# Checkerboard Paths: A Combinatorial Exercise Kyle Pena, Messiah College.

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Checkerboard Paths: A Combinatorial Exercise Kyle Pena, Messiah College

Problem Statement “On a four by four checkerboard we place 8 white and 8 black pieces at random. What is the probability that a ‘path’ of adjacent pieces connects any corner to its respective diagonal corner?” A possible arrangement

Examples of Paths A Path in BlackA Path in White A Configuration of Pieces Containing Two Possible Paths

Simplifications Simplification 1: Consider paths of only one color Bijection: Ф(black)=white #blackpaths = #whitepaths, and so (#blackpaths) + (#whitepaths) = 2*(#whitepath) Multiply by two “at the end”

Simplifications (cont’d) Simplification 2: Only LL to UR Bijection: Rotate board clockwise 90 degrees. By similar argument, 2x

Simplifications (cont’d ◦ cont’d) Simplification 3: Only seven pieces at a time (we consider the eighth later). The process by which we account for the eighth piece here is a little more involved…

Finding The Number of Paths The Easy Way To recap: ONLY WHITE PIECES ONLY LL TO UR ONLY SEVEN PIECES Original Problem (easy to count)(difficult to count) We shall now count those arrangements of seven white pieces which form paths from the lower left to the upper right corner.

Using Pascal’s Triangle 1 11 211 33 6 1 4 10 1 5 4 5 1 1 11 201566 11 R=5 R=1 R=2 R=3 R=4 R=0 R=6 6 C 4 = 20 rckrck

Counts lattice paths (inductively) 1 1 1 1 1 2 3 4 1 3 6 10 1 4 20

Accounting for the Eighth Piece A harrowing task! (Caveats!) 20*9 = 180 arrangements? Wrong! (Presence of duplicated arrangements) Observe: =

Counting Duplicated Arrangements Double countings of arrangements occur at corners

Counting Corners (via Pascal’s Triangle) Two kinds of corners: 1 1 1 1 1 2 3 4 1 3 6 10 1 4 20 3 1 3 * 1 = 3

Counting Corners (via Pascal’s Triangle) Two kinds of corners: 1 1 1 1 1 2 3 4 1 3 6 10 1 4 20 3 * 1 = 3 3 1 3 + 3 = 6 corners

20 minutes later… 1 corner3 corners6 corners0 corners 3 corners 6 corners 0 corners 5 corners 6 corners 5 corners 3 corners 6 corners 3 corners 1 corner

And the results are… 60 corners over all 20 paths, that is, 30 duplicated arrangements…

De-simplifying. 180-30=150 eight piece arrangements (simplification 3) Other pair of diagonals: 150*2 = 300 arrangements (simplification 2) Other color: 300*2 = 600 arrangements containing corner to diagonal corner paths (simplification 1)

Calculating Probability (#viable arrangements)/(#total arrangements) Total arrangements easily computed by counting ways we can place eight white pieces on a 4 by 4 board That is, 4*4 C 8 16 C 8 = 12840. 600/12840 = 15/321 =  0.0467, or approximately 4.7%. We can think of this as just shy of a 1 in 20 chance. We’re finished!

Easily Generalized… To a variety of tasks… - Counting m+n-1 piece paths on an m×n checkerboard: If m≥n then #paths = m+n-1 C n If n>m then #paths = m+n-1 C m

Easily Generalized (Counting Corners) Total number of corners over all m+n-1 piece paths on an m×n checkerboard #paths =  i=2 to n  j=1 to m-1 [( i-1 C j )*( m-i C n-j-1 )] +  i=1 to m-1  j=2 to n [( i C j-1 )*( m-i-1 C n-j )]

Generalizations… Visit home.messiah.edu/~kp1197, or google “Kyle Pena” (I happen to be the internet’s most famous Kyle Pena) for generalizations to n dimensional lattices of arbitrary size.home.messiah.edu/~kp1197

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