Looking Glass Zoo

Here is a puzzle of Carrollian style from 1936. It was proposed by Sir Arthur Eddington and published in the New Statesman and Nation magazine [1].

The puzzle

I took some nephews and nieces to the Zoo, and we halted at a cage marked

  • Tovus Slithius, male and female.
  • Beregovus Mimsius, male and female.
  • Rathus Momus, male and female.
  • Jabberwockius Vulgaris, male and female.

The eight animals were asleep in a row, and the children began to guess which was which. “That one at the end is Mr. Tove.” “No, no! It’s Mrs. Jabberwock,” and so on. I suggested that they should each write down the names in order from left to right, and offered a prize to the one who got most names right.

As the four species were easily distinguished, no mistake would arise in pairing the animals; naturally a child who identified one animal as Mr. Tove identified the other animal of the same species as Mrs. Tove.

The keeper, who consented to judge the lists, scrutinised them carefully. “Here’s a queer thing. I take two of the lists, say, John’s and Mary’s. The animal which John supposes to be the animal which Mary supposes to be Mr. Tove is the animal which Mary supposes to be the animal which John supposes to be Mrs. Tove. It is just the same for every pair of lists, and for all four species.

“Curiouser and curiouser! Each boy supposes Mr. Tove to be the animal which he supposes to be Mr. Tove; but each girl supposes Mr. Tove to be the animal which she supposes to be Mrs Tove. And similarly for the other animals. I mean, for instance, that the animal Mary calls Mr. Tove is really Mrs. Rathe, but the animal she calls Mrs. Rathe is really Mrs. Tove.”

“It seems a little involved,” I said, “but I suppose it is a remarkable coincidence.”

“Very remarkable,” replied Mr. Dodgson (whom I had supposed to be the keeper) “and it could not have happened if you had brought any more children.”

How many nephews and nieces were there? Was the winner a boy or a girl? And how many names did the winner get right?

A solution

A possible solution involves a clever application of matrices in which each child’s list is evaluated in a 4×4 matrix in the following way. We label each species with a number: 1 for Tovus, 2 for Beregovus, 3 for Rathus, and 4 for Jabberwockius. Then a guess is nothing, but a permutation of the numbers 1, 2, 3, 4. For example, the guess which gets the Beregovus(2) right, but supposes a Tovus(1) to be a Rathus(3), a Rathus(3) to be a Jabberwockius(4) and a Jabberwockius(4) to be a Tovus(1) can be written as {\sigma: 1\mapsto 3}, {2\mapsto 2}, {3\mapsto 4}, {4\mapsto 1}. Thus a so-called permutation matrix can be constructed whose only non-zero elements are {A_{j\sigma(j)}}, {j=1,2,3,4}, so it looks something like this

\displaystyle \begin{pmatrix} 0&0&\ast&0\\ 0&\ast&&0\\ 0&0&0&\ast\\ \ast&0&0&0 \end{pmatrix}.

In addition, we must indicate whether the genders are guessed correctly, or not. This can be achieved by setting {A_{j\sigma(j)}} to {+1}, if the child got the genders right and {-1}, if he/she didn’t. Let assume that in the previous example the gender of the Tovuses and Beregovus were guessed right, then we get the matrix

\displaystyle \begin{pmatrix} 0&0&+1&0\\ 0&+1&&0\\ 0&0&0&-1\\ -1&0&0&0 \end{pmatrix}.

Let {A_1,\dots,A_b,\dots,A_{b+g}} denote the matrices made from the lists as explained above. The first {b} matrices are of the boys, and the rest of matrices belongs to the girls.

Now, it’s time to decrypt the keeper’s observations. Mr. Dodgson’s first clue simply means that any pair of different matrices anti-commute, i.e.

\displaystyle A_iA_j+A_jA_i=0\quad (i\neq j).\qquad\qquad (\text{1st clue})

Let’s translate into matrix form the second clue about the matrices of boys and girls. It states that any matrix belonging to a boy has a square equal to the identity matrix, and any girl’s matrix has square equal to minus the identity, that is

\displaystyle A_i^2=I,\quad i=1,\dots,b,\qquad\qquad (\text{2nd clue - boys})


\displaystyle A_i^2=-I,\quad i=b+1,\dots,b+g.\qquad\qquad (\text{2nd clue - girls})

These matrix equations are very important. It was solved in 1928 by Dirac, who found a set of matrices satisfying these conditions and used it to construct his relativistic equation for the electron, which led him to predict the existence of positron (discovered in 1932). The famous Dirac equation can be written as

\displaystyle (i\hbar\gamma^\mu\partial_\mu-mc)\psi=0,

where now we should concentrate on the objects {\gamma^\mu}. These are the {4\times 4} matrices in question. They have the form

\displaystyle \begin{gathered} \gamma^0=\begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix},\quad \gamma^1=\begin{pmatrix} 0&0&0&1\\ 0&0&1&0\\ 0&-1&0&0\\ -1&0&0&0 \end{pmatrix},\quad \gamma^2=\begin{pmatrix} 0&0&0&-i\\ 0&0&i&0\\ 0&i&0&0\\ -i&0&0&0 \end{pmatrix},\\ \gamma^3=\begin{pmatrix} 0&0&1&0\\ 0&0&0&-1\\ -1&0&0&0\\ 0&1&0&0 \end{pmatrix},\quad \gamma^5=\begin{pmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{pmatrix}, \end{gathered}

(The index 5 is due to the fact that physicists used to call {\gamma^0} to {\gamma^4}.)

To answer Eddington’s question, there were {5} children, {3} boys ({A_1=\gamma^0}, {A_2=\gamma^3}, {A_3=\gamma^5}) and {2} girls ({A_4=\gamma^1}, {A_5=\gamma^2}), the winner was a boy (with matrix {A_1=\gamma^0}), who guessed {4} animals correctly in total, both in species and gender.


  1. A. Eddington, in Caliban’s Puzzle Column by H. Phillips, New Statesman and Nation, December 19, 1936 and January 9, 1937.
  2. D.H. Sattinger and O.L. Weaver, Lie Groups and Algebras with Applications to Physics, Geometry, and Mechanics. Applied Mathematical Sciences 61, Springer, 1986.

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