Towards decidability of conjugacy of pairs and triplets Benny George - - PowerPoint PPT Presentation

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Towards decidability of conjugacy of pairs and triplets Benny George - - PowerPoint PPT Presentation

Sep 09, 2023 149 likes •775 views

Introduction Characterizing conjugacy Conclusion Towards decidability of conjugacy of pairs and triplets Benny George K benny@tcs.tifr.res.in Tata Institute of Fundamental Research Mumbai (This is joint work with Samrith Ram,Dept. Of

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Introduction Characterizing conjugacy Conclusion

Towards decidability of conjugacy of pairs and triplets

Benny George K benny@tcs.tifr.res.in

Tata Institute of Fundamental Research Mumbai (This is joint work with Samrith Ram,Dept. Of Mathematics,IITB)

Third Indian Conference on Logic and its Applications

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 1 / 16

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Introduction Characterizing conjugacy Conclusion

Outline

1

Introduction

2

Characterizing conjugacy

3

Conclusion

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 2 / 16

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SLIDE 3

Introduction Characterizing conjugacy Conclusion

Problem Statement

Given languages X and Y when does the equation XZ = ZY have solutions ?

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 3 / 16

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Introduction Characterizing conjugacy Conclusion

Problem Statement

Given languages X and Y when does the equation XZ = ZY have solutions ? X = {ab, abab}, Y = {ba, bababa}

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 3 / 16

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SLIDE 5

Introduction Characterizing conjugacy Conclusion

Problem Statement

Given languages X and Y when does the equation XZ = ZY have solutions ? X = {ab, abab}, Y = {ba, bababa} Z = {a, aba, ababa, . . .}

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 3 / 16

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Introduction Characterizing conjugacy Conclusion

Problem Statement

Given languages X and Y when does the equation XZ = ZY have solutions ? X = {ab, abab}, Y = {ba, bababa} Z = {a, aba, ababa, . . .} The general problem is fairly complicated.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 3 / 16

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SLIDE 7

Introduction Characterizing conjugacy Conclusion

Problem Statement

Given languages X and Y when does the equation XZ = ZY have solutions ? X = {ab, abab}, Y = {ba, bababa} Z = {a, aba, ababa, . . .} The general problem is fairly complicated. So we restrict our attention to |X| = 2 and |Y | = 3

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 3 / 16

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Introduction Characterizing conjugacy Conclusion

Terminology

A is a finite alphabet. A+ (resp. A∗)is the free semigroup (resp. monoid) generated by A Elements of A∗ are called words. Subsets of A∗ are called languages. Lower cases letters x, y, z, . . . denotes words and upper case letters X, Y , Z . . . denotes languages. Lower case letters a, b, c, . . . are used for constants. 1 denotes the empty word. |w| is length of word w. For two words u and v, u is a prefix(suffix, factor) of v if there exists x, y such that v = ux(v = xu, v = xuy). A word is called primitive if it is not of the form wk for k > 1.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 4 / 16

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SLIDE 9

Introduction Characterizing conjugacy Conclusion

Terminology

A is a finite alphabet. A+ (resp. A∗)is the free semigroup (resp. monoid) generated by A Elements of A∗ are called words. Subsets of A∗ are called languages. Lower cases letters x, y, z, . . . denotes words and upper case letters X, Y , Z . . . denotes languages. Lower case letters a, b, c, . . . are used for constants. 1 denotes the empty word. |w| is length of word w. For two words u and v, u is a prefix(suffix, factor) of v if there exists x, y such that v = ux(v = xu, v = xuy). A word is called primitive if it is not of the form wk for k > 1.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 4 / 16

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SLIDE 10

Introduction Characterizing conjugacy Conclusion

Terminology

A is a finite alphabet. A+ (resp. A∗)is the free semigroup (resp. monoid) generated by A Elements of A∗ are called words. Subsets of A∗ are called languages. Lower cases letters x, y, z, . . . denotes words and upper case letters X, Y , Z . . . denotes languages. Lower case letters a, b, c, . . . are used for constants. 1 denotes the empty word. |w| is length of word w. For two words u and v, u is a prefix(suffix, factor) of v if there exists x, y such that v = ux(v = xu, v = xuy). A word is called primitive if it is not of the form wk for k > 1.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 4 / 16

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SLIDE 11

Introduction Characterizing conjugacy Conclusion

Terminology

A is a finite alphabet. A+ (resp. A∗)is the free semigroup (resp. monoid) generated by A Elements of A∗ are called words. Subsets of A∗ are called languages. Lower cases letters x, y, z, . . . denotes words and upper case letters X, Y , Z . . . denotes languages. Lower case letters a, b, c, . . . are used for constants. 1 denotes the empty word. |w| is length of word w. For two words u and v, u is a prefix(suffix, factor) of v if there exists x, y such that v = ux(v = xu, v = xuy). A word is called primitive if it is not of the form wk for k > 1.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 4 / 16

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SLIDE 12

Introduction Characterizing conjugacy Conclusion

Terminology

A is a finite alphabet. A+ (resp. A∗)is the free semigroup (resp. monoid) generated by A Elements of A∗ are called words. Subsets of A∗ are called languages. Lower cases letters x, y, z, . . . denotes words and upper case letters X, Y , Z . . . denotes languages. Lower case letters a, b, c, . . . are used for constants. 1 denotes the empty word. |w| is length of word w. For two words u and v, u is a prefix(suffix, factor) of v if there exists x, y such that v = ux(v = xu, v = xuy). A word is called primitive if it is not of the form wk for k > 1.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 4 / 16

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SLIDE 13

Introduction Characterizing conjugacy Conclusion

Terminology

A is a finite alphabet. A+ (resp. A∗)is the free semigroup (resp. monoid) generated by A Elements of A∗ are called words. Subsets of A∗ are called languages. Lower cases letters x, y, z, . . . denotes words and upper case letters X, Y , Z . . . denotes languages. Lower case letters a, b, c, . . . are used for constants. 1 denotes the empty word. |w| is length of word w. For two words u and v, u is a prefix(suffix, factor) of v if there exists x, y such that v = ux(v = xu, v = xuy). A word is called primitive if it is not of the form wk for k > 1.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 4 / 16

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SLIDE 14

Introduction Characterizing conjugacy Conclusion

Terminology

A is a finite alphabet. A+ (resp. A∗)is the free semigroup (resp. monoid) generated by A Elements of A∗ are called words. Subsets of A∗ are called languages. Lower cases letters x, y, z, . . . denotes words and upper case letters X, Y , Z . . . denotes languages. Lower case letters a, b, c, . . . are used for constants. 1 denotes the empty word. |w| is length of word w. For two words u and v, u is a prefix(suffix, factor) of v if there exists x, y such that v = ux(v = xu, v = xuy). A word is called primitive if it is not of the form wk for k > 1.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 4 / 16

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SLIDE 15

Introduction Characterizing conjugacy Conclusion

Motivation

Language equations are natural generalizations of word equations. Word equations have been extensively investigated. Language equations are also fairly natural objects. They are commonly used to describe state systems, context free grammars etc. Conjugacy equation is a simple language equation, namely XZ = ZY but not much is known about this simple looking equation over languages.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 5 / 16

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SLIDE 16

Introduction Characterizing conjugacy Conclusion

Motivation

Language equations are natural generalizations of word equations. Language equations Word equations Word equations have been extensively investigated. Language equations are also fairly natural objects. They are commonly used to describe state systems, context free grammars etc. Conjugacy equation is a simple language equation, namely XZ = ZY but not much is known about this simple looking equation over languages.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 5 / 16

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SLIDE 17

Introduction Characterizing conjugacy Conclusion

Motivation

Language equations are natural generalizations of word equations. Language equations {a}X + {b}Y = Z Word equations w = sxxt Word equations have been extensively investigated. Language equations are also fairly natural objects. They are commonly used to describe state systems, context free grammars etc. Conjugacy equation is a simple language equation, namely XZ = ZY but not much is known about this simple looking equation over languages.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 5 / 16

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SLIDE 18

Introduction Characterizing conjugacy Conclusion

Motivation

Language equations are natural generalizations of word equations. Language equations {a}X + {b}Y = Z XYX = YXY Word equations w = sxxt xy = yx Word equations have been extensively investigated. Language equations are also fairly natural objects. They are commonly used to describe state systems, context free grammars etc. Conjugacy equation is a simple language equation, namely XZ = ZY but not much is known about this simple looking equation over languages.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 5 / 16

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SLIDE 19

Introduction Characterizing conjugacy Conclusion

Motivation

Language equations are natural generalizations of word equations. Language equations {a}X + {b}Y = Z XYX = YXY Word equations w = sxxt xy = yx Word equations have been extensively investigated. Language equations are also fairly natural objects. They are commonly used to describe state systems, context free grammars etc. Conjugacy equation is a simple language equation, namely XZ = ZY but not much is known about this simple looking equation over languages.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 5 / 16

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SLIDE 20

Introduction Characterizing conjugacy Conclusion

Motivation

Language equations are natural generalizations of word equations. Language equations {a}X + {b}Y = Z XYX = YXY Word equations w = sxxt xy = yx Word equations have been extensively investigated. Language equations are also fairly natural objects. They are commonly used to describe state systems, context free grammars etc. Conjugacy equation is a simple language equation, namely XZ = ZY but not much is known about this simple looking equation over languages.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 5 / 16

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SLIDE 21

Introduction Characterizing conjugacy Conclusion

Motivation

Language equations are natural generalizations of word equations. Language equations {a}X + {b}Y = Z XYX = YXY Word equations w = sxxt xy = yx Word equations have been extensively investigated. Language equations are also fairly natural objects. They are commonly used to describe state systems, context free grammars etc. Conjugacy equation is a simple language equation, namely XZ = ZY but not much is known about this simple looking equation over languages.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 5 / 16

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SLIDE 22

Introduction Characterizing conjugacy Conclusion

Previous Works

Word equations are decidable. G.S.Makanin (1977) Maximal conjugator problem. Michal Kunc (2005) Conjugacy of pairs. Cassaigne,Karhum¨ aki, Maˇ nuch (2001) Conjugacy of finite biprefix codes. Cassaigne, Salmela, Karhum¨ aki (2007)

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 6 / 16

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SLIDE 23

Introduction Characterizing conjugacy Conclusion

Previous Works

Word equations are decidable. G.S.Makanin (1977) Maximal conjugator problem. Michal Kunc (2005) Conjugacy of pairs. Cassaigne,Karhum¨ aki, Maˇ nuch (2001) Conjugacy of finite biprefix codes. Cassaigne, Salmela, Karhum¨ aki (2007)

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 6 / 16

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SLIDE 24

Introduction Characterizing conjugacy Conclusion

Previous Works

Word equations are decidable. G.S.Makanin (1977) Maximal conjugator problem. Michal Kunc (2005) Conjugacy of pairs. Cassaigne,Karhum¨ aki, Maˇ nuch (2001) Conjugacy of finite biprefix codes. Cassaigne, Salmela, Karhum¨ aki (2007)

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 6 / 16

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SLIDE 25

Introduction Characterizing conjugacy Conclusion

Previous Works

Word equations are decidable. G.S.Makanin (1977) Maximal conjugator problem. Michal Kunc (2005) Conjugacy of pairs. Cassaigne,Karhum¨ aki, Maˇ nuch (2001) Conjugacy of finite biprefix codes. Cassaigne, Salmela, Karhum¨ aki (2007)

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 6 / 16

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Introduction Characterizing conjugacy Conclusion

Our Contribution

We partially characterize conjugacy when one set is a two element set and the other is a three element set. The characterization we obtain actually holds even when one set is a two element set and the other element is any finite set.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 7 / 16

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Introduction Characterizing conjugacy Conclusion

Our Contribution

We partially characterize conjugacy when one set is a two element set and the other is a three element set. The characterization we obtain actually holds even when one set is a two element set and the other element is any finite set.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 7 / 16

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Introduction Characterizing conjugacy Conclusion

Outline

1

Introduction

2

Characterizing conjugacy

3

Conclusion

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 8 / 16

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SLIDE 29

Introduction Characterizing conjugacy Conclusion

Conjugacy of words

What are all the possible solutions of xz = zy when x, y, z are words? Note that x and y must be of same length and that xnz = zyn for all n. Thus x = (αβ)n , y = (βα)n and z = (αβ)mα In the above equation we can choose αβ to be a primitive word.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 9 / 16

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Introduction Characterizing conjugacy Conclusion

Conjugacy of words

What are all the possible solutions of xz = zy when x, y, z are words? Note that x and y must be of same length and that xnz = zyn for all n. Thus x = (αβ)n , y = (βα)n and z = (αβ)mα In the above equation we can choose αβ to be a primitive word.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 9 / 16

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SLIDE 31

Introduction Characterizing conjugacy Conclusion

Conjugacy of words

What are all the possible solutions of xz = zy when x, y, z are words? Note that x and y must be of same length and that xnz = zyn for all n. Thus x = (αβ)n , y = (βα)n and z = (αβ)mα In the above equation we can choose αβ to be a primitive word.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 9 / 16

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SLIDE 32

Introduction Characterizing conjugacy Conclusion

Conjugacy of words

What are all the possible solutions of xz = zy when x, y, z are words? Note that x and y must be of same length and that xnz = zyn for all n. Thus x = (αβ)n , y = (βα)n and z = (αβ)mα In the above equation we can choose αβ to be a primitive word.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 9 / 16

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Introduction Characterizing conjugacy Conclusion

Sub problems of 2,3 conjugacy

We may assume that X = {x1, x2}, Y = {y1, y2, y3}. So we have the following sub problems Elements in X are of equal length Elements in X are of unequal length (i.e |x1| < |x2|)

No element in Y of size x2 There are elements in Y of size x2

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 10 / 16

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SLIDE 34

Introduction Characterizing conjugacy Conclusion

Sub problems of 2,3 conjugacy

We may assume that X = {x1, x2}, Y = {y1, y2, y3}. So we have the following sub problems Elements in X are of equal length Elements in X are of unequal length (i.e |x1| < |x2|)

No element in Y of size x2 There are elements in Y of size x2

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 10 / 16

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SLIDE 35

Introduction Characterizing conjugacy Conclusion

Sub problems of 2,3 conjugacy

We may assume that X = {x1, x2}, Y = {y1, y2, y3}. So we have the following sub problems Elements in X are of equal length Elements in X are of unequal length (i.e |x1| < |x2|)

No element in Y of size x2 There are elements in Y of size x2

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 10 / 16

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SLIDE 36

Introduction Characterizing conjugacy Conclusion

Sub problems of 2,3 conjugacy

We may assume that X = {x1, x2}, Y = {y1, y2, y3}. So we have the following sub problems Elements in X are of equal length Elements in X are of unequal length (i.e |x1| < |x2|)

No element in Y of size x2 There are elements in Y of size x2

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 10 / 16

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SLIDE 37

Introduction Characterizing conjugacy Conclusion

Sub problems of 2,3 conjugacy

We may assume that X = {x1, x2}, Y = {y1, y2, y3}. So we have the following sub problems Elements in X are of equal length Elements in X are of unequal length (i.e |x1| < |x2|)

No element in Y of size x2 There are elements in Y of size x2

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 10 / 16

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SLIDE 38

Introduction Characterizing conjugacy Conclusion

Conjugacy when X is a uniform set.

Y must have two elements of the same length as x1. Z must make X and the above two element subset Y1 conjugates. Other elements of Y must be “insertable”. Thus we have the following characterization. X = {pu, pv}, Y1 = {up, vp} and Z = {pu, pv}ip (or the symmetric case) Since zy3 = xiz′, we know y3 ∈ (up + vp)k

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 11 / 16

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SLIDE 39

Introduction Characterizing conjugacy Conclusion

Conjugacy when X is a uniform set.

Y must have two elements of the same length as x1. Z must make X and the above two element subset Y1 conjugates. Other elements of Y must be “insertable”. Thus we have the following characterization. X = {pu, pv}, Y1 = {up, vp} and Z = {pu, pv}ip (or the symmetric case) Since zy3 = xiz′, we know y3 ∈ (up + vp)k

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 11 / 16

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SLIDE 40

Introduction Characterizing conjugacy Conclusion

Conjugacy when X is a uniform set.

Y must have two elements of the same length as x1. Z must make X and the above two element subset Y1 conjugates. Other elements of Y must be “insertable”. Thus we have the following characterization. X = {pu, pv}, Y1 = {up, vp} and Z = {pu, pv}ip (or the symmetric case) Since zy3 = xiz′, we know y3 ∈ (up + vp)k

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 11 / 16

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SLIDE 41

Introduction Characterizing conjugacy Conclusion

Conjugacy when X is a uniform set.

Y must have two elements of the same length as x1. Z must make X and the above two element subset Y1 conjugates. Other elements of Y must be “insertable”. Thus we have the following characterization. X = {pu, pv}, Y1 = {up, vp} and Z = {pu, pv}ip (or the symmetric case) Since zy3 = xiz′, we know y3 ∈ (up + vp)k

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 11 / 16

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SLIDE 42

Introduction Characterizing conjugacy Conclusion

Conjugacy when X is a uniform set.

Y must have two elements of the same length as x1. Z must make X and the above two element subset Y1 conjugates. Other elements of Y must be “insertable”. Thus we have the following characterization. X = {pu, pv}, Y1 = {up, vp} and Z = {pu, pv}ip (or the symmetric case) Since zy3 = xiz′, we know y3 ∈ (up + vp)k

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 11 / 16

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SLIDE 43

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

Since XZ = ZY , X nZ = ZY n Observe that there is a least sized element in Y . Call it y1 Observe that there is a least sized element in Z too. Call it z1 z1yiy1yk

1 = xi1 . . . xik+2zi

z1y1yiyk

1 = xj1 . . . xjk+2zj

Thus every element in Y of length less than x2 commutes with y1

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 12 / 16

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SLIDE 44

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

Since XZ = ZY , X nZ = ZY n Observe that there is a least sized element in Y . Call it y1 Observe that there is a least sized element in Z too. Call it z1 z1yiy1yk

1 = xi1 . . . xik+2zi

z1y1yiyk

1 = xj1 . . . xjk+2zj

Thus every element in Y of length less than x2 commutes with y1

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 12 / 16

slide-45
SLIDE 45

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

Since XZ = ZY , X nZ = ZY n Observe that there is a least sized element in Y . Call it y1 Observe that there is a least sized element in Z too. Call it z1 z1yiy1yk

1 = xi1 . . . xik+2zi

z1y1yiyk

1 = xj1 . . . xjk+2zj

Thus every element in Y of length less than x2 commutes with y1

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 12 / 16

slide-46
SLIDE 46

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

Since XZ = ZY , X nZ = ZY n Observe that there is a least sized element in Y . Call it y1 Observe that there is a least sized element in Z too. Call it z1 z1yiy1yk

1 = xi1 . . . xik+2zi

z1y1yiyk

1 = xj1 . . . xjk+2zj

Thus every element in Y of length less than x2 commutes with y1

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 12 / 16

slide-47
SLIDE 47

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

Since XZ = ZY , X nZ = ZY n Observe that there is a least sized element in Y . Call it y1 Observe that there is a least sized element in Z too. Call it z1 z1yiy1yk

1 = xi1 . . . xik+2zi

z1y1yiyk

1 = xj1 . . . xjk+2zj

Thus every element in Y of length less than x2 commutes with y1

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 12 / 16

slide-48
SLIDE 48

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

Since XZ = ZY , X nZ = ZY n Observe that there is a least sized element in Y . Call it y1 Observe that there is a least sized element in Z too. Call it z1 z1yiy1yk

1 = xi1 . . . xik+2zi

z1y1yiyk

1 = xj1 . . . xjk+2zj

Thus every element in Y of length less than x2 commutes with y1

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 12 / 16

slide-49
SLIDE 49

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

x1x2xk

1 z1 = yr1 . . . yrk+2zr

x2x1xk

1 z1 = ys1 . . . ysk+2zs

x1 and x2 commute with each other. Words which commute with each other are powers of a word p. Thus words in Y are powers of a conjugate word p′. So X = {(αβ)i1, (αβ)i2} and Y = {(βα)i1, (βα)i2, (βα)i3}

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 13 / 16

slide-50
SLIDE 50

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

x1x2xk

1 z1 = yr1 . . . yrk+2zr

x2x1xk

1 z1 = ys1 . . . ysk+2zs

x1 and x2 commute with each other. Words which commute with each other are powers of a word p. Thus words in Y are powers of a conjugate word p′. So X = {(αβ)i1, (αβ)i2} and Y = {(βα)i1, (βα)i2, (βα)i3}

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 13 / 16

slide-51
SLIDE 51

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

x1x2xk

1 z1 = yr1 . . . yrk+2zr

x2x1xk

1 z1 = ys1 . . . ysk+2zs

x1 and x2 commute with each other. Words which commute with each other are powers of a word p. Thus words in Y are powers of a conjugate word p′. So X = {(αβ)i1, (αβ)i2} and Y = {(βα)i1, (βα)i2, (βα)i3}

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 13 / 16

slide-52
SLIDE 52

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

x1x2xk

1 z1 = yr1 . . . yrk+2zr

x2x1xk

1 z1 = ys1 . . . ysk+2zs

x1 and x2 commute with each other. Words which commute with each other are powers of a word p. Thus words in Y are powers of a conjugate word p′. So X = {(αβ)i1, (αβ)i2} and Y = {(βα)i1, (βα)i2, (βα)i3}

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 13 / 16

slide-53
SLIDE 53

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

x1x2xk

1 z1 = yr1 . . . yrk+2zr

x2x1xk

1 z1 = ys1 . . . ysk+2zs

x1 and x2 commute with each other. Words which commute with each other are powers of a word p. Thus words in Y are powers of a conjugate word p′. So X = {(αβ)i1, (αβ)i2} and Y = {(βα)i1, (βα)i2, (βα)i3}

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 13 / 16

slide-54
SLIDE 54

Introduction Characterizing conjugacy Conclusion

Conjugacy when there are no elements in Y of length x2

x1x2xk

1 z1 = yr1 . . . yrk+2zr

x2x1xk

1 z1 = ys1 . . . ysk+2zs

x1 and x2 commute with each other. Words which commute with each other are powers of a word p. Thus words in Y are powers of a conjugate word p′. So X = {(αβ)i1, (αβ)i2} and Y = {(βα)i1, (βα)i2, (βα)i3}

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 13 / 16

slide-55
SLIDE 55

Introduction Characterizing conjugacy Conclusion

Outline

1

Introduction

2

Characterizing conjugacy

3

Conclusion

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 14 / 16

slide-56
SLIDE 56

Introduction Characterizing conjugacy Conclusion

Open Problems

2,3 conjugacy is not yet full characterised. We think that 2,3 and 2,n, n ≥ 3 is essentially the same. How about undecidability in the general case? Defect theorems for Language equations? Shapes of (maximal)conjugators.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 15 / 16

slide-57
SLIDE 57

Introduction Characterizing conjugacy Conclusion

Open Problems

2,3 conjugacy is not yet full characterised. We think that 2,3 and 2,n, n ≥ 3 is essentially the same. How about undecidability in the general case? Defect theorems for Language equations? Shapes of (maximal)conjugators.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 15 / 16

slide-58
SLIDE 58

Introduction Characterizing conjugacy Conclusion

Open Problems

2,3 conjugacy is not yet full characterised. We think that 2,3 and 2,n, n ≥ 3 is essentially the same. How about undecidability in the general case? Defect theorems for Language equations? Shapes of (maximal)conjugators.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 15 / 16

slide-59
SLIDE 59

Introduction Characterizing conjugacy Conclusion

Open Problems

2,3 conjugacy is not yet full characterised. We think that 2,3 and 2,n, n ≥ 3 is essentially the same. How about undecidability in the general case? Defect theorems for Language equations? Shapes of (maximal)conjugators.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 15 / 16

slide-60
SLIDE 60

Introduction Characterizing conjugacy Conclusion

Open Problems

2,3 conjugacy is not yet full characterised. We think that 2,3 and 2,n, n ≥ 3 is essentially the same. How about undecidability in the general case? Defect theorems for Language equations? Shapes of (maximal)conjugators.

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 15 / 16

slide-61
SLIDE 61

Introduction Characterizing conjugacy Conclusion

THANK YOU Questions?

B.G.K TIFR Conjugacy of pairs and triplets ICLA’09 16 / 16