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Handbook of Constraint Programming 10.5.1, 10.5.2

Presented by: Shant KarakashianSymmetries in CP, Sprint 2010

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Outline• Symmetry Breaking During Search (SBDS)

– Example: 8-Queens– Adding Constraints– Correctness – Implementations– Problems

• Symmetry Breaking via Dominance Detection (SBDD)– Main Idea– Example: Add Different – Technique: No-goods & Dominance– Implementation of Dominance Checks– Propagation– Scheduling Checks– Comparing SBDS & SBDD

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Symmetry Breaking During Search

• Symmetry-excluding search trees introduced by [Backofen and Will '99]

• Gent and Smith '00 described in more detail the implementation of the technique using the name “Symmetry Breaking During Search” (SBDS)

• Basic idea: add constraints to a problem so that after backtracking from a search decision no symmetric equivalent of that decision is ever allowed

• Dynamic technique: cannot add the constraints until search decision is made

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SBDS Example: 8-Queens

• 8 Queens positioned on 8x8 board• No two queens on the same row, column, or

diagonal• Each queen (Q[i], i=1,..8) belongs to a row• Each queen can be on a column j=1,..,8 (Q[i]=j)• Symmetries of the problem: – r90, r180, r270, x, y, d1, d2

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Adding Constraints at the 1st Level

• Q[1]=2: no constraints added

• After backtracking from Q[1]=2:– Assert Q[1]≠2– Never try any state

containing a symmetric equivalent to Q[1]=2

– Add constraints (Q[1]≠2)g

Q[1]=2 Q[1]≠2

(var ≠ val)g (Q[1] ≠2)r90 ≡ Q[2] ≠8(Q[1] ≠2)r180 ≡ Q[8] ≠7(Q[1] ≠2)r270 ≡ Q[7] ≠1(Q[1] ≠2)x ≡ Q[1] ≠7(Q[1] ≠2)y ≡ Q[8] ≠2(Q[1] ≠2)d1 ≡ Q[2] ≠1(Q[1] ≠2)d2 ≡ Q[7] ≠8

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Adding Constraints at the 2nd Level• Q[2]=4: no constraints added• Q[2] ≠4: add constraint if the corresponding symmetry is not

broken• For symmetry r90, add (Q[2]=4)r90

– only if r90 is not already broken– i.e. if (Q[2]=8) ≡ (Q[1]=2)r90

• Add constraints of the form:• (A → var≠val) → (A → var≠val)g

– e.g. (Q[1]=2 → Q[2]≠4) → (Q[1]=2 → Q[2]≠4)r90 – (Q[1]=2 → Q[2]≠4)r90 ≡ (Q[2]=8 → Q[4]≠7)

Q[1]=2Q[1]≠2

Q[2]=4 Q[2]≠4

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Constraints skipped at the 2nd Level

• Some symmetries are already broken:– e.g. (Q[1]=2 → Q[2]≠4)x → (Q[1]=2 → Q[2]≠4)x – (Q[1]=2 → Q[2]≠4)x ≡ (Q[1]=7 → Q[2]≠5)

• No solution containing Q[1]=2 can have the symmetries: x, y, d1 or d2– Problem constraints break these symmetries

Q[1]=2Q[1]≠2

Q[2]=4 Q[2]≠4

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Constraints Added to the Example

Q[1]=2 Q[1]≠2

Q[2]=4 Q[2]≠4

(var≠val)g (Q[1]≠2)r90 ≡ Q[2]≠8(Q[1]≠2)r180 ≡ Q[8]≠7(Q[1]≠2)r270 ≡ Q[7]≠1(Q[1]≠2)x ≡ Q[1]≠7(Q[1]≠2)y ≡ Q[8]≠2(Q[1]≠2)d1 ≡ Q[2]≠1(Q[1]≠2)d2 ≡ Q[7]≠8

(A → var≠val) → (A → var≠val)g (Q[1]=2 → Q[2]≠4) → (Q[1]=2 → Q[2]≠4)r90 ≡ (Q[2]=8 → Q[4]≠7)(Q[1]=2 → Q[2]≠4) → (Q[1]=2 → Q[2]≠4)r180 ≡ (Q[8]=7 → Q[7]≠5)(Q[1]=2 → Q[2]≠4) → (Q[1]=2 → Q[2]≠4)r270 ≡ (Q[7]=1 → Q[5]≠2)

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Example Contradicting the Notation in Figure 10.10 (page 352)

Q[1]=1 Q[1]≠1

Q[2]=5 Q[2]≠5

Consider 5-queens problem. <q1,q2,q3,q4,q5> = <1,3,5,2,4> is a solution.Applying the notation usein in Figure 10.10 for the assignment Q[1]=1 we get:(Q[1]=1 → (Q[2]≠5)r90 ≡ (Q[5]≠4)Notice that this constraint eliminates the solution <1,3,5,2,4>, before any solution symmetric to it is encountered.

This constraint must be specified as follows according to the expression (10.2):(Q[1]=1 → Q[2]≠5) → (Q[1]=1 → Q[2]≠5)r90 ≡ (Q[1]=5 → Q[5]≠4)

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Constraints Added by SBDS

• When constraints are added at the point of backtracking from a choice (var ≠ val):

(A → var≠val) → (A → var≠val)g – A is true and (var ≠ val) is true

• Can simplify the constraint to:Ag → (var≠val)g

• Ag ensures that only unbroken symmetries are considered

• Practical implementations do not add all the constraints

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Correctness of SBDS

• Backofen and Will '99 proved that this method:– Explores full non-symmetric search space– No solutions can be completely missed– No returned two solutions can be equivalent

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Implementations of SBDS

• Number of implementations of SBDS have been provided

• ECLiPSe is publicly available implementation [Petrie ’05]

• Demands the implementation of a function for each symmetry

• User has to identify and implement each symmetry

• Used successfully despite the difficulties

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Problems with SBDS

If the number of symmetries in a problem is large: • Large number of symmetry functions has to be

specified• In the worst case, too many functions to be

successfully compiled• Feasible to use the method in its entirety only

with problems with small symmetry groups• Choose only a subset of the symmetry groups for

problems with large symmetry groups

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Symmetry Breaking via Dominance Detection

• SBDD break symmetries during search• Checks every node being explored if it is

symmetrically equivalent to one already explored

• If equivalent, prunes the branch• Developed independently by Focacci & Milano

‘01 and Schamberger & Sellmann ‘01

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SBDD Main Idea

• Need to store the whole explored tree?• Explored tree is exponentially sized • Only need to store the roots of fully explored

subtrees• Search backtracked from an equivalent node: – Either visited the equivalent node– Deduced no need to visit

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SBDD Example: All Different

• Variables: v1, v2, v3

• All different constraints• Domains: {1,2,3,4}• 24 solutions • Any variable

permutation is a symmetric solution

• Nodes 7 & 3 are symmetric solutions

0:root

1:v1=1

2:v2=2

3:v3=3 4:v3≠3

8:v3≠2

6:v2=3

7:v3=2

5:v2≠2

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No-Good Nodes

• Based on the notion of no-goods: prune if equivalent to a no-good

• No-goods are the roots of completely traversed maximal subtrees by depth first search

• Node v is a no-good w.r.t. n if there exists an ancestor na of n s.t. v is the left child of na and v is not an ancestor of n

• All solutions search: – Fully explored trees may contain solutions– Avoid searching symmetric equivalent subtrees

n

r

a

v

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Dominance

• P(v): the path from the root of the search tree to a no-good node v

• S(n): When search at node n, set of variables whose domains are reduced to a singleton

• Node n is dominated if there exists a no-good v w.r.t. n and a symmetry g s.t. (P(v))g (\subseteq) S(n)

• v dominates n

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SBDD Technique

• Never generate the children of dominated nodes

• Example:– P(2)={v1=1, v2=2}

– (P(2))g={v1=1, v3=2}

– S(7)={v1=1, v3=2, v2=3}– Node 7 dominated by

no-good 2

0:root

1:v1=1

2:v2=2

3:v3=3 4:v3≠3

8:v3≠2

6:v2=3

7:v3=2

5:v2≠2

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Implementation of Dominance Checks

• SBDD requires a problem specific function:f(v,n) → {true, false}

• For each pair v & n, the search for g amounts to solving a sub graph isomorphism problem

• NP-complete problems at each node but good results obtained using the technique

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Implementation of Dominance Checks 2

Three broad techniques for implementation:1. Implement a dominance checker specific for

a particular problem2. Construct a constraint encoding of the

dominance problem (problem specific)3. Use computational group theory

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SBDD and Propagation

• Failed dominance checks can result in propagation:– Dominance check reports the missing variable-

value pair (v,x) to make the node dominated– Can remove x from the domain of v and propagate

the result

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Scheduling Dominance Checks

Dominance check:• Does not need to be done at every node of

the search tree• Sufficient if only the leaf nodes are checked• Tradeoff between cost of checks and savings in

search time

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Comparing SBDS & SBDD• Both do not require symmetries to be variable symmetries as in lex-

leader method• The representative solution in each equivalent class of solutions is

always the leftmost one in the traversed search tree for both • Both Respect variable and value ordering heuristics• SBDS places constraints to stop search before reaching

symmetrically equivalent nodes • SBDD prunes search nodes that are symmetrically equivalent to

visited nodes• Set of acceptable solutions is the same for both• SBDD can outperform SBDS in many problems since it does not

have to post large number of constraints