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This is cool. I never actually realized that magic squares exhausted all of the sums to 15; in other words, there is no triplet which sums to 15 and is not alre
by jjnoakes 8y ago
This is cool. I never actually realized that magic squares exhausted all of the sums to 15; in other words, there is no triplet which sums to 15 and is not already a row, column, or diagonal. No false positives, so to speak.
I wonder if this is true of larger magic squares.
- Sharlin 8y agoIt is not. There are indeed exactly 8 ways to pick three elements from the set [1,9] such that their sum equals 15. But for 4x4, there are 86 (!) ways to pick four elements from [1, 15] so that their sum is 34 (the magic constant for 4x4), whereas only ten of the combinations are used in construction of the square. Some Python: from collections import Counter from itertools import combinations def number_of_sums(n, M): combs = combinations(range(1,n**2+1), n) sums = (sum(c) for c in combs) return Counter(sums)[M] print(number_of_sums(3, 15)) print(number_of_sums(4, 34))