Inspired by Marilyn vos Savant's using an arithmetic mean where a geometric mean was clearly called for last weekend (did anyone else catch that?) I decided to look into the optimum solution of the advance tournament scheduling problem for various power means. Power means of n elements include: -1: Harmonic mean, the reciprocal of the arithmetic mean of the reciprocals 0: Geometric mean, the nth root of the product 1: Arithmetic mean, the sum divided by n 2: Quadratic mean (RMS), the square root of the arithmetic mean of the squares The power doesn't have to be an integer. It can be any real number. (For 0, it's taken as the limit when the power approaches 0.) (The power can presumably be a complex number too, but I'll save that for another day.) Here's the result: 2E6BB72 44444444 n:2520 min:18 max:28 0,0,27,175,339,496,370,308,267,243,162,82,51 22.4476 -16.023502 20.896856 0775CF5 55444433 n:10080 min:18 max:28 0,0,182,788,1032,1248,1678,1144,1406,1144,836,440,182 22.7560 -14.684336 21.008307 05F7D78 54444443 n:20160 min:18 max:28 0,0,668,1134,1786,2078,2856,3100,2806,2452,1866,1008,406 22.9565 -9.398537 21.600937 077BE78 44444444 n:2520 min:17 max:28 0,45,98,37,166,264,243,406,360,404,263,176,58 23.3071 -8.666145 21.729642 03FDE78 44444444 n:35 min:16 max:28 1,0,0,2,1,0,4,4,7,3,9,3,1 24.0000 There are five different best solutions, depending on the power. I show the powers at which best solution changed, and what the power mean was at that power. Of course at that power it was a tie between the above and the below graph. For instance 2E6BB72 is the best solution for powers below (more negative than) -16.023502, and 0775CF5 is the best solution for powers above it but below -14.684336. At -16.023502 itself, 20.896856 is the mean for both 2E6BB72 and 0775CF5. (See my July 21 message for how to turn those hexadecimal numbers into the graphs they represent, and what the meaning of the stuff after those hexadecimal numbers is.) It's interesting that 03FDE78, K4,4, is the best for all powers above -8.666145, including the harmonic, geometric, arithmetic, and quadratic means. It's also 7-way-tied for the highest median, 24, and has the unique highest mode, 26. I checked all powers from -200 to +200. Beyond that, I was plagued by underflow, overflow, and precision-loss errors. (Try raising 28 to the 200th power on your scientific calculator, see how far you get.) But I'm convinced that the bottom solution goes all the way down and the top solution goes all the way up. In the limit of high powers, the mean is the maximum value, which in this case is a 1281-way tie for 28. In the case of a tie, it's whichever set has the highest proportion of 28s. And K4,4 does. It may not look like it, since only 1 of the 35 K4,4 graphs has 28, and other graphs have as many as 768 28s. But that's 768 out of 40320, which is less than 1/35. In the limit of low (very negative) powers, the mean is the minimum value, which in this case is a 76-way tie for 18. In the case of a tie, it's whichever set has the lowest proportion of 18s. And, sure enough, 2E6BB72 does. As for the *worst* graph, for any power mean it's 2DDEFC0 55555511 n:28 min:16 max:16 28,0,0,0,0,0,0,0,0,0,0,0,0 16.0000 Since all values are 16, all power means are 16 (as are the median and the mode). It's interesting that the worst graph and the best (for most purposes) graph have the two lowest values of n. Now I'm wondering about the general case of some finite quantity of finite sets of positive integers. As the power is gradually ramped up, can which set has the highest power mean occur more than once? Obviously yes, since, given the two sets {2,9} and {1,6,7,8} the former excels for very low and very high powers, and the latter is best for powers in between. (They're tied at a power of 1. At what other power are they tied?) It there any limit to how many times two sets can trade places as best? How about three or more sets?