Quandles and the Towers of Hanoi

Transcription

1 The the Bob Laboratory for Algebraic and Symbolic Computation Reem-Kayden Center for Science and Computation Bard College Annandale-on-Hudson, NY August 8, 2011

2 The The

3 The Figure Eight (4 1 ) Knot The

4 Knot A knot quandle is a system of equations relating the arcs of a knot, one for each crossing. a b c The a a b = c b a/b = c c

5 Figure Eight 4 1 The Q(4 1 ) = 0, 1, 2, 3 0/2 = 1, 1 3 = 2, 2/0 = 3, 3 1 = 0

6 Type I Reidemeister Move The x y x x = y = x x

7 Type II Reidemeister Move The y x y z w (x y)/y = z/y = w = x x

8 Type III Reidemeister Move x y w z s v x y t z r u The (x z) (y z) = w s = v = u = t z = (x y) z

9 The Definition A quandle (Q,, /) is a set Q along with a binary operations and / on Q satisfying: I x(x x = x); IIa xy((x/y) y = x); IIb xy((x y)/y = x); and III xyz((x y) z = (x z) (y z)). Since uniquely determines /, we can limit our mention of /. [Note: Axioms I, IIa, and IIb constitute the theory of idempotent, right quasigroups.]

10 Unary Quandle The Table: Unary Quandle U 3

11 Latin Quandle The Table: Latin (Quasigroup) Quandle

12 Basic Elements of a CSP The A, a finite domain; V = {v 1, v 2,..., v n,...},a countable collection of variables; and Γ, which is a collection of relations R A n for various positive integers n.

13 Example: Sat The Question Does there exists a truth assignment for the proposition below? α = ( v 1 v 2 ) (v 3 v 2 ) The domain is easy: A = {0, 1}

14 Constraints The Definition (Constraint) A constraint over Γ is a pair (v i1, v i2,..., v im ), R, where R is a relation in Γ of arity m. α = ( v 1 v 2 ) (v 3 v 2 ) translates to the following constraints: and C 1 = (v 1, v 2 ), {(0, 0), (0, 1), (1, 1)} C 2 = (v 3, v 2 ), {(0, 0), (1, 0), (1, 1)}.

15 CSP(Γ) The Definition (CSP(Γ)) CSP(Γ) is the combinatorial decision problem with the following components. Instance: A triple I = (V, A, C) where C is a finite set of constraints over Γ and V is a finite subset of V. Solution: A function θ : V A such that for every constraint (v 1, v 2,..., v m ), R C, (θ(v 1 ), θ(v 2 ),..., θ(v m )) R.

16 Tractable and NP-Complete CSP The Definition Let Γ be a collection of relations over a finite domain A. CSP(Γ) is tractable if CSP(Γ ) is in P for all finite Γ Γ. CSP(Γ) is NP-complete if CSP(Γ ) is NP-complete for some finite Γ Γ.

17 Complexity of CSP s The Tractable: 2-Color 2-Sat NP-Complete: k-color for k > 2 k-sat for k > 2 Schedule n-queens

18 CSP(Q) The Definition (Subpower) Recall that a subpower of a quandle Q is a subquandle of Q n. Let Sub(Q) stand for the class of subpowers of Q. Definition (CSP(Q)) CSP(Q) = CSP(Γ) where Γ = Sub(Q). Definition Q is NP-complete if CSP(Q) is NP-complete and is tractable if CSP(Q) is tractable.

19 U 2 The Theorem U 2 is NP-complete. Proof Table: U 2 Sub(U 2 ) includes all relations over {0, 1}. So 3-Sat is a finite, NP-complete subproblem of CSP(Q) where Q = U 2.

20 Connectedness The Definition (Connected) A quandle Q is connected if its right Cayley graph for is connected. Proposition A quandle Q is disconnected iff there exists a surjective homomorphism h : Q U 2. Definition (Totally Connected) A quandle Q is totally connected if every element of Sub(Q) is connected.

21 Homomorphic Images The Theorem Suppose h : Q Q is a surjective quandle homomorphism. If Q is tractable, so is Q. If Q is NP-complete, so is Q. Corollary Disconnected quandles are NP-complete.

22 Subpowers The Theorem Suppose Q Sub(Q). If Q is tractable, so is Q. If Q is NP-complete, so is Q. Corollary If Q is not totally connected, then it is NP-complete. If Q is not totally connected, then U 2 appears in its variety.

23 Merling Terms The Definition A Merling term t(x, y) for a quandle Q satisfies the following 1 t(x, y) = y in Q; and 2 t(x, y) = x in U 2. Note: If such a term exists for Q, then Q and U 2 have independent varieties. Proposition Every totally connected quandle has a Merling term. Why? F (2, Q) is a subpower of Q, so it is connected. t(x, y) is the path from x to y in F (2, Q).

24 From Merling to Malcev The Lemma If a quandle Q has a Merling term, then it also has a Mal cev term. This follows from a two-stage transformation of t(x, y): 1 Construct term s(x, y, z) via selective substitution. 2 Use right cancellation to form m(x, y, z). This transformation is demonstrated on the term t(x, y) = ((x (y x)) (x y)) (y x).

25 s(x, y, z) The To construct s(x, y, z) from t(x, y): Replace all instances of y with z, and Replace every instance of x except the first with y. For example, t(x, y) = ((x (y x)) (x y)) (y x) yields s(x, y, z) = ((x (z y)) (y z)) (z y).

26 Properties of s(x, y, z) The Hence, if t(x, y) = y, then s(x, x, y) = t(x, y) = y, s(x, y, y) = ((x (y y)) (y y)) (y y) = ((x y) y) y.

27 m(x, y, z) The To determine m(x, y, z) from s(x, y, z), use right cancellation to unravel s(x, y, y). For example, for let s(x, y, y) = ((x y) y) y, m(x, y, z) = ((s(x, y, z)/z)/z)/z.

28 m(x, y, z) is Mal cev The If t(x, y) = y, then m(x, x, y) = ((s(x, x, y)/y)/y)/y = ((y/y)/y)/y = y, m(x, y, y) = ((s(x, y, y)/y)/y)/y = (((((x y) y) y)/y)/y)/y = x.

29 Malv cev Terms and Tractability The Theorem (Bulatov,Dalmau) If an algebra has a Mal cev term, then it is tractable. Corollary If a quandle Q is totally connected, then it is tractable.

30 The Theorem Every quandle that is not NP-complete is tractable. In fact, since right self distributivity is never employed, this result extends as follows. Theorem Every idempotent, right quasigroup that is not NP-complete is tractable.

31 Types 4 and 5 Omitted The Theorem (Hobby) The pseudo variety of a finite quandle omits types 4 and 5. The proof does use right, self distributivity and so does not apply to idempotent, right quasigroups.

32 The Specific: Let Q = A R be a finitely presented quandle. Does there exist an algorithm that decides whether two expressions q 1 and q 2 over the generators A represent the same element of Q? General: Does there exist a general algorithm over all quandles that takes A R as input and then proceeds with a correct decision method? In groups the specific version depends on the group, so the general algorithm does not exist.

33 Conjugation The Given a group G, define a quandle structure Conj(G) via and g h = h 1 gh g/h = hgh 1. A conjugation quandle is a quandle Q that embeds into Conj(G) for some G.

34 Conj(G) Naively The Given a finitely presented group G = A W with undecidable word problem, is Conj(G) a finitely presented quandle with undecidable work problem? Conj(G) is not necessarily finitely presented. Conj(G) does not faithfully reflect G (Groups omit type 1!). Conj(G) is not sufficiently general among quandles.

35 Lopsided Quandle The The following quandle cannot arise through conjugation Table: Lopsided Quandle

36 Inner Automorphisms The Let Q be a quandle. For q Q define r q, R q : Q Q by right translation as follows r q (p) = p q and R q (p) = p/q.

37 Inner Automorphisms The r q is always a permutation: R q = r 1 q. r q is a quandle homomorphism: r q (a b) = (a b) q = (a q) (b q) = r q (a) r q (b). Let Inn(Q) be the group generated by the permutations {r q q Q}. We will show that Inn(Q) is general among groups.

38 The Quandle The Given a finitely presented group G = A W, we wish to construct a finitely presented quandle = A R W.

39 Example: G = C 2 The Consider the following group presentation: C 2 = a aa. Let x be a fresh generator and Q C2 be the following quandle presentation: Q C2 = a, x (x a) a = x, (a a) a = a, or, rather, Q C2 = a, x x aa = x, a aa = a.

40 The Quandle The Given G = A W, let x be a new generator not in A and let = A x R W where A x = A {x} and R W = {a w = a a A x, w W }.

41 The Formal Expression q g The Given a quandle expression q over A x and a group expression g over A, define the quandle expression q g over A x by structural induction: q e = q, q e 1 = q, q a = q a = r a (q), for a A, q a 1 = q/a = R a (q) for a A, q g 1g 2 = (q g 1 ) g 2, q (g 1g 2 ) 1 = (q g 1 2 ) g 1 1.

42 Main Result The Theorem Given a finitely presented group G, Corollary G g = e iff x g = x. The word problem for finitely presented quandles is, in general, undecidable. Proof. Let G be Novikov s group (1952), then must have undecidable word problem.

43 Forward Direction The Proposition If G g = e then x g = x. Define ρ : A Inn( ) by ρ a = r a. Then ρ extends to a group homomorphism ρ : FGA Inn( ). Furthermore, for any h G and q, ρ h (q) = q h.

44 Forward Direction (Continued) The In particular, for each w W and a A x, so It follows that ρ w (a) = a w = a = id(a) Inn( ) ρ w = id. W ker(ρ). Let π G : FGA G be the canonical projection. Recall that ker(π G ) is the smallest normal subgroup containing W. Hence, ker(π G ) ker(ρ).

45 Forward Direction (Continued) The This means that ρ defines a homomorphism of type ρ : G Inn( ). Consequently, if G g = e then Inn( ) ρ g = id so that x g = ρ g (x) = id(x) = x.

46 Backward Direction The Proposition If x g = x then G g = e. First, consider the following finitely presented group: G x = A x C W, where C W = {w 1 aw = a a A x, w W }. The proposition is a consequence of the following: Fact 1: If x g = x then G x g 1 xg = x. Fact 2: if G x g 1 xg = x then G g = e.

47 Fact 1 The It is left to the audience to verify that for any h FGA and a A x, Conj(G x ) a h = a iff G x h 1 ah = a. In particular, for all w W and a A x, Conj(G x ) a w = a, so that the relations in R W hold in Conj(G x ). Therefore, there exists a quandle homomorphism φ : Conj(G x ) such that φ(a) = a for all a A x.

48 Fact 1 (Continued) The Then if x g = x the quandle homomorphism φ ensures Conj(G x ) x g = x. It follows that G x g 1 xg = x.

49 Fact 2 The Let the free product of G and x. G[x] = A x W = G x, Since the relations C W hold in G[x], there exists a unique homomorphism χ : G x G[x] such that χ(a) = a for all a A x.

50 Fact 2 (Continued) The Moreover, the following square commutes: χ π x = ι G π G where π x and π G are the canonical homomorphisms that fix A.

51 Fact 2 (Continued) The Consequently, if G x g 1 xg = x then, via χ, G[x] g 1 xg = x. However, since G[x] = G x, G[x] g = e, which, since G is included in G[x], means that G g = e.

52 Main Result The Theorem Given a finitely presented group G, Corollary G g = e iff x g = x. The word problem for finitely presented quandles is, in general, undecidable. Proof. Let G be Novikov s group (1952), then must have undecidable word problem.

53 R, a Term Rewriting System The ι rules: x x x x/x x ρ rules: (x y)/y x (x/y) y x δ rules: x (y z) ((x/z) y) z x (y/z) ((x z) y)/z x/(y z) ((x/z)/y) z x/(y/z) ((x z)/y)/z)

54 Basics The Definition Let t and t be terms over {, /}. We say t t if t rewrites to t in one step, t t if t rewrites to t in zero or more steps, and t + t if t rewrites to t in one or more steps.

55 The Definition A term t is a normal form if whenever t t, t is identical to t. Theorem (Golbus, Gutierrez, McGrail (GGM)) The R is strongly normalizing (SN). That is, every infinite rewrite sequence t 0 t 1... includes a normal form.

56 The Proof. (Sketch) Assume without loss of generality that t i + t i+1 for each i. Define a function m from terms to the natural numbers as follows: { 0, if t is a variable; m(t) = (1) 1 + m(t 1 ) + 3m(t 2 ), if t = t 1 o t 2. Then {m(t i ) i N} is a subset of N with no least element.

57 The Theorem (GGM) The R is confluent. That is whenever t u and t v there exists a term s with u w and v w. Notes: If S is SN and confluent, it has unique normal forms. So R has unique normal forms.

58 R Decides Q The Theorem (GGM) The R decides the pure equational theory of Q. That is, the following are equivalent for any two terms t and s: t = s is a theorem of Q t and s have the same normal form r over R.

59 The The There are n disks, each of unique size. There are k pegs. The disks are initially stacked on a source peg with smaller disks stacked directly on larger disks. The disks must all be moved to a target peg. Only one disk may be moved at one time. One may never stack a larger disk on a smaller disk.

60 and the The The encoded into : Initial Term y (x 3 (x 2 x 1 )) Normal Form ((((((y/x 1 )/x 2 ) x 1 ) x 3 )/x 1 ) x 2 ) x 1 Interpret both terms above in terms of number of pegs. Interpret t = ((y/(x 2 x 1 )) x 3 ) (x 2 x 1 ) as a four-peg solution.

61 Grand Daddy Term The Definition For n = 1, 2,..., define the term t n as follows: t 1 = x 1 ; and t n+1 = x n+1 t n. Definition For n = 1, 2,..., define g n as below: g 1 = y x 1 ; and g n+1 = ((y/t n ) x n+1 ) t n.

62 Solutions as Descendants The Definition A nice solution to the n-disk puzzle that uses no more than n pegs corresponds to some term s with g n s. Question If g n s, does s correspond to some solution?

63 The The Thank you!!! Any questions?