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

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
2
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

3
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 lt 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
• 221 2 3 0
• 221 2 4 0
• 221 3 4 0

1in3 has number 22 M 1 !2 !3 !4 satisfies all
4
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
• 221 2 3 0
• 221 2 4 0 in MAX-CSP(22,?).
  Highest value for ?
• 221 3 4 0
• 22 2 3 4 0

1in3 has number 22
5
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 Bobs profit.
• Take turns.

R2, Bob
R1, Anna
100 R1, 0 R2
100 R2, 0 R1
6
Protocol choice

• Randomly choose R1 and R2 (independently) between
  1 and 255 (Throw two dice choosing relations).

7
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
  Bobs profit?

How well does MAX-SAT (e.g., yices, ubcsat) or
MAX-CSP solvers do on symmetric instances ???
8
Game strategy in a nutshell

• Choose GR1,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.

9
Our approach by ExampleSAT 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 payoff find maximum
assignment
10
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.

appmean approximation of the mean (k variables
true)
11
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) gt 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 Hk1, absH1(mb1) gt f
• MAYBE, the closer absH1(mb1) comes to f, the
  better
• NO, UP or clause learning

absH abstract representation of H
12

• 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

8/9
7/9
abstract representation
3/9
H
0 1 2 3 4 5
6
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
6/7 8/9
5/77/9
H0
3/75/9
maximum assignment away from max bias blurry
0 1 2 3 4
5
13

• 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

8/9
7/9
3/8
H
0 1 2 3 4 5
6
clearly above 3/4
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
7/88/9
6/87/9
H1
maximum assignment away from max bias blurry
2/73/8
0 1 2 3 4
5
14
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
8/9
7/9
abstract representation guarantees 7/9
H
7/8 8/9
6/78/9
6/8 7/9
5/77/9
abstract representation guarantees 7/9
abstract representation guarantees 8/9
H0
UBCSAT
H1
NEVER GOES DOWN DERANDOMIZATION
15
rank 2 10 1 or(1) 7 1 2 or(!1 !2)
4/6
4/6
10 1 0 10 2 0 10 3 0 7 1 2 0 7 1 3
0 7 2 3 0
3/6
3/6
abstract representation guarantees 0.625 6
3.75 4 satisfied.
Evergreen game G 10,7 How do you choose
a CSP(G)-instance to minimize payoff? 0.618
0 1 2
3
4/6
4/6
4/6
4/6
5 1 0 10 2 0 10 3 0 13 1 2 0 13 1 3
0 7 2 3 0
3/6
3/6
rank 2 5 1 or(!1) 13 1 2 or(1 !2)
The effect of n-map
16
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?

17
Where we are

• Introduction
• Look-forward
• Look-backward
• SPOT how to use the look-ahead polynomials
  together with superresolution.

18
Look Forward

• Why?
• To make informed decisions
• To play the Evergreen game
• How?
• Abstract representation based on look-ahead
  polynomials

19
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.

20
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
to look-ahead polynomial 3p(1-p)2 B1,3(p)
(Bernstein)
21
3p(1-p)2 for MAX-CSP(22)
22
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.

23
The general case MAX-CSP(G)
G R1, , tR(F) fraction of constraints in
F that use R.
appSATR(x) over all R is a super set of the
Bernstein polynomials (computer graphics,
weighted sum of Bernstein polynomials)
x p
24
Rational Bezier Curves
25
Bernstein Polynomials
http//graphics.idav.ucdavis.edu/education/CAGDNot
es/Bernstein-Polynomials.pdf
26
all the appSATR(x) polynomials
27
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?

28
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)
• 221 2 3 0
• 221 2 4 0
• 221 3 4 0
• 22 2 3 4 0

29
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, egt0.
1
not always (solid)
u critical transition point
always (fluid)
0
30
The Magic Number

• u 4/9

31
3p(1-p)2 for MAX-CSP(22)
32
Produce the Magic Number

• Use an optimally biased coin
• 1/3 in this case
• In general min max problem

33
The 22 reductionsNeeded for implementation
1,0
2,0
22
60
240
3,0
2,1
3,1
1,1
3,0
2,0
3
15
255
3,1
2,1
22 is expanded into 6 additional relations.
0
34
The 22 N-MappingsNeeded for implementation
1
41
134
2
2
0
0
1
1
22
73
146
104
2
2
0
0
97
148
1
22 is expanded into 7 additional relations.
35
General Dichotomy Theorem
MAX-CSP(G,f) For each finite set G of relations
closed under renaming there exists an algebraic
number tG For f tG MAX-CSP(G,f) has
polynomial solution For f tG e MAX-CSP(G,f)
is NP-complete, egt0.
1
hard (solid) NP-complete
polynomial solution Use optimally biased
coin. Derandomize. P-Optimal.
tG critical transition point
easy (fluid) Polynomial
0
due to Lieberherr/Specker (1979, 1982)
implications for the Evergreen game? Are you a
better player?
36
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.

37
General Dichotomy Theorem(Discussion)
MAX-CSP(G,f) For each finite set G of relations
closed under renaming there exists an algebraic
number tG For f tG MAX-CSP(G,f) has
polynomial solution For f tG e MAX-CSP(G,f)
is NP-complete, egt0.
1
hard (solid), NP-complete exponential,
super-polynomial proofs ??? relies on clause
learning
tG critical transition point
easy (fluid), Polynomial (finding an
assignment) constant proofs (done statically
using look-ahead polynomials) no clause learning
0
38
min max problem
sat(H,M) fraction of satisfied constraints in
CSP(G)-instance H by assignment M
tG min max
sat(H,M)
all (0,1) assignments M
all CSP(G) instances H
39
Problem reductions are the key

• Solution to simpler problem implies solution to
  original problem.

40
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
max sat(H,M,n)
all (0,1) assignments M to n variables
n to infinity
all SYMMETRIC CSP(G) -instances H with n
variables
41
Reduction achieved

• Instead of minimizing over all constraint systems
  it is sufficient to minimize over the symmetric
  constraint systems.

42
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.

43
Symmetric the worst-case
n variables n! permutations
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
. .
44
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)
all (0,1) assignments M to n variables where
the first k variables are set to 1
n to infinity
all SYMMETRIC CSP(G) -instances H with n
variables
45
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.

46
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47
The Game Evergreen(r,m) for Boolean MAX-CSP(G),
rgt1,mgt0

• Two players They agree on a protocol P1 to
  choose a set of m relations of rank r.
• 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 2m(1000 choose r)
  constraints and gives it to player 2 (1 second
  limit).
• Bob gets paid the fraction of constraints she can
  satisfy in H (100 seconds limit).
• Take turns (go to 1).

48
For Evergreen(3,2)
100 R1, 0 R2
100 R2, 0 R1
49
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.

50
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
100 R1, 0 R2
100 R2, 0 R1
51
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).

52
Protocol choice variant 3

• Randomly choose R1 and R2 (independently) between
  1 and 255 (Throw two dice).

53
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
54
For Evergreen(3,2)
Tells us how to mix the two relations
100 R1, 0 R2
100 R2, 0 R1
55
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).

56
Game strategy in a nutshell

• Anna Best Choose tG instance
• Bob Gets paid tG
• etc.

57
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/resea
  rch/demeter/papers/publications.html

58
Additional Information

• Useful unpublished paper on look-ahead
  polynomials http//www.ccs.neu.edu/research/demet
  er/biblio/partial-sat-II.html
• Technical report on the topic of this talk
  http//www.ccs.neu.edu/research/demeter/biblio/POp
  tMAXCSP.html

59
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?

60
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.

61
Polynomials for rank 3

• x3 x2 x1 x0 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)2x3-2x2x
• maximum at x1/3 1/3(2/3)24/27

Check 2 and 4 are the same
62
Polynomials for rank 3

• x3 x2 x1 x0 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)2x3-2x2x
• maximum at x1/3 1/3(2/3)24/27

63
Recall

• (fg) fg fg
• (f2) 2f f
• For relation 2
• x(1-x)2 (1-x)2 x2(1-x)(-1) (1-x)(1-3x)
• x1 is minimum
• x1/3 is maximum
• value at maximum 4/27

64
Harold

• concern intension, extension query, predicate
• extension intension(software)
• Harold Ossher confirmed pointcuts

65
The Game Evergreen(r,m) for Boolean MAX-CSP(G),
rgt1,mgt0

• Two players They agree on a protocol P1 to
  choose a set of m relations of rank r.
• 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 2m(1000 choose r)
  constraints and gives it to Bob (1 second limit).
  
• Bob gets paid by Anna the fraction of constraints
  he can satisfy in H (100 seconds limit).
• Take turns (go to 1).

66
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).
• Dont choose two odd numbers All false would
  satisfy all constraints.
• Dont choose both numbers above 128 All true
  would satisfy all constraints.

67
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.

68
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-1e which
  is enough to guarantee a maximum assignment.
• what if we also allow n-maps?

69
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.

70
Exploring the search space

• Look-ahead polynomials dont 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.

71
Crosscutting the search space
current
by look-ahead
even better
better
by search
best
72
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.

73
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

74
Never require closed under renaming?

• symmetric formulas dont 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.

75
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).

76
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.

77
Question for Daniel

• p(x) t1p1(x)t2p2(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?

78
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.

79
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 assignment and proof that it is
maximum
80
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.

81
SAT Rank 2 instance
rank 2 10 1 or(1) 5 1 or(!1)
14 1 2 or(1 2) 7 1 3 or(!1 !3)
N1 !2 !3 4 5 !6 unsat1/9 FN -gt D
UP 1 !3 !5 4 6FN -gt SSR
Restart F5(1)N -gt UP !1 2 !4 !6 5
3F5(1)N -gt SSR !1 2 !4 !6 5 3F5(1),0()N -gt
Finale end

• 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

82
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), dont mention second
  argument
• 12(1) or(1) or(1,), 10(2,1) 12(1,2)
• 0() empty clause

83
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84
UP / D
85
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.)

86
mean versus appmean

• mean does less averaging, so it is preferred?
• appmean looks at the neighborhood of xn

87
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.

88
The Game EvergreenTM(r,m) for Boolean
MAX-CSP(G), rgt1,mgt0

• Two players They agree on a protocol P1 to
  choose a set of m relations of rank r.
• 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 2m(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).

TM true maximum
89
EvergreenTM versus Evergreen

• Anna can try to create instances that are hard to
  solve by Bobs 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.

90
Using Mathematica
15 1 0 !1 238 1 2 0 1 or 2

• Combine215, 238
• t1(1-x)-t2(-2x)x
• DDCombine215, 238,x,x
• -2t2 is negative must be a maximum
• SolveDCombine215, 238,x0,x
• x -t12t2/2t2
• RootsOf215,238/.t2-gt(1-t1)/.t1-gt1/5(5-sqrt(5)

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

92
Mathematica

• IncludeItr_, n_ ModFloorr/n,
  2AppSATr_ SimplifyIncludeItr,
  1x0((1 - x))((3 - 0)) ((   IncludeItr,
  2 IncludeItr, 4 IncludeItr, 16))     
       x1((1 - x))((3 - 1)) ((IncludeItr, 8
         IncludeItr, 32 IncludeItr, 64)) 
               x2((1 - x))((3 - 2))
  IncludeItr,               128x3((1 -
  x))((3 - 3))Combine2r1_, r2_
  t1AppSATr1 t2AppSATr2RootsOf2r1_, r2_
     ReplaceAllCombine2r1, r2,
  SolveDCombine2r1, r2, x 0,
  xSolve2r1_, r2_ For i 1, i lt       
   Length RootsOf2r1, r2, i, PrintMinimize
  RootsOf2r1, r2           i,  0 lt t1 lt 1, 0
  lt t2 lt 1, t1 t2 1, t1, t2