Matpic
mathematics, in pictures
Theme

Decided by exhaustion — page 10

Questions with finitely many cases, settled by going through all of them — and what changes when a claim about every argument becomes a count.
A random walk across the regions to the one that satisfies every clause. Four overlapping ellipses with a dot in each of their sixteen regions, one marked as the only assignment satisfying the clauses, and a path of arrows from a random starting region to it, each arrow crossing one ellipse. Logic

A walk that beats trying everything

To decide whether clauses of three letters can all be satisfied, the obvious method tries all 2ⁿ assignments. Uwe Schöning's method, from 1999, starts at a random assignment and wanders: pick a clause that is false, flip one of its letters at random, and repeat three times as many times as there are letters. A single try usually fails, but it succeeds with chance at least about (3/4)ⁿ, so about (4/3)ⁿ tries are enough — and the reason is a walk on a line that goes the wrong way two times in three.

 Class numbers of x² + ny², with Euler's idoneal numbers marked. A scatter of class number against n up to 2000, on a logarithmic scale, with the 65 idoneal numbers marked on the power-of-two levels. Number

Euler's sixty-five convenient numbers

For some n, whether a prime can be written as x² + ny² is settled by its remainder on division by 4n alone, the way a prime's remainder on division by four settles whether it is a sum of two squares. Euler found sixty-five such n, from 1 to 1,848, called them convenient, and used the largest to prove that 18,518,809 is prime. Every one of them is a number whose class group has no element of order more than two — and whether the list is complete is still not known.

 A single table for 9 guests over 4 nights. Small circles of 9 guests, one per night, each showing the night's seating as a closed zigzag path; every pair of guests is adjacent in exactly one of them. Probability

Every pair side by side, once

Seat an odd number of guests at round tables for as many nights as it takes, the same table sizes every night, so that every two guests sit side by side on exactly one night. For a single table a zigzag turned a notch each night does it for any number of guests. For other table plans the answer is almost always yes — and for six guests at two tables of three, nine at tables of four and five, and eleven at three, three and five, an exhaustive search proves it is no.

All themes