Playground

Search versus construction

Pick an order n. The search below tries to build every (v, k, 1) difference set with v = n² + n + 1 points and k = n + 1 members, one member at a time, backing up whenever a difference repeats. It runs in a background thread, so the page stays responsive, and each set it finds is checked against the Singer sets for the same order.

The search only looks for sets that contain 0 and 1. That loses nothing: the difference 1 occurs exactly once, as a − b = 1 for one pair, and sliding the set back by b gives a copy that contains 0 and 1. So each family of translates shows up exactly once. The completeness conjecture of Lesson 17 predicts that every set found is of Singer type: a Singer set up to sliding, or a mirror image of one.

What to try

• n = 6 finishes at once with nothing found. The Bruck–Ryser theorem already says so: 6 leaves remainder 2 when divided by 4 and is not a sum of two squares.
• n = 10 passes Bruck–Ryser (10 = 1² + 3²), and the search rules it out by brute force in about a second. Ruling out every plane of order 10, cyclic or not, took Lam, Thiel, and Swiercz years of computer time (1989): a cyclic plane is far more constrained.
• n = 11 and 12 take seconds to minutes, and the work grows about ninefold with each step in n. Singer's construction, in the last box, stays far below a millisecond: it writes the set down instead of searching for it.