Permutation Groups symmetric group S:all permutations r:列m on-to cyclic group Cn:rotations π=(012…n-1)π(i)=(i+1)modn (012…n-1)》generated by(012..n-1) Dihedral group D:rotations reflections p()=(n-1)-i generated by (012...n-1)and p
Permutation Groups symmetric group cyclic group Dihedral group Sn Cn Dn : all permutations : rotations : rotations & reflections ⇡ : [n] 1-1 ! on-to [n] h(012 ··· n 1)i generated by (012 ··· n 1) ⇡ = (012 ··· n 1) ⇡(i)=(i + 1) mod n generated by (012 ··· n 1) and ⇢(i)=(n 1) i ⇢
Group Action configuration x:[n][m] X (m]linl permutation π：m1m group G on-to (πox)(i)=x(π() group action o:G×X→X ·associativity:（(π·o)ox=To(oox) ·identity::eor=r
Group Action 0 1 2 3 4 5 0 1 2 3 4 5 configuration x : [n] ! [m] permutation ⇡ : [n] 1-1 ! on-to [n] group action X = [m] [n] group G : G ⇥ X ! X • associativity: • identity: (⇡ · ) x = ⇡ ( x) e x = x (⇡ x)(i) = x(⇡(i))
Graph Isomorphism (Gl)Problem input:two undirected graphs G and H output:G≈H? p n 6) vertices 8 7 ^b Gl is in NP,but is NOT known to be in P or NPC trivial algorithm:O(n!)time Babai-Luks'83:20(vmlogn)time Babai 2015:a quasi-polynomial time algorithm! npolylog(n)=2polylog(n)
Graph Isomorphism (GI) Problem ? • GI is in NP, but is NOT known to be in P or NPC • trivial algorithm: O(n!) time • Babai-Luks ’83: time n vertices Babai 2015: a quasi-polynomial time algorithm! npolylog(n) = 2polylog(n) 2O( pn log n) ⇠ = input: two undirected graphs G and H output: G ⇠ = H ?
String Isomorphism (SI) input:two strings x,y:[n][m] a permutation group G Sn output::x≈cy?(3o∈Gs.t.oox=y)
input: two strings a permutation group output: String Isomorphism (SI) ⇠ = ? x, y : [n] ! [m] G ✓ Sn x ⇠ =G y? (9 2 G s.t. x = y)
String Isomorphism (SI) input:two strings a,y [n][m] a permutation group G Sn output:x≥cy?(3o∈Gs.t.oox=y) a graph X(V,E)is a string :() →0， 1:edge 0:no edge positions←→vertex-.pairs all permutations on V induces a permutation group on Johnson group:SCS() on vertex pairs two graphs≈iff their string versions兰sgy
String Isomorphism (SI) a graph X(V,E) is a string x : ✓V 2 ◆ ! {0, 1} all permutations on V induces a permutation group on on vertex pairs two graphs X ⇠ = Y iff their string versions x ⇠ =S(2) V y positions ⟷ vertex-pairs 1: edge 0: no edge Johnson group: input: two strings a permutation group output: x, y : [n] ! [m] G ✓ Sn x ⇠ =G y? (9 2 G s.t. x = y) S(2) V ⇢ S( V 2 ) ✓V 2 ◆