#OK to post homework
#Yuxuan Yang, April 15th, Assignment 22
with(combinat):

#1
#Suppose j is the largest number such that ai[j]+j<ai[i], then because of the disjoint and complete cover property, we know that j+i=ai[i]
#At the same time, ai[j]+j is really close to ai[i], then we have j*alpha+j=i*alpha and j+i=i*alpha, which leads to alpha=Golden Ratio.

#4
LP:=proc(G) local s,i,lag:
s:={}:lag:=LaG(G):
for i from 1 to nops(G) do
 if lag[i]=0 then s:=s union {i}: fi:
od:
s:
end:


#5
AvNuLP:=proc(n,p,K) local i,G,lp:
lp:=0:
for i from 1 to K do
G:=RDG(n,p):
lp:=lp+nops(LP(G)):
od:
evalf(lp/K):
end:
#seq(AvNuLP(10, p, 10000), p in {1/5, 2/5, 3/5, 4/5})
#5.502600000, 3.790600000, 2.778200000, 2.026700000
#K=10000 is too much for my computer. I used K=1000 instead.
#seq(AvNuLP(50, p, 1000), p in {1/5, 2/5, 3/5, 4/5})
#11.40600000, 6.719000000, 4.461000000, 2.973000000
#14.33000000, 7.976000000, 5.219000000, 3.387000000



#Maple code #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])<n do
Hope:=OneStep(Hope):
od:

[seq(Hope[3][i],i=1..n)]:
end:
   





#LaG([{2, 4}, {4, 5}, {4}, {5}, {}]);