A notch below the monolength category in elegance and challenge to constructors is what we may call the bilength puzzle -- one in which each answer has one of exactly two specified lengths. In the common situations where monolengths are impossible, we will now show that bilengths are always possible. In such a case, the average answer length, w, is not an integer. That is, when you divide q by a there will be positive integers n and r, called the quotient and remainder respectively, such that q = na + r and 1 ≤ r ≤ a - 1. With a little mathematical sleight of hand, we can write q = na + r as q = n(a - r + r) + r = n(a - r) + nr + r = n(a - r) + (n + 1)r, which says that you can form a bilength puzzle with a - r answers of length n and r answers of length n + 1. This completes the proof, but we should note that the pair n, n + 1 of lengths is not necessarily unique; there may well be other possibilities as the following example will show.

Example. In my puzzle, bgc, the quotation has q = 253 letters while the acrostic has a = 33 letters. Dividing 253 by 33, we find that 253 = 7 × 33 + 22 (a = 33, r = 22) so as shown above, we could have 11 7-letter answers and 22 8-letter answers, which is what bgc actually has. But we could also have written 253 = 7(22 + 11) + 22 = 7 × 22 + 7 × 11 + 2 × 11 = 7 × 22 + 9 × 11, so another possibility would have been to have 11 words of length 9 and 22 of length 7.

Of course, we could similarly define terms such as trilength, quadlength, etc., but it soon becomes ridiculous. After all, every crostic is k-length for some k. At the opposite extreme from monolength would be a puzzle in which no two answers have the same length--they all have different lengths. Given an acrostic length a and assuming each answer must have at least 3 letters, the least possible quote length is q = 3 + 4 + 5 + × × × + (a + 2) = (a + 2)(a + 3)/2 - 3 = a(a + 5)/2. If answers are to have no more than 20 letters, then a ≤ 18. With a = 18, you’d have to have q = 207 because you’d have to include all the lengths 3, 4, 5, . . ., 20. With a = 15, you’d have to have 187 ≤ q ≤ 195. It seems that opportunities for totally-mixed-length crostics are rare in practice. Construction would be difficult and hardly seems worthwhile, though as Sue remarked, it would be tempting to try to construct one in triangular form, i.e., having answers of consecutive lengths in order from the top down.