But Shaun’s already proved P equals NP. Can’t we move onto quantified SAT? Woof!

But Shaun’s already proved P equals NP. Can’t we move onto quantified SAT?

Woof!

Set Variables

using as an example

Steiner Triples

Set Variables

using as an example

Steiner Triples

Steiner Kirkman

… and BIBD’s, 0/1 encoding, symmetry breaking

See • choco manual 1.4.2, page 14 and 15• CSPLib prob028• choco manual 6.3, pages 64 and 65• choco manual 8.56 lex (constraint)

Represented with two bounds

kernel: intersection of all possible sets

envelope: the union of all possible sets upper bound

lower bound

We will do this by example, solving a problem

We are given a set S of size n. Produce n.(n-1)/6 triples (subsets of S of size 3) such that given any pair of triples their intersection is of size at most 1

Equivalently, every possible pair of elements in S occurs in onlyone of the n.(n-1)/6 triples

Given a set of size n, how many triples are we to produce?

Equivalently, every possible pair of elements in S occurs in onlyone of the triples

How many pairs are there? n.(n-1)/2

Every triple contains 3 pairs Therefore n.(n-1)/6 triples required

We are given a set S of size n. Produce n.(n-1)/6 triples (subsets of S of size 3) such that given any pair of triples their intersection is of size at most 1

A triple is also called a “block” and the entries of a block a “point”

This is then a BIBD (balanced incomplete block design)

Applications: design of experiments, testing, …

Modelling the Steiner Triple problem using Set Variables

Given a set of size n, how many triples are we to produce?

Equivalently, every possible pair of elements in S occurs in onlyone of the triples

How many pairs are there? n.(n-1)/2

Every triple contains 3 pairs Therefore n.(n-1)/6 triples required

Compile and Run

An alternative representation

0/1 Array

m=7

block

point

Sum of a column = 3, intersection between two columns is at most 1

1 1 1

1 1 1

1 1 1

1 1 1

1 1 1

1 1 1

1 1 1

Compile & Run

For both models try n=1, n=7, n=9

Show effect of symmetry breaking and search over only decision variable

How about we do the following:

Generate all triples <i,j,k> where 0 ≤ i < j < k < m and pick the ones we want

Have a zero/one IntegerVariable for each triple, in an arrayA, such that A[i][j][k] is 1 iff we select triple <i,j,k>

Flatten the array A into a one-D vector, say v?

Constrain v such that it sums to m.(m-1)/6

For every pair of triples that match on two indices make thesum of their corresponding constrained integer variables be at most 1 (i.e. select at most 1 of these triples)

Is this dumb?

Fix i and j: Sum of A[i][j][*] ≤ 1

Fix i and k: Sum of A[i][*][k] ≤ 1

Fix j and k: Sum of A[*][j][k] ≤ 1

But we need to ensure that the sum of the above sums is also at most 1

For every pair of triples that match on two indices make thesum of their corresponding constrained integer variables be at most 1 (i.e. select at most 1 of these triples)

The age of stupid

The age of stupid

The age of stupid

The age of stupid

The age of stupid

The age of stupid

The age of stupid

The age of stupid

The age of stupid

Compile & Run

An array, m by m, where pair[i][j] is an integer

pair[i][j] = k means that pair (i,j) is in kth block

Proposed by Chris Unsworth

Location, location, location

pair[i][j] = k iff points i and j are in block[k]

The number of points in a block is 3

Proposed by Chris Unsworth

Location, location, location

Constrain pair[i][j] = k iff block[k][i] = 1 and b[k][j] = 1

i and j are in the kth triple

if and only if kth triple contains i

and kth triple contains j

public class SteinerTriple03 { public static void main(String[] args) {

CPModel model = new CPModel(); int m = I nteger.parseInt(args[0]); int n = m*(m- 1)/6;

I ntegerVariable[][] block = makeIntVarArray("block",n,m,0,1); / / block[i][j] = 1 <- > j is in ith block I ntegerVariable[][] pair = new IntegerVariable[m][m]; / / pair[i][j] = k <- > (i,j) is in block[k]

for (int i=0;i<m- 1;i++) for (int j=i+1;j<m;j++) pair[i][j] = makeIntVar("pair_"+ i +"_"+ j,0,n- 1);

/ / / / one variable for each pair (i,j) where i<j / /

for (int i=0;i<n;i++) model.addConstraint(eq(sum(block[i]),3));

/ / / / there are 3 points in a block / /

for (int i=0;i<m- 1;i++) for (int j=i+1;j<m;j++) for (int k=0;k<n;k++)

model.addConstraint(ifOnlyI f(eq(pair[i][j],k),and(eq(block[k][i],1),eq(block[k][j],1)))); / / / / pair[i][j] = k <- > block[k][i] = 1 && block[k][j] = 1 / /





































CPSolver sol = new CPSolver(); sol.read(model);

I ntDomainVar[] decision = new IntDomainVar[m*(m- 1)/2]; for (int i=0,k=0;i<m- 1;i++)

for (int j=i+1;j<m;j++,k++) decision[k] = sol.getVar(pair[i][j]);

sol.setVarIntSelector(new StaticVarOrder(decision)); / / / / decision variables are the pairs / /

System.out.println(sol.solve()); sol.printRuntimeSatistics(); for (int i=0;i<n;i++){

for (int j=0;j<m;j++) if(sol.getVar(block[i][j]).getVal() == 1) System.out.print(j +" "); for (int j=0;j<m;j++) System.out.print(sol.getVar(block[i][j]).getVal()); System.out.println();

} }

CPSolver sol = new CPSolver(); sol.read(model);

I ntDomainVar[] decision = new IntDomainVar[m*(m- 1)/2]; for (int i=0,k=0;i<m- 1;i++)

for (int j=i+1;j<m;j++,k++) decision[k] = sol.getVar(pair[i][j]);

sol.setVarIntSelector(new StaticVarOrder(decision)); / / / / decision variables are the pairs / /

System.out.println(sol.solve()); sol.printRuntimeSatistics(); for (int i=0;i<n;i++){

for (int j=0;j<m;j++) if(sol.getVar(block[i][j]).getVal() == 1) System.out.print(j +" "); for (int j=0;j<m;j++) System.out.print(sol.getVar(block[i][j]).getVal()); System.out.println();

} }

Compile & Run

More generally Steiner Triple is a BIBD

So?

1. Set variables2. Steiner triples3. Various models (four!)4. Symmetry breaking with

lex5. Dare to be stupid

But Shaun’s already proved P equals NP. Can’t we move onto quantified SAT? Woof!

Documents

But Shaun’s already proved P equals NP. Can’t we move onto quantified SAT? Woof!