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The problem of female students in Kirkman
This is a planning problem and can be solved by exhaustive method. Just list all the combinations and remove the unqualified combinations.

Set up 15 girls and divide them into 5 teams. Three people in each team always have X kinds of points (I won't count the number of X's), and then choose seven combinations from X to test whether each combination meets the requirements (provided that everyone in these seven groups appears 1 time, and the people met in each group are different from those met in other groups).

Mathematical solution, introduced the solution of British priest Andrew Frost.

Let 15 girls be represented by the following symbols: X, A 1, A2, B 1, B2, C 1, C2, D 1, D2, E 1, E2, F65438. Arrange them in seven rows, with five threesomes (* * * fifteen people) every day, so that X is at the front of the seven rows: (x, a 1, a2); (x,b 1,B2); (x,c 1,C2); (x,d 1,D2); (x,e 1,E2); (x,f 1,F2); (x,g 1,g2)。

So we only need to allocate 14 elements, and then in each line, we will have a group of three people, that is, we will combine the seven elements with the following label, A, B, C, D, E, F and G, and fill them in each line, but each letter is only allowed twice. that is

Sunday: (x, a, a), (b, d, f), (b, e, g), (c, d, g), (c, e, f);

Monday: (x, b, b), (a, b, e), (a, f, g), (c, d, g), (c, e, f);

Tuesday: (x, c, c), (a, d, e), (a, f, g), (b, d, f), (b, e, g);

Wednesday: (x, d, d), (a, b, c), (a, f, g), (b, e, g), (c, e, f);

Thursday: (X, E, E), (A, B, C), (A, F, G), (B, D, F), (C, D, G)

Friday: (x, f, f), (a, b, c), (a, d, e), (b, e, g), (c, d, g);

Saturday: (x, g, g), (a, b, c), (a, d, e), (b, d, f), (c, e, f)

Now fill in the subscript. If there are two identical letters in the same line, for example, in the third line, BDF and Berg, B appear twice, you can mark b 1, B2 with different footprints; If every "threesome" has two footprints, it is the same strain, and other threesome groups can no longer use it; If the footprint is not determined by two principles, it is determined as 1. Get a solution:

Sunday: (x, a 1, a2), (b 1, d 1, f 1), (b2, e 1, g 1), (c/kloc.

Monday: (x, b 1, b2), (a 1, b2, e2), (a2, f2, g2), (c 1, d 1, g 1), (c2

Tuesday: (x, c 1, c2), (a 1, d 1, e 1), (a2, f 1, g 1), (b/kloc.

Wednesday: (x, d 1, d2), (a 1, b2, c2), (a2, f2, g 1), (b2, e 1, g2), (c 1.

Thursday: (x, e 1, e2), (a 1, b 1, c 1), (a2, f 1, g2), (b2, d 1.

Friday: (x, f 1, f2), (a 1, b2, c 1), (a2, d2, e 1), (b 1, e2, g/kloc-.

Saturday: (x, g 1, g2), (a 1, b 1, c2), (a2, d 1, e2), (b2, d2, f 1), (c)