Transcript The Evergreen Project: The Promise of Polynomials to Boost CSP/SAT Solvers*

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
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Two objectives
• I want you to become
– better writers of MAX-SAT/MAX-CSP solvers
• better decision making
• crosscutting exploration of search space
– better players of the Evergreen game
• the game reveals what the polynomials can do
• iterated game of base zero-sum game:
– Together: choose domain
– Anna: choose instance (min .max. possible loss)
– Bob: solve instance, Anna pays Bob satisfaction fraction
• perfect information 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
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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
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Evergreen(3,2) game
Anna and Bob: They agree on a protocol P1 to
choose a set of 2 relations (=G) of rank 3.
– Anna chooses CSP(G)-instance H (limited).
– Bob solves H and gets paid by Anna the fraction
that Bob satisfies in H.
• This gives nice control to Anna. Anna will choose an
instance that will minimize Bob’s profit.
– Take turns.
R2, Bob
R1, Anna
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100% R2, 0% R1
100% R1, 0% R2
Protocol choice
• Randomly choose R1 and R2
(independently) between 1 and 255
(Throw two dice choosing relations).
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Tell me
• How would you react as Anna?
– 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 Bob use to
maximize its payoff?
• Should any MAX-CSP solver be able to
maximize Bob’s profit? How well does MAX-SAT (e.g., yices,
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ubcsat)
or MAX-CSP solvers do on symmetric 7
instances ???
Game strategy in a nutshell
• Choose G={R1,R2} randomly.
• Anna chooses instance so that payoff is
minimized.
• Bob finds solution so that payoff is
maximized (Solve MAX-CSP(G))
• Take turns: Choose G= … Bob chooses …
• Requires thorough understanding of MAXCSP(G) problem domain. Requires an
excellent MAX-CSP(G) solver.
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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
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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.
2. Anna constructs a CSP(G) instance H with 1000
variables and at most 2*m*(1000 choose r)
constraints and gives it to player 2 (1 second limit).
3. Bob 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
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Evergreen(3,2) protocol
possibilities
• Variant 1
– Player Bob chooses both relations G
– Player Anna chooses CSP(G) instance H.
– Player Bob solves H and gets paid by Anna.
• This gives too much control to Bob. Bob can
choose two odd relations which guarantees him a
pay of 1 independent of how Anna chooses the
instance H.
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Evergreen(3,2) protocol
possibilities
• Variant 2:
– Anna chooses a relation R1 (e.g. 22).
– Bob chooses a relation R2.
– Anna chooses CSP(G) instance H.
– Bob solves H and gets paid by Anna.
R2, Bob
R1, Anna
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100% R2, 0% R1
100% R1, 0% R2
Problem with variant 2
• Anna can just ignore relation R2.
• Gives Anna too much control because the
payoff for Bob depends only on R1 chosen
by Anna (and the quality of the solver that
Bob uses).
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Protocol choice: variant 3
• Randomly choose R1 and R2
(independently) between 1 and 255
(Throw two dice).
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Tell me
• How would you react as Anna?
– 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
• Anna: Best: Choose tG instance
• Bob: 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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Check: 2 and 4 are the same
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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 (same as 2)
• 0 1 -2 1
5
For 4: 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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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. Anna constructs a CSP(G) instance H with 1000
variables and at most 2*m*(1000 choose r)
constraints and gives it to Bob (1 second limit).
3. Bob gets paid by Anna the fraction of constraints he
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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How to play Evergreen(3,2)
• G = {R1, R2} is given (by some protocol)
• Anna: compute t=(t1, t2) so that
max appmeant(x) for x in [0,1] is minimum.
Construct a symmetric instance SYMG(t)
H.
• Bob: Solves H.
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Question
• For any G and any CSP(G)-instance H, is
there a weight assignment to the
constraints of H so that the look-ahead
polynomial absH has its maximum not at 0
or 1 and guarantees a maximum
assignment for H without weights?
– the polynomial might guarantee maximum1+e which is enough to guarantee a maximum
assignment.
– what if we also allow n-maps?
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Absolute P-optimality
• Bringing max to boundary is polynomial.
• Bringing max away from boundary using weights? What
is the complexity.
• Definition: ImproveLookAhead(G,H,N): Given G, a
CSP(instance) H and an assignment N for H. Is there an
assignment that satisfies at least laH,N(mb) + 1. mb =
maximum bias.
• Assume G sufficiently closed.
• Theorem: [Absolute P-optimality]
ImproveLookAhead(G,H,N) is NP-hard iff MAX-CSP(G)
is NP-hard.
• Warning: ImproveAllZero(G,H) is NP-hard iff MAXCSP(G) is NP-hard.
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Exploring the search space
• Look-ahead polynomials don’t eliminate
parts of the search space.
• They crosscut the search space early in
the search process. Whenever the lookahead polynomial guarantees more than
the currently best assignment, we can cut
across the search space but might have to
get back to the part we jumped over.
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Crosscutting the search space
current
by look-ahead
even
better
better
by search
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best
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Early better than later
• Look-ahead polynomials are more useful early in
the search.
• Later in the search the maximum will be at 0 or
1.
• Look-ahead polynomials will make mistakes
which are compensated by superresolvents.
• Superresolvents cut part of the search space
and they help the look-ahead polynomials to
eliminate the mistakes.
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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
– improve assignments: closed under n-maps
and reductions
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Never require closed under
renaming?
• symmetric formulas don’t require it? They
do? Consider
2: 1 2 3 0
2: 1 2 4 0
2: 1 3 4 0
2: 2 3 4 0
is not symmetric. {1 !2 !3 4} does not satisfy
all, only ¾. {!1 2 3 !4} only satisfies ¼.
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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 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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SAT Rank 2 instance
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)
=F
find maximum assignment
and proof that it is maximum
79
Solution Strategy
• The MAX-CSP transition system gives
many options:
– Choose initial assignment. Has significant
impact on length of proof. Best to start with a
maximum assignment.
– variable ordering. Irrelevant because start
with maximum assignment.
– value ordering: Also irrelevant.
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14: 1 2 = or(1 2)
7: 1 3 = or(!1 !3)
SAT Rank 2 instance
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
rank 2
10: 1 = or(1)
5: 1 = or(!1)
N={1 !2 !3 4 5 !6}
unsat=1/9
{}|F|{}|N -> D UP*
{1* !3 !5 4 6}|F|{}|N -> SSR Restart
{}|F|5(1)|N -> UP*
{!1 2 !4 !6 5 3}F|5(1)|N -> SSR
{!1 2 !4 !6 5 3}F|5(1),0()|N -> Finale
end
81
Rank 2 relations
ba
1 00 0
0
2 01 1
0
4 10 0
1
8 11 1
1
10 12
10(1) = or(1) = or(*,1), don’t mention second
argument
12(1) = or(1) = or(1,*), 10(2,1) = 12(1,2)
0() = empty clause
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UP / D
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Variable ordering
• maximizes likelihood that look-ahead
polynomials make correct decisions
• Finds variable where look-ahead
polynomials give the strongest indication
– even if the look-ahead polynomial chooses
the wrong mb, the decision might still be right
– what is better:
• laH1(mb1) is max
• | laH1(mb1) - laH0(mb0) | is max (is more instance
specific. Will adapt to superresolvents.)
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mean versus appmean
• mean does less averaging, so it is
preferred?
• appmean looks at the neighborhood of x*n
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Derandomization
• In a perfectly symmetric CSP(G) instance,
it is sufficient to try any assignment with k
ones for k from 0 to n to achieve the
maximum (tG).
• But in a non-symmetric instance, we need
derandomization to achieve tG and
superresolution to achieve the maximum.
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The Game EvergreenTM(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.
2.
3.
4.
The players use P1 to choose a set G of m relations of rank r.
Anna constructs a CSP(G) instance H with 1000 variables and
at most 2*m*(1000 choose r) constraints and gives it to Bob (1
second limit). Anna knows the maximum assignment and has
a proof for maximality but keeps it secret until Bob gives
response.
Bob gets paid the fraction of constraints he can satisfy in H
relative to the maximum number that can be satisfied (100
seconds limit).
Take turns (go to 1).
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TM = true maximum
88
EvergreenTM versus Evergreen
• Anna can try to create
instances that are hard to
solve by Bob’s solver.
• If Bob has a perfect
solver, he will be paid 1.0.
• The game depends a lot
on the solver quality.
• Incomplete information
(maximum assignment is
kept secret).
• Challenge for Anna to find
instance where maximum
is known with short proof.
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• Anna can control the
maximum Bob is paid
assuming a perfect
solver.
• Bob may be paid little
even with a perfect
solver.
• The game depends less
on solver quality.
• Complete information.
89
15: 1 0 = !1
238: 1 2 0 = 1 or 2
Using Mathematica
• Combine2[15, 238]
– t1(1-x)-t2(-2+x)*x
• D[D[Combine2[15, 238],x],x]
– -2*t2 is negative: must be a maximum
• Solve[D[Combine2[15, 238],x]=0,x]
– x = -t1+2*t2/2*t2
• RootsOf2[15,238]/.t2->(1-t1)/.t1->1/5(5sqrt(5)
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Mathematica
• Solve2(15, 238)
– ½ (sqrt(5)-1)
– t1 = 1-1/sqrt(5)
– t2 = 1/sqrt(5)
• Solve2(22, 22)
– 4/9
– t1 = ½
– t2 = ½
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Mathematica
IncludeIt[r_, n_] := Mod[Floor[r/n], 2];
AppSAT[r_] := Simplify[IncludeIt[r, 1]*x^0*((1 - x))^((3 - 0)) + ((
IncludeIt[r, 2] + IncludeIt[r, 4] + IncludeIt[r, 16]))*
x^1*((1 - x))^((3 - 1)) + ((IncludeIt[r, 8] +
IncludeIt[r, 32] + IncludeIt[r, 64]))*
x^2*((1 - x))^((3 - 2)) + IncludeIt[r,
128]*x^3*((1 - x))^((3 - 3))];
Combine2[r1_, r2_] := t1*AppSAT[r1] + t2*AppSAT[r2];
RootsOf2[r1_, r2_] :=
ReplaceAll[Combine2[r1, r2], Solve[D[Combine2[r1, r2], x] == 0,
x]];
Solve2[r1_, r2_] := For [i = 1, i <=
Length[ RootsOf2[r1, r2]], i++, Print[Minimize[{ RootsOf2[r1,
r2][[
i]], 0 < t1 < 1, 0 < t2 < 1, t1 + t2 == 1}, {t1, t2}]]];
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92