The Evergreen Project:
The Promise of Polynomials
to Boost CSP/SAT Solvers*
Karl J. Lieberherr
Northeastern University
Boston
joint work with Ahmed Abdelmeged, Christine Hang and Daniel Rinehart
UBC March 2007
Title inspired by a paper by Carla Gomes / David Shmoys
1
Two objectives
• I want you to become
– better writers of MAX-SAT/MAX-CSP solvers
• UBCSAT (Tompkins/Hoos) can it be improved by
what you learn today?
– better players of the Evergreen game
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Introduction
• Boolean MAX-CSP(G) for rank d, G = set of relations of
rank d
– Input
•
•
•
•
Input = Bag of Constraint = CSP(G) instance
Constraint = Relation + Set of Variable
Relation = int. // Relation number < 2 ^ (2 ^ d) in G
Variable = int
– Output
• (0,1) assignment to variables which maximizes the number of
satisfied constraints.
• Example Input: G = {22} of rank 3. H =
– 22:1 2 3
0
– 22:1 2 4 0
– 22:1 3 4 0
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1in3 has number 22
M = {1 !2 !3 !4} satisfies all
3
Variation
MAX-CSP(G,f):
Given a CSP(G) instance H expressed in n variables which
may assume only the values 0 or 1, find an assignment
to the n variables which satisfies at least the fraction f of
the constraints in H.
Example: G = {22} of rank 3
MAX-CSP({22},f): H =
22:1 2 3
22:1 2
4
22:1
3 4
22: 2 3 4
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0
0
0
0
in MAX-CSP({22},?). Highest value for ?
1in3 has number 22
4
P1 = you
P2 = I
Evergreen(3,2) game
Two players: They agree on a protocol P1 to
choose a set of 2 relations (=G) of rank 3.
– Player P1 chooses CSP(G)-instance H (limited).
– Player P2 solves H and gets paid by P1 the
fraction that P2 satisfies in H.
• This gives nice control to P1. P1 will choose an
instance that will minimize P2’s profit.
– Take turns.
R2, P2
R1, P1
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100% R2, 0% R1
100% R1, 0% R2
Protocol choice
• Randomly choose symmetric R1 and
symmetric R2 (independently) between 1
and 255 (Throw two dice choosing
symmetric relations).
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Tell me
• How would you react as player P1?
– The relations 22 and 22 have been chosen.
– You must create a CSP({22}) instance with
1000 variables in which only the smallest
possible fraction can be satisfied.
– What kind of instance will this be?
• What kind of algorithm should P2 use to
maximize its payoff?
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Game strategy in a nutshell
• Choose G={R1,R2} randomly (both
symmetric).
• P1 chooses instance so that payoff is
minimized.
• P2 finds solution so that payoff is
maximized (Solve MAX-CSP(G))
• Take turns: Choose G= …
• P2. Choose instance so that payoff is
minimized.
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Our approach by Example:
SAT Rank 2 example
14 : 1 2
14 :
34
14 :
56
7: 1 3
7: 1
5
7:
3 5
7: 2 4
7: 2
6
7:
4
6
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0
0
0
0
0
0
0
0
0
14: 1 2 = or(1 2)
7: 1 3 = or(!1 !3)
=H
Evergreen game:
maximize payoff
find maximum assignment
9
excellent peripheral vision
0
1
2
3
4
5
6
=k
8/9
7/9
Blurry vision
What do we learn from the abstract representation absH?
• set 1/3 of the variables to true (maximize).
• the best assignment will satisfy at least 7/9 constraints.
• very useful but the vision is blurry in the “middle”.
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appmean = approximation of the mean (k variables true)
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Our approach by Example
• Given a CSP(G)-instance H and an assignment
N which satisfies fraction f in H
– Is there an assignment that satisfies more than f?
• YES (we are done), absH(mb) > f
• MAYBE, The closer absH() comes to f, the better
– Is it worthwhile to set a certain literal k to 1 so that we
can reach an assignment which satisfies more than f
• YES (we are done), H1 = Hk=1, absH1(mb1) > f
• MAYBE, the closer absH1(mb1) comes to f, the better
• NO, UP or clause learning
absH= abstract representation of H
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14 : 1 2
14 :
14 :
7: 1
7: 1
7:
7: 2
7: 2
7:
14 : 2
14 :
14 :
7: 1
7: 1
7:
7: 2
7: 2
7:
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56
3
5
5
3
4
6
6
4
34
56
3
5
5
3
4
4
6
6
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
8/9
7/9
abstract representation
H
0
1
2
3/9
3
4
5
6
6/7 = 8/9
5/7=7/9
H0
3/7=5/9
maximum assignment away
from max bias: blurry
0
1
2
3
4
5
12
14 : 1 2
14 :
34
14 :
56
7: 1 3
7: 1
5
7:
3 5
7: 2 4
7: 2
6
7:
4
6
14 : 1 2
14 :
34
14 :
56
7:
3
7:
5
7:
3 5
7: 2 4
7: 2
6
7:
4
6
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0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
8/9
7/9
3/8
H
0
1
2
3
clearly
above
3/4
4
5
6
7/8=8/9
6/8=7/9
H1
maximum assignment away
from max bias: blurry
2/7=3/8
0
1
2
3
4
5
13
14 : 1 2
14 :
14 :
7: 1
7: 1
7:
7: 2
7: 2
7:
0
34
0
56 0
3
0
5
0
3 5
0
4
0
6 0
4
6 0
8/9
7/9
abstract representation
guarantees 7/9
H
7/8 = 8/9
6/7=8/9
6/8 = 7/9
5/7=7/9
abstract representation
guarantees 7/9
abstract representation
guarantees 8/9
H0
NEVER GOES DOWN: DERANDOMIZATION
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UBCSAT
H1
14
4/6
4/6
3/6
3/6
abstract representation guarantees
0.625 * 6 = 3.75: 4 satisfied.
0
1
2
rank 2
10: 1 = or(1)
7: 1 2 = or(!1 !2)
10 : 1 0
10 : 2 0
10 : 3 0
7:120
7:130
7:230
Evergreen game: G = {10,7}
How do you choose a
CSP(G)-instance to minimize
payoff? 0.618 …
3
4/6
4/6
4/6
rank 2
5: 1 = or(!1)
13: 1 2 = or(1 !2)
5:10
10 : 2 0
10 : 3 0
13 : 1 2 0
13 : 1 3 0
7:230
4/6
3/6
3/6
The effect of n-map
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First Impression
• The abstract representation = look-ahead
polynomials seems useful for guiding the
search.
• The look-ahead polynomials give us
averages: the guidance can be misleading
because of outliers.
• But how can we compute the look-ahead
polynomials? How do the polynomials help
play the Evergreen(3,2) game?
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Where we are
•
•
•
•
Introduction
Look-forward
Look-backward
SPOT: how to use the look-ahead
polynomials together with superresolution.
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Look Forward
• Why?
– To make informed decisions
– To play the Evergreen game
• How?
– Abstract representation based on look-ahead
polynomials
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Look-ahead Polynomial
(Intuition)
• The look-ahead polynomial computes the
expected fraction of satisfied constraints
among all random assignments that are
produced with bias p.
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Consider an instance: 40 variables,
1000 constraints (1in3)
1, …
22:
679
22:
,40
0
12
27
38 0
Abstract representation:
reduce the instance to
look-ahead polynomial
2
3p(1-p) = B1,3(p) (Bernstein)
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3p(1-p)2 for MAX-CSP({22})
Fraction of constraints that
are guaranteed to be
satisfied
1in3
0.5
0.4
0.3
0.2
0.1
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Coin bias (Probability of setting a variable to true)
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Look-ahead Polynomial
(Definition)
• H is a CSP(G) instance.
• N is an arbitrary assignment.
• The look-ahead polynomial laH,N(p)
computes the expected fraction of satisfied
constraints of H when each variable in N is
flipped with probability p.
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The general case MAX-CSP(G)
G = {R1, … }, tR(F) = fraction of constraints in F that use R.
x=p
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appSATR(x) over all R is a
super set of the Bernstein polynomials
(computer graphics, weighted sum of Bernstein polynomials)
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Rational Bezier Curves
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Bernstein Polynomials
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http://graphics.idav.ucdavis.edu/education/CAGDNotes/Bernstein-Polynomials.pdf
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all the appSATR(x) polynomials
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Look-ahead Polynomial in Action
• Focus on purely mathematical question first
• Algorithmic solution will follow
• Mathematical question: Given a CSP(G)
instance. For which fractions f is there
always an assignment satisfying fraction f of
the constraints? In which constraint systems
is it impossible to satisfy many constraints?
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Remember?
MAX-CSP(G,f):
Given a CSP(G) instance H expressed in n variables which
may assume only the values 0 or 1, find an assignment
to the n variables which satisfies at least the fraction f of
the constraints in H.
Example: G = {22} of rank 3
MAX-CSP({22},f):
22:1 2 3
22:1 2
4
22:1
3 4
22: 2 3 4
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0
0
0
0
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Mathematical Critical Transition
Point
MAX-CSP({22},f):
For f ≤ u: problem has always a solution
For f ≥ u + e: problem has not always a solution, e>0.
1
not always (solid)
u = critical transition point
always (fluid)
0
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The Magic Number
• u = 4/9
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3p(1-p)2 for MAX-CSP({22})
Fraction of constraints that
are guaranteed to be
satisfied
1in3
0.5
0.4
0.3
0.2
0.1
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Coin bias (Probability of setting a variable to true)
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Produce the Magic Number
• Use an optimally biased coin
– 1/3 in this case
• In general: min max problem
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The 22 reductions:
Needed for implementation
1,0
22
2,0
60
240
3,0
2,1
3,1
1,1
2,0
3
15
2,1
22 is expanded into 6 additional
relations.
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3,0
255
3,1
0
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The 22 N-Mappings:
Needed for implementation
41
0
22
1
73
1
2
0
134
146
2
1
104
2
2
0
0
97
1
148
22 is expanded into 7 additional
relations.
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General Dichotomy Theorem
MAX-CSP(G,f): For each finite set G of relations closed
under renaming there exists an algebraic number tG :
1
For f ≤ tG: MAX-CSP(G,f) has polynomial solution
For f ≥ tG+ e: MAX-CSP(G,f) is NP-complete, e>0.
hard (solid)
NP-complete
tG = critical transition point
0
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polynomial solution:
Use optimally biased coin.
Derandomize.
P-Optimal.
easy (fluid)
Polynomial
due to Lieberherr/Specker (1979, 1982)
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implications for the Evergreen game? Are you a better player?
Context
• Ladner [Lad 75]: if P !=NP, then there are
decision problems in NP that are neither
NP-complete, nor they belong to P.
• Conceivable that MAX-CSP(G,f) contains
problems of intermediate complexity.
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General Dichotomy Theorem
(Discussion)
MAX-CSP(G,f): For each finite set G of relations closed
under renaming there exists an algebraic number tG
1
For f ≤ tG: MAX-CSP(G,f) has polynomial solution
For f ≥ tG+ e: MAX-CSP(G,f) is NP-complete, e>0.
hard (solid), NP-complete
exponential, super-polynomial proofs ???
relies on clause learning
tG = critical transition point
0
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easy (fluid), Polynomial (finding an assignment)
constant proofs (done statically using look-ahead polynomials)
no clause learning
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min max problem
sat(H,M) = fraction of satisfied constraints in
CSP(G)-instance H by assignment M
tG = min
all CSP(G)
instances H
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max
sat(H,M)
all (0,1) assignments M
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Problem reductions are the key
• Solution to simpler problem implies
solution to original problem.
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min max problem
sat(H,M,n) = fraction of satisfied constraints in
CSP(G)-instance H by assignment M with n variables.
tG = lim
min
n to
all SYMMETRIC
infinity
CSP(G) -instances
H with n variables
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max
sat(H,M,n)
all (0,1) assignments M
to n variables
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Reduction achieved
• Instead of minimizing over all constraint
systems it is sufficient to minimize over the
symmetric constraint systems.
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Reduction
• Symmetric case is the worst-case: If in a
symmetric constraint system the fraction f
of constraints can be satisfied, then in any
constraint system the fraction f can be
satisfied.
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Symmetric the worst-case
n variables
n! permutations
.
.
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If in the big system the
fraction f is satisfied,
then there must
be a least one small system
where the fraction f is satisfied
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min max problem
sat(H,M,n) = fraction of satisfied constraints in
system S by assignment I
tG = lim
min
max
sat(H,M,n)
n to
all SYMMETRIC all (0,1) assignments M
infinity CSP(G) -instances to n variables where the
H with n variables first k variables are set
to 1
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Observations
• The look-ahead polynomial look-forward
approach has not been used in state-ofthe-art MAX-SAT and Boolean MAX-CSP
solvers.
• Often a fair coin is used. The optimally
biased coin is often significantly better.
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The Game Evergreen(r,m) for
Boolean MAX-CSP(G), r>1,m>0
Two players: They agree on a protocol P1 to
choose a set of m relations of rank r.
1. The players use P1 to choose a set G of m relations
of rank r closed under renamings.
2. Player 1 constructs a CSP(G) instance H with 1000
variables and at most 2*(1000 choose r) constraints
and gives it to player 2 (1 second limit).
3. Player 2 gets paid the fraction of constraints she
can satisfy in H (100 seconds limit).
4. Take turns (go to 1).
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For
Evergreen(3,2)
100% R2, 0% R1
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100% R1, 0% R2
48
Evergreen(3,2) protocol
possibilities
• Variant 1
– Player P2 chooses both relations G
– Player P1 chooses CSP(G) instance H.
– Player P2 solves H and gets paid by P1.
• This gives too much control to P2. P2 can choose
two odd relations which guarantees P2 a pay of 1
independent of how P1 chooses the instance H.
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Evergreen(3,2) protocol
possibilities
• Variant 2:
– Player P1 chooses a relation R1 (e.g. 22).
– Player P2 chooses a relation R2.
– Player P1 chooses CSP(G) instance H.
– Player P2 solves H and gets paid by P1.
• This gives nice control to P2. P2 will choose a relation
R2 that will maximize P2’s profit.
R2, P2
R1, P1
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100% R2, 0% R1
100% R1, 0% R2
Problem with variant 2
• P1 can just ignore relation R2.
• Gives P1 too much control because the
payoff for P2 depends only on R1 chosen
by P1.
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Protocol choice: variant 3
• Randomly choose symmetric R1 and
symmetric R2 (independently) between 1
and 255 (Throw two dice).
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Tell me
• How would you react as player P1?
– The relations 22 and 22 have been chosen.
– You must create a CSP({22}) instance with
1000 variables in which only the smallest
possible fraction can be satisfied.
– What kind of instance will this be? 4/9
symmetric instance with (1000 choose 3) constraints
• What kind of algorithm should P1 use to
maximize its payoff? compute optimal k +
best MAX-CSP solver.
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For
Evergreen(3,2)
Tells us how to
mix the two
relations
100% R2, 0% R1
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100% R1, 0% R2
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Role of tG in the
Evergreen(3,2) game
• 1: Instance construction: Choose a
CSP(G) instance so that only the fraction
tG can be satisfied: symmetric formula.
• 2: Choose an algorithm so that at least the
fraction tG is satisfied. (2 gets paid tG from
1).
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Game strategy in a nutshell
• 1. Best: Choose tG instance
• 2. Best: Get’s paid tG
• etc.
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Additional Information
• Rich literature on clause learning in SAT
and CSP solver domain. Superresolution
is the most general form of clause learning
with restarts.
• Papers on look-ahead polynomials and
superresolution:
http://www.ccs.neu.edu/research/demeter/
papers/publications.html
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Additional Information
• Useful unpublished paper on look-ahead
polynomials:
http://www.ccs.neu.edu/research/demeter/
biblio/partial-sat-II.html
• Technical report on the topic of this talk:
http://www.ccs.neu.edu/research/demeter/
biblio/POptMAXCSP.html
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Future work
• Exploring best combination of look-forward
and look-back techniques.
• Find all maximum-assignments or
estimate their number.
• Robustness of maximum assignments.
• Are our MAX-CSP solvers useful for
reasoning about biological pathways?
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Conclusions
• Presented SPOT, a family of MAX-CSP
solvers based on look-ahead polynomials
and non-chronological backtracking.
• SPOT has a desirable property: P-optimal.
• SPOT can be implemented very efficiently.
• Preliminary experimental results are
encouraging. A lot more work is needed to
assess the practical value of the lookahead polynomials.
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Polynomials for rank 3
• x^3 x^2 x^1 x^0 relation
• -1 3 -3 1
1
• 1 -2 1 0
2
• 0 1 -2 1
3
• 1 -2 1 0
4
• 0 1 -2 1
5
For 2: x*(1-x)2=x3-2x2+x
maximum at x=1/3; 1/3*(2/3)2=4/27
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Recall
• (f*g)’ = f’*g + f*g’
• (f2)’ = 2*f * f’
• For relation 2:
– x*(1-x)2 = (1-x)2 + x*2(1-x)*(-1)= (1-x)(1-3x)
– x=1 is minimum
– x=1/3 is maximum
– value at maximum: 4/27
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Check: 2 and 4 should be the same
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Harold
•
•
•
concern: intension, extension: query, predicate
extension = intension(software)
Harold Ossher: confirmed pointcuts
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The Game Evergreen(r,m) for
Boolean MAX-CSP(G), r>1,m>0
Two players: They agree on a protocol P1 to
choose a set of m relations of rank r.
1. The players use P1 to choose a set G of m relations
of rank r.
2. Player 1 constructs a CSP(G) instance H with 1000
variables and at most 2*(1000 choose r) constraints
and gives it to player 2 (1 second limit).
3. Player 2 gets paid by player 1 the fraction of
constraints she can satisfy in H (100 seconds limit).
4. Take turns (go to 1).
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Evergreen(3,2)
• Rank 3: Represent relations by the integer
corresponding to the truth table in
standard sorted order 000 – 111.
• choose relations between 1 and 254
(exclude 0 and 255).
• Don’t choose two odd numbers: All false
would satisfy all constraints.
• Don’t choose both numbers above 128: All
true would satisfy all constraints.
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The Game Evergreen(r,m) for
Boolean MAX-CSP(G), r>1,m>0
Two players: They agree on a protocol P1 to
choose a set of m relations of rank r.
1. The players use P1 to choose a set G of m relations
of rank r.
2. Player 1 constructs a CSP(G) instance H with 1000
variables and at most 2*(1000 choose r) constraints
and gives it to player 2 (1 second limit).
3. Player 2 gets paid by player 1 the fraction of
constraints she can satisfy in H (100 seconds limit).
4. Take turns (go to 1).
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Requirements for algorithms and
properties to work
• Relative P-optimality
• Absolute P-optimality
– G needs to be closed under renaming and
reductions and n-maps
• Look-ahead polynomials
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What happens during the solution
process
• Maximum of polynomial will be at the
boundary, say 0. Can be achieved in P.
Notice folding effect.
• Many superresolvents will be learned until
better assignment is found.
• Most constraints use an odd relation, a
few an even relation (if many constraints
can be satisfied).
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What happens …
• Because the polynomial only depends on
a few numbers, it is not sensitive to the
detailed properties of the instance.
• But if one variable has a visible bias
towards either 1 or 0, polynomials might
detect it.
• Adjust the weight of the constraints to
bring the maximum of the polynomial into
the middle so that abs(mb) increases.
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Question for Daniel
•
•
•
•
p(x) = t1*p1(x)+t2*p2(x)
mb at 0
p(mb)
perturb t1,t2 so that p(x) gets a higher
maximum. The fraction of t1 should go up
if R1 is an unsatisfied relation.
• How high can we bring the fraction of
satisfied constraints this way?
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Question
• Does this solve the original problem?
• If we get all all satisfied, yes.
• Can force that, by deleting all but one
unsatisfied and adding them later on.
• Are forced to work with many relations.
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