Knight's Tour - Oregon State University

Case 2: Assume that we start on v 1 . C 1 =1+b E 12c, C 2 = d E 12e+b E 22c, C 3 = d E 22e+b E 32c,and C 4 = d E 32e.A: Assume we end on v 1 . Therefore E 1 ,E 2 ,E 3 are even so the floors and the ceilingscan be dropped from the above equations. C 3 = C 4 =⇒ E 22+ E 32= E 32=⇒ E 2 =0. →←.B: Assume we end on v 4 . Therefore E 1 ,E 2 ,E 3 are odd so C 1 =1+ E 1−1= E 1+1,2 2C 2 = E 1+1+ E 2−1= E 1+E 2, C2 2 2 3 = E 2+1+ E 3−1= E 2+E 3,andC2 2 2 4 = E 3+1. C2 1 = C 2 =⇒ E 1 =E 3 and C 3 = C 4 =⇒ E 2 =1.This is what we wanted to prove. You must start on v 1 and the edge between v 2 andv 3 must be used exactly once. It is clear that this one use of the edge between v 2 and v 3 mustoccur after the counts of v 1 and v 2 are completed. In other words, you must tour all thegrey squares and then tour all the white squares.3.2.1 ToursLemma 3.11 There is no tour on a 4 × 4 board.Proof. Because the 4 × 4 board is symmetric, we can assume that we start on a greysquare. Using theorem 3.10 we need to complete all the grey squares before jumping to awhitesquare. Imaginethatatourhadfinishedthegreysquares. Wewillshowthatfromany position on the white squares there cannot be a tour. This is because there are twodisjoint cycles. Since there is no possible way to jump from one cycle to the next, there arealways four white squares left over.4 × 4 boardLemma 3.12 Thegreysquarescanbetouredona4 × m board for m ≥ 5.Proof. To construct a tour of the grey squares, we use 1 beginning board, followed by 0or more middle boards, and finally 1 ending board. Then all the sub-boardsareconnectedasshowninthefigure. The tour begins in the upper-left and the square that the tourfinishes on is determined by whether m is even or odd.143