IEOR 269, Spring 2010 Integer Programming and Combinatorial ...

IEOR269 notes, Prof. Hochbaum, 2010 41
19 The Chinese Checkerboard Problem and a First Look at Cutting
Planes
In this problem 10 we are given a standard Chinese checkerboard and three different types of diamond
tiles (Figure 15). Each type of tile has a prescribed orientation, and each tile covers exactly four
circles on the board. The Chinese checkerboard problem is a special case of the Set Packing problem
and asks:
What is the maximum number of diamonds that can be packed on a Chinese checkerboard
such that no two diamonds overlap or share a circle?
19.1 Problem Setup
Let D be the collection of diamonds, and let O be a set containing all the pairs of diamonds that
overlap; that is, (d, d ′ ) ∈ O implies that diamonds d and d ′ have a circle in common. For every
d ∈ D, define a decision variable x d as follows.
x d =
{ 1 if diamond d is selected
0 if diamond d is not selected
To determine the total number of decision variables, we consider all possible placements for each
type of diamond. Starting with the yellow diamond, we sweep each circle on the checkerboard and
increment our count if a yellow diamond can be placed with its upper vertex at that circle. Figure
16 illustrates this process; a yellow diamond can be “hung” from every black circle in the image,
and there are 88 total black circles. Using the same procedure, we find that there are 88 legal
placements for green diamonds and another 88 for yellow diamonds. This brings the total number
of decision variables to 264.
19.2 The First Integer Programming Formulation
We can formulate the problem as the following integer program, which is also known as the set
packing problem.
max
∑
d∈D
x d
subject to x d + x d ′ ≤ 1 for ( d, d ′) ∈ O
x d ∈ {0, 1}
for d ∈ D
The optimal objective value for this IP is 27, but the optimal objective for the LP relaxation is
132. This is a huge difference!
19.3 An Improved ILP Formulation
As it turns out, we can do much better by reformulating the constraints to make them tighter. We
begin by observing that for many circles on the board, there are 12 different diamonds competing
for that space. (See Figure 17.)
10 Thanks to Professor Jim Orlin for the use of his slides and to Professor Andreas Schulz for this example.