#OK to post homework
#Charles Kenney, March 15, 2020, Assignment 15

#1. The results of running NickAv(4,4,1.3,1.0,1000,10000,E,E,M) 10 times:
#9.9328, 11.9492, 10.8632, 10.7632, 10.8384, 
#11.0156, 11.4548, 11.0688, 10.2564, 10.4040
#These results vary from one another a good amount. Could it be that
#The algorithm is sensitive to the starting random matrix?

#2. The results of running NickAv(40,40,1.3,1.0,1000,10000,E,E,M) 10 times:
#822.9956, 909.2212, 860.5820, 864.0736, 888.7244, 
#875.4416, 883.7940, 936.1312, 906.6236, 868.9172
#In absolute terms, these results are even farther from one another than
#the results for NickAv(4,4,...), but in relative terms they
#are similarly spaced.


#3. NickAvONE(M,N,x,y,K1,K2,G,En,Ma) is just like NickAv(...)
#except that instead of inputting a random matrix, it starts with
#a matrix of all 1s
NickAvONE:=proc(M,N,x,y,K1,K2,G,En,Ma) local A,i,su,j:
A:=[seq([seq(1,j=1..N)],i=1..M)]:
for i from 1 to K1 do
 A:=OSM(A,M,N,x,y):
od:
su:=0:
for i from 1 to K2 do
 A:=OSM(A,M,N,x,y):
 su:=su+subs({En=Ene(A),Ma=Mag(A)},G):
od:
evalf(su/K2):
end:
#Now the outputs from running NickAvONE(4,4,1.3,1.0,1000,10000,E,E,M)
#ten times are:
#10.3772, 10.9248, 9.8596, 11.4092, 10.1216, 
#11.0996, 10.7924, 10.7076, 10.7196, 11.8880
#These do not seem to be much more clustered than the outputs
#which began with a random matrix. Curious!
#Here are the outputs from running 
#NickAvONE(40,40,1.3,1.0,1000,10000,E,E,M) 10 times:
#1372.4368, 1340.6256, 1351.5400, 1423.3656, 1369.7848,
#1296.9268, 1309.1656, 1398.4184, 1448.8112, 1365.8460
#Again, the clustering is about the same as before.



#NickAvALT(M,N,x,y,K1,K2,G,En,Ma) is just like NickAv(...)
#except that instead of inputting a random matrix, it starts with
#a matrix alternating +/- 1s
NickAvALT:=proc(M,N,x,y,K1,K2,G,En,Ma) local A,i,su,j:
A:=[seq([seq((-1)^(i+j),j=1..N)],i=1..M)]:
for i from 1 to K1 do
 A:=OSM(A,M,N,x,y):
od:
su:=0:
for i from 1 to K2 do
 A:=OSM(A,M,N,x,y):
 su:=su+subs({En=Ene(A),Ma=Mag(A)},G):
od:
evalf(su/K2):
end:
#The outputs from running NickAvALT(4,4,1.3,1.0,1000,10000,E,E,M)
#ten times are:
#10.7172, 10.6196, 10.0492, 10.4244, 11.0508,
#10.8988, 10.6592, 10.0556, 10.3104, 11.6188

#And from NickAvALT(40,40,1.3,1.0,1000,10000,E,E,M):
#807.4344, 838.7960, 869.7960, 873.6140, 871.3292,
#892.9936, 810.8612, 871.3244, 824.1748, 852.0464




#4. Shrink(A) starts with a 3^k by 3^k matrix of {1,-1} 
#as always in a list of lists, and outputs
#a 3^(k-1) by 3^(k-1) matrix according to majority rule.
#First, we need a procedure that outputs a full list
#of the neighbors of a point in a matrix (with no wrap-around,
#for simplicity)

#ThreeBy inputs a matrix A and a position c=[i,j],
#outputs a list of the 9 nearest neighbors of c,
#which form a 3x3 box centered at c.
ThreeBy:=proc(A,c) local i,j:
i:=c[1]: j:=c[2]:
[A[i-1][j-1],A[i-1][j],A[i-1][j+1],A[i][j-1],A[i][j],
A[i][j+1],A[i+1][j-1],A[i+1][j],A[i+1][j+1]]:
end:

#MajorRul inputs a {+1,-1} list of length 9,
#outputs a 1 if a majority of entries are +1,
#and a -1 otherwise
MajorRul:=proc(L) local f:
f:=add(L):
if f>0 then
1:
else
-1:
fi:
end:

Shrink:=proc(A) local i,j,k:
k:=log[3](nops(A)):
[seq([seq(MajorRul(ThreeBy(A,[3*j-1,3*i-1])),i=1..3^(k-1))],j=1..3^(k-1))]:
end: