For each graph topology, and each of the 8! orderings of competitor strength, it calculates how many of the (8 choose 2)=28 pairwise judgements can be derived from just the head-to-head results from a tournament represented by the input graph. This gives us a number from 0 to 28 (in this case, from 16 to 28 because there are 16 edges in each graph so we know those will be good). Then it writes a distribution. So in the example you quoted, of the 8! orderings, 1728 gave 18 bits of results, 2352 gave 19 bits of results, etc. up to 840 orderings gave all 28 bits of results (or a full ordering of all competitors). On Wed, Jun 19, 2019 at 12:22 PM Dan Asimov <dasimov@earthlink.net> wrote:
Tomas, please remind me of just what the program is computing — thanks.
—Dan
----- Okay, I wrote and ran the code for the 8/4 case over the six graphs and this is what I got. I'm a bit surprised we get so much information . . . the graphs are in the order I gave the graph list before. TL;DR: the pretty graph we've been talking about is, indeed, the best one. ... ... Average bits 22.6464 18:1728 19:2352 20:3504 21:7200 22:4152 23:6528 24:5712 25:3576 26:3120 27:1608 28:840 ... ... -----
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