd cbaa cbab cbac cbad cbba cbbb cbbc cbbd cbca cbcb cbcc cbcd cbda cbdb cbdc cbdd ccaa ccab ccac ccad ccba ccbb ccbc ccbd ccca cccb cccc cccd ccda ccdb ccdc ccdd cdaa cdab cdac cdad cdba cdbb cdbc cdbd cdca cdcb cdcc cdcd cdda cddb cddc cddd daaa daab daac daad daba dabb dabc dabd daca dacb dacc dacd dada dadb dadc dadd dbaa dbab dbac dbad dbba dbbb dbbc dbbd dbca dbcb dbcc dbcd dbda dbdb dbdc dbdd dcaa dcab dcac dcad dcba dcbb dcbc dcbd dcca dccb dccc dccd dcda dcdb dcdc dcdd ddaa ddab ddac ddad ddba ddbb ddbc ddbd ddca ddcb ddcc ddcd ddda dddb dddc dddd

Example: given the input efgh (or any combination thereof of the four characters e,f,g,h), there is an equal chance of outputting:

e f g h ee ef eg eh fe ff fg fh ge gf gg gh he hf hg hh eee eef eeg eeh efe eff efg efh ege egf egg egh ehe ehf ehg ehh fee fef feg feh ffe fff ffg ffh fge fgf fgg fgh fhe fhf fhg fhh gee gef geg geh gfe gff gfg gfh gge ggf ggg ggh ghe ghf ghg ghh hee hef heg heh hfe hff hfg hfh hge hgf hgg hgh hhe hhf hhg hhh eeee eeef eeeg eeeh eefe eeff eefg eefh eege eegf eegg eegh eehe eehf eehg eehh efee efef efeg efeh effe efff effg effh efge efgf efgg efgh efhe efhf efhg efhh egee egef egeg egeh egfe egff egfg egfh egge eggf eggg eggh eghe eghf eghg eghh ehee ehef eheg eheh ehfe ehff ehfg ehfh ehge ehgf ehgg ehgh ehhe ehhf ehhg ehhh feee feef feeg feeh fefe feff fefg fefh fege fegf fegg fegh fehe fehf fehg fehh ffee ffef ffeg ffeh fffe ffff fffg fffh ffge ffgf ffgg ffgh ffhe ffhf ffhg ffhh fgee fgef fgeg fgeh fgfe fgff fgfg fgfh fgge fggf fggg fggh fghe fghf fghg fghh fhee fhef fheg fheh fhfe fhff fhfg fhfh fhge fhgf fhgg fhgh fhhe fhhf fhhg fhhh geee geef geeg geeh gefe geff gefg gefh gege gegf gegg gegh gehe gehf gehg gehh gfee gfef gfeg gfeh gffe gfff gffg gffh gfge gfgf gfgg gfgh gfhe gfhf gfhg gfhh ggee ggef ggeg ggeh ggfe ggff ggfg ggfh ggge gggf gggg gggh gghe gghf gghg gghh ghee ghef gheg gheh ghfe ghff ghfg ghfh ghge ghgf ghgg ghgh ghhe ghhf ghhg ghhh heee heef heeg heeh hefe heff hefg hefh hege hegf hegg hegh hehe hehf hehg hehh hfee hfef hfeg hfeh hffe hfff hffg hffh hfge hfgf hfgg hfgh hfhe hfhf hfhg hfhh hgee hgef hgeg hgeh hgfe hgff hgfg hgfh hgge hggf hggg hggh hghe hghf hghg hghh hhee hhef hheg hheh hhfe hhff hhfg hhfh hhge hhgf hhgg hhgh hhhe hhhf hhhg hhhh

Example: given the input ijkl (or any combination thereof of the four characters i,j,k,l), there is an equal chance of outputting:

i j k l ii ij ik il ji jj jk jl ki kj kk kl li lj lk ll iii iij iik iil iji ijj ijk ijl iki ikj ikk ikl ili ilj ilk ill jii jij jik jil jji jjj jjk jjl jki jkj jkk jkl jli jlj jlk jll kii kij kik kil kji kjj kjk kjl kki kkj kkk kkl kli klj klk kll lii lij lik lil lji ljj ljk ljl lki lkj lkk lkl lli llj llk lll iiii iiij iiik iiil iiji iijj iijk iijl iiki iikj iikk iikl iili iilj iilk iill ijii ijij ijik ijil ijji ijjj ijjk ijjl ijki ijkj ijkk ijkl ijli ijlj ijlk ijll ikii ikij ikik ikil ikji ikjj ikjk ikjl ikki ikkj ikkk ikkl ikli iklj iklk ikll ilii ilij ilik ilil ilji iljj iljk iljl ilki ilkj ilkk ilkl illi illj illk illl jiii jiij jiik jiil jiji jijj jijk jijl jiki jikj jikk jikl jili jilj jilk jill jjii jjij jjik jjil jjji jjjj jjjk jjjl jjki jjkj jjkk jjkl jjli jjlj jjlk jjll jkii jkij jkik jkil jkji jkjj jkjk jkjl jkki jkkj jkkk jkkl jkli jklj jklk jkll jlii jlij jlik jlil jlji jljj jljk jljl jlki jlkj jlkk jlkl jlli jllj jllk jlll kiii kiij kiik kiil kiji kijj kijk kijl kiki kikj kikk kikl kili kilj kilk kill kjii kjij kjik kjil kjji kjjj kjjk kjjl kjki kjkj kjkk kjkl kjli kjlj kjlk kjll kkii kkij kkik kkil kkji kkjj kkjk kkjl kkki kkkj kkkk kkkl kkli kklj kklk kkll klii klij klik klil klji kljj kljk kljl klki klkj klkk klkl klli kllj kllk klll liii liij liik liil liji lijj lijk lijl liki likj likk likl lili lilj lilk lill ljii ljij ljik ljil ljji ljjj ljjk ljjl ljki ljkj ljkk ljkl ljli ljlj ljlk ljll lkii lkij lkik lkil lkji lkjj lkjk lkjl lkki lkkj lkkk lkkl lkli lklj lklk lkll llii llij llik llil llji lljj lljk lljl llki llkj llkk llkl llli lllj lllk llll

Every individual character within the string has equal probability as well.

For example, given an output of aaab, this does not mean that a has more probability of occuring, rather a and b have equal probability of occurence.

Another example, given an output of aaabeeeee, this does not mean that e has the highest probability of occuring, rather a and b and e all have equal probability of occurence.

The input is guranteed to have at least two distinct characters.

code-golf string random

edited yesterday

asked yesterday

Flog Edoc

413

New contributor

edited yesterday

asked yesterday

Flog Edoc

413

New contributor

edited yesterday

asked yesterday

Flog Edoc

413

New contributor

asked yesterday

Flog Edoc

413

asked yesterday

Flog Edoc

413

New contributor

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

$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– DJMcMayhem♦
yesterday

$begingroup$
@EmbodimentofIgnorance Thanks, fixed.
$endgroup$
– Flog Edoc
yesterday

$begingroup$
Is the input guaranteed to have at least two distinct characters?
$endgroup$
– Neil
yesterday

1

$begingroup$
@Flog Edoc: (2) There is a real difference between 'will contain at least two distinct characters' and 'all characters will be distinct'. Which do you intend?
$endgroup$
– Chas Brown
23 hours ago

1

$begingroup$
do you mean given an *input* of aaab? Can we have examples that aren't isomorphic to each other?
$endgroup$
– Jo King
22 hours ago

|
show 9 more comments

$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– DJMcMayhem♦
yesterday

$begingroup$
@EmbodimentofIgnorance Thanks, fixed.
$endgroup$
– Flog Edoc
yesterday

$begingroup$
Is the input guaranteed to have at least two distinct characters?
$endgroup$
– Neil
yesterday

1

$begingroup$
@Flog Edoc: (2) There is a real difference between 'will contain at least two distinct characters' and 'all characters will be distinct'. Which do you intend?
$endgroup$
– Chas Brown
23 hours ago

1

$begingroup$
do you mean given an *input* of aaab? Can we have examples that aren't isomorphic to each other?
$endgroup$
– Jo King
22 hours ago

Comments are not for extended discussion; this conversation has been moved to chat.

– DJMcMayhem♦
yesterday

@EmbodimentofIgnorance Thanks, fixed.

– Flog Edoc
yesterday

Is the input guaranteed to have at least two distinct characters?

– Neil
yesterday

@Flog Edoc: (2) There is a real difference between 'will contain at least two distinct characters' and 'all characters will be distinct'. Which do you intend?

– Chas Brown
23 hours ago

do you mean given an *input* of aaab? Can we have examples that aren't isomorphic to each other?

– Jo King
22 hours ago

|
show 9 more comments

15 Answers
15

active

oldest

votes

Jelly, 10 5 bytes

ṗJẎQX

A monadic Link accepting a list of characters which yields a list of characters.

Try it online!

How?

ṗJẎQX - Link list of characters    e.g.  aabc

 J    - range of length                  [1,2,3,4]

ṗ     - Cartesian power (vectorises)     [[a,a,b,c],[aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc],[aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc],[aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

  Ẏ   - tighten                          [a,a,b,c,aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

   Q  - de-duplicate                     [a,b,c,aa,ab,ac,ba,bb,bc,ca,cb,cc,aaa,aab,aac,aba,abb,abc,aca,acb,acc,baa,bab,bac,bba,bbb,bbc,bca,bcb,bcc,caa,cab,cac,cba,cbb,cbc,cca,ccb,ccc,aaaa,aaab,aaac,aaba,aabb,aabc,aaca,aacb,aacc,abaa,abab,abac,abba,abbb,abbc,abca,abcb,abcc,acaa,acab,acac,acba,acbb,acbc,acca,accb,accc,baaa,baab,baac,baba,babb,babc,baca,bacb,bacc,bbaa,bbab,bbac,bbba,bbbb,bbbc,bbca,bbcb,bbcc,bcaa,bcab,bcac,bcba,bcbb,bcbc,bcca,bccb,bccc,caaa,caab,caac,caba,cabb,cabc,caca,cacb,cacc,cbaa,cbab,cbac,cbba,cbbb,cbbc,cbca,cbcb,cbcc,ccaa,ccab,ccac,ccba,ccbb,ccbc,ccca,cccb,cccc]

    X - uniform random choice

edited yesterday

answered yesterday

Jonathan Allan

53.7k535173

$begingroup$
That fraction can be $frac1{2^{L+1}-2}$... ;P
$endgroup$
– Erik the Outgolfer
yesterday

add a comment |

05AB1E, 6 bytes

ā€ã˜ÙΩ

Try it online!

Explanation

ā       # push [1 ... len(input)]

 €ã     # apply repeated cartesian product on each and input

   ˜    # flatten

    Ù   # remove duplicates

     Ω  # pick random string

edited 20 hours ago

answered yesterday

Emigna

47.4k433144

add a comment |

Brachylog, 8 bytes

⊇ᶠbṛ;?ṛw

Try it online!

Prints output directly instead of constraining the output variable, because for some reason this actually fails without the w at the end. Originally this used ≠ and the ∈& backtracking hack I used to implement bogosort, but I realized that it would be far saner to just use b instead.

       w    Print

      ṛ     a random one of

     ?      the input

    ;       or

   ṛ        a random element of

 ᶠ          every

⊇           sublist of

            the input

  b         except the first one (which would be the input).

answered yesterday

Unrelated String

1,571312

add a comment |

Perl 6, 35 bytes

{set($_,[X~] $_ xx$_).pick}o*.comb

Try it online!

Explanation:

{                          }o*.comb  # Anonymous code block

        [ X~] $_ xx$_  # Find all combinations with repetition of the input

                      # And most of the prefixes

     $_,               # Add the singular characters

 set(                )       # Remove duplicates

                      .pick  # And pick a random one

edited 20 hours ago

answered yesterday

Jo King

26.5k364130

add a comment |

Python 2, 111 121 bytes

lambda s:choice(g(list(set(s)),len(s)))

from random import *

g=lambda s,n:s+(n>1and[c+k for k in g(s,n-1)for c in s]or)

Try it online!

Takes any iterable (list, set, string, etc.) of either distinct characters or non-distinct characters as input. Outputs a non-empty string with the required distribution.

g is a recursive function to generate a list of compliant strings.

edited 21 hours ago

answered yesterday

Chas Brown

5,1691523

add a comment |

Japt, 10 9 bytes

-1 byte from @Shaggy

M¬©Uª¯UÊö

M¬©Uª¯UÊö       Full Program

M¬              Random number between 1 and 0 (50% => 1, 50% => 0)            

  ©             If 1 then        

    U           Output the whole string        

     ª          Else

      ¯         Get a sub string from 0 to

         ö      A positive random number lower than

       UÊ       The length of the string

Try it online!

Japt, 9 bytes

Removes chars at random position. Done by @Shaggy

M¬©Uª®p2ö

Try it online!

edited yesterday

answered yesterday

Luis felipe De jesus Munoz

5,71821671

1

$begingroup$
Does this work for abcd --> ac?
$endgroup$
– AdmBorkBork
yesterday

$begingroup$
@AdmBorkBork No, it just returns a substring of the original. Since it is Removing a non-zero random number of characters from it. I assume it is valid
$endgroup$
– Luis felipe De jesus Munoz
yesterday

1

$begingroup$
9 bytes?
$endgroup$
– Shaggy
yesterday

1

$begingroup$
Or remove letters at random for the same byte count: tio.run/##y0osKPn/3/fQmkMrQw@tOrSuwOjwtv//lZzzU1IV3PNz0pQA
$endgroup$
– Shaggy
yesterday

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

|
show 1 more comment

Wolfram Language (Mathematica), 41 bytes

""<>#&@*RandomChoice@*Subsets@*Characters

Try it online!

answered 18 hours ago

attinat

4797

$begingroup$
Wha is this part? ""<>#&@
$endgroup$
– Jonah
7 hours ago

add a comment |

C (gcc), 68 65 bytes

f(s,i)char*s;{for(;i<strlen(s);i++)rand()&1?putc(s[i],stdout):0;}

Try it online!

edited yesterday

answered yesterday

Natural Number Guy

1516

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

1

$begingroup$
If the op changed the rules that's a pity.
$endgroup$
– Natural Number Guy
16 hours ago

add a comment |

Java, 106 bytes

void k(String a){a.chars().filter(value->Math.random()>.5).forEach(value->System.out.print((char)value));}

answered yesterday

Ilya Gazman

479313

$begingroup$
Given a string like abc, there is no chance of the string cba appearing. Next, you can turn the function into a lambda, and you do not need to name all the lambda variables value, one letter names suffice. Something like: a->a.chars().filter(value->Math.random()>.5) since you do not need to print your output, though you still need to fix the first problem I listed
$endgroup$
– Embodiment of Ignorance
yesterday

add a comment |

R, 72 bytes

sample(t<-unique(s<-scan(,"")),sample(n<-length(s),p=length(t)^(1:n)),T)

Try it online!

Input and output are vectors of characters. All possible outputs have equal probability.

Explanation (ungolfed version):

s<-scan(,"")                         # takes input as vector of characters

n<-length(s)

t<-unique(s)

sample(t,                           # sample uniformly at random from the list of unique characters

       sample(n,p=length(t)^(1:n)), # with length k in 1..n chosen randomly such that P[k] is proportional to length(t)^k

       T)                           # the sampling is with replacement

edited 20 hours ago

answered 21 hours ago

Robin Ryder

5317

add a comment |

Charcoal, 34 bytes

≔Φθ⁼κ⌕θιη≔⊕‽ΣＥθＸＬη⊕κζＷζ«§ηζ≔÷⊖ζＬηζ

Try it online! Link is to verbose version of code. Explanation:

≔Φθ⁼κ⌕θιη

Extract the unique characters of the input. Let's call the number of unique characters n.

≔⊕‽ΣＥθＸＬη⊕κζ

Calculate the number of combinations for each possible length and take the sum. Then, pick a random number between 1 and this number (inclusive). This ensures that all combinations are equally likely (within the accuracy of the random number generator).

Ｗζ«§ηζ≔÷⊖ζＬηζ

Convert the number into bijective base n, using the unique characters as the digits.

answered 18 hours ago

Neil

82.6k745179

add a comment |

MATLAB / Octave, 110 bytes

Choose a random permutation of a subset of the input letters (with repetitions), whose random length is based on the probability of generating a word with that length.

@(a)a(randi(numel(a),[find(cumsum(numel(a).^[1:numel(a)])>=randi(sum(numel(a).^[1:numel(a)])),1,'first'),1]));

Try it online!

answered 16 hours ago

PieCot

97959

add a comment |

T-SQL, 222 bytes

This creates all combinations of each unique character with recursive sql, then picks a random row from the distinct combinations.

DECLARE @ varchar(max)='T-SQL';



WITH C as(SELECT DISTINCT substring(@,number+1,1)x

FROM spt_values

WHERE'P'=type and len(@)>number),D

as(SELECT x y

FROM c UNION ALL

SELECT y+x

FROM C JOIN D

ON len(y)<len(@))SELECT top 1*FROM D

GROUP BY y

ORDER BY newid()

Note the online version will always give the same result unlike MS-SQL Studio Management. This is because newid() always returns the same value in the online testing. This should work in Studio Management.

Try it online ungolfed version

edited 9 hours ago

answered 19 hours ago

t-clausen.dk

2,074314

add a comment |

Python 2, 124 bytes

lambda s:g(list(set(s)),len(s))

from random import*

g=lambda s,n:choice(s)+(random()*~-len(s)**(n)>~-len(s)and g(s,n-1)or'')

Try it online!

A different approach: instead of first constructing a list of all compliant strings and choosing one at random, this approach randomly decides to continue extending the string or stopping.

This again has to deal with the exacting input requirements, at a cost of 31 bytes; but the function of interest g takes a list s of characters to be used, and an integer n which is the maximum length of the returned string.

As an example, consider the set of characters ['a','b'] and suppose we want to generate random strings of length 1, 2, or 3.

Then if we randomly choose 'a' as the starting character, the possible strings which could be returned are:

a

 aa

  aaa

  aab

 ab

  aba

  abb

shown above in 'tree' form, which is a total of 1 + 2 + 2*2 = 2^0 + 2^1 + 2^2 = 7 strings. So if we generate a string a; then 1/7 of the time we should stop and return 'a', and 6/7 of the time we should add further to the string - then the distribution will be uniform in the way desired.

More generally, if n is maximum length of the string and x is the number of characters in the set, then the number of strings starting with some given character is going to be:

$$h(n)=sum_{i=0}^{n-1} x^i$$
$$h(n)=1+x+x^2+x^3...+x^{n-1}$$

For x>1 (guaranteed by OP's rule 'at least two distinct...'), we have:

$$h(n)(x-1)=(1+x+x^2+x^3...+x^{n-1})(x-1)=x^n-1$$
$$h(n)=frac{x^n-1}{x-1}$$

So to decide whether we should continue extending our string, let p be a random number in [0,1); then we should recurse only if any of these equivalent statements are true:

$$p>frac{1}{h(n)}$$
$$p>frac{x-1}{x^n-1}$$
$$p(x^n-1)>x-1$$

of which g is the golfed implementation.

answered 8 hours ago

Chas Brown

5,1691523

add a comment |

J, 27 bytes

[:(?@#{])@;[:,@{&.>#<@#"{<

Try it online!

standard formatting

[: (?@# { ])@; [: ,@{&.> # <@#"1 _ <

explanation

Eg, for the string 'abc' we first use #<@#"{< to create:

┌─────┬─────────┬─────────────┐

│┌───┐│┌───┬───┐│┌───┬───┬───┐│

││abc│││abc│abc│││abc│abc│abc││

│└───┘│└───┴───┘│└───┴───┴───┘│

└─────┴─────────┴─────────────┘

We then take the cartesian product of each of these {, flatten the results, and finally remove the outer boxing ;.

(?@#{])@ takes a random result from that list.

Note: TIO will return the same result every time, but in a normal J REPL it will be random.

edited 7 hours ago

answered 9 hours ago

Jonah

2,5711017

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "200"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

Flog Edoc is a new contributor. Be nice, and check out our Code of Conduct.

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f182735%2fto-string-or-not-to-string%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

15 Answers
15

active

oldest

votes

15 Answers
15

active

oldest

votes

Jelly, 10 5 bytes

ṗJẎQX

A monadic Link accepting a list of characters which yields a list of characters.

Try it online!

How?

ṗJẎQX - Link list of characters    e.g.  aabc

 J    - range of length                  [1,2,3,4]

ṗ     - Cartesian power (vectorises)     [[a,a,b,c],[aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc],[aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc],[aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

  Ẏ   - tighten                          [a,a,b,c,aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

   Q  - de-duplicate                     [a,b,c,aa,ab,ac,ba,bb,bc,ca,cb,cc,aaa,aab,aac,aba,abb,abc,aca,acb,acc,baa,bab,bac,bba,bbb,bbc,bca,bcb,bcc,caa,cab,cac,cba,cbb,cbc,cca,ccb,ccc,aaaa,aaab,aaac,aaba,aabb,aabc,aaca,aacb,aacc,abaa,abab,abac,abba,abbb,abbc,abca,abcb,abcc,acaa,acab,acac,acba,acbb,acbc,acca,accb,accc,baaa,baab,baac,baba,babb,babc,baca,bacb,bacc,bbaa,bbab,bbac,bbba,bbbb,bbbc,bbca,bbcb,bbcc,bcaa,bcab,bcac,bcba,bcbb,bcbc,bcca,bccb,bccc,caaa,caab,caac,caba,cabb,cabc,caca,cacb,cacc,cbaa,cbab,cbac,cbba,cbbb,cbbc,cbca,cbcb,cbcc,ccaa,ccab,ccac,ccba,ccbb,ccbc,ccca,cccb,cccc]

    X - uniform random choice

edited yesterday

answered yesterday

Jonathan Allan

53.7k535173

$begingroup$
That fraction can be $frac1{2^{L+1}-2}$... ;P
$endgroup$
– Erik the Outgolfer
yesterday

add a comment |

Jelly, 10 5 bytes

ṗJẎQX

A monadic Link accepting a list of characters which yields a list of characters.

Try it online!

How?

ṗJẎQX - Link list of characters    e.g.  aabc

 J    - range of length                  [1,2,3,4]

ṗ     - Cartesian power (vectorises)     [[a,a,b,c],[aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc],[aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc],[aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

  Ẏ   - tighten                          [a,a,b,c,aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

   Q  - de-duplicate                     [a,b,c,aa,ab,ac,ba,bb,bc,ca,cb,cc,aaa,aab,aac,aba,abb,abc,aca,acb,acc,baa,bab,bac,bba,bbb,bbc,bca,bcb,bcc,caa,cab,cac,cba,cbb,cbc,cca,ccb,ccc,aaaa,aaab,aaac,aaba,aabb,aabc,aaca,aacb,aacc,abaa,abab,abac,abba,abbb,abbc,abca,abcb,abcc,acaa,acab,acac,acba,acbb,acbc,acca,accb,accc,baaa,baab,baac,baba,babb,babc,baca,bacb,bacc,bbaa,bbab,bbac,bbba,bbbb,bbbc,bbca,bbcb,bbcc,bcaa,bcab,bcac,bcba,bcbb,bcbc,bcca,bccb,bccc,caaa,caab,caac,caba,cabb,cabc,caca,cacb,cacc,cbaa,cbab,cbac,cbba,cbbb,cbbc,cbca,cbcb,cbcc,ccaa,ccab,ccac,ccba,ccbb,ccbc,ccca,cccb,cccc]

    X - uniform random choice

edited yesterday

answered yesterday

Jonathan Allan

53.7k535173

$begingroup$
That fraction can be $frac1{2^{L+1}-2}$... ;P
$endgroup$
– Erik the Outgolfer
yesterday

add a comment |

Jelly, 10 5 bytes

ṗJẎQX

A monadic Link accepting a list of characters which yields a list of characters.

Try it online!

How?

ṗJẎQX - Link list of characters    e.g.  aabc

 J    - range of length                  [1,2,3,4]

ṗ     - Cartesian power (vectorises)     [[a,a,b,c],[aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc],[aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc],[aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

  Ẏ   - tighten                          [a,a,b,c,aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

   Q  - de-duplicate                     [a,b,c,aa,ab,ac,ba,bb,bc,ca,cb,cc,aaa,aab,aac,aba,abb,abc,aca,acb,acc,baa,bab,bac,bba,bbb,bbc,bca,bcb,bcc,caa,cab,cac,cba,cbb,cbc,cca,ccb,ccc,aaaa,aaab,aaac,aaba,aabb,aabc,aaca,aacb,aacc,abaa,abab,abac,abba,abbb,abbc,abca,abcb,abcc,acaa,acab,acac,acba,acbb,acbc,acca,accb,accc,baaa,baab,baac,baba,babb,babc,baca,bacb,bacc,bbaa,bbab,bbac,bbba,bbbb,bbbc,bbca,bbcb,bbcc,bcaa,bcab,bcac,bcba,bcbb,bcbc,bcca,bccb,bccc,caaa,caab,caac,caba,cabb,cabc,caca,cacb,cacc,cbaa,cbab,cbac,cbba,cbbb,cbbc,cbca,cbcb,cbcc,ccaa,ccab,ccac,ccba,ccbb,ccbc,ccca,cccb,cccc]

    X - uniform random choice

edited yesterday

answered yesterday

Jonathan Allan

53.7k535173

Jelly, 10 5 bytes

ṗJẎQX

A monadic Link accepting a list of characters which yields a list of characters.

Try it online!

How?

ṗJẎQX - Link list of characters    e.g.  aabc

 J    - range of length                  [1,2,3,4]

ṗ     - Cartesian power (vectorises)     [[a,a,b,c],[aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc],[aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc],[aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

  Ẏ   - tighten                          [a,a,b,c,aa,aa,ab,ac,aa,aa,ab,ac,ba,ba,bb,bc,ca,ca,cb,cc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,aaa,aaa,aab,aac,aaa,aaa,aab,aac,aba,aba,abb,abc,aca,aca,acb,acc,baa,baa,bab,bac,baa,baa,bab,bac,bba,bba,bbb,bbc,bca,bca,bcb,bcc,caa,caa,cab,cac,caa,caa,cab,cac,cba,cba,cbb,cbc,cca,cca,ccb,ccc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,aaaa,aaaa,aaab,aaac,aaaa,aaaa,aaab,aaac,aaba,aaba,aabb,aabc,aaca,aaca,aacb,aacc,abaa,abaa,abab,abac,abaa,abaa,abab,abac,abba,abba,abbb,abbc,abca,abca,abcb,abcc,acaa,acaa,acab,acac,acaa,acaa,acab,acac,acba,acba,acbb,acbc,acca,acca,accb,accc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,baaa,baaa,baab,baac,baaa,baaa,baab,baac,baba,baba,babb,babc,baca,baca,bacb,bacc,bbaa,bbaa,bbab,bbac,bbaa,bbaa,bbab,bbac,bbba,bbba,bbbb,bbbc,bbca,bbca,bbcb,bbcc,bcaa,bcaa,bcab,bcac,bcaa,bcaa,bcab,bcac,bcba,bcba,bcbb,bcbc,bcca,bcca,bccb,bccc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,caaa,caaa,caab,caac,caaa,caaa,caab,caac,caba,caba,cabb,cabc,caca,caca,cacb,cacc,cbaa,cbaa,cbab,cbac,cbaa,cbaa,cbab,cbac,cbba,cbba,cbbb,cbbc,cbca,cbca,cbcb,cbcc,ccaa,ccaa,ccab,ccac,ccaa,ccaa,ccab,ccac,ccba,ccba,ccbb,ccbc,ccca,ccca,cccb,cccc]

   Q  - de-duplicate                     [a,b,c,aa,ab,ac,ba,bb,bc,ca,cb,cc,aaa,aab,aac,aba,abb,abc,aca,acb,acc,baa,bab,bac,bba,bbb,bbc,bca,bcb,bcc,caa,cab,cac,cba,cbb,cbc,cca,ccb,ccc,aaaa,aaab,aaac,aaba,aabb,aabc,aaca,aacb,aacc,abaa,abab,abac,abba,abbb,abbc,abca,abcb,abcc,acaa,acab,acac,acba,acbb,acbc,acca,accb,accc,baaa,baab,baac,baba,babb,babc,baca,bacb,bacc,bbaa,bbab,bbac,bbba,bbbb,bbbc,bbca,bbcb,bbcc,bcaa,bcab,bcac,bcba,bcbb,bcbc,bcca,bccb,bccc,caaa,caab,caac,caba,cabb,cabc,caca,cacb,cacc,cbaa,cbab,cbac,cbba,cbbb,cbbc,cbca,cbcb,cbcc,ccaa,ccab,ccac,ccba,ccbb,ccbc,ccca,cccb,cccc]

    X - uniform random choice

edited yesterday

answered yesterday

Jonathan Allan

53.7k535173

edited yesterday

answered yesterday

Jonathan Allan

53.7k535173

answered yesterday

Jonathan Allan

53.7k535173

answered yesterday

Jonathan Allan

53.7k535173

$begingroup$
That fraction can be $frac1{2^{L+1}-2}$... ;P
$endgroup$
– Erik the Outgolfer
yesterday

add a comment |

$begingroup$
That fraction can be $frac1{2^{L+1}-2}$... ;P
$endgroup$
– Erik the Outgolfer
yesterday

That fraction can be $frac1{2^{L+1}-2}$... ;P

– Erik the Outgolfer
yesterday

add a comment |

05AB1E, 6 bytes

ā€ã˜ÙΩ

Try it online!

Explanation

ā       # push [1 ... len(input)]

 €ã     # apply repeated cartesian product on each and input

   ˜    # flatten

    Ù   # remove duplicates

     Ω  # pick random string

edited 20 hours ago

answered yesterday

Emigna

47.4k433144

add a comment |

05AB1E, 6 bytes

ā€ã˜ÙΩ

Try it online!

Explanation

ā       # push [1 ... len(input)]

 €ã     # apply repeated cartesian product on each and input

   ˜    # flatten

    Ù   # remove duplicates

     Ω  # pick random string

edited 20 hours ago

answered yesterday

Emigna

47.4k433144

add a comment |

05AB1E, 6 bytes

ā€ã˜ÙΩ

Try it online!

Explanation

ā       # push [1 ... len(input)]

 €ã     # apply repeated cartesian product on each and input

   ˜    # flatten

    Ù   # remove duplicates

     Ω  # pick random string

edited 20 hours ago

answered yesterday

Emigna

47.4k433144

05AB1E, 6 bytes

ā€ã˜ÙΩ

Try it online!

Explanation

ā       # push [1 ... len(input)]

 €ã     # apply repeated cartesian product on each and input

   ˜    # flatten

    Ù   # remove duplicates

     Ω  # pick random string

edited 20 hours ago

answered yesterday

Emigna

47.4k433144

edited 20 hours ago

answered yesterday

Emigna

47.4k433144

answered yesterday

Emigna

47.4k433144

answered yesterday

Emigna

47.4k433144

add a comment |

Brachylog, 8 bytes

⊇ᶠbṛ;?ṛw

Try it online!

       w    Print

      ṛ     a random one of

     ?      the input

    ;       or

   ṛ        a random element of

 ᶠ          every

⊇           sublist of

            the input

  b         except the first one (which would be the input).

answered yesterday

Unrelated String

1,571312

add a comment |

Brachylog, 8 bytes

⊇ᶠbṛ;?ṛw

Try it online!

       w    Print

      ṛ     a random one of

     ?      the input

    ;       or

   ṛ        a random element of

 ᶠ          every

⊇           sublist of

            the input

  b         except the first one (which would be the input).

answered yesterday

Unrelated String

1,571312

add a comment |

Brachylog, 8 bytes

⊇ᶠbṛ;?ṛw

Try it online!

       w    Print

      ṛ     a random one of

     ?      the input

    ;       or

   ṛ        a random element of

 ᶠ          every

⊇           sublist of

            the input

  b         except the first one (which would be the input).

answered yesterday

Unrelated String

1,571312

Brachylog, 8 bytes

⊇ᶠbṛ;?ṛw

Try it online!

       w    Print

      ṛ     a random one of

     ?      the input

    ;       or

   ṛ        a random element of

 ᶠ          every

⊇           sublist of

            the input

  b         except the first one (which would be the input).

answered yesterday

Unrelated String

1,571312

answered yesterday

Unrelated String

1,571312

answered yesterday

Unrelated String

1,571312

answered yesterday

Unrelated String

1,571312

add a comment |

Perl 6, 35 bytes

{set($_,[X~] $_ xx$_).pick}o*.comb

Try it online!

Explanation:

{                          }o*.comb  # Anonymous code block

        [ X~] $_ xx$_  # Find all combinations with repetition of the input

                      # And most of the prefixes

     $_,               # Add the singular characters

 set(                )       # Remove duplicates

                      .pick  # And pick a random one

edited 20 hours ago

answered yesterday

Jo King

26.5k364130

add a comment |

Perl 6, 35 bytes

{set($_,[X~] $_ xx$_).pick}o*.comb

Try it online!

Explanation:

{                          }o*.comb  # Anonymous code block

        [ X~] $_ xx$_  # Find all combinations with repetition of the input

                      # And most of the prefixes

     $_,               # Add the singular characters

 set(                )       # Remove duplicates

                      .pick  # And pick a random one

edited 20 hours ago

answered yesterday

Jo King

26.5k364130

add a comment |

Perl 6, 35 bytes

{set($_,[X~] $_ xx$_).pick}o*.comb

Try it online!

Explanation:

{                          }o*.comb  # Anonymous code block

        [ X~] $_ xx$_  # Find all combinations with repetition of the input

                      # And most of the prefixes

     $_,               # Add the singular characters

 set(                )       # Remove duplicates

                      .pick  # And pick a random one

edited 20 hours ago

answered yesterday

Jo King

26.5k364130

Perl 6, 35 bytes

{set($_,[X~] $_ xx$_).pick}o*.comb

Try it online!

Explanation:

{                          }o*.comb  # Anonymous code block

        [ X~] $_ xx$_  # Find all combinations with repetition of the input

                      # And most of the prefixes

     $_,               # Add the singular characters

 set(                )       # Remove duplicates

                      .pick  # And pick a random one

edited 20 hours ago

answered yesterday

Jo King

26.5k364130

edited 20 hours ago

answered yesterday

Jo King

26.5k364130

answered yesterday

Jo King

26.5k364130

answered yesterday

Jo King

26.5k364130

add a comment |

Python 2, 111 121 bytes

lambda s:choice(g(list(set(s)),len(s)))

from random import *

g=lambda s,n:s+(n>1and[c+k for k in g(s,n-1)for c in s]or)

Try it online!

Takes any iterable (list, set, string, etc.) of either distinct characters or non-distinct characters as input. Outputs a non-empty string with the required distribution.

g is a recursive function to generate a list of compliant strings.

edited 21 hours ago

answered yesterday

Chas Brown

5,1691523

add a comment |

Python 2, 111 121 bytes

lambda s:choice(g(list(set(s)),len(s)))

from random import *

g=lambda s,n:s+(n>1and[c+k for k in g(s,n-1)for c in s]or)

Try it online!

Takes any iterable (list, set, string, etc.) of either distinct characters or non-distinct characters as input. Outputs a non-empty string with the required distribution.

g is a recursive function to generate a list of compliant strings.

edited 21 hours ago

answered yesterday

Chas Brown

5,1691523

add a comment |

Python 2, 111 121 bytes

lambda s:choice(g(list(set(s)),len(s)))

from random import *

g=lambda s,n:s+(n>1and[c+k for k in g(s,n-1)for c in s]or)

Try it online!

Takes any iterable (list, set, string, etc.) of either distinct characters or non-distinct characters as input. Outputs a non-empty string with the required distribution.

g is a recursive function to generate a list of compliant strings.

edited 21 hours ago

answered yesterday

Chas Brown

5,1691523

Python 2, 111 121 bytes

lambda s:choice(g(list(set(s)),len(s)))

from random import *

g=lambda s,n:s+(n>1and[c+k for k in g(s,n-1)for c in s]or)

Try it online!

Takes any iterable (list, set, string, etc.) of either distinct characters or non-distinct characters as input. Outputs a non-empty string with the required distribution.

g is a recursive function to generate a list of compliant strings.

edited 21 hours ago

answered yesterday

Chas Brown

5,1691523

edited 21 hours ago

answered yesterday

Chas Brown

5,1691523

answered yesterday

Chas Brown

5,1691523

answered yesterday

Chas Brown

5,1691523

add a comment |

Japt, 10 9 bytes

-1 byte from @Shaggy

M¬©Uª¯UÊö

M¬©Uª¯UÊö       Full Program

M¬              Random number between 1 and 0 (50% => 1, 50% => 0)            

  ©             If 1 then        

    U           Output the whole string        

     ª          Else

      ¯         Get a sub string from 0 to

         ö      A positive random number lower than

       UÊ       The length of the string

Try it online!

Japt, 9 bytes

Removes chars at random position. Done by @Shaggy

M¬©Uª®p2ö

Try it online!

edited yesterday

answered yesterday

Luis felipe De jesus Munoz

5,71821671

1

$begingroup$
Does this work for abcd --> ac?
$endgroup$
– AdmBorkBork
yesterday

$begingroup$
@AdmBorkBork No, it just returns a substring of the original. Since it is Removing a non-zero random number of characters from it. I assume it is valid
$endgroup$
– Luis felipe De jesus Munoz
yesterday

1

$begingroup$
9 bytes?
$endgroup$
– Shaggy
yesterday

1

$begingroup$
Or remove letters at random for the same byte count: tio.run/##y0osKPn/3/fQmkMrQw@tOrSuwOjwtv//lZzzU1IV3PNz0pQA
$endgroup$
– Shaggy
yesterday

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

|
show 1 more comment

Japt, 10 9 bytes

-1 byte from @Shaggy

M¬©Uª¯UÊö

M¬©Uª¯UÊö       Full Program

M¬              Random number between 1 and 0 (50% => 1, 50% => 0)            

  ©             If 1 then        

    U           Output the whole string        

     ª          Else

      ¯         Get a sub string from 0 to

         ö      A positive random number lower than

       UÊ       The length of the string

Try it online!

Japt, 9 bytes

Removes chars at random position. Done by @Shaggy

M¬©Uª®p2ö

Try it online!

edited yesterday

answered yesterday

Luis felipe De jesus Munoz

5,71821671

1

$begingroup$
Does this work for abcd --> ac?
$endgroup$
– AdmBorkBork
yesterday

$begingroup$
@AdmBorkBork No, it just returns a substring of the original. Since it is Removing a non-zero random number of characters from it. I assume it is valid
$endgroup$
– Luis felipe De jesus Munoz
yesterday

1

$begingroup$
9 bytes?
$endgroup$
– Shaggy
yesterday

1

$begingroup$
Or remove letters at random for the same byte count: tio.run/##y0osKPn/3/fQmkMrQw@tOrSuwOjwtv//lZzzU1IV3PNz0pQA
$endgroup$
– Shaggy
yesterday

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

|
show 1 more comment

Japt, 10 9 bytes

-1 byte from @Shaggy

M¬©Uª¯UÊö

M¬©Uª¯UÊö       Full Program

M¬              Random number between 1 and 0 (50% => 1, 50% => 0)            

  ©             If 1 then        

    U           Output the whole string        

     ª          Else

      ¯         Get a sub string from 0 to

         ö      A positive random number lower than

       UÊ       The length of the string

Try it online!

Japt, 9 bytes

Removes chars at random position. Done by @Shaggy

M¬©Uª®p2ö

Try it online!

edited yesterday

answered yesterday

Luis felipe De jesus Munoz

5,71821671

Japt, 10 9 bytes

-1 byte from @Shaggy

M¬©Uª¯UÊö

M¬©Uª¯UÊö       Full Program

M¬              Random number between 1 and 0 (50% => 1, 50% => 0)            

  ©             If 1 then        

    U           Output the whole string        

     ª          Else

      ¯         Get a sub string from 0 to

         ö      A positive random number lower than

       UÊ       The length of the string

Try it online!

Japt, 9 bytes

Removes chars at random position. Done by @Shaggy

M¬©Uª®p2ö

Try it online!

edited yesterday

answered yesterday

Luis felipe De jesus Munoz

5,71821671

edited yesterday

answered yesterday

Luis felipe De jesus Munoz

5,71821671

answered yesterday

Luis felipe De jesus Munoz

5,71821671

answered yesterday

Luis felipe De jesus Munoz

5,71821671

1

$begingroup$
Does this work for abcd --> ac?
$endgroup$
– AdmBorkBork
yesterday

$begingroup$
@AdmBorkBork No, it just returns a substring of the original. Since it is Removing a non-zero random number of characters from it. I assume it is valid
$endgroup$
– Luis felipe De jesus Munoz
yesterday

1

$begingroup$
9 bytes?
$endgroup$
– Shaggy
yesterday

1

$begingroup$
Or remove letters at random for the same byte count: tio.run/##y0osKPn/3/fQmkMrQw@tOrSuwOjwtv//lZzzU1IV3PNz0pQA
$endgroup$
– Shaggy
yesterday

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

|
show 1 more comment

1

$begingroup$
Does this work for abcd --> ac?
$endgroup$
– AdmBorkBork
yesterday

$begingroup$
@AdmBorkBork No, it just returns a substring of the original. Since it is Removing a non-zero random number of characters from it. I assume it is valid
$endgroup$
– Luis felipe De jesus Munoz
yesterday

1

$begingroup$
9 bytes?
$endgroup$
– Shaggy
yesterday

1

$begingroup$
Or remove letters at random for the same byte count: tio.run/##y0osKPn/3/fQmkMrQw@tOrSuwOjwtv//lZzzU1IV3PNz0pQA
$endgroup$
– Shaggy
yesterday

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

Does this work for abcd --> ac?

– AdmBorkBork
yesterday

@AdmBorkBork No, it just returns a substring of the original. Since it is Removing a non-zero random number of characters from it. I assume it is valid

– Luis felipe De jesus Munoz
yesterday

9 bytes?

– Shaggy
yesterday

Or remove letters at random for the same byte count: tio.run/##y0osKPn/3/fQmkMrQw@tOrSuwOjwtv//lZzzU1IV3PNz0pQA

– Shaggy
yesterday

I don't think this is correct anymore

– Jo King
yesterday

|
show 1 more comment

Wolfram Language (Mathematica), 41 bytes

""<>#&@*RandomChoice@*Subsets@*Characters

Try it online!

answered 18 hours ago

attinat

4797

$begingroup$
Wha is this part? ""<>#&@
$endgroup$
– Jonah
7 hours ago

add a comment |

Wolfram Language (Mathematica), 41 bytes

""<>#&@*RandomChoice@*Subsets@*Characters

Try it online!

answered 18 hours ago

attinat

4797

$begingroup$
Wha is this part? ""<>#&@
$endgroup$
– Jonah
7 hours ago

add a comment |

Wolfram Language (Mathematica), 41 bytes

""<>#&@*RandomChoice@*Subsets@*Characters

Try it online!

answered 18 hours ago

attinat

4797

Wolfram Language (Mathematica), 41 bytes

""<>#&@*RandomChoice@*Subsets@*Characters

Try it online!

answered 18 hours ago

attinat

4797

answered 18 hours ago

attinat

4797

answered 18 hours ago

attinat

4797

answered 18 hours ago

attinat

4797

$begingroup$
Wha is this part? ""<>#&@
$endgroup$
– Jonah
7 hours ago

add a comment |

$begingroup$
Wha is this part? ""<>#&@
$endgroup$
– Jonah
7 hours ago

Wha is this part? ""<>#&@

– Jonah
7 hours ago

add a comment |

C (gcc), 68 65 bytes

f(s,i)char*s;{for(;i<strlen(s);i++)rand()&1?putc(s[i],stdout):0;}

Try it online!

edited yesterday

answered yesterday

Natural Number Guy

1516

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

1

$begingroup$
If the op changed the rules that's a pity.
$endgroup$
– Natural Number Guy
16 hours ago

add a comment |

C (gcc), 68 65 bytes

f(s,i)char*s;{for(;i<strlen(s);i++)rand()&1?putc(s[i],stdout):0;}

Try it online!

edited yesterday

answered yesterday

Natural Number Guy

1516

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

1

$begingroup$
If the op changed the rules that's a pity.
$endgroup$
– Natural Number Guy
16 hours ago

add a comment |

C (gcc), 68 65 bytes

f(s,i)char*s;{for(;i<strlen(s);i++)rand()&1?putc(s[i],stdout):0;}

Try it online!

edited yesterday

answered yesterday

Natural Number Guy

1516

C (gcc), 68 65 bytes

f(s,i)char*s;{for(;i<strlen(s);i++)rand()&1?putc(s[i],stdout):0;}

Try it online!

edited yesterday

answered yesterday

Natural Number Guy

1516

edited yesterday

answered yesterday

Natural Number Guy

1516

answered yesterday

Natural Number Guy

1516

answered yesterday

Natural Number Guy

1516

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

1

$begingroup$
If the op changed the rules that's a pity.
$endgroup$
– Natural Number Guy
16 hours ago

add a comment |

$begingroup$
I don't think this is correct anymore
$endgroup$
– Jo King
yesterday

1

$begingroup$
If the op changed the rules that's a pity.
$endgroup$
– Natural Number Guy
16 hours ago

I don't think this is correct anymore

– Jo King
yesterday

If the op changed the rules that's a pity.

– Natural Number Guy
16 hours ago

add a comment |

Java, 106 bytes

void k(String a){a.chars().filter(value->Math.random()>.5).forEach(value->System.out.print((char)value));}

answered yesterday

Ilya Gazman

479313

$begingroup$
Given a string like abc, there is no chance of the string cba appearing. Next, you can turn the function into a lambda, and you do not need to name all the lambda variables value, one letter names suffice. Something like: a->a.chars().filter(value->Math.random()>.5) since you do not need to print your output, though you still need to fix the first problem I listed
$endgroup$
– Embodiment of Ignorance
yesterday

add a comment |

Java, 106 bytes

void k(String a){a.chars().filter(value->Math.random()>.5).forEach(value->System.out.print((char)value));}

answered yesterday

Ilya Gazman

479313

$begingroup$
Given a string like abc, there is no chance of the string cba appearing. Next, you can turn the function into a lambda, and you do not need to name all the lambda variables value, one letter names suffice. Something like: a->a.chars().filter(value->Math.random()>.5) since you do not need to print your output, though you still need to fix the first problem I listed
$endgroup$
– Embodiment of Ignorance
yesterday

add a comment |

Java, 106 bytes

void k(String a){a.chars().filter(value->Math.random()>.5).forEach(value->System.out.print((char)value));}

answered yesterday

Ilya Gazman

479313

Java, 106 bytes

void k(String a){a.chars().filter(value->Math.random()>.5).forEach(value->System.out.print((char)value));}

answered yesterday

Ilya Gazman

479313

answered yesterday

Ilya Gazman

479313

answered yesterday

Ilya Gazman

479313

answered yesterday

Ilya Gazman

479313

$begingroup$
Given a string like abc, there is no chance of the string cba appearing. Next, you can turn the function into a lambda, and you do not need to name all the lambda variables value, one letter names suffice. Something like: a->a.chars().filter(value->Math.random()>.5) since you do not need to print your output, though you still need to fix the first problem I listed
$endgroup$
– Embodiment of Ignorance
yesterday

add a comment |

$begingroup$
Given a string like abc, there is no chance of the string cba appearing. Next, you can turn the function into a lambda, and you do not need to name all the lambda variables value, one letter names suffice. Something like: a->a.chars().filter(value->Math.random()>.5) since you do not need to print your output, though you still need to fix the first problem I listed
$endgroup$
– Embodiment of Ignorance
yesterday

Given a string like abc, there is no chance of the string cba appearing. Next, you can turn the function into a lambda, and you do not need to name all the lambda variables value, one letter names suffice. Something like: a->a.chars().filter(value->Math.random()>.5) since you do not need to print your output, though you still need to fix the first problem I listed

– Embodiment of Ignorance
yesterday

add a comment |

R, 72 bytes

sample(t<-unique(s<-scan(,"")),sample(n<-length(s),p=length(t)^(1:n)),T)

Try it online!

Input and output are vectors of characters. All possible outputs have equal probability.

Explanation (ungolfed version):

s<-scan(,"")                         # takes input as vector of characters

n<-length(s)

t<-unique(s)

sample(t,                           # sample uniformly at random from the list of unique characters

       sample(n,p=length(t)^(1:n)), # with length k in 1..n chosen randomly such that P[k] is proportional to length(t)^k

       T)                           # the sampling is with replacement

edited 20 hours ago

answered 21 hours ago

Robin Ryder

5317

add a comment |

R, 72 bytes

sample(t<-unique(s<-scan(,"")),sample(n<-length(s),p=length(t)^(1:n)),T)

Try it online!

Input and output are vectors of characters. All possible outputs have equal probability.

Explanation (ungolfed version):

s<-scan(,"")                         # takes input as vector of characters

n<-length(s)

t<-unique(s)

sample(t,                           # sample uniformly at random from the list of unique characters

       sample(n,p=length(t)^(1:n)), # with length k in 1..n chosen randomly such that P[k] is proportional to length(t)^k

       T)                           # the sampling is with replacement

edited 20 hours ago

answered 21 hours ago

Robin Ryder

5317

add a comment |

R, 72 bytes

sample(t<-unique(s<-scan(,"")),sample(n<-length(s),p=length(t)^(1:n)),T)

Try it online!

Input and output are vectors of characters. All possible outputs have equal probability.

Explanation (ungolfed version):

s<-scan(,"")                         # takes input as vector of characters

n<-length(s)

t<-unique(s)

sample(t,                           # sample uniformly at random from the list of unique characters

       sample(n,p=length(t)^(1:n)), # with length k in 1..n chosen randomly such that P[k] is proportional to length(t)^k

       T)                           # the sampling is with replacement

edited 20 hours ago

answered 21 hours ago

Robin Ryder

5317

R, 72 bytes

sample(t<-unique(s<-scan(,"")),sample(n<-length(s),p=length(t)^(1:n)),T)

Try it online!

Input and output are vectors of characters. All possible outputs have equal probability.

Explanation (ungolfed version):

s<-scan(,"")                         # takes input as vector of characters

n<-length(s)

t<-unique(s)

sample(t,                           # sample uniformly at random from the list of unique characters

       sample(n,p=length(t)^(1:n)), # with length k in 1..n chosen randomly such that P[k] is proportional to length(t)^k

       T)                           # the sampling is with replacement

edited 20 hours ago

answered 21 hours ago

Robin Ryder

5317

edited 20 hours ago

answered 21 hours ago

Robin Ryder

5317

answered 21 hours ago

Robin Ryder

5317

answered 21 hours ago

Robin Ryder

5317

add a comment |

Charcoal, 34 bytes

≔Φθ⁼κ⌕θιη≔⊕‽ΣＥθＸＬη⊕κζＷζ«§ηζ≔÷⊖ζＬηζ

Try it online! Link is to verbose version of code. Explanation:

≔Φθ⁼κ⌕θιη

Extract the unique characters of the input. Let's call the number of unique characters n.

≔⊕‽ΣＥθＸＬη⊕κζ

Ｗζ«§ηζ≔÷⊖ζＬηζ

Convert the number into bijective base n, using the unique characters as the digits.

answered 18 hours ago

Neil

82.6k745179

add a comment |

Charcoal, 34 bytes

≔Φθ⁼κ⌕θιη≔⊕‽ΣＥθＸＬη⊕κζＷζ«§ηζ≔÷⊖ζＬηζ

Try it online! Link is to verbose version of code. Explanation:

≔Φθ⁼κ⌕θιη

Extract the unique characters of the input. Let's call the number of unique characters n.

≔⊕‽ΣＥθＸＬη⊕κζ

Ｗζ«§ηζ≔÷⊖ζＬηζ

Convert the number into bijective base n, using the unique characters as the digits.

answered 18 hours ago

Neil

82.6k745179

add a comment |

Charcoal, 34 bytes

≔Φθ⁼κ⌕θιη≔⊕‽ΣＥθＸＬη⊕κζＷζ«§ηζ≔÷⊖ζＬηζ

Try it online! Link is to verbose version of code. Explanation:

≔Φθ⁼κ⌕θιη

Extract the unique characters of the input. Let's call the number of unique characters n.

≔⊕‽ΣＥθＸＬη⊕κζ

Ｗζ«§ηζ≔÷⊖ζＬηζ

Convert the number into bijective base n, using the unique characters as the digits.

answered 18 hours ago

Neil

82.6k745179

Charcoal, 34 bytes

≔Φθ⁼κ⌕θιη≔⊕‽ΣＥθＸＬη⊕κζＷζ«§ηζ≔÷⊖ζＬηζ

Try it online! Link is to verbose version of code. Explanation:

≔Φθ⁼κ⌕θιη

Extract the unique characters of the input. Let's call the number of unique characters n.

≔⊕‽ΣＥθＸＬη⊕κζ

Ｗζ«§ηζ≔÷⊖ζＬηζ

Convert the number into bijective base n, using the unique characters as the digits.

answered 18 hours ago

Neil

82.6k745179

answered 18 hours ago

Neil

82.6k745179

answered 18 hours ago

Neil

82.6k745179

answered 18 hours ago

Neil

82.6k745179

add a comment |

MATLAB / Octave, 110 bytes

Choose a random permutation of a subset of the input letters (with repetitions), whose random length is based on the probability of generating a word with that length.

@(a)a(randi(numel(a),[find(cumsum(numel(a).^[1:numel(a)])>=randi(sum(numel(a).^[1:numel(a)])),1,'first'),1]));

Try it online!

answered 16 hours ago

PieCot

97959

add a comment |

MATLAB / Octave, 110 bytes

Choose a random permutation of a subset of the input letters (with repetitions), whose random length is based on the probability of generating a word with that length.

@(a)a(randi(numel(a),[find(cumsum(numel(a).^[1:numel(a)])>=randi(sum(numel(a).^[1:numel(a)])),1,'first'),1]));

Try it online!

answered 16 hours ago

PieCot

97959

add a comment |

MATLAB / Octave, 110 bytes

Choose a random permutation of a subset of the input letters (with repetitions), whose random length is based on the probability of generating a word with that length.

@(a)a(randi(numel(a),[find(cumsum(numel(a).^[1:numel(a)])>=randi(sum(numel(a).^[1:numel(a)])),1,'first'),1]));

Try it online!

answered 16 hours ago

PieCot

97959

MATLAB / Octave, 110 bytes

Choose a random permutation of a subset of the input letters (with repetitions), whose random length is based on the probability of generating a word with that length.

@(a)a(randi(numel(a),[find(cumsum(numel(a).^[1:numel(a)])>=randi(sum(numel(a).^[1:numel(a)])),1,'first'),1]));

Try it online!

answered 16 hours ago

PieCot

97959

answered 16 hours ago

PieCot

97959

answered 16 hours ago

PieCot

97959

answered 16 hours ago

PieCot

97959

add a comment |

T-SQL, 222 bytes

This creates all combinations of each unique character with recursive sql, then picks a random row from the distinct combinations.

DECLARE @ varchar(max)='T-SQL';



WITH C as(SELECT DISTINCT substring(@,number+1,1)x

FROM spt_values

WHERE'P'=type and len(@)>number),D

as(SELECT x y

FROM c UNION ALL

SELECT y+x

FROM C JOIN D

ON len(y)<len(@))SELECT top 1*FROM D

GROUP BY y

ORDER BY newid()

Try it online ungolfed version

edited 9 hours ago

answered 19 hours ago

t-clausen.dk

2,074314

add a comment |

T-SQL, 222 bytes

This creates all combinations of each unique character with recursive sql, then picks a random row from the distinct combinations.

DECLARE @ varchar(max)='T-SQL';



WITH C as(SELECT DISTINCT substring(@,number+1,1)x

FROM spt_values

WHERE'P'=type and len(@)>number),D

as(SELECT x y

FROM c UNION ALL

SELECT y+x

FROM C JOIN D

ON len(y)<len(@))SELECT top 1*FROM D

GROUP BY y

ORDER BY newid()

Try it online ungolfed version

edited 9 hours ago

answered 19 hours ago

t-clausen.dk

2,074314

add a comment |

T-SQL, 222 bytes

This creates all combinations of each unique character with recursive sql, then picks a random row from the distinct combinations.

DECLARE @ varchar(max)='T-SQL';



WITH C as(SELECT DISTINCT substring(@,number+1,1)x

FROM spt_values

WHERE'P'=type and len(@)>number),D

as(SELECT x y

FROM c UNION ALL

SELECT y+x

FROM C JOIN D

ON len(y)<len(@))SELECT top 1*FROM D

GROUP BY y

ORDER BY newid()

Try it online ungolfed version

edited 9 hours ago

answered 19 hours ago

t-clausen.dk

2,074314

T-SQL, 222 bytes

This creates all combinations of each unique character with recursive sql, then picks a random row from the distinct combinations.

DECLARE @ varchar(max)='T-SQL';



WITH C as(SELECT DISTINCT substring(@,number+1,1)x

FROM spt_values

WHERE'P'=type and len(@)>number),D

as(SELECT x y

FROM c UNION ALL

SELECT y+x

FROM C JOIN D

ON len(y)<len(@))SELECT top 1*FROM D

GROUP BY y

ORDER BY newid()

Try it online ungolfed version

edited 9 hours ago

answered 19 hours ago

t-clausen.dk

2,074314

edited 9 hours ago

answered 19 hours ago

t-clausen.dk

2,074314

answered 19 hours ago

t-clausen.dk

2,074314

answered 19 hours ago

t-clausen.dk

2,074314

add a comment |

Python 2, 124 bytes

lambda s:g(list(set(s)),len(s))

from random import*

g=lambda s,n:choice(s)+(random()*~-len(s)**(n)>~-len(s)and g(s,n-1)or'')

Try it online!

A different approach: instead of first constructing a list of all compliant strings and choosing one at random, this approach randomly decides to continue extending the string or stopping.

As an example, consider the set of characters ['a','b'] and suppose we want to generate random strings of length 1, 2, or 3.

Then if we randomly choose 'a' as the starting character, the possible strings which could be returned are:

a

 aa

  aaa

  aab

 ab

  aba

  abb

More generally, if n is maximum length of the string and x is the number of characters in the set, then the number of strings starting with some given character is going to be:

$$h(n)=sum_{i=0}^{n-1} x^i$$
$$h(n)=1+x+x^2+x^3...+x^{n-1}$$

For x>1 (guaranteed by OP's rule 'at least two distinct...'), we have:

$$h(n)(x-1)=(1+x+x^2+x^3...+x^{n-1})(x-1)=x^n-1$$
$$h(n)=frac{x^n-1}{x-1}$$

So to decide whether we should continue extending our string, let p be a random number in [0,1); then we should recurse only if any of these equivalent statements are true:

$$p>frac{1}{h(n)}$$
$$p>frac{x-1}{x^n-1}$$
$$p(x^n-1)>x-1$$

of which g is the golfed implementation.

answered 8 hours ago

Chas Brown

5,1691523

add a comment |

Python 2, 124 bytes

lambda s:g(list(set(s)),len(s))

from random import*

g=lambda s,n:choice(s)+(random()*~-len(s)**(n)>~-len(s)and g(s,n-1)or'')

Try it online!

A different approach: instead of first constructing a list of all compliant strings and choosing one at random, this approach randomly decides to continue extending the string or stopping.

As an example, consider the set of characters ['a','b'] and suppose we want to generate random strings of length 1, 2, or 3.

Then if we randomly choose 'a' as the starting character, the possible strings which could be returned are:

a

 aa

  aaa

  aab

 ab

  aba

  abb

More generally, if n is maximum length of the string and x is the number of characters in the set, then the number of strings starting with some given character is going to be:

$$h(n)=sum_{i=0}^{n-1} x^i$$
$$h(n)=1+x+x^2+x^3...+x^{n-1}$$

For x>1 (guaranteed by OP's rule 'at least two distinct...'), we have:

$$h(n)(x-1)=(1+x+x^2+x^3...+x^{n-1})(x-1)=x^n-1$$
$$h(n)=frac{x^n-1}{x-1}$$

So to decide whether we should continue extending our string, let p be a random number in [0,1); then we should recurse only if any of these equivalent statements are true:

$$p>frac{1}{h(n)}$$
$$p>frac{x-1}{x^n-1}$$
$$p(x^n-1)>x-1$$

of which g is the golfed implementation.

answered 8 hours ago

Chas Brown

5,1691523

add a comment |

Python 2, 124 bytes

lambda s:g(list(set(s)),len(s))

from random import*

g=lambda s,n:choice(s)+(random()*~-len(s)**(n)>~-len(s)and g(s,n-1)or'')

Try it online!

A different approach: instead of first constructing a list of all compliant strings and choosing one at random, this approach randomly decides to continue extending the string or stopping.

As an example, consider the set of characters ['a','b'] and suppose we want to generate random strings of length 1, 2, or 3.

Then if we randomly choose 'a' as the starting character, the possible strings which could be returned are:

a

 aa

  aaa

  aab

 ab

  aba

  abb

More generally, if n is maximum length of the string and x is the number of characters in the set, then the number of strings starting with some given character is going to be:

$$h(n)=sum_{i=0}^{n-1} x^i$$
$$h(n)=1+x+x^2+x^3...+x^{n-1}$$

For x>1 (guaranteed by OP's rule 'at least two distinct...'), we have:

$$h(n)(x-1)=(1+x+x^2+x^3...+x^{n-1})(x-1)=x^n-1$$
$$h(n)=frac{x^n-1}{x-1}$$

So to decide whether we should continue extending our string, let p be a random number in [0,1); then we should recurse only if any of these equivalent statements are true:

$$p>frac{1}{h(n)}$$
$$p>frac{x-1}{x^n-1}$$
$$p(x^n-1)>x-1$$

of which g is the golfed implementation.

answered 8 hours ago

Chas Brown

5,1691523

Python 2, 124 bytes

lambda s:g(list(set(s)),len(s))

from random import*

g=lambda s,n:choice(s)+(random()*~-len(s)**(n)>~-len(s)and g(s,n-1)or'')

Try it online!

A different approach: instead of first constructing a list of all compliant strings and choosing one at random, this approach randomly decides to continue extending the string or stopping.

As an example, consider the set of characters ['a','b'] and suppose we want to generate random strings of length 1, 2, or 3.

Then if we randomly choose 'a' as the starting character, the possible strings which could be returned are:

a

 aa

  aaa

  aab

 ab

  aba

  abb

More generally, if n is maximum length of the string and x is the number of characters in the set, then the number of strings starting with some given character is going to be:

$$h(n)=sum_{i=0}^{n-1} x^i$$
$$h(n)=1+x+x^2+x^3...+x^{n-1}$$

For x>1 (guaranteed by OP's rule 'at least two distinct...'), we have:

$$h(n)(x-1)=(1+x+x^2+x^3...+x^{n-1})(x-1)=x^n-1$$
$$h(n)=frac{x^n-1}{x-1}$$

So to decide whether we should continue extending our string, let p be a random number in [0,1); then we should recurse only if any of these equivalent statements are true:

$$p>frac{1}{h(n)}$$
$$p>frac{x-1}{x^n-1}$$
$$p(x^n-1)>x-1$$

of which g is the golfed implementation.

answered 8 hours ago

Chas Brown

5,1691523

answered 8 hours ago

Chas Brown

5,1691523

answered 8 hours ago

Chas Brown

5,1691523

answered 8 hours ago

Chas Brown

5,1691523

add a comment |

J, 27 bytes

[:(?@#{])@;[:,@{&.>#<@#"{<

Try it online!

standard formatting

[: (?@# { ])@; [: ,@{&.> # <@#"1 _ <

explanation

Eg, for the string 'abc' we first use #<@#"{< to create:

┌─────┬─────────┬─────────────┐

│┌───┐│┌───┬───┐│┌───┬───┬───┐│

││abc│││abc│abc│││abc│abc│abc││

│└───┘│└───┴───┘│└───┴───┴───┘│

└─────┴─────────┴─────────────┘

We then take the cartesian product of each of these {, flatten the results, and finally remove the outer boxing ;.

(?@#{])@ takes a random result from that list.

Note: TIO will return the same result every time, but in a normal J REPL it will be random.

edited 7 hours ago

answered 9 hours ago

Jonah

2,5711017

add a comment |

J, 27 bytes

[:(?@#{])@;[:,@{&.>#<@#"{<

Try it online!

standard formatting

[: (?@# { ])@; [: ,@{&.> # <@#"1 _ <

explanation

Eg, for the string 'abc' we first use #<@#"{< to create:

┌─────┬─────────┬─────────────┐

│┌───┐│┌───┬───┐│┌───┬───┬───┐│

││abc│││abc│abc│││abc│abc│abc││

│└───┘│└───┴───┘│└───┴───┴───┘│

└─────┴─────────┴─────────────┘

We then take the cartesian product of each of these {, flatten the results, and finally remove the outer boxing ;.

(?@#{])@ takes a random result from that list.

Note: TIO will return the same result every time, but in a normal J REPL it will be random.

edited 7 hours ago

answered 9 hours ago

Jonah

2,5711017

add a comment |

J, 27 bytes

[:(?@#{])@;[:,@{&.>#<@#"{<

Try it online!

standard formatting

[: (?@# { ])@; [: ,@{&.> # <@#"1 _ <

explanation

Eg, for the string 'abc' we first use #<@#"{< to create:

┌─────┬─────────┬─────────────┐

│┌───┐│┌───┬───┐│┌───┬───┬───┐│

││abc│││abc│abc│││abc│abc│abc││

│└───┘│└───┴───┘│└───┴───┴───┘│

└─────┴─────────┴─────────────┘

We then take the cartesian product of each of these {, flatten the results, and finally remove the outer boxing ;.

(?@#{])@ takes a random result from that list.

Note: TIO will return the same result every time, but in a normal J REPL it will be random.

edited 7 hours ago

answered 9 hours ago

Jonah

2,5711017

J, 27 bytes

[:(?@#{])@;[:,@{&.>#<@#"{<

Try it online!

standard formatting

[: (?@# { ])@; [: ,@{&.> # <@#"1 _ <

explanation

Eg, for the string 'abc' we first use #<@#"{< to create:

┌─────┬─────────┬─────────────┐

│┌───┐│┌───┬───┐│┌───┬───┬───┐│

││abc│││abc│abc│││abc│abc│abc││

│└───┘│└───┴───┘│└───┴───┴───┘│

└─────┴─────────┴─────────────┘

We then take the cartesian product of each of these {, flatten the results, and finally remove the outer boxing ;.

(?@#{])@ takes a random result from that list.

Note: TIO will return the same result every time, but in a normal J REPL it will be random.

edited 7 hours ago

answered 9 hours ago

Jonah

2,5711017

edited 7 hours ago

answered 9 hours ago

Jonah

2,5711017

answered 9 hours ago

Jonah

2,5711017

answered 9 hours ago

Jonah

2,5711017

add a comment |

Flog Edoc is a new contributor. Be nice, and check out our Code of Conduct.

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

Flog Edoc is a new contributor. Be nice, and check out our Code of Conduct.

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

To string or not to string

15 Answers 15

Jelly, 10 5 bytes

How?

05AB1E, 6 bytes

Brachylog, 8 bytes

Perl 6, 35 bytes

Explanation:

Python 2, 111 121 bytes

Japt, 10 9 bytes

Japt, 9 bytes

Wolfram Language (Mathematica), 41 bytes

C (gcc), 68 65 bytes

Java, 106 bytes

R, 72 bytes

Charcoal, 34 bytes

MATLAB / Octave, 110 bytes

T-SQL, 222 bytes

Python 2, 124 bytes

J, 27 bytes

standard formatting

explanation

Your Answer

Sign up or log in

Post as a guest

Post as a guest

15 Answers 15

15 Answers 15

Jelly, 10 5 bytes

How?

Jelly, 10 5 bytes

How?

Jelly, 10 5 bytes

How?

Jelly, 10 5 bytes

How?

05AB1E, 6 bytes

05AB1E, 6 bytes

05AB1E, 6 bytes

05AB1E, 6 bytes

Brachylog, 8 bytes

Brachylog, 8 bytes

Brachylog, 8 bytes

Brachylog, 8 bytes

Perl 6, 35 bytes

Explanation:

Perl 6, 35 bytes

Explanation:

Perl 6, 35 bytes

Explanation:

Perl 6, 35 bytes

Explanation:

Python 2, 111 121 bytes

Python 2, 111 121 bytes

Python 2, 111 121 bytes

Python 2, 111 121 bytes

Japt, 10 9 bytes

Japt, 9 bytes

Japt, 10 9 bytes

Japt, 9 bytes

Japt, 10 9 bytes

Japt, 9 bytes

Japt, 10 9 bytes

Japt, 9 bytes

Wolfram Language (Mathematica), 41 bytes

Wolfram Language (Mathematica), 41 bytes

Wolfram Language (Mathematica), 41 bytes

Wolfram Language (Mathematica), 41 bytes

C (gcc), 68 65 bytes

C (gcc), 68 65 bytes

C (gcc), 68 65 bytes

C (gcc), 68 65 bytes

Java, 106 bytes

Java, 106 bytes

Java, 106 bytes

Java, 106 bytes

R, 72 bytes

R, 72 bytes

R, 72 bytes

R, 72 bytes

15 Answers
15

15 Answers
15

15 Answers
15