ontain 5, 5, 7, and 9 letters. A has a non-palindrome
line (the word being BOY), and the general solution for such cases, where
the line contains 2_n_ + 1 letters, is 4(2^_n_ - 1). Where the line is a
single palindrome, with its middle letter in the centre, as in B, the
general formula is [4(2^_n_ - 1)]^{2}. This is the form of the
Rat-catcher's Puzzle, and therefore the expression that I have given
above. In cases C and D we have double palindromes, but these two
represent very different types. In C, where the line contains 4^n-1
letters, the general expression is 4^(2^{2_n_}-2). But D is by far the
most difficult case of all.

I had better here state that in the diamonds under consideration (i.) no
diagonal readings are allowed--these have to be dealt with specially in
cases where they are possible and admitted; (ii.) readings may start
anywhere; (iii.) readings may go backwards and forwards, using letters
more than once in a single reading, but not the same letter twice in
immediate succession. This last condition will be understood if the
reader glances at C, where it is impossible to go forwards and backwards
in a reading without repeating the first O touched--a proceeding which I
have said is not allowed. In the case D it is very different, and this is
what accounts for its greater difficulty. The formula for D is this:

[Illustration:

(_n_+5)x2^{2_n_+2} + (2^{_n_+2}x(1x3x5x7
. . . . . (2n-1)) / _n_) - 2^{_n_+4} - 8 ]

where the number of letters in the line is 4_n_+1. In the example given
there are therefore 400 readings for _n_ = 2.

See also Nos. 256, 257, and 258 in _A. in M._

[Illustration

A

Y
YOY
YOBOY
YOY
Y

B
L
LEL
LEVEL
LEL
L

C

N
NON
NOOON
NOONOON
NOOON
NON
N

D

L
LEL
LEVEL
LEVEVEL
LEVELEVEL
LEVEVEL
LEVEL
LEL
L ]

31.--_The Manciple's Puzzle._

The simple Ploughman, who was so ridiculed for his opinion, was perfectly
correct: the Miller should receive seven pieces of money, and the Weaver
only one. As all three ate equal shares of the bread, it should be
evident that each ate 8/3 of a loaf. Therefore, as the Miller provided
15/3 and ate 8/3, he contributed 7/3 to the Manciple's meal; whereas the
Weaver provided 9/3, ate 8/3, and contributed only 1/3. Therefore, since
they contributed to the Manciple in the proportion of 7 to 1, they must
di