Permutations without repeating myself












0














This may be a quite simple question, but there we go.



Imagine that I've got a sequence $s = [s_1, s_2, dots, s_n]$. I want to generate programatically a new random permutation $s'$ of $s$ so that $forall i in [1,n],; s[i] neq s'[i]$, of, what is the same, none of the values of the new permutation coincide 1-1 with the values of the previous one.



For instance, for $s = [1,2,3,4]$, $s' = [2,3,4,1]$ is a desired permutation because $s'$ is a permutation of $s$ and $2 neq 1$, $2neq 3$, and so on.



I know that the Fiser-Yates produces random permutation, but you cannot force the requirement that $s[i] neq s'[i]$ for all $i$'s.



Thank you.



Edit: apparently I'm asking for how to generate a derangement with uniform probability distribution.






algorithms permutations







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  • Two sequences can be different while representing the same permutation. I am not sure about what you are asking.
    – nicomezi
    Nov 29 '18 at 13:13










  • So the question is how to generate a derangement with uniform probability distribution?
    – Peter Taylor
    Nov 29 '18 at 14:21










  • @PeterTaylor exactly
    – JnxF
    Nov 29 '18 at 14:22
















0














This may be a quite simple question, but there we go.



Imagine that I've got a sequence $s = [s_1, s_2, dots, s_n]$. I want to generate programatically a new random permutation $s'$ of $s$ so that $forall i in [1,n],; s[i] neq s'[i]$, of, what is the same, none of the values of the new permutation coincide 1-1 with the values of the previous one.



For instance, for $s = [1,2,3,4]$, $s' = [2,3,4,1]$ is a desired permutation because $s'$ is a permutation of $s$ and $2 neq 1$, $2neq 3$, and so on.



I know that the Fiser-Yates produces random permutation, but you cannot force the requirement that $s[i] neq s'[i]$ for all $i$'s.



Thank you.



Edit: apparently I'm asking for how to generate a derangement with uniform probability distribution.






algorithms permutations







share|cite|improve this question
























  • Two sequences can be different while representing the same permutation. I am not sure about what you are asking.
    – nicomezi
    Nov 29 '18 at 13:13










  • So the question is how to generate a derangement with uniform probability distribution?
    – Peter Taylor
    Nov 29 '18 at 14:21










  • @PeterTaylor exactly
    – JnxF
    Nov 29 '18 at 14:22














0












0








0







This may be a quite simple question, but there we go.



Imagine that I've got a sequence $s = [s_1, s_2, dots, s_n]$. I want to generate programatically a new random permutation $s'$ of $s$ so that $forall i in [1,n],; s[i] neq s'[i]$, of, what is the same, none of the values of the new permutation coincide 1-1 with the values of the previous one.



For instance, for $s = [1,2,3,4]$, $s' = [2,3,4,1]$ is a desired permutation because $s'$ is a permutation of $s$ and $2 neq 1$, $2neq 3$, and so on.



I know that the Fiser-Yates produces random permutation, but you cannot force the requirement that $s[i] neq s'[i]$ for all $i$'s.



Thank you.



Edit: apparently I'm asking for how to generate a derangement with uniform probability distribution.






algorithms permutations







share|cite|improve this question















This may be a quite simple question, but there we go.



Imagine that I've got a sequence $s = [s_1, s_2, dots, s_n]$. I want to generate programatically a new random permutation $s'$ of $s$ so that $forall i in [1,n],; s[i] neq s'[i]$, of, what is the same, none of the values of the new permutation coincide 1-1 with the values of the previous one.



For instance, for $s = [1,2,3,4]$, $s' = [2,3,4,1]$ is a desired permutation because $s'$ is a permutation of $s$ and $2 neq 1$, $2neq 3$, and so on.



I know that the Fiser-Yates produces random permutation, but you cannot force the requirement that $s[i] neq s'[i]$ for all $i$'s.



Thank you.



Edit: apparently I'm asking for how to generate a derangement with uniform probability distribution.






algorithms permutations




algorithms permutations






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 14:23







JnxF

















asked Nov 29 '18 at 13:03









JnxFJnxF

1,0691821




1,0691821












  • Two sequences can be different while representing the same permutation. I am not sure about what you are asking.
    – nicomezi
    Nov 29 '18 at 13:13










  • So the question is how to generate a derangement with uniform probability distribution?
    – Peter Taylor
    Nov 29 '18 at 14:21










  • @PeterTaylor exactly
    – JnxF
    Nov 29 '18 at 14:22


















  • Two sequences can be different while representing the same permutation. I am not sure about what you are asking.
    – nicomezi
    Nov 29 '18 at 13:13










  • So the question is how to generate a derangement with uniform probability distribution?
    – Peter Taylor
    Nov 29 '18 at 14:21










  • @PeterTaylor exactly
    – JnxF
    Nov 29 '18 at 14:22
















Two sequences can be different while representing the same permutation. I am not sure about what you are asking.
– nicomezi
Nov 29 '18 at 13:13




Two sequences can be different while representing the same permutation. I am not sure about what you are asking.
– nicomezi
Nov 29 '18 at 13:13












So the question is how to generate a derangement with uniform probability distribution?
– Peter Taylor
Nov 29 '18 at 14:21




So the question is how to generate a derangement with uniform probability distribution?
– Peter Taylor
Nov 29 '18 at 14:21












@PeterTaylor exactly
– JnxF
Nov 29 '18 at 14:22




@PeterTaylor exactly
– JnxF
Nov 29 '18 at 14:22










2 Answers
2






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One way to generate a random derangement $sigma$ of a set $S$ of $n$ letters is based on the recursion $!n = (n-1)(!(n-1) + !(n-2)$. First compute the derangement numbers $!k$ for $1 le k le n$.

Choose $sigma(s_1)$ uniformly from $S backslash {s_1}$. With probability $!(n-2)/(!(n-1)+!(n-2))$, let $sigma(sigma(s_1)) = s_1$, and the rest of $sigma$ is a derangement of the remaining $n-2$
letters $S backslash {s_1, sigma(s_1)}$. Otherwise, recursively take a derangement of
$tau$ of $S backslash {s_1}$; let $sigma(tau^{-1}(sigma(s_1))) = s_1$, otherwise $sigma(i) = tau(i)$.






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    0














    Generate a random derangement (that's a permutation with no fixed points), then multiply it by $s$ to obtain $s'$.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      1














      One way to generate a random derangement $sigma$ of a set $S$ of $n$ letters is based on the recursion $!n = (n-1)(!(n-1) + !(n-2)$. First compute the derangement numbers $!k$ for $1 le k le n$.

      Choose $sigma(s_1)$ uniformly from $S backslash {s_1}$. With probability $!(n-2)/(!(n-1)+!(n-2))$, let $sigma(sigma(s_1)) = s_1$, and the rest of $sigma$ is a derangement of the remaining $n-2$
      letters $S backslash {s_1, sigma(s_1)}$. Otherwise, recursively take a derangement of
      $tau$ of $S backslash {s_1}$; let $sigma(tau^{-1}(sigma(s_1))) = s_1$, otherwise $sigma(i) = tau(i)$.






      share|cite|improve this answer


























        1














        One way to generate a random derangement $sigma$ of a set $S$ of $n$ letters is based on the recursion $!n = (n-1)(!(n-1) + !(n-2)$. First compute the derangement numbers $!k$ for $1 le k le n$.

        Choose $sigma(s_1)$ uniformly from $S backslash {s_1}$. With probability $!(n-2)/(!(n-1)+!(n-2))$, let $sigma(sigma(s_1)) = s_1$, and the rest of $sigma$ is a derangement of the remaining $n-2$
        letters $S backslash {s_1, sigma(s_1)}$. Otherwise, recursively take a derangement of
        $tau$ of $S backslash {s_1}$; let $sigma(tau^{-1}(sigma(s_1))) = s_1$, otherwise $sigma(i) = tau(i)$.






        share|cite|improve this answer
























          1












          1








          1






          One way to generate a random derangement $sigma$ of a set $S$ of $n$ letters is based on the recursion $!n = (n-1)(!(n-1) + !(n-2)$. First compute the derangement numbers $!k$ for $1 le k le n$.

          Choose $sigma(s_1)$ uniformly from $S backslash {s_1}$. With probability $!(n-2)/(!(n-1)+!(n-2))$, let $sigma(sigma(s_1)) = s_1$, and the rest of $sigma$ is a derangement of the remaining $n-2$
          letters $S backslash {s_1, sigma(s_1)}$. Otherwise, recursively take a derangement of
          $tau$ of $S backslash {s_1}$; let $sigma(tau^{-1}(sigma(s_1))) = s_1$, otherwise $sigma(i) = tau(i)$.






          share|cite|improve this answer












          One way to generate a random derangement $sigma$ of a set $S$ of $n$ letters is based on the recursion $!n = (n-1)(!(n-1) + !(n-2)$. First compute the derangement numbers $!k$ for $1 le k le n$.

          Choose $sigma(s_1)$ uniformly from $S backslash {s_1}$. With probability $!(n-2)/(!(n-1)+!(n-2))$, let $sigma(sigma(s_1)) = s_1$, and the rest of $sigma$ is a derangement of the remaining $n-2$
          letters $S backslash {s_1, sigma(s_1)}$. Otherwise, recursively take a derangement of
          $tau$ of $S backslash {s_1}$; let $sigma(tau^{-1}(sigma(s_1))) = s_1$, otherwise $sigma(i) = tau(i)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 14:48









          Robert IsraelRobert Israel

          319k23208458




          319k23208458























              0














              Generate a random derangement (that's a permutation with no fixed points), then multiply it by $s$ to obtain $s'$.






              share|cite|improve this answer


























                0














                Generate a random derangement (that's a permutation with no fixed points), then multiply it by $s$ to obtain $s'$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Generate a random derangement (that's a permutation with no fixed points), then multiply it by $s$ to obtain $s'$.






                  share|cite|improve this answer












                  Generate a random derangement (that's a permutation with no fixed points), then multiply it by $s$ to obtain $s'$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 13:09









                  Alexander BursteinAlexander Burstein

                  1,129217




                  1,129217






























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