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This Python 3 program constructively produces possible tables that satisfy the constraints of the problem. It runs in 51ms.

I’ve assumed that Alan (A) and David (D) are men and Barbara (B) and Christine (C) are women.

Solution:There were 2 men and 4 women in the tournament.Here’s a diagram of the possible matches played, a square with a 1 in it indicates that the player in that row beat the player in that column:

Analytically we can see that

mmen playT(m – 1)matches amongst themselves, and each of them must be won by a man.We are told that there are at least 2 men, and one of them wins a match against a woman. So there can be at most

m – 1matches amongst the men.So, we are looking for

mwherem ≥ 2andT(m – 1) ≤ m – 1.The only possible value is

m = 2.So the two men are

AandD, andDbeatsAwhen they play. They each lose the rest of their matches.The only remaining variable is

w, the number of women(w ≥ 2).There are

T(w + 1)matches in total, 2 of them won by men and the remainingT(w + 1) – 2must be 1 more than a multiple ofw.So, in order for

kto be a whole number it follows thatwmust be a divisor of 4(w ≥ 2). The only possible values are 2 and 4.Hence there are 2 men and 4 women (6 players in total). The men win 1 game each, the women win 3 games each, except for Christine who wins 4.

It is now easy to construct a set of possible matches, as follows:

If the men are

A,D, and the women areB,C,X,Y. Then:AbeatB, and loses the rest of his matches (D,C,X,Y). (1 win).DbeatA, and loses the rest of his matches (B,C,X,Y). (1 win).Xalready has 2 wins (againstAandD), let’s suppose they also beatY, and lose againstBandCto give the required 3 wins.Yhas 2 wins (againstAandD), and lost againstX. So needs one more win againstBorC. Let’s sayYbeatsBfor 3 wins.Bhas wins againstDandX, and has lost toAandY, so needs to win againstCto get 3 wins.All matches are now specified and

Chas wins againstA,D,X,Yand lost againstBto give 4 wins, as required.