As you said you don't mind a non-optimal solution, I though I would re-use your initial function, and add a way to find a good starting arrangement for your initial list s
Your initial function:
def pigeon_hole(s):
a = [[], [], []]
sum_a = [0, 0, 0]
for x in s:
i = sum_a.index(min(sum_a))
sum_a[i] += x
a[i].append(x)
return map(sum, a)
This is a way to find a sensible initial ordering for your list, it works by creating rotations of your list in sorted and reverse sorted order. The best rotation is found by minimizing the standard deviation, once the list has been pigeon holed:
def rotate(l):
l = sorted(l)
lr = l[::-1]
rotation = [np.roll(l, i) for i in range(len(l))] + [np.roll(lr, i) for i in range(len(l))]
blocks = [pigeon_hole(i) for i in rotation]
return rotation[np.argmin(np.std(blocks, axis=1))] # the best rotation
import random
print pigeon_hole(rotate([random.randint(0, 20) for i in range(20)]))
# Testing with some random numbers, these are the sums of the three sub lists
>>> [64, 63, 63]
Although this could be optimized further it is quite quick taking 0.0013s for 20 numbers. Doing a quick comparison with @Mo Tao's answer, using a = rotate(range(1, 30))
# This method
a = rotate(range(1, 30))
>>> [[29, 24, 23, 18, 17, 12, 11, 6, 5], [28, 25, 22, 19, 16, 13, 10, 7, 4, 1], [27, 26, 21, 20, 15, 14, 9, 8, 3, 2]]
map(sum, a)
# Sum's to [145, 145, 145] in 0.002s
# Mo Tao's method
>>> [[25, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1], [29, 26, 20, 19, 18, 17, 16], [28, 27, 24, 23, 22, 21]]
# Sum's to [145, 145, 145] in 1.095s
This method also seems to find the optimal solution in many cases, although this probably wont hold for all cases. Testing this implementation 500 times using a list of 30 numbers against Mo Tao's answer, and comparing if the sub-lists sum to the same quantity:
c = 0
for i in range(500):
r = [random.randint(1, 10) for j in range(30)]
res = pigeon_hole(rotate(r))
d, e = sorted(res), sorted(tao(r)) # Comparing this to the optimal solution by Mo Tao
if all([k == kk] for k, kk in zip(d, e)):
c += 1
memory = {}
best_f = pow(sum(s), 3)
best_state = None
>>> 500 # (they do)
I thought I would provide an update with a more optimized version of my function here:
def rotate2(l):
# Calculate an acceptable minimum stdev of the pigeon holed list
if sum(l) % 3 == 0:
std = 0
else:
std = np.std([0, 0, 1])
l = sorted(l, reverse=True)
best_rotation = None
best_std = 100
for i in range(len(l)):
rotation = np.roll(l, i)
sd = np.std(pigeon_hole(rotation))
if sd == std:
return rotation # If a min stdev if found
elif sd < best_std:
best_std = sd
best_rotation = rotation
return best_rotation
The main change is that the search for a good ordering stops once a suitable rotation has been found. Also only the reverse sorted list is searched which doesnt appear to alter the result. Timing this with
print timeit.timeit("rotate2([random.randint(1, 10) for i in range(30)])", "from __main__ import rotate2, random", number=1000) / 1000.
results in a large speed up. On my current computer rotate takes about 1.84ms and rotate2 takes about 0.13ms, so about a 14x speed-up. For comparison גלעד ברקן 's implementation took about 0.99ms on my machine.
