k******a
发帖数: 2436 | 31 define functions:
f1 (x, y) = three games: x1-y2, x3-y4, x5-y6
f2 (x, y) = three games: x1-y3, x2-y5, x4-y6
f3 (x, y) = three games: x1-y4, x2-y6, x3-y5
f4 (x, y) = three games: x1-y5, x2-y4, x3-y6
f5 (x, y) = three games: x1-y6, x2-y3, x4-y5
Week 1: f1(A, A), f1(B, B), f1(C, C)
Week 2: f2(A, A), f2(B, B), f2(C, C)
Week 3: f3(A, A), f3(B, B), f3(C, C)
Week 4: f4(A, A), f4(B, B), f4(C, C)
Week 5: f5(A, A), f5(B, B), f5(C, C)
Week 6: same as week 1
Week 7: same as week 2
Week 8: same as week 3
Week 9: same as week 4
Week 10: same as week 5
Week 11: f1(A, B), f1(B, C), f1(C, A)
Week 12: f2(A, B), f2(B, C), f2(C, A)
Week 13: f3(A, B), f3(B, C), f3(C, A)
Week 14: f4(A, B), f4(B, C), f4(C, A)
Week 15: f5(A, B), f5(B, C), f5(C, A)
Week 16: f1(B, A), f1(C, B), f1(A, C)
Week 17: f2(B, A), f2(C, B), f2(A, C)
Week 18: f3(B, A), f3(C, B), f3(A, C)
Week 19: f4(B, A), f4(C, B), f4(A, C)
Week 20: f5(B, A), f5(C, B), f5(A, C)
proof: I hope I did not make a mistake, in function f1 to f5, no two teams
with the same index number can meet, and the same pair of index number of a
game occur and only occur once. The proof is apparent.
【在 C****c 的大作中提到】
: A1-A6
: B1-B6
: C1-C6
: 3个组，每个组6个队伍
: 组内打双循环
: 组外打单循环，但是A1不和B1,C1打，只打B2-B6和C2-C6
: 这样算下来一共180场比赛，要求20周搞定，每周9场
: 我用matlab一个18*18*20的矩阵算的
: 结果发现竟然填不满
: 然后我现在想证明这种东西不一定有解
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