[seqfan] Re: the revolver sequence
Max Alekseyev
maxale at gmail.com
Thu Sep 24 00:58:14 CEST 2009
Hi Tanya,
I've got different values starting with a(9) = 13, a(10) = 32 etc.
Could you please double check your values or lists all 18 ways of
loading a revolver with n=9 chambers?
Thanks,
Max
On Wed, Sep 23, 2009 at 5:06 PM, Tanya Khovanova
<mathoflove-seqfan at yahoo.com> wrote:
> Hello all,
>
> I just submitted a sequence related to the revolver puzzle:
> http://blog.tanyakhovanova.com/?p=174
>
> I generalized it to revolvers having n chambers.
>
> I think there should be more sequences: for example, the one when it is preferable to spin and the one when the chances are the same.
>
> Also, my programs are usually not very efficient. It will be nice if someone can add more terms to this sequence.
>
> %I A165646
> %S A165646 0,0,0,0,2,3,4,5,18,43,60,84,294,472,724
> %N A165646 The revolver sequence.
> %C A165646 a(n) is in how many ways you can load a revolver with n chambers, so that if a person survives after the first shot, he/she has better chances of survival for the second shot if the shooter continues rather than spins. a(n) <= A000031(n) as A000031(n) counts the number of possible revolver loadings.
> %H A165646 Tanya Khovanova, <a href="http://blog.tanyakhovanova.com/?p=174">Heard on the Street</a>
> %e A165646 Based on the famous interview question where the revolver has six chambers and the shooter loads two adjacent bullets. As the answer to this question is to continue, a(6) must be at least 1.
> %t A165646 << Combinatorica` colors[x_] := Table[PadLeft[Table[1, {n, i}], x], {i, 0, x}] continueBetter[list_] := (len = Length[list]; c0 = 0; c00 = 0; rotateN = 0; While[rotateN < len, newList = RotateLeft[list, rotateN]; If[newList[[1]] == 0, c0++; If[newList[[2]] == 0, c00++]]; rotateN++]; If[c0 > 0, c00/c0 > c0/len, False]) continueNum[num_] := (neckNum = Length[colors[num]]; ans = 0; count = 1; While[count <= neckNum, ans = ans + Length[Select[ListNecklaces[num, colors[num][[count]], Cyclic], continueBetter[#] &]]; count++]; ans) Table[continueNum[n], {n, 2, 15}]
> %Y A165646 A000031
> %K A165646 more,nonn
> %O A165646 1,5
> %A A165646 Tanya Khovanova (tanyakh(AT)yahoo.com), Sep 23 2009
>
>
>
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