4
$\begingroup$

The following code produces lists of sublists with maximum length 5 that shuffle zeros around an ordered list (here {1,2}). It produces all permutations while keeping the ordering of the list {1,2} within the permutations:

mylst[K_]:=Select[Drop[Tuples[{0,1,2},K],1],Total[#]==3&&Count[#,1]==1&&#[[Position[#,x_/;!TrueQ[x==0],{1},1,Heads->False][[1,1]]]]!=2&]
mylst /@ Range[5]

Each of mylst[K] generates $K(K-1)/2$ terms.

Is there a better way to code mylst?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

Just for fun, here is the pattern based version

mylst2[K_] := ReplaceList[
  ConstantArray[0, K],
  {a___, x_, b___, y_, c___} :> {a, 1, b, 2, c}
  ]
Improve this answer
$\endgroup$
10
  • $\begingroup$ Thanks for the simplified version! Seems really compact way to do the same task! $\endgroup$
    – Chen Stats Yu
    Jan 9, 2015 at 13:26
  • 1
    $\begingroup$ @ChenStatsYu Surprising! Rule based solutions often have poor complexity, I guess in this case it works... $\endgroup$
    – C. E.
    Jan 9, 2015 at 13:38
  • 1
    $\begingroup$ mylst2[200]; // AbsoluteTiming (yours) VS mylst3[200]; // AbsoluteTiming (Carlo's), it's about {0.078000, Null} VS {0.780001, Null}. Nearly 10 times. $\endgroup$
    – Chen Stats Yu
    Jan 9, 2015 at 13:46
  • 1
    $\begingroup$ No, I think your solution is better Pickett. I forfeit! $\endgroup$
    – Carlo
    Jan 9, 2015 at 13:55
  • 1
    $\begingroup$ @ChenStatsYu It isn't. $\endgroup$
    – C. E.
    Jan 9, 2015 at 16:12
5
$\begingroup$ 
mylst2[K_] := Map[
    ReplacePart[#, FirstPosition[#, 2] -> 1] &,
    Permutations[PadRight[{2, 2}, K]]
]

This might not be what you want for K == 0. But it has much better complexity (quadratic vs exponential).

Improve this answer
$\endgroup$
1
  • $\begingroup$ That will do! I dont really use the case $K=0$. $\endgroup$
    – Chen Stats Yu
    Jan 9, 2015 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.