UBC March 2007 1

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

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 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 3MAX-CSP({22},f): H =

22:1 2 3 0 22:1 2 4 0 in MAX-CSP({22},?). Highest value for ?22:1 3 4 022: 2 3 4 0 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.R1, Anna

R2, Bob

100% R1, 0% R2100% R2, 0% R1

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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,

ubcsat)or MAX-CSP solvers do on symmetric instances ???

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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 MAX-

CSP(G) problem domain. Requires an excellent MAX-CSP(G) solver.

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Our approach by Example:SAT Rank 2 example

14 : 1 2 014 :       3 4 014 :           5 6 0  7 : 1    3            0 7 : 1        5      0 7 :       3   5      0 7 :   2    4         0 7 :   2         6   0 7 :         4   6   0

14: 1 2 = or(1 2) 7: 1 3 = or(!1 !3)

= H

Evergreen game:maximize payofffind maximum assignment

UBC March 2007 10appmean = approximation of the mean (k variables true)

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”.

excellent peripheral vision

0 1 2 3 4 5 6 = k

8/9

7/9

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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 learningabsH= abstract representation of H

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14 : 1 2 014 :       3 4 014 :           5 6 0  7 : 1    3            0 7 : 1        5      0 7 :       3   5      0 7 :   2     4          0 7 :   2         6   0 7 :         4   6   0

14 : 2 014 :       3 4 014 :           5 6 0  7 : 1    3            0 7 : 1        5      0 7 :       3   5      0 7 :   2     4          0 7 :   2         6   0 7 :         4   6   0

8/9

6/7 = 8/9

3/7=5/9

3/9H

H0

abstract representation

0 1 2 3 4 5 6

0 1 2 3 4 5

maximum assignment away from max bias: blurry

7/9

5/7=7/9

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14 : 1 2 014 :       3 4 014 :           5 6 0  7 : 1    3            0 7 : 1        5      0 7 :       3   5      0 7 :   2     4          0 7 :   2         6   0 7 :         4   6   0

14 : 1 2 014 :       3 4 014 :           5 6 0  7 :     3            0 7 :         5      0 7 :       3   5      0 7 :   2     4          0 7 :   2         6   0 7 :         4   6   0

8/9

7/8=8/9

6/8=7/9

H

H1

3/8

2/7=3/8

0 1 2 3 4 5 6

0 1 2 3 4 5

maximum assignment away from max bias: blurry

7/9

clearlyabove 3/4

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8/97/9

14 : 1 2 014 :       3 4 014 :           5 6 0  7 : 1    3            0 7 : 1        5      0 7 :       3   5      0 7 :   2     4          0 7 :   2         6   0 7 :         4   6   0

6/7=8/9

5/7=7/9

7/8 = 8/9

6/8 = 7/9

abstract representationguarantees 7/9

abstract representationguarantees 7/9 abstract representation

guarantees 8/9

H

H0H1NEVER GOES DOWN: DERANDOMIZATION

UBCSAT

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10 : 1 010 : 2 010 : 3 0 7 : 1 2 0 7 : 1 3 0 7 : 2 3 0

rank 210: 1 = or(1) 7: 1 2 = or(!1 !2)

5 : 1 010 : 2 010 : 3 013 : 1 2 013 : 1 3 0 7 : 2 3 0

rank 25: 1 = or(!1)13: 1 2 = or(1 !2)

4/6 4/6

3/6 3/6

abstract representation guarantees0.625 * 6 = 3.75: 4 satisfied.

4/6 4/6

3/6 3/6

4/6 4/6

0 1 2 3

The effect of n-map

Evergreen game: G = {10,7}How do you choose aCSP(G)-instance to minimizepayoff? 0.618 …

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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, … ,40

22: 6 7 9 0

22: 12 27 38 0

Abstract representation:reduce the instance tolook-ahead polynomial 3p(1-p)2 = B1,3(p) (Bernstein)

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1in3

0

0.1

0.2

0.3

0.4

0.5

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)

Fra

ctio

n o

f co

nst

rain

ts t

hat

ar

e g

uar

ante

ed t

o b

e sa

tisf

ied

3p(1-p)2 for MAX-CSP({22})

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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 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

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 3MAX-CSP({22},f):

22:1 2 3 0 22:1 2 4 0 22:1 3 4 022: 2 3 4 0

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Mathematical Critical Transition Point

MAX-CSP({22},f):

For f ≤ u: problem has always a solutionFor f ≥ u + : problem has not always a solution,

u critical transition point

always (fluid)

not always (solid)

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The Magic Number

• u = 4/9

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1in3

0

0.1

0.2

0.3

0.4

0.5

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)

Fra

ctio

n o

f co

nst

rain

ts t

hat

ar

e g

uar

ante

ed t

o b

e sa

tisf

ied

3p(1-p)2 for MAX-CSP({22})

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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

22 60

3

240

15 255

0

1,0

1,1

2,1

2,0

3,03,1

3,0

3,1

2,0

2,1

22 is expanded into 6 additionalrelations.

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The 22 N-Mappings:Needed for implementation

22

41

73

134

97

146

148

0

2

1

22 is expanded into 7 additionalrelations.

1040

21

0

2

1

0

2

1

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General Dichotomy Theorem

MAX-CSP(G,f): For each finite set G of relations closedunder renaming there exists an algebraic number tG :

For f ≤ tG: MAX-CSP(G,f) has polynomial solutionFor f ≥ tG+ : MAX-CSP(G,f) is NP-complete,

tG critical transition pointeasy (fluid)Polynomial

hard (solid)NP-complete

due to Lieberherr/Specker (1979, 1982)

polynomial solution:Use optimally biased coin.Derandomize.P-Optimal.

implications for the Evergreen game? Are you a better player?

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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 closedunder renaming there exists an algebraic number tG

For f ≤ tG: MAX-CSP(G,f) has polynomial solutionFor f ≥ tG+ : MAX-CSP(G,f) is NP-complete,

tG critical transition pointeasy (fluid), Polynomial (finding an assignment)constant proofs (done statically using look-ahead polynomials)no clause learning

hard (solid), NP-completeexponential, super-polynomial proofs ???relies on clause learning

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min max problem

tG = min max sat(H,M)

all (0,1) assignments M

all CSP(G)instances H

sat(H,M) = fraction of satisfied constraints in CSP(G)-instance H by assignment 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

tG = lim min max sat(H,M,n)

all (0,1) assignments Mto n variables

n toinfinity

sat(H,M,n) = fraction of satisfied constraints inCSP(G)-instance H by assignment M with n variables.

all SYMMETRIC CSP(G) -instances H with 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 variablesn! permutations

If in the big system thefraction f is satisfied, then there mustbe a least one small systemwhere the fraction f is satisfied

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min max problem

tG = lim min max sat(H,M,n)

all (0,1) assignments Mto n variables where thefirst k variables are setto 1

n toinfinity

sat(H,M,n) = fraction of satisfied constraints in system S by assignment I

all SYMMETRIC CSP(G) -instances H with n variables

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Observations

• The look-ahead polynomial look-forward approach has not been used in state-of-the-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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ForEvergreen(3,2)

100% R1, 0% R2100% R2, 0% R1

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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.

R1, Anna R2, Bob

100% R1, 0% R2100% R2, 0% R1

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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

• What kind of algorithm should P1 use to maximize its payoff? compute optimal k + best MAX-CSP solver.

symmetric instance with (1000 choose 3) constraints

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ForEvergreen(3,2)

100% R1, 0% R2100% R2, 0% R1

Tells us how tomix the tworelations

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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 look-ahead 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 5For 2: x*(1-x)2=x3-2x2+xmaximum at x=1/3; 1/3*(2/3)2=4/27

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 5For 4: x*(1-x)2=x3-2x2+xmaximum 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 maximum-

1+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 MAX-CSP(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 look-ahead 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 spacecurrent

better

best

by look-ahead

evenbetter

by search

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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 2 014 :       3 4 014 :           5 6 0  7 : 1    3            0 7 : 1        5      0 7 :       3   5      0 7 :   2    4         0 7 :   2         6   0 7 :         4   6   0

14: 1 2 = or(1 2) 7: 1 3 = or(!1 !3)

= F

find maximum assignmentand proof that it is maximum

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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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SAT Rank 2 instance

14 : 1 2 014 :       3 4 014 :           5 6 0  7 : 1    3            0 7 : 1        5      0 7 :       3   5      0 7 :   2    4         0 7 :   2         6   0 7 :         4   6   0

14: 1 2 = or(1 2) 7: 1 3 = or(!1 !3)

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 -> Finaleend

rank 210: 1 = or(1) 5: 1 = or(!1)

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Rank 2 relations

ba1 00 0 02 01 1 04 10 0 18 11 1 1 10 1210(1) = or(1) = or(*,1), don’t mention second

argument12(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. 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). Anna knows the maximum assignment and has a proof for maximality but keeps it secret until Bob gives response.

3. Bob gets paid the fraction of constraints he can satisfy in H relative to the maximum number that can be satisfied (100 seconds limit).

4. Take turns (go to 1).

TM = true maximum

UBC March 2007 89

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.

• 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.

UBC March 2007 90

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(5-sqrt(5)

15: 1 0 = !1238: 1 2 0 = 1 or 2

UBC March 2007 91

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

UBC March 2007 92

MathematicaIncludeIt[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}]]];