Thank you Dave. This does it but slowly. takes every subset of the
list a of squares, and then gets a 'partition' that will work, many
are very inefficient (with lots of 1s).
any hints about how to speed up ?
def subset(x):
for z in range(1,2**len(x)):
q=bin(z)
subs=[]
for dig in range(len(q)):
if q[dig]=='1':
subs.append(x[dig])
yield subs
def bin(x):
q=""
while x>=1:
q+=str(x%2)
x//=2
return q
def squ(z,b):
if z==0:
return 0
for x in b:
if z>=x:
return x,squ(z-x,b)
def flatten(lst):
for elem in lst:
if type(elem) in (tuple, list):
for i in flatten(elem):
yield i
else:
yield elem
sizelim=150
a=[x**2 for x in range(int(sizelim**0.5),1,-1)]
q,r=[],[]
for aa in range(sizelim):
r.append([])
for xx in range(1,sizelim):
for z in subset(a):
q=[]
z.append(1)
for rr in flatten(squ(xx,z)):
if rr !=0:
q.append(rr)
item=[len(q),q]
if item not in r[xx]:
r[xx].append(item)
r[xx].sort()
for eee in r:
if eee:
print r.index(eee),eee[0:3]
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