Lecture 22:
CP Models for
Sports Scheduling
© J. Christopher Beck 2008
1
Outline




Arc consistency and generalized arc
consistency
Recall the problem (again)
Step 1: Two CP Models
Step 2: A better CP Model
© J. Christopher Beck 2008
2
Readings

P Ch 10.3
© J. Christopher Beck 2008
3
Arc Consistency


Fundamental notion in CP!
Given: c1(v1,v2)



a binary constraint
e.g., v1 < v2
Given: D1 = D2 = {0, 1, …, 5}
© J. Christopher Beck 2008
4
Arc Consistency

c1 is arc consistent iff


for all values d1 є D1 there exists a value
d2 є D2 such that c1(v1=d1,v2=d2)  T
And similarly for all values d2 є D2
V1
<
{0,1,2,3,4,5}
© J. Christopher Beck 2008
V2
{0,1,2,3,4,5}
5
Generalized Arc Consistency
(GAC)


Given: c1(v1,..., vm)
C1 is GAC iff


for all variables di, for all values di є Di
there exists a tuple of values [dj є Dj], j≠i
such that C1(vi=di,[vj=dj])  T
E.g., c1(v1,v2,v3,v4)

for every value in d1 є D1 there must be a
triple [d2 є D2, d3 є D3, d4 є D4] s.t.
c1(v1=d1, v2=d2, v3=d3, v4=d4)  T
© J. Christopher Beck 2008
6
All-Diff vs. Clique of ≠



Logically, all-diff(v1, v2, …, vn) =def
vi ≠ vj for 1 ≤ i < j ≤ n
D1=D2=D3={1,3}
Establish AC (or GAC) for


v1 ≠ v2, v1 ≠ v3, v2 ≠ v3
all-diff(v1,v2,v3)
© J. Christopher Beck 2008
7
Generalizing All-Diff



All-Diff say that each value can appear
at most once
#1: each value must appear a given
(specific) number of times
#2: Global Cardinality Constraint (GCC):
each value must appear a given
(variable) number of times

e.g., between 2 and 5 times
© J. Christopher Beck 2008
8
The Global Cardinality
Constraint

gcc(card, value, base)




card – array of variables
value – array of values
base – array of variables
value[i] is taken by card[i] elements of
base
gcc([{0,1}, {1,3}], [0, 1], [w1, …, wn])
card
© J. Christopher Beck 2008
value
base
9
Question from a
Previous Quiz
gcc([C0, C1, C2, C3, C4], [0, 1, 2, 3, 4]
[W0, W1, W2, W3, W4, W5, W6, W7, W8, W9])
The initial domain of each W variable is {0,1,2,3,4}
The initial domains of the cardinality variables are:
Which of the following assignments of the W variables
satisfies the gcc-constraint?
C0 0,2,3,4
C1 2,3
C2 1,3
C3 1
C4 0,1
W0
W1
W2
W3
W4
W5
W6
W7
W8
W9
1
4
0
1
1
0
2
2
0
1
3
2
0
0
2
2
1
3
1
2
1
1
3
0
2
1
0
3
2
0
2
0
1
4
0
0
0
2
1
1
1
4
2
2
5
3
1
1
0
0
4
1
2
2
2
Question from a Quiz
gcc([C0, C1, C2, C3, C4], [0, 1, 2, 3, 4]
[W0, W1, W2, W3, W4, W5, W6, W7, W8, W9])
The initial domain of each W variable is {0,1,2,3,4}
The initial domains of the cardinality variables are:
Which of the following assignments of the W variables
satisfies the gcc-constraint?
C0 0,2,3,4
C1 2,3
C2 1,3
C3 1
C4 0,1
W0
W1
W2
W3
W4
W5
W6
W7
W8
W9
1
4
0
1
1
0
2
2
0
1
3
2
0
0
2
2
1
3
1
2
1
1
3
0
2
1
0
3
2
0
2
0
1
4
0
0
0
2
1
1
1
4
2
2
5
3
1
1
0
0
4
1
2
2
2
GAC on All-Diff & GCC

There exists a polynomial time
algorithm to enforce GAC on all-diff and
GCC

otherwise if wouldn’t be very useful
© J. Christopher Beck 2008
12
3-Step Approach

Step 1: HAPs

Find at least n HAPs




HAP – string of H, A, B
Find a set of n consistent HAPs
Step 2: Assign games to HAPs
Step 3: Assign teams to HAPs
© J. Christopher Beck 2008
14
Constraint Programming
Models


Double RR with odd n
Break is now 3 consecutive Hs or As



No breaks are allowed
H or B in at least 2 of first 4 rounds
No team can be away in both of the
final 2 games
© J. Christopher Beck 2008
15
Step 1: CP Model 1

ht, at, bt – 0,1 variables denoting home,
away, bye for a given team in slot t
ht  at  bt  1
t  1,...,2n
at  at 1  at  2  2 t  1,...,2n  2
ht  ht 1  ht  2  2 t  1,...,2n  2
a2 n 1  a2 n  1
4
a
i
2
i 1
© J. Christopher
Beck 2008
No breaks
No team away for both
of final 2 games
H or B in at least 2 of first 4 slots
16
Step 1: CP Model 2

gt є {H,A,B} – variable denoting home,
away, bye for a given team in slot t
gcc([{0,1,2}, {0,1,2}], [H, A], [gt, gt+1, gt+2])
t = 1, …, 2n-2
(g2n ≠ A) OR (g2n-1 ≠ A)
No breaks
No team away for both
of final 2 games
gcc([{0,1,2}], [A], [g1, g2, g3 , g4])
H or B in at least 2 of first 4 slots
© J. Christopher Beck 2008
17
Step 1: CP Model 2

Some additional constraints
gcc([{n-1}, {n-1}, {2}], [H, A, B], [g1, …, g2n])
2 Bs, Equal # As & Hs
Let t1, … tn, be the HAP for one slot (across all teams)
gcc([{(n-1)/2}, {(n-1)/2}, {1}], [H, A, B], [t1, …, tn])
1 Bs, Equal # As & Hs
© J. Christopher Beck 2008
18
Step 1: CP Model


With either of these models, you can
then easily generate all possible HAPs
You could also add a cost function to
generate only “good” HAPs
© J. Christopher Beck 2008
19
Step 2: Assign Games


Given a set of consistent HAPs
Devise a CP model to assign games

The one in the book is overly complicated!
slot
1
2
3
4
5
6
7
8
9
10
Team a
H
A
H
A
B
H
A
H
A
B
Team b
H
B
A
H
A
H
B
A
H
A
Team c
A
A
B
H
H
A
A
B
H
H
Team d
B
H
H
A
A
B
H
H
A
A
Team e
A
H
A
B
H
A
H
A
B
H
© J. Christopher Beck 2008
20
Step 2: CP Model

Let P be a n * 2n matrix of patterns



P(i,j) = {H,A,B} – The team assigned to
pattern i plays P(i,j) in time slot j
This is given as input from Step 1!
Let G be a n * 2n matrix


Each element of G has initial domain [0..n]
G(i,j) is the “team” that team i will play in
time slot j
Really the pattern that
© J. Christopher Beck 2008
pattern i will play in slot j
22
P(i,j) & G(i,j)
P
G
H
A
H
A
B
A
H
A
H
B
H
B
A
H
A
A
B
H
A
H
A
A
B
H
H
H
H
B
A
A
B
H
H
A
A
B
A
A
H
H
A
H
A
B
H
H
A
H
B
A
0..5
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
0..5
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
0..5
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
0..5
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
0..5
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
© J. Christopher Beck 2008
23
Preprocess P(i,j) & G(i,j)
P
G
H
A
H
A
B
A
H
A
H
B
H
B
A
H
A
A
B
H
A
H
A
A
B
H
H
H
H
B
A
A
B
H
H
A
A
B
A
A
H
H
A
H
A
B
H
H
A
H
B
A
3,5
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
3,5
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
1,2
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
0
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
1,2
0..5 0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
© J. Christopher Beck 2008
24
Preprocess P(i,j) & G(i,j)
P
G
H
A
H
A
B
A
H
A
H
B
H
B
A
H
A
A
B
H
A
H
A
A
B
H
H
H
H
B
A
A
B
H
H
A
A
B
A
A
H
H
A
H
A
B
H
H
A
H
B
A
3,5
4,5
0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
3,5
0
0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
1,2
4,5
0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
0
1,3
0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
1,2
1,3
0..5 0..5 0..5 0..5
0..5 0..5 0..5
0..5
© J. Christopher Beck 2008
25
Preprocess P(i,j) & G(i,j)
P
G
H
A
H
A
B
A
H
A
H
B
H
B
A
H
A
A
B
H
A
H
A
A
B
H
H
H
H
B
A
A
B
H
H
A
A
B
A
A
H
H
A
H
A
B
H
H
A
H
B
A
3,5
4,5
2,5
2,3
0
3,5
4,5
2,5
2,3
0
3,5
0
1,4
1,4
3,5
3,5
0
1,4
1,4
3,5
1,2
4,5
0
1,4
2,4
1,2
4,5
0
1,4
2,4
0
1,3
2,5
2,3
3,5
0
1,3
2,5
2,3
3,5
1,2
1,3
1,4
0
2,4
1,2
1,3
1,4
0
2,4
© J. Christopher Beck 2008
26
Constraints

(G(i,j) = k) ↔ (G(k,j) = i)


G
for all i,k s.t. P(i,j) ≠ B, P(k,j) ≠ B
for all j
3,5
4,5
2,5
2,3
0
3,5
4,5
2,5
2,3
0
3,5
0
1,4
1,4
3,5
3,5
0
1,4
1,4
3,5
1,2
4,5
0
1,4
2,4
1,2
4,5
0
1,4
2,4
0
1,3
2,5
2,3
3,5
0
1,3
2,5
2,3
3,5
1,2
1,3
1,4
0
2,4
1,2
1,3
1,4
0
2,4
© J. Christopher Beck 2008
27
Constraints


P
all-diff(G(1,1), G(1,3), G(1,7), G(1,9))
all-diff(G(1,2), G(1,4), G(1,6), G(1,8))
H
A
H
A
B
A
H
A
H
B
H
B
A
H
A
A
B
H
A
H
A
A
B
H
H
H
H
B
A
A
B
H
H
A
A
B
A
A
H
H
A
H
A
B
H
H
A
H
B
A
© J. Christopher Beck 2008
28
All-Diff for Row 2?
P
H
A
H
A
B
A
H
A
H
B
H
B
A
H
A
A
B
H
A
H
A
A
B
H
H
H
H
B
A
A
B
H
H
A
A
B
A
A
H
H
A
H
A
B
H
H
A
H
B
A
© J. Christopher Beck 2008
29