Word equations are combinatorial equalities between strings of symbols, variables and functions, which can be used to model problems in a wide range of domains. While some complexity results for the solving of specific classes of equations are known, currently there does not exist a systematic equation solver. We present in this paper a reduction of the problem of solving word equations to Boolean satisfiability, and describe the implementation of a general-purpose tool that leverages existing SAT solvers for this purpose. Our solver will prove useful in the resolution of word equations, and in the computer-based exploration of various combinatorial conjectures.

1. Solving Equations on Words with Morphisms
and Antimorphisms
A. Blondin Mass´e, S. Gaboury, S. Hall´e and M. Larouche
Laboratoire d’informatique formelle
Universit´e du Qu´ebec `a Chicoutimi
Chicoutimi, Canada
Language and Automata Theory and Applications
(LATA 2014)
March 13th, 2014
Madrid, Spain
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 1 / 25

2. Outline
1. Introduction
2. Words equations
3. From equations to graphs
4. Conclusion
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 2 / 25

3. Outline
1. Introduction
2. Words equations
3. From equations to graphs
4. Conclusion
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 3 / 25

4. Distinct squares conjecture
Conjecture (Fraenkel and Simpson, 1998)
A word of length n contains less than n distinct squares.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 4 / 25

5. Distinct squares conjecture
Conjecture (Fraenkel and Simpson, 1998)
A word of length n contains less than n distinct squares.
The 8 squares of the word w = 0000110110101 of length
13 are
00, 11, 0000, 0101, 1010, 011011, 101101, 110110.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 4 / 25

6. Distinct squares conjecture
Conjecture (Fraenkel and Simpson, 1998)
A word of length n contains less than n distinct squares.
The 8 squares of the word w = 0000110110101 of length
13 are
00, 11, 0000, 0101, 1010, 011011, 101101, 110110.
We look at the solutions of the equations
x1u2
1y1 = x2u2
2y2 = . . . = xku2
kyk,
with various lenghts of the xi’s and the ui’s.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 4 / 25

7. Tilings
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 5 / 25

8. Tilings
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 5 / 25

9. Tilings
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 5 / 25

10. Tilings
On the Freeman chain code F = {0, 1, 2, 3} :
(xyˆxˆy)2
= pztˆzˆts,
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 5 / 25

11. Tilings
On the Freeman chain code F = {0, 1, 2, 3} :
(xyˆxˆy)2
= pztˆzˆts,
It is easier to solve the equation on the turns alphabet
{R, L, F} :
(xLyLˆxLˆyL)2
= uzLtLˆzLˆtLv.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 5 / 25

12. Solving equations on words
Makanin showed that the problem of solving equations on
words with constants and variables is decidable
(1977) ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 6 / 25

13. Solving equations on words
Makanin showed that the problem of solving equations on
words with constants and variables is decidable
(1977) ;
Plandowski showed that the problem is in fact in
PSPACE (1999) ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 6 / 25

14. Solving equations on words
Makanin showed that the problem of solving equations on
words with constants and variables is decidable
(1977) ;
Plandowski showed that the problem is in fact in
PSPACE (1999) ;
Abdulrab propose a Lisp implementation of a solver in
1990 (not available anymore) ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 6 / 25

15. Solving equations on words
Makanin showed that the problem of solving equations on
words with constants and variables is decidable
(1977) ;
Plandowski showed that the problem is in fact in
PSPACE (1999) ;
Abdulrab propose a Lisp implementation of a solver in
1990 (not available anymore) ;
More recenly, tools such as Hampi, Omega and Stranger
deal with string constraints and regular expressions ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 6 / 25

16. Solving equations on words
Makanin showed that the problem of solving equations on
words with constants and variables is decidable
(1977) ;
Plandowski showed that the problem is in fact in
PSPACE (1999) ;
Abdulrab propose a Lisp implementation of a solver in
1990 (not available anymore) ;
More recenly, tools such as Hampi, Omega and Stranger
deal with string constraints and regular expressions ;
None handles morphisms and antimorphisms.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 6 / 25

17. Outline
1. Introduction
2. Words equations
3. From equations to graphs
4. Conclusion
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 7 / 25

18. Basics
Alphabet Σ, free monoid Σ∗, ε is the empty word ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 8 / 25

19. Basics
Alphabet Σ, free monoid Σ∗, ε is the empty word ;
A morphism is a map ϕ : Σ∗ → Σ∗ such that
ϕ(uv) = ϕ(u)ϕ(v) ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 8 / 25

20. Basics
Alphabet Σ, free monoid Σ∗, ε is the empty word ;
A morphism is a map ϕ : Σ∗ → Σ∗ such that
ϕ(uv) = ϕ(u)ϕ(v) ;
An antimorphism is a map ϕ : Σ∗ → σ∗ such that
ϕ(uv) = ϕ(v)ϕ(u)
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 8 / 25

21. Basics
Alphabet Σ, free monoid Σ∗, ε is the empty word ;
A morphism is a map ϕ : Σ∗ → Σ∗ such that
ϕ(uv) = ϕ(u)ϕ(v) ;
An antimorphism is a map ϕ : Σ∗ → σ∗ such that
ϕ(uv) = ϕ(v)ϕ(u)
Morphism and antimorphisms are completely determined
by their action on single letters ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 8 / 25

22. Basics
Alphabet Σ, free monoid Σ∗, ε is the empty word ;
A morphism is a map ϕ : Σ∗ → Σ∗ such that
ϕ(uv) = ϕ(u)ϕ(v) ;
An antimorphism is a map ϕ : Σ∗ → σ∗ such that
ϕ(uv) = ϕ(v)ϕ(u)
Morphism and antimorphisms are completely determined
by their action on single letters ;
If ϕ is an antimorphism, then ϕ = · · ϕ , where ϕ is a
morphism ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 8 / 25

23. Basics
Alphabet Σ, free monoid Σ∗, ε is the empty word ;
A morphism is a map ϕ : Σ∗ → Σ∗ such that
ϕ(uv) = ϕ(u)ϕ(v) ;
An antimorphism is a map ϕ : Σ∗ → σ∗ such that
ϕ(uv) = ϕ(v)ϕ(u)
Morphism and antimorphisms are completely determined
by their action on single letters ;
If ϕ is an antimorphism, then ϕ = · · ϕ , where ϕ is a
morphism ;
A morphism (or antimorphism) is called k-uniform if
ϕ(a) = k for every a ∈ Σ ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 8 / 25

24. Basics
Alphabet Σ, free monoid Σ∗, ε is the empty word ;
A morphism is a map ϕ : Σ∗ → Σ∗ such that
ϕ(uv) = ϕ(u)ϕ(v) ;
An antimorphism is a map ϕ : Σ∗ → σ∗ such that
ϕ(uv) = ϕ(v)ϕ(u)
Morphism and antimorphisms are completely determined
by their action on single letters ;
If ϕ is an antimorphism, then ϕ = · · ϕ , where ϕ is a
morphism ;
A morphism (or antimorphism) is called k-uniform if
ϕ(a) = k for every a ∈ Σ ;
We then write |ϕ| = k.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 8 / 25

25. Words equations systems
Systems of words equations are represented by
Σ the set of constants ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 9 / 25

26. Words equations systems
Systems of words equations are represented by
Σ the set of constants ;
V the set of variables ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 9 / 25

27. Words equations systems
Systems of words equations are represented by
Σ the set of constants ;
V the set of variables ;
M the set of morphisms ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 9 / 25

28. Words equations systems
Systems of words equations are represented by
Σ the set of constants ;
V the set of variables ;
M the set of morphisms ;
A the set of antimorphisms ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 9 / 25

29. Words equations systems
Systems of words equations are represented by
Σ the set of constants ;
V the set of variables ;
M the set of morphisms ;
A the set of antimorphisms ;
a list of words equations Li = Ri (i = 1, 2, . . . , m).
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 9 / 25

30. Symbolic expressions
A symbolic expression (or simply expression) is either
ε, a constant, or a variable ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 10 / 25

31. Symbolic expressions
A symbolic expression (or simply expression) is either
ε, a constant, or a variable ;
v1v2 · · · vk, where vi is itself an expression ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 10 / 25

32. Symbolic expressions
A symbolic expression (or simply expression) is either
ε, a constant, or a variable ;
v1v2 · · · vk, where vi is itself an expression ;
ϕ(v), where v is an expression and ϕ ∈ M ∪ A ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 10 / 25

33. Example
Take
Σ = {a, b};
V = {x, y};
M = {ϕ}, ϕ : a → b, b → a;
A = { · }
Then
abxy = ϕ(x)yab
is a word equation.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 11 / 25

34. Assignments and solutions
An assignment is a map I : V → Σ∗ ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 12 / 25

35. Assignments and solutions
An assignment is a map I : V → Σ∗ ;
Is is naturally extended as follows
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 12 / 25

36. Assignments and solutions
An assignment is a map I : V → Σ∗ ;
Is is naturally extended as follows
1. I(ε) = ε ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 12 / 25

37. Assignments and solutions
An assignment is a map I : V → Σ∗ ;
Is is naturally extended as follows
1. I(ε) = ε ;
2. I(u) = u, for u ∈ Σ ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 12 / 25

38. Assignments and solutions
An assignment is a map I : V → Σ∗ ;
Is is naturally extended as follows
1. I(ε) = ε ;
2. I(u) = u, for u ∈ Σ ;
3. I(u1 · u2 · · · uk) = I(u1) · I(u2) · · · I(uk), if ui is an
expression (i = 1, 2, . . . , k) ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 12 / 25

39. Assignments and solutions
An assignment is a map I : V → Σ∗ ;
Is is naturally extended as follows
1. I(ε) = ε ;
2. I(u) = u, for u ∈ Σ ;
3. I(u1 · u2 · · · uk) = I(u1) · I(u2) · · · I(uk), if ui is an
expression (i = 1, 2, . . . , k) ;
4. I(ϕ(u)) = ϕ(I(u)), if ϕ ∈ M ∪ A and u is an
expression.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 12 / 25

40. Assignments and solutions
An assignment is a map I : V → Σ∗ ;
Is is naturally extended as follows
1. I(ε) = ε ;
2. I(u) = u, for u ∈ Σ ;
3. I(u1 · u2 · · · uk) = I(u1) · I(u2) · · · I(uk), if ui is an
expression (i = 1, 2, . . . , k) ;
4. I(ϕ(u)) = ϕ(I(u)), if ϕ ∈ M ∪ A and u is an
expression.
Let L = R be an equation ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 12 / 25

41. Assignments and solutions
An assignment is a map I : V → Σ∗ ;
Is is naturally extended as follows
1. I(ε) = ε ;
2. I(u) = u, for u ∈ Σ ;
3. I(u1 · u2 · · · uk) = I(u1) · I(u2) · · · I(uk), if ui is an
expression (i = 1, 2, . . . , k) ;
4. I(ϕ(u)) = ϕ(I(u)), if ϕ ∈ M ∪ A and u is an
expression.
Let L = R be an equation ;
A solution of L = R is an assigment S : V → Σ∗such that
S(L) = S(R).
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 12 / 25

42. Bounded lengths
Since solving general equations on words is very hard, we
need to restrict the context ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 13 / 25

43. Bounded lengths
Since solving general equations on words is very hard, we
need to restrict the context ;
To simplify, we assume that all lenghts of variables are
known ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 13 / 25

44. Bounded lengths
Since solving general equations on words is very hard, we
need to restrict the context ;
To simplify, we assume that all lenghts of variables are
known ;
Let λ : V → N, where λ(v) is the length of the variable v ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 13 / 25

45. Bounded lengths
Since solving general equations on words is very hard, we
need to restrict the context ;
To simplify, we assume that all lenghts of variables are
known ;
Let λ : V → N, where λ(v) is the length of the variable v ;
Naturally,
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 13 / 25

46. Bounded lengths
Since solving general equations on words is very hard, we
need to restrict the context ;
To simplify, we assume that all lenghts of variables are
known ;
Let λ : V → N, where λ(v) is the length of the variable v ;
Naturally,
1. λ(ε) = 0 ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 13 / 25

47. Bounded lengths
Since solving general equations on words is very hard, we
need to restrict the context ;
To simplify, we assume that all lenghts of variables are
known ;
Let λ : V → N, where λ(v) is the length of the variable v ;
Naturally,
1. λ(ε) = 0 ;
2. λ(u) = 1 for u ∈ Σ ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 13 / 25

48. Bounded lengths
Since solving general equations on words is very hard, we
need to restrict the context ;
To simplify, we assume that all lenghts of variables are
known ;
Let λ : V → N, where λ(v) is the length of the variable v ;
Naturally,
1. λ(ε) = 0 ;
2. λ(u) = 1 for u ∈ Σ ;
3. λ(u1 · u2 · · · uk) = k
i=1 λ(ui), if ui is an expression for
i = 1, 2, . . . , k ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 13 / 25

49. Bounded lengths
Since solving general equations on words is very hard, we
need to restrict the context ;
To simplify, we assume that all lenghts of variables are
known ;
Let λ : V → N, where λ(v) is the length of the variable v ;
Naturally,
1. λ(ε) = 0 ;
2. λ(u) = 1 for u ∈ Σ ;
3. λ(u1 · u2 · · · uk) = k
i=1 λ(ui), if ui is an expression for
i = 1, 2, . . . , k ;
4. λ(ϕ(u)) = |ϕ| · λ(u), if ϕ is uniform and u is an
expression.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 13 / 25

50. Is the problem difficult ?
Theorem
Let Σ, V, M and A be some sets of constants, variables,
morphisms and antimorphisms ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 14 / 25

51. Is the problem difficult ?
Theorem
Let Σ, V, M and A be some sets of constants, variables,
morphisms and antimorphisms ;
Let L = R be an equation on these four sets ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 14 / 25

52. Is the problem difficult ?
Theorem
Let Σ, V, M and A be some sets of constants, variables,
morphisms and antimorphisms ;
Let L = R be an equation on these four sets ;
Assume that all variables are of known length and that
|L| = |R| = n ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 14 / 25

53. Is the problem difficult ?
Theorem
Let Σ, V, M and A be some sets of constants, variables,
morphisms and antimorphisms ;
Let L = R be an equation on these four sets ;
Assume that all variables are of known length and that
|L| = |R| = n ;
Then the problem of determining if some solution
S : V → Σ∗ exists for L = R is NP-complete.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 14 / 25

54. Proof of NP-completeness
Clearly in NP ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 15 / 25

55. Proof of NP-completeness
Clearly in NP ;
We reduce 3-IN-1-SAT to the problem ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 15 / 25

56. Proof of NP-completeness
Clearly in NP ;
We reduce 3-IN-1-SAT to the problem ;
Consider an instance of 3-IN-1-SAT :
x =
m
i=1
(xi1 ∨ xi2 ∨ xi3),
where xij ∈ {b1, b1, b2, b2, . . . , bn, bn}.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 15 / 25

57. Proof of NP-completeness
Clearly in NP ;
We reduce 3-IN-1-SAT to the problem ;
Consider an instance of 3-IN-1-SAT :
x =
m
i=1
(xi1 ∨ xi2 ∨ xi3),
where xij ∈ {b1, b1, b2, b2, . . . , bn, bn}.
We construct an instance of words equation as follows.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 15 / 25

58. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

59. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

60. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Lengths :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

61. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Lengths :
λ(ci) = 3 for i = 1, 2, . . . , m ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

62. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Lengths :
λ(ci) = 3 for i = 1, 2, . . . , m ;
λ(bi) = λ(bi) = 1 for i = 1, 2, . . . , n ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

63. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Lengths :
λ(ci) = 3 for i = 1, 2, . . . , m ;
λ(bi) = λ(bi) = 1 for i = 1, 2, . . . , n ;
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

64. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Lengths :
λ(ci) = 3 for i = 1, 2, . . . , m ;
λ(bi) = λ(bi) = 1 for i = 1, 2, . . . , n ;
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equations :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

65. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Lengths :
λ(ci) = 3 for i = 1, 2, . . . , m ;
λ(bi) = λ(bi) = 1 for i = 1, 2, . . . , n ;
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equations :
c1c2 · · · cm = x11x12x13x21 · · · xm3.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

66. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Lengths :
λ(ci) = 3 for i = 1, 2, . . . , m ;
λ(bi) = λ(bi) = 1 for i = 1, 2, . . . , n ;
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equations :
c1c2 · · · cm = x11x12x13x21 · · · xm3.
ϕ(ci) = 1, for i = 1, 2, . . . , m ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

67. Reduction
Constants : Σ = {0, 1} ∪ {$1, $2, . . . , $m+n}
Variables : V = {c1, c2, . . . , cm} ∪ {b1, b1, b2, b2, . . . , bn, bn}
Lengths :
λ(ci) = 3 for i = 1, 2, . . . , m ;
λ(bi) = λ(bi) = 1 for i = 1, 2, . . . , n ;
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equations :
c1c2 · · · cm = x11x12x13x21 · · · xm3.
ϕ(ci) = 1, for i = 1, 2, . . . , m ;
ϕ(bibi) = 1, for i = 1, 2, . . . , n ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 16 / 25

68. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

69. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

70. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

71. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

72. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
=
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

73. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)
=
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

74. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)
=
1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

75. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1
=
1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

76. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1
=
1 $1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

77. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)
=
1 $1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

78. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)
=
1 $1 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

79. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2
=
1 $1 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

80. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2
=
1 $1 1 $2
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

81. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)
=
1 $1 1 $2
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

82. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)
=
1 $1 1 $2 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

83. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3
=
1 $1 1 $2 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

84. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3
=
1 $1 1 $2 1 $3
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

85. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1)
=
1 $1 1 $2 1 $3
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

86. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1)
=
1 $1 1 $2 1 $3 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

87. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4
=
1 $1 1 $2 1 $3 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

88. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4
=
1 $1 1 $2 1 $3 1 $4
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

89. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2)
=
1 $1 1 $2 1 $3 1 $4
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

90. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2)
=
1 $1 1 $2 1 $3 1 $4 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

91. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2) $5
=
1 $1 1 $2 1 $3 1 $4 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

92. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2) $5
=
1 $1 1 $2 1 $3 1 $4 1 $5
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

93. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2) $5 ϕ(c3)
=
1 $1 1 $2 1 $3 1 $4 1 $5
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

94. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2) $5 ϕ(c3)
=
1 $1 1 $2 1 $3 1 $4 1 $5 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

95. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2) $5 ϕ(c3) $6
=
1 $1 1 $2 1 $3 1 $4 1 $5 1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

96. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2) $5 ϕ(c3) $6
=
1 $1 1 $2 1 $3 1 $4 1 $5 1 $6
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

97. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2) $5 ϕ(c3) $6 c1 c2 c3
=
1 $1 1 $2 1 $3 1 $4 1 $5 1 $6
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

98. Reduction (cont.)
Consider the boolean expression
x = (b1 ∨ ¬b2 ∨ b3) ∧ (¬b1 ∨ ¬b3 ∨ b4) ∧ (¬b2 ∨ b3 ∨ ¬b4).
Constants : Σ = {0, 1} ∪ {$1, $2, $3, $4, $5, $6}
Variables : V = {c1, c2, c3} ∪ {b1, b1, b2, b2, b3, b3, b4, b4}
Morphism : ϕ(0) = ε ; ϕ(1) = 1 ;
Equation :
ϕ(b1b1)$1 ϕ(b2b2)$2 ϕ(b3b3)$3 ϕ(c1) $4 ϕ(c2) $5 ϕ(c3) $6 c1 c2 c3
=
1 $1 1 $2 1 $3 1 $4 1 $5 1 $6 b1b2b3b1b3b4b2b3b4
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 17 / 25

99. Outline
1. Introduction
2. Words equations
3. From equations to graphs
4. Conclusion
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 18 / 25

100. Example 1 (only with words)
Consider the length 20 equation
uuv = wwx
where |u| = 4, |v| = 12, |w| = 6, |x| = 8.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 19 / 25

101. Example 1 (only with words)
Consider the length 20 equation
uuv = wwx
where |u| = 4, |v| = 12, |w| = 6, |x| = 8.
We create the following graph :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 19 / 25

102. Example 1 (only with words)
Consider the length 20 equation
uuv = wwx
where |u| = 4, |v| = 12, |w| = 6, |x| = 8.
We create the following graph :
e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12 e13 e14 e15 e16 e17 e18 e19 e20
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 19 / 25

103. Example 1 (only with words)
Consider the length 20 equation
uuv = wwx
where |u| = 4, |v| = 12, |w| = 6, |x| = 8.
We create the following graph :
e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12 e13 e14 e15 e16 e17 e18 e19 e20
u1 u2 u3 u4
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 19 / 25

104. Example 1 (only with words)
Consider the length 20 equation
uuv = wwx
where |u| = 4, |v| = 12, |w| = 6, |x| = 8.
We create the following graph :
e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12 e13 e14 e15 e16 e17 e18 e19 e20
u1 u2 u3 u4
w1 w2 w3 w4 w5 w6
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 19 / 25

105. Example 1 (only with words)
Consider the length 20 equation
uuv = wwx
where |u| = 4, |v| = 12, |w| = 6, |x| = 8.
We create the following graph :
u1, u3, e1, e3, e5, e7, e9, e11, w1, w3, w5
u2, u4, e2, e4, e6, e8, e10, e12, w2, w4, w6
e13
e14
e15
e16
e17
e18
e19
e20
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 19 / 25

106. Example 2 (tilings)
Next, consider the length 36 equation
(xLyLˆxLˆyL)2
= uzLtLˆzLˆtLv,
where |x| = 3, |y| = 4, |z| = 3, |t| = 4, |u| = 3, |v| = 15.
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 20 / 25

107. Example 2 (tilings)
Next, consider the length 36 equation
(xLyLˆxLˆyL)2
= uzLtLˆzLˆtLv,
where |x| = 3, |y| = 4, |z| = 3, |t| = 4, |u| = 3, |v| = 15.
We create the following graph :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 20 / 25

108. Example 2 (tilings)
Next, consider the length 36 equation
(xLyLˆxLˆyL)2
= uzLtLˆzLˆtLv,
where |x| = 3, |y| = 4, |z| = 3, |t| = 4, |u| = 3, |v| = 15.
We create the following graph :
e1, e19, x1, ˆt3
e6, e24, y2, z3
e10, e28, ˆx1, t3 e15, e33, ˆy2, ˆz3
L, e3, e4, e7, e9,
e12, e13, e16,
e18, x3, z1, y3,
t2, ˆx3, ˆz1, ˆy3, ˆt2
¯·¯·
¯· ¯·
e2, e20, x2, ˆt4
e8, e26, y4, t1
e14, e32, ˆy1, ˆz2
e5, e23, y1, z2
e17, e35, ˆy4, ˆt1
e11, e29, ˆx2, t4
¯·¯·
¯·
¯· ¯·
¯·
e2, e14, e17, e20,
e32, e35, x2, ˆy1,
ˆy4, ˆt4, ˆz2, ˆt1
e5, e8, e11, e23,
e26, e29, y1, y4,
ˆx2, z2, t1, t4
¯·
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 20 / 25

109. Example 2 (tilings)
Next, consider the length 36 equation
(xLyLˆxLˆyL)2
= uzLtLˆzLˆtLv,
where |x| = 3, |y| = 4, |z| = 3, |t| = 4, |u| = 3, |v| = 15.
We create the following graph :
e2, e20, x2, ˆt4
e8, e26, y4, t1
e14, e32, ˆy1, ˆz2
e5, e23, y1, z2
e17, e35, ˆy4, ˆt1
e11, e29, ˆx2, t4
¯·¯·
¯·
¯· ¯·
¯·
R, e1, e6, e10,
e15, e19, e24,
e28, e33, x1,
y2, ˆx1, ˆy2,
ˆt3, z3, t3, ˆz3
L, e3, e4, e7, e9,
e12, e13, e16,
e18, x3, z1, y3,
t2, ˆx3, ˆz1, ˆy3, ˆt2
¯·
e2, e14, e17, e20,
e32, e35, x2, ˆy1,
ˆy4, ˆt4, ˆz2, ˆt1
e5, e8, e11, e23,
e26, e29, y1, y4,
ˆx2, z2, t1, t4
¯·
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 20 / 25

110. Reduction rules
Vertices with unlabelled edges are merged :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

111. Reduction rules
Vertices with unlabelled edges are merged :
u v
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

112. Reduction rules
Vertices with unlabelled edges are merged :
u v u, v
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

113. Reduction rules
Vertices with unlabelled edges are merged :
u v u, v
Constants are propagated :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

114. Reduction rules
Vertices with unlabelled edges are merged :
u v u, v
Constants are propagated :
a u
ϕ
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

115. Reduction rules
Vertices with unlabelled edges are merged :
u v u, v
Constants are propagated :
a u
ϕ
a u, ϕ(a)
ϕ
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

116. Reduction rules
Vertices with unlabelled edges are merged :
u v u, v
Constants are propagated :
a u
ϕ
a u, ϕ(a)
ϕ
Morphisms are functions :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

117. Reduction rules
Vertices with unlabelled edges are merged :
u v u, v
Constants are propagated :
a u
ϕ
a u, ϕ(a)
ϕ
Morphisms are functions :
u
v1
v2
ϕ
ϕ
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

118. Reduction rules
Vertices with unlabelled edges are merged :
u v u, v
Constants are propagated :
a u
ϕ
a u, ϕ(a)
ϕ
Morphisms are functions :
u
v1
v2
ϕ
ϕ
u v1, v2
ϕ
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 21 / 25

119. Reduction rules (cont.)
Morphisms can be combined :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 22 / 25

120. Reduction rules (cont.)
Morphisms can be combined :
u1
u2
u3
u4
ϕ1
ϕ2
ϕ3
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 22 / 25

121. Reduction rules (cont.)
Morphisms can be combined :
u1
u2
u3
u4
ϕ1
ϕ2
ϕ3
u1
u2
u3
u4
ϕ1
ϕ2
ϕ3
ϕ2 ◦ ϕ1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 22 / 25

122. Reduction rules (cont.)
Morphisms can be combined :
u1
u2
u3
u4
ϕ1
ϕ2
ϕ3
u1
u2
u3
u4
ϕ1
ϕ2
ϕ3
ϕ2 ◦ ϕ1
ϕ3 ◦ ϕ2
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 22 / 25

123. Reduction rules (cont.)
Morphisms can be combined :
u1
u2
u3
u4
ϕ1
ϕ2
ϕ3
u1
u2
u3
u4
ϕ1
ϕ2
ϕ3
ϕ2 ◦ ϕ1
ϕ3 ◦ ϕ2
ϕ3 ◦ ϕ2 ◦ ϕ1
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 22 / 25

124. Outline
1. Introduction
2. Words equations
3. From equations to graphs
4. Conclusion
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 23 / 25

125. Concluding remarks
The approach based on graphs was compared with one
using directly boolean constraint programming ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 24 / 25

126. Concluding remarks
The approach based on graphs was compared with one
using directly boolean constraint programming ;
In almost all cases, it performed better, especially when
there are morphisms and antimorphisms ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 24 / 25

127. Concluding remarks
The approach based on graphs was compared with one
using directly boolean constraint programming ;
In almost all cases, it performed better, especially when
there are morphisms and antimorphisms ;
Equations considered in this talk :
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 24 / 25

128. Concluding remarks
The approach based on graphs was compared with one
using directly boolean constraint programming ;
In almost all cases, it performed better, especially when
there are morphisms and antimorphisms ;
Equations considered in this talk :
have fixed lengths ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 24 / 25

129. Concluding remarks
The approach based on graphs was compared with one
using directly boolean constraint programming ;
In almost all cases, it performed better, especially when
there are morphisms and antimorphisms ;
Equations considered in this talk :
have fixed lengths ;
deal only with uniform morphisms and
antimorphisms ;
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 24 / 25

130. Concluding remarks
The approach based on graphs was compared with one
using directly boolean constraint programming ;
In almost all cases, it performed better, especially when
there are morphisms and antimorphisms ;
Equations considered in this talk :
have fixed lengths ;
deal only with uniform morphisms and
antimorphisms ;
How could it be extended to the non-uniform case ?
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 24 / 25

131. Questions ?
Merci de m’avoir ´ecout´e
Thank you for your attention
Les agradezco su atenci´on
Blondin Mass´e et al. (LIF, UQAC) March 13th, 2014 25 / 25