Trip Technician wrote:
On 21 Apr, 14:46, MRAB <goo...@mrabarnett.plus.com> wrote:
Trip Technician wrote:
Thank you Dave. This does it but slowly. takes every subset of the
list a ofsquares, 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]
Even this code doesn't find them all! For 135 it finds [49, 49, 36, 1],
[81, 25, 25, 4] and [81, 36, 9, 9], but not [121, 9, 4, 1].- Hide quoted text -
- Show quoted text -
blowed if i know why that is !
I think I might have cracked it:
import math
def sumsq(n):
if n == 0:
return [[]]
root = int(math.sqrt(n))
square = root ** 2
sums = [[square] + s for s in sumsq(n - square)]
while root > 1:
root -= 1
square = root ** 2
if square < n // len(sums[0]):
break
more_sums = [[square] + s for s in sumsq(n - square)]
if len(more_sums[0]) == len(sums[0]):
sums.extend(more_sums)
return sums
for n in range(1, 150):
# Find all the possible sums.
sums = sumsq(n)
# Create a set of the unique combinations.
sums = set(tuple(sorted(s, reverse=True)) for s in sums)
# Convert back to a list of lists.
sums = [list(s) for s in sorted(sums, reverse=True)]
print n, sums
--
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