Computational Complexity of Semigroup Properties Trevor Jack Joint - - PowerPoint PPT Presentation

Computational Complexity of Semigroup Properties

Trevor Jack Joint work with Peter Mayr

Trevor Jack Semigroup Complexity August 7, 2018 1 / 16

Regularity Preliminaries

Notation and Regularity Problem

Transformation Semigroups [n] := {1, ..., n} Tn is the semigroup of all unary functions on [n] S ≤ Tn

Trevor Jack Semigroup Complexity August 7, 2018 2 / 16

Regularity Preliminaries

Notation and Regularity Problem

Transformation Semigroups [n] := {1, ..., n} Tn is the semigroup of all unary functions on [n] S ≤ Tn General Inquiry: Given generators a1, . . . , ak ∈ Tn, what is the complexity

• f verifying certain properties about S = a1, . . . , an within:

P ⊆ NP ⊆ PSPACE ⊆ EXPTIME?

Trevor Jack Semigroup Complexity August 7, 2018 2 / 16

Regularity Preliminaries

Notation and Regularity Problem

Transformation Semigroups [n] := {1, ..., n} Tn is the semigroup of all unary functions on [n] S ≤ Tn General Inquiry: Given generators a1, . . . , ak ∈ Tn, what is the complexity

• f verifying certain properties about S = a1, . . . , an within:

P ⊆ NP ⊆ PSPACE ⊆ EXPTIME? Definition b ∈ Tn is regular in S if for some s ∈ S, bsb = b.

Trevor Jack Semigroup Complexity August 7, 2018 2 / 16

Regularity Preliminaries

Notation and Regularity Problem

Transformation Semigroups [n] := {1, ..., n} Tn is the semigroup of all unary functions on [n] S ≤ Tn General Inquiry: Given generators a1, . . . , ak ∈ Tn, what is the complexity

• f verifying certain properties about S = a1, . . . , an within:

P ⊆ NP ⊆ PSPACE ⊆ EXPTIME? Definition b ∈ Tn is regular in S if for some s ∈ S, bsb = b. RegularElement Input: a1, ..., ak, b ∈ Tn Output: Is b regular in a1, ..., ak?

Trevor Jack Semigroup Complexity August 7, 2018 2 / 16

Regularity Theorems and Proofs

RegularElement Theorem and Proof

Theorem RegularElement is PSPACE-Complete.

Trevor Jack Semigroup Complexity August 7, 2018 3 / 16

Regularity Theorems and Proofs

RegularElement Theorem and Proof

Theorem RegularElement is PSPACE-Complete. Definition A deterministic finite automata (DFA) has:

1 a set of states Z with a start state and an accept state; and 2 a set of transformations Σ, which map states to states. Trevor Jack Semigroup Complexity August 7, 2018 3 / 16

Regularity Theorems and Proofs

RegularElement Theorem and Proof

Theorem RegularElement is PSPACE-Complete. Definition A deterministic finite automata (DFA) has:

1 a set of states Z with a start state and an accept state; and 2 a set of transformations Σ, which map states to states.

The proof uses the following PSPACE-complete problem (Kozen, 1970): Finite Automata Intersection (FAI) Input: DFA’s A1, ..., Aℓ with shared transitions Σ Output: Whether there is w ∈ Σ∗ accepted by each Ai.

Trevor Jack Semigroup Complexity August 7, 2018 3 / 16

Regularity Theorems and Proofs

Proof Sketch

Proof. Given DFAs A1, ..., Aℓ with sets of states Z1, ..., Zℓ and shared transitions Σ, define the following transformation semigroup: Transformed Set: Z = ℓ

i=1 Zi along with new state 0.

Generators: Σ defined naturally on Z and fixing 0. Add generator h that sends accept states to start states and sends every other state to 0. Then h is regular in this semigroup iff there is a w ∈ Σ∗ accepted by each A1, ..., Aℓ. Hence, RegularElement is PSPACE-hard. RegularElement is in NPSPACE because we can nondeterministically guess the generators that produce an s satisfying bsb = b. So, by Savitch’s Theorem, RegularElement is in PSPACE, and thus PSPACE-complete.

Trevor Jack Semigroup Complexity August 7, 2018 4 / 16

Regularity Theorems and Proofs

Regular Semigroup

Open Problem How hard is it to check that every element in S is regular?

Trevor Jack Semigroup Complexity August 7, 2018 5 / 16

Regularity Theorems and Proofs

Regular Semigroup

Open Problem How hard is it to check that every element in S is regular? A semigroup is completely regular if each element generates a subgroup. Theorem Determining if a1, . . . , ak ≤ Tn is completely regular is in P. Proof requires use of ”transformation graphs”

Trevor Jack Semigroup Complexity August 7, 2018 5 / 16

Transformation Graphs Introduction

Model Checking

Fix u, v semigroup words over variables z1, . . . , zm

Trevor Jack Semigroup Complexity August 7, 2018 6 / 16

Transformation Graphs Introduction

Model Checking

Fix u, v semigroup words over variables z1, . . . , zm Model(u ≈ v) Input: a1, ..., ak ∈ Tn Output: Whether a1, ..., ak models u(z1, ..., zm) ≈ v(z1, ..., zm).

Trevor Jack Semigroup Complexity August 7, 2018 6 / 16

Transformation Graphs Introduction

Model Checking

Fix u, v semigroup words over variables z1, . . . , zm Model(u ≈ v) Input: a1, ..., ak ∈ Tn Output: Whether a1, ..., ak models u(z1, ..., zm) ≈ v(z1, ..., zm). Example: Band Identity z1z1 ≈ z1

Trevor Jack Semigroup Complexity August 7, 2018 6 / 16

Transformation Graphs Introduction

Model Checking

Fix u, v semigroup words over variables z1, . . . , zm Model(u ≈ v) Input: a1, ..., ak ∈ Tn Output: Whether a1, ..., ak models u(z1, ..., zm) ≈ v(z1, ..., zm). Example: Band Identity z1z1 ≈ z1 Theorem Model(u ≈ v) is in P.

Trevor Jack Semigroup Complexity August 7, 2018 6 / 16

Transformation Graphs Notation

Notation

Let W be the set of all prefixes of u and v including the empty word 1. For x ∈ [n], s1, . . . , sm ∈ S define evaluations, e(x, s1, . . . , sm): W → [n], w → xw(s1, . . . , sm). E(W , S) := {e(x, s1, . . . , sm) : x ∈ [n], s1, . . . , sm ∈ S} ⊆ [n]W . Then S models u ≈ v iff f (u) = f (v) for all f ∈ E(W , S).

Trevor Jack Semigroup Complexity August 7, 2018 7 / 16

Transformation Graphs Notation

Notation

Let W be the set of all prefixes of u and v including the empty word 1. For x ∈ [n], s1, . . . , sm ∈ S define evaluations, e(x, s1, . . . , sm): W → [n], w → xw(s1, . . . , sm). E(W , S) := {e(x, s1, . . . , sm) : x ∈ [n], s1, . . . , sm ∈ S} ⊆ [n]W . Then S models u ≈ v iff f (u) = f (v) for all f ∈ E(W , S). Example: Band Identity E(W , S) := {(x, xs, xs2) : x ∈ [n], s ∈ S}

Trevor Jack Semigroup Complexity August 7, 2018 7 / 16

Transformation Graphs Proof

Lemmas

Lemma Let S = a1, ..., ak ⊆ Tn, d ∈ N, and f ∈ [n]d. Then fS can be enumerated in O(ndk) time.

Trevor Jack Semigroup Complexity August 7, 2018 8 / 16

Transformation Graphs Proof

Lemmas

Lemma Let S = a1, ..., ak ⊆ Tn, d ∈ N, and f ∈ [n]d. Then fS can be enumerated in O(ndk) time. Definition The degree-d transformation graph of S = a1, ..., ak is G d = (V , E) having vertices V = [n]d and edges E = {(x, y) ∈ V 2 : ∃i ∈ [k](xai = y}, where S acts on [n]d component-wise. Enumerate fS using depth-first search algorithm. There are a maximum of ndk edges and the algorithm traverses each once, hence O(ndk) time.

Trevor Jack Semigroup Complexity August 7, 2018 8 / 16

Transformation Graphs Proof

Lemmas

Lemma Let S = a1, ..., ak ⊆ Tn, d ∈ N, and f ∈ [n]d. Then fS can be enumerated in O(ndk) time. Definition The degree-d transformation graph of S = a1, ..., ak is G d = (V , E) having vertices V = [n]d and edges E = {(x, y) ∈ V 2 : ∃i ∈ [k](xai = y}, where S acts on [n]d component-wise. Enumerate fS using depth-first search algorithm. There are a maximum of ndk edges and the algorithm traverses each once, hence O(ndk) time. Lemma Let f ∈ [n]W . Then f ∈ E(W , S) iff ∀i ∈ [m] ∃g ∈ fS ∀wzi ∈ W : f (wzi) = g(w).

Trevor Jack Semigroup Complexity August 7, 2018 8 / 16

Transformation Graphs Extensions

Extension of Model(u ≈ v) strategy

We now return to the problem of determining if S := a1, . . . , ak ≤ Tn is completely regular. Lemma a ∈ Tn generates a subgroup iff a|Im(a) is a permutation. [Proof of lemma omitted]

Trevor Jack Semigroup Complexity August 7, 2018 9 / 16

Transformation Graphs Extensions

Extension of Model(u ≈ v) strategy

We now return to the problem of determining if S := a1, . . . , ak ≤ Tn is completely regular. Lemma a ∈ Tn generates a subgroup iff a|Im(a) is a permutation. [Proof of lemma omitted] Let W = {1, z, z2} and define: e((x, y), s) : W → [n]2, w → (x, y)w(s) E(W , S) := {e((x, y), s) : (x, y) ∈ [n]2, s ∈ S} ⊆ [n]6 Then every element of S permutes its images iff for every f ∈ E(W , S), f (z) ∈ {(x, x) : x ∈ [n]} ⇒ f (z2) ∈ {(x, x) : x ∈ [n]}.

Trevor Jack Semigroup Complexity August 7, 2018 9 / 16

Transformation Graphs Extensions

Quasi-Identities

Open Problem: Quasi-Identities Complexity of whether S models u1(z1, ..., zm) ≈ v1(z1, ..., zm) ⇒ u2(z1, ..., zm) ≈ v2(z1, ..., zm)?

Trevor Jack Semigroup Complexity August 7, 2018 10 / 16

Transformation Graphs Extensions

Quasi-Identities

Open Problem: Quasi-Identities Complexity of whether S models u1(z1, ..., zm) ≈ v1(z1, ..., zm) ⇒ u2(z1, ..., zm) ≈ v2(z1, ..., zm)? Model(z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u ≈ v)

Input: a1, ..., ak ∈ Tn Output: Whether a1, ..., ak models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u ≈ v

Trevor Jack Semigroup Complexity August 7, 2018 10 / 16

Transformation Graphs Extensions

Quasi-Identities

Open Problem: Quasi-Identities Complexity of whether S models u1(z1, ..., zm) ≈ v1(z1, ..., zm) ⇒ u2(z1, ..., zm) ≈ v2(z1, ..., zm)? Model(z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u ≈ v)

Input: a1, ..., ak ∈ Tn Output: Whether a1, ..., ak models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u ≈ v

Idempotent Quasi-Identity Examples

1 idempotents are central: z1 ≈ z2

1 ⇒ z1z2 ≈ z2z1

2 idempotents commute: z1 ≈ z2

1, z2 ≈ z2 2 ⇒ z1z2 ≈ z2z1

3 Clifford semigroup (completely regular and idempotents commute) 4 composition of idempotents are idempotent:

z1 ≈ z2

1, z2 ≈ z2 2 ⇒ z1z2 ≈ (z1z2)2

Trevor Jack Semigroup Complexity August 7, 2018 10 / 16

Transformation Graphs Extensions

Idempotent Quasi-Identity Problems are in P.

Theorem Model(z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u ≈ v) is in P.

Trevor Jack Semigroup Complexity August 7, 2018 11 / 16

Transformation Graphs Extensions

Idempotent Quasi-Identity Problems are in P.

Theorem Model(z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u ≈ v) is in P.

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Trevor Jack Semigroup Complexity August 7, 2018 11 / 16

Transformation Graphs Extensions

Idempotent Quasi-Identity Problems are in P.

Theorem Model(z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u ≈ v) is in P.

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

To obtain the Theorem from the Lemma, let W ′ := W ∪ {wz2

i : wzi ∈ W , 1 ≤ i ≤ ℓ}. Enumerate E(W ′, S) and check

if each f ∈ E(W ′, S) satisfies: [∀i ∈ [ℓ], ∀wzi ∈ W : f (wzi) = f (wz2

i )] ⇒ f (u) = f (v).

Trevor Jack Semigroup Complexity August 7, 2018 11 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (2) ⇒ (1)

Trevor Jack Semigroup Complexity August 7, 2018 12 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (2) ⇒ (1) Let s1, . . . , sℓ be idempotent.

Trevor Jack Semigroup Complexity August 7, 2018 12 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (2) ⇒ (1) Let s1, . . . , sℓ be idempotent. Then for each x ∈ [n], i ∈ [ℓ], and wzi ∈ W : xw(s1, . . . , sm)si = xw(s1, . . . , sm)s2

i .

Trevor Jack Semigroup Complexity August 7, 2018 12 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (2) ⇒ (1) Let s1, . . . , sℓ be idempotent. Then for each x ∈ [n], i ∈ [ℓ], and wzi ∈ W : xw(s1, . . . , sm)si = xw(s1, . . . , sm)s2

i .

Thus, by (2), xu(s1, . . . , sm) = xv(s1, . . . , sm) for every x ∈ [n].

Trevor Jack Semigroup Complexity August 7, 2018 12 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (1) ⇒ (2)

Trevor Jack Semigroup Complexity August 7, 2018 13 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (1) ⇒ (2) Pick any x ∈ [n], s1, . . . , sm ∈ S satisfying: ∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si = xw(s1, . . . , sm)s2

i .

Trevor Jack Semigroup Complexity August 7, 2018 13 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (1) ⇒ (2) Pick any x ∈ [n], s1, . . . , sm ∈ S satisfying: ∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si = xw(s1, . . . , sm)s2

i .

Then xw(s1, . . . , sm)si = xw(s1, . . . , sm)ski

i

for each wzi ∈ W and ki ∈ N.

Trevor Jack Semigroup Complexity August 7, 2018 13 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (1) ⇒ (2) Pick any x ∈ [n], s1, . . . , sm ∈ S satisfying: ∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si = xw(s1, . . . , sm)s2

i .

Then xw(s1, . . . , sm)si = xw(s1, . . . , sm)ski

i

for each wzi ∈ W and ki ∈ N. For each i ∈ [ℓ], let ki ∈ N satisfy ski

i

= (ski

i )2. By induction on the length

• f u: xu(s1, . . . , sm) = xu(sk1

1 , . . . , skℓ ℓ , . . . , sm).

Trevor Jack Semigroup Complexity August 7, 2018 13 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (1) ⇒ (2) Pick any x ∈ [n], s1, . . . , sm ∈ S satisfying: ∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si = xw(s1, . . . , sm)s2

i .

Then xw(s1, . . . , sm)si = xw(s1, . . . , sm)ski

i

for each wzi ∈ W and ki ∈ N. For each i ∈ [ℓ], let ki ∈ N satisfy ski

i

= (ski

i )2. By induction on the length

• f u: xu(s1, . . . , sm) = xu(sk1

1 , . . . , skℓ ℓ , . . . , sm).

Similarly: xv(s1, . . . , sm) = xv(sk1

1 , . . . , skℓ ℓ , . . . , sm).

Trevor Jack Semigroup Complexity August 7, 2018 13 / 16

Transformation Graphs Extensions

Sketch of Lemma Proof

Lemma Let W be the set of prefixes of u, v ∈ {z1, . . . , zm}∗, and S ≤ Tn. TFAE:

1 S models z1 ≈ z2

1, . . . , zℓ ≈ z2 ℓ ⇒ u(z1, . . . , zm) ≈ v(z1, . . . , zm).

2 ∀x ∈ [n], ∀s1, . . . , sm ∈ S : [∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si =

xw(s1, . . . , sm)s2

i ] ⇒ xu(s1, . . . , sm) = xv(s1, . . . , sm).

Proof Sketch: (1) ⇒ (2) Pick any x ∈ [n], s1, . . . , sm ∈ S satisfying: ∀i ∈ [ℓ], ∀wzi ∈ W : xw(s1, . . . , sm)si = xw(s1, . . . , sm)s2

i .

Then xw(s1, . . . , sm)si = xw(s1, . . . , sm)ski

i

for each wzi ∈ W and ki ∈ N. For each i ∈ [ℓ], let ki ∈ N satisfy ski

i

= (ski

i )2. By induction on the length

• f u: xu(s1, . . . , sm) = xu(sk1

1 , . . . , skℓ ℓ , . . . , sm).

Similarly: xv(s1, . . . , sm) = xv(sk1

1 , . . . , skℓ ℓ , . . . , sm).

By (1), xu(sk1

1 , . . . , skℓ ℓ , . . . , sm) = xv(sk1 1 , . . . , skℓ ℓ , . . . , sm)

Trevor Jack Semigroup Complexity August 7, 2018 13 / 16

Transformation Graphs Extensions

Quantified Identities

Open Problem: Quantified Identities Complexity of whether S models ∃z1, ..., zℓ∀zℓ+1, ..., zm(u1(z1, ..., zm) ≈ v1(z1, ..., zm))?

Trevor Jack Semigroup Complexity August 7, 2018 14 / 16

Transformation Graphs Extensions

Quantified Identities

Open Problem: Quantified Identities Complexity of whether S models ∃z1, ..., zℓ∀zℓ+1, ..., zm(u1(z1, ..., zm) ≈ v1(z1, ..., zm))? Examples: ∃z1(z1z2 ≈ z1) (left zero) and ∃z1(z2z1 ≈ z1) (right zero).

Trevor Jack Semigroup Complexity August 7, 2018 14 / 16

Transformation Graphs Extensions

Quantified Identities

Open Problem: Quantified Identities Complexity of whether S models ∃z1, ..., zℓ∀zℓ+1, ..., zm(u1(z1, ..., zm) ≈ v1(z1, ..., zm))? Examples: ∃z1(z1z2 ≈ z1) (left zero) and ∃z1(z2z1 ≈ z1) (right zero). Theorem Determining if a transformation semigroup has a left zero is in P.

Trevor Jack Semigroup Complexity August 7, 2018 14 / 16

Transformation Graphs Extensions

Quantified Identities

Open Problem: Quantified Identities Complexity of whether S models ∃z1, ..., zℓ∀zℓ+1, ..., zm(u1(z1, ..., zm) ≈ v1(z1, ..., zm))? Examples: ∃z1(z1z2 ≈ z1) (left zero) and ∃z1(z2z1 ≈ z1) (right zero). Theorem Determining if a transformation semigroup has a left zero is in P. Theorem Determining if a transformation semigroup has a right zero is in P.

Trevor Jack Semigroup Complexity August 7, 2018 14 / 16

Transformation Graphs Nilpotence

Nilpotence

0 ∈ S is called a zero if s0 = 0 = 0s for all s ∈ S. A semigroup S containing a zero is called nilpotent if Sd = {0} for some d ∈ N. Theorem Determining if S = a1, ..., ak ⊆ Tn is nilpotent is in P.

Trevor Jack Semigroup Complexity August 7, 2018 15 / 16

Transformation Graphs Nilpotence

Nilpotence

0 ∈ S is called a zero if s0 = 0 = 0s for all s ∈ S. A semigroup S containing a zero is called nilpotent if Sd = {0} for some d ∈ N. Theorem Determining if S = a1, ..., ak ⊆ Tn is nilpotent is in P. Lemma S is nilpotent iff it has a zero element, 0, and the graph (V , E) V := [n] \ Im(0) E := {(x, y) ∈ V 2 | xai = y for some i ∈ [k])} is acyclic.

Trevor Jack Semigroup Complexity August 7, 2018 15 / 16

Transformation Graphs Open Problems

Open Problems

Open Problem: Semigroup Regularity How hard is it to check that every element in S is regular?

Trevor Jack Semigroup Complexity August 7, 2018 16 / 16

Transformation Graphs Open Problems

Open Problems

Open Problem: Semigroup Regularity How hard is it to check that every element in S is regular? Open Problem: Quasi-Identities Complexity of whether S models u1(z1, ..., zm) ≈ v1(z1, ..., zm) ⇒ u2(z1, ..., zm) ≈ v2(z1, ..., zm)?

Trevor Jack Semigroup Complexity August 7, 2018 16 / 16

Transformation Graphs Open Problems

Open Problems

Open Problem: Semigroup Regularity How hard is it to check that every element in S is regular? Open Problem: Quasi-Identities Complexity of whether S models u1(z1, ..., zm) ≈ v1(z1, ..., zm) ⇒ u2(z1, ..., zm) ≈ v2(z1, ..., zm)? Open Problem: Quantified Identities Complexity of whether S models ∃z1, ..., zℓ∀zℓ+1, ..., zm(u1(z1, ..., zm) ≈ v1(z1, ..., zm))?

Trevor Jack Semigroup Complexity August 7, 2018 16 / 16