Help23:=proc(): print(` GenS(G) `): end: #GenS(G): Inputs a directed graph (w/o cycles) and outputs a list of "generations" #L[1] : the set of vertices that are sinks (no childre), and L[i] the set of vertices #all whose children are in L[1] union L[i-1] GenS:=proc(G) local n,T,i,j,NYD,AD,j1: n:=nops(G): NYD:={seq(i,i=1..n)}: T[1]:={}: for i from 1 to n do if G[i]={} then T[1]:=T[1] union {i}: fi: od: NYD:=NYD minus T[1]: AD:=T[1]: for j from 2 while NYD<>{} do T[j]:={}: for i in NYD do if G[i] minus AD={} then T[j]:=T[j] union {i}: fi: od: AD:=AD union T[j]: NYD:=NYD minus T[j]: od: [seq(T[j1],j1=1..j-1)]: end: #DZ:=[{3,4,5},{3,4,5},{6},{},{},{}]; #GenS(DZ); # [{4, 5, 6}, {3}, {1, 2}] #LaGn(G): Same as LaG(G) from C22.txt, but hopefully better (at least more adaptable for the Centipede games) LaGn:=proc(G) local n,L,T,i,j,s: n:=nops(G): L:=GenS(G): for i in L[1] do T[i]:=0: od: for j from 2 to nops(L) do for i in L[j] do if member(0,{seq(T[s], s in G[i])}) then T[i]:=1: else T[i]:=0: fi: od: od: [seq(T[i],i=1..n)]: end: Els:=proc(): [{2, 6}, {3, 7}, {4, 8}, {5, 9}, {}, {}, {}, {}, {}], {[5, [4, 3]], [6, [2, -1] ], [7, [2, 1]], [8, [4, 1]], [9, [3, 6]]}: end: ###STUFF FROM C22.txt #C22.txt: April 11, 2022 Help22:=proc(): print(`mex(S), ai(i), aiC(i), RDG(n,p), Parents(G), OneStep0(G,AL,T) , OneStep1(G,AL,T), OneStep(G,AL,T), LaG(G) `): end: #mex(S): inputs a set of NON-NEGATIVE integers and outputs the smallest #non-neg. intger NOT in the set S, i.e. min({0,1,2,3,...} minus S) mex:=proc(S) local i: for i from 0 while member(i,S) do od: i: end: #ai(i): the a-component in the losing position for Wythoff's game [a_i,i+a_i] ai:=proc(i) local j: option remember: if i=0 then RETURN(0): else mex({seq(ai(j),j=0..i-1),seq(ai(j)+j,j=0..i-1)}): fi: end: #aiC(i) : Faster way to compute ai(i) using Wythof's formula trunc(phi*i): aiC:=proc(i) trunc((1+sqrt(5))/2*i): end: #RDG(n,p): inputs a pos. integer n and a RATIONAL number p=a/b between 0 and 1 #and outputs a random directed graph w/o cycles such tha the prob. of #an edge is p RDG:=proc(n,p) local a,b,L,i,j,S,ra: a:=numer(p): b:=denom(p): ra:=rand(1..b): L:=[]: for i from 1 to n do S:={}: for j from i+1 to n do if ra()<=a then S:=S union {j}: fi: od: L:=[op(L),S]: od: L: end: #Parents(G): inputs a directed graph on {1,...,n} where n:=nops(G) such that G[i] is the set of children of i (the set of outgoing neihbors) #outputs the list whose i-th component is the set of "parents" of i, i.e. those j such that i belongs to G[j]. For example #Parents([{2,3},{3},{}]); should be [{},{1},{1,2}] Parents:=proc(G) local n,i,j,P: option remember: n:=nops(G): #We initialize all the parents-sets to be empty for j from 1 to n do P[j]:={}: od: for i from 1 to n do for j in G[i] do P[j]:=P[j] union {i}: od: od: #For each vertex i (we look for its children) and append i to the set of parents of each such child [seq(P[j],j=1..n)]: end: #OneStep0(G,AL,T): Given a graph G, with a set of already labelled vertices AL and a table T of lables #performs one step in the labelling algorithm, the implication that if a vertex is labeled 0 then all its parents are labeled 1 #(since there exists is a legal move that will make the opponent lose) OneStep0:=proc(G,AL,T) local AL1,n,T1,i,j,P: AL1:=AL: n:=nops(G): P:=Parents(G): T1:=T: for i in AL do if T1[i]=0 then for j in P[i] do T1:=[op(1..j-1,T1),1,op(j+1..n,T1)]: AL1:=AL1 union {j}: od: fi: od: RETURN(G,AL1,T1): end: #OneStep1(G,AL,T): Given a graph G, with a set of already labelled vertices AL and a table T of lables #performs one step in the labelling algorithm, that if all the children of a vertex are already labeld 1 then #it is labeled 0 (since whatever the player can do it would be winning for the opponent) OneStep1:=proc(G,AL,T) local n,T1,i,j: n:=nops(G): T1:=T: for i from 1 to n do if not member(i,AL) then if G[i] minus AL={} and {seq(T1[j],j in G[i])}={1} then T1:=[op(1..i-1,T1),0,op(i+1..n,T1)]: RETURN(G,AL union {i},T1): fi: fi: od: G,AL,T1: end: #OneStep(G,AL,T): Given a graph G, with a set of already labelled vertices AL and a table T of lables #performs one step in the labelling algorithm, that if all the children of a vertex are already labeld 1 then #it is labeled 0 (since whatever the player can do it would be winning for the opponent) OneStep:=proc(G,AL,T) local n,Hope: n:=nops(G): if nops(AL)=n then RETURN(G,AL,op(T)): fi: Hope:=OneStep0(G,AL,T): if nops(Hope[2])>nops(AL) then RETURN(Hope): fi: Hope:=OneStep1(G,AL,T): if nops(Hope[2])>nops(AL) then RETURN(Hope): fi: FAIL: end: #LaG(G) LaG:=proc(G) local n,i,AL,T,Hope: n:=nops(G): T:=[(-1)$n]: AL:={}: for i from 1 to n do if G[i]={} then T:=[op(1..i-1,T),0,op(i+1..n,T)]: AL:=AL union {i}: fi: od: Hope:=G,AL,T: while nops(Hope[2])