This is Problem 208 "Looking-glass Zoo" in Question Time, Problems for a Brainy Day by Hubert Phillips, Farrar & Rinehart, New York, 1938, which is an American reprint of Hubert Phillips' fifth collection of puzzles from New Statesman (where he published under the name "Caliban") and News Chronicle (where he published under the name "Dogberry"). Many of these puzzles were submitted by readers, and this one was submitted by Sir Arthur Eddington. Sir Arthur also submitted a solution using 4x4 matrices, using a theorem of his that there are exactly five such matrices which anticommute. A separate solution using the roots of minus one was submitted pseudonymously by "Champ." We give yet a third solution below.

Given that there is at least one boy and one girl (John and Mary are mentioned) then the answer is that there were 3 nephews and 2 nieces, the winner was a boy who got 4 right.

Number the animals 1 through 8, such that the females are even and the males are odd, with members of the same species consecutive; i.e. 1 is Mr. Tove, 2 Mrs. Tove, etc.

Then each childs guesses can be represented by a permutation. I use the standard notation of a permutation as a set of orbits. For example: (1 3 5)(6 8) means 1 -> 3, 3 -> 5, 5 -> 1, 6 -> 8, 8 -> 6 and 2,4,7 are unchanged.

1? Let P be any child's guesses. Then P(mate(i)) = mate(P(i)).

2? If Q is another child's guesses, then P,Q? = T, where P,Q? is the commutator of P and Q (P composed with Q composed with P inverse composed with Q inverse) and T is the special permutation (1 2) (3 4) (5 6) (7 8) that just swaps each animal with its spouse.

3? If P represents a boy, then P*P = I (I use * for composition, and I for the identity permutation: (1)(2)(3)(4)(5)(6)(7)(8)

4? If P represents a girl, then P*P = T.

1? and 4? together mean that all girl's guesses must be of the form
(A B C D) (E F G H) where A and C are mates, as are B & D, E & F G & H.

So without loss of generality let Mary = (1 3 2 4) (5 7 6 8) Without too much effort we see that the only possibilities for other girls "compatible" with Mary (I use compatible to mean the relation

expressed in 2?) are
g1: (1 5 2 6) (3 8 4 7) g2: (1 6 2 5) (3 7 4 8) g3: (1 7 2 8) (3 5 4 6) g4: (1 8 2 7) (3 6 4 5)

Note that g1 is incompatible with g2 and g3 is incompatible with g4. Thus no 4 of Mary and g1-4 are mutually compatible. Thus there are at most three girls: Mary, g1 and g3 (without loss of generality)

By 1? and 3?, each boy must be represented as a product of transpostions and/or singletons: e.g. (1 3) (2 4) (5) (6) (7) (8) or (1) (2) (3 4) (5 8) (6 7).

Let J represent John's guesses and consider J(1). If J(1) = 1, then J(2) = 2 (by 1?) using 2? and Mary J(3) = 4, J(4) = 3, and g1 & J => J(5) = 6, J(6) = 5, & g3 & J => J(8) = 7 J(7) = 8 i.e. J = (1)(2)(3 4)(5 6)(7 8). But the J,Mary? <> T. In fact, we can see that J must have no fixed points, J(i) <> i for all i, since there is nothing special about i = 1.

If J(1) = 2, then we get from Mary that J(3) = 3. contradiction.

If J(1) = 3, then J(2) = 4, J(3) = 1, J(4) = 2 (from Mary) =>

J(5) = 7, J(6) = 8, J(7) = 5, J(8) = 6 => J = (1 3)(2 4)(5 7)(6 8) (from g1)

But then J is incompatible with g3.

A similar analysis shows that J(1) cannot be 4,5,6,7 or 8; i.e. no J can be compatible with all three girls. So without loss of generality, throw away g3.

We have Mary = (1 3 2 4) (5 7 6 8)

g1 = (1 5 2 6) (3 8 4 7)

The following are the only possible boy guesses which are compatible with

both of these
B1: (1)(2)(3 4)(5 6)(7)(8) B2: (1 2)(3)(4)(5)(6)(7 8) B3: (1 3)(2 4)(5 7)(6 8) B4: (1 4)(2 3)(5 8)(6 7) B5: (1 5)(2 6)(3 8)(4 7) B6: (1 6)(2 5)(3 7)(4 8)

Note that B1 & B2 are incombatible, as are B3 & B4, B5 & B6, so at most three of them are mutually compatible. In fact, Mary, g1, B1, B3 and B5 are all mutually compatible (as are all the other possibilities you can get by choosing either B1 or B2, B3 or B4, B5 or B6). So if there are 2 girls there can be 3 boys, but no more, and we have already eliminated the case of 3 girls and 1 boy.

The only other possibility to consider is whether there can be 4 or more boys and 1 girl. Suppose there are Mary and 4 boys. Each boy must map 1 to a different digit or they would not be mutually compatible. For example if b1 and b2 both map 1 to 3, then they both map 3 to 1 (since a boy's map consists of transpositions), so both b1*b2 and b2*b1 map 1 to 1. Furthermore, b1 and b2 cannot map 1 onto spouses. For example, if b1(1) = a and b is the spouse of a, then b1(2) = b. If b2(1) = b, then b2(2) = a. Then b1*b2(1) = b1(b) = 2 and b2*b1(1) = b2(a) = 2 (again using the fact that boys are all transpostions). Thus the four boys must be:

B1: (1)(2)... or (1 2).... B2: (1 3)... or (1 4) ... B3: (1 5) ... or (1 6) ... B4: (1 7) ... or (1 8) ...

Consider B4. The only permutation of the form (1 7)... which is compatible

with Mary ( (1 3 2 4) (5 7 6 8) ) is
(1 7)(2 8)(3 5)(4 6)
The only (1 8)... possibility is
(1 8)(2 7)(3 6)(4 5)

Suppose B4 = (1 7)(2 8)(3 5)(4 6)

If B3 starts (1 5), it must be (1 5)(2 6)(3 8)(4 7) to be compatible with B4. This is compatible with Mary also.

Assuming this and B2 starts with (1 3) we get B2 = (1 3)(2 4)(5 8)(6 7) in order to be compatible with B4. But then B2*B3 and B3*B2 moth map 1 to 8. I.e. no B2 is mutually compatible with B3 & B4.

Similarly if B2 starts with (1 4) it must be (1 4)(2 3)(5 7)(6 8) to work with B4, but this doesn't work with B3.

Likewise B3 starting with (1 6) leads to no possible B2 and the identical reasoning eliminates B4 = (1 8)...

So no B4 is possible!

I.e at most 3 boys are mutually compatiblw with Mary, so 2 girls & 3 boys is optimal.

Thus:

Mary = (1 3 2 4) (5 7 6 8) Sue = (1 5 2 6) (3 8 4 7) John = (1)(2)(3 4)(5 6)(7)(8) Bob = (1 3)(2 4)(5 7)(6 8) Jim = (1 5)(2 6)(3 8)(4 7)

is one optimal solution, with the winner being John (4 right: 1 2 7 & 8)

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