pythoncombinationssubset-sumlexicographicrecursion-schemes

# Generate lexicographic series efficiently in Python

I want to generate a lexicographic series of numbers such that for each number the sum of digits is a given constant. It is somewhat similar to 'subset sum problem'. For example if I wish to generate 4-digit numbers with sum = 3 then I have a series like:

[3 0 0 0]

[2 1 0 0]

[2 0 1 0]

[2 0 0 1]

[1 2 0 0] ... and so on.

I was able to do it successfully in Python with the following code:

``````import numpy as np

M = 4 # No. of digits
N = 3 # Target sum

a = np.zeros((1,M), int)
b = np.zeros((1,M), int)

a[0][0] = N
jj = 0

while a[jj][M-1] != N:
ii = M-2
while a[jj][ii] == 0:
ii = ii-1
kk = ii
if kk > 0:
b[0][0:kk-1] = a[jj][0:kk-1]
b[0][kk] = a[jj][kk]-1
b[0][kk+1] = N - sum(b[0][0:kk+1])
b[0][kk+2:] = 0
a = np.concatenate((a,b), axis=0)
jj += 1

for ii in range(0,len(a)):
print a[ii]

print len(a)
``````

I don't think it is a very efficient way (as I am a Python newbie). It works fine for small values of M and N (<10) but really slow beyond that. I wish to use it for M ~ 100 and N ~ 6. How can I make my code more efficient or is there a better way to code it?

Solution

• Very effective algorithm adapted from Jorg Arndt book "Matters Computational"
(Chapter `7.2 Co-lexicographic order for compositions into exactly k parts`)

``````n = 4
k = 3

x = [0] * n
x[0] = k

while True:
print(x)
v = x[-1]
if (k==v ):
break
x[-1] = 0
j = -2
while (0==x[j]):
j -= 1
x[j] -= 1
x[j+1] = 1 + v

[3, 0, 0, 0]
[2, 1, 0, 0]
[2, 0, 1, 0]
[2, 0, 0, 1]
[1, 2, 0, 0]
[1, 1, 1, 0]
[1, 1, 0, 1]
[1, 0, 2, 0]
[1, 0, 1, 1]
[1, 0, 0, 2]
[0, 3, 0, 0]
[0, 2, 1, 0]
[0, 2, 0, 1]
[0, 1, 2, 0]
[0, 1, 1, 1]
[0, 1, 0, 2]
[0, 0, 3, 0]
[0, 0, 2, 1]
[0, 0, 1, 2]
[0, 0, 0, 3]
``````

Number of compositions and time on seconds for plain Python (perhaps numpy arrays are faster) for n=100, and k = 2,3,4,5 (2.8 ghz Cel-1840)

``````2  5050 0.040000200271606445
3  171700 0.9900014400482178
4  4421275 20.02204465866089
5  91962520 372.03577995300293
I expect time  2 hours for 100/6 generation
``````

Same with numpy arrays (`x = np.zeros((n,), dtype=int)`) gives worse results - but perhaps because I don't know how to use them properly

``````2  5050 0.07999992370605469
3  171700 2.390003204345703
4  4421275 54.74532389640808
``````

Native code (this is Delphi, C/C++ compilers might optimize better) generates 100/6 in 21 seconds

``````3  171700  0.012
4  4421275  0.125
5  91962520  1.544
6  1609344100 20.748
``````

Cannot go sleep until all measurements aren't done :)

MSVS VC++: 18 seconds! (O2 optimization)

``````5  91962520 1.466
6  1609344100 18.283
``````

So 100 millions variants per second. A lot of time is wasted for checking of empty cells (because fill ratio is small). Speed described by Arndt is reached on higher k/n ratios and is about 300-500 millions variants per second:

``````n=25, k=15 25140840660 60.981  400 millions per second
``````