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I've been thinking about sorting $6$ elements with the minimal possible number of comparisons. I can do it in $10$ comparisons but I've no idea if this is optimal. Or is there a better algorithm ?

Algorithm
1. Sort $a_1, a_2, a_3$ and $a_4, a_5, a_6$.
Number of comparisons: $3+3=6$.
2. Merge two subarrays.
Number of comparisons: $3 + 3 - 2 = 4$.
Total number of comparisons: $6 + 4 = 10$.

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    $\begingroup$ Closely related question. In my answer there, I give a reference to the relevant section in TAoCP. What research have you done? $\endgroup$
    – Raphael
    Oct 3, 2015 at 10:48
  • $\begingroup$ How does step 2 take only 4 comparisons? ​ ​ $\endgroup$
    – user12859
    Oct 3, 2015 at 12:04
  • $\begingroup$ You are right, I am wrong.... How to fix it ? $\endgroup$
    – user40545
    Oct 3, 2015 at 22:31
  • $\begingroup$ See, for example, Daniel Fischer's answer in Exactly how many comparisons does merge sort make?. $\endgroup$
    – greybeard
    Oct 4, 2015 at 7:01
  • $\begingroup$ Ok, I know that I was wrong. I think about fixing my algorithm. $\endgroup$
    – user40545
    Oct 4, 2015 at 8:24

1 Answer 1

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According to A036604, 10 comparisons are indeed optimal. The link probably contains a citation to a paper proving this.

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  • $\begingroup$ thanks, my algorithm is ok ? $\endgroup$
    – user40545
    Oct 3, 2015 at 8:19
  • $\begingroup$ @user40545 I don't know, but I believe you can verify that yourself. $\endgroup$ Oct 3, 2015 at 8:21
  • $\begingroup$ Ok, hovewer it is sufficient to analyze decison tree to prove optimality ? ($log(6!)\ge 10$) $\endgroup$
    – user40545
    Oct 3, 2015 at 8:22
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    $\begingroup$ @user40545 You can do this calculation and see for yourself. Try to be more independent. $\endgroup$ Oct 3, 2015 at 8:24
  • $\begingroup$ I thnik you are right, I am too dependent. $\endgroup$
    – user40545
    Oct 3, 2015 at 8:30

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