I think it will be well to give here a formula for the general solution of each of the four most common forms of the diamond-letter puzzle. By the word "line" I mean the complete diagonal. Thus in A, B, C, and D, the lines respectively contain 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)×2^{2_n_+2} + (2^{_n_+2}×(1×3×5×7 . . . . . (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 divide the eight pieces of money in the same proportion. PUZZLING TIMES AT SOLVAMHALL CASTLE _SIR HUGH EXPLAINS HIS PROBLEMS_ The friends of Sir Hugh de Fortibus were so perplexed over many of his strange puzzles that at a gathering of his kinsmen and retainers he undertook to explain his posers. [Illustration]