I'd like to go through all n-digit numbers such that second digit of the number is always lower or equal to the first, third is lower or equal to the second etc. I can get this by writing a horrible code such as:

```for i in range(10):
for j in range(i+1):
for k in range(j+1):```

etc., but with 10-digit numbers my code starts looking horrible, and also that's a lot of writing, and indentation get horrible if I want to commend few of those. Is there a nice, concise way of getting this?

Edit: just so that people know why I'm bothering with this, https://projecteuler.net/problem=74 has me check numbers from 1 to one milion. Unfortunately, It's not as straightforward as I thought -- numbers with leading zeros are treated differently than the ones with zeros inside, so some additional magic had to be performed. Anyway, thanks to all for insightful suggestions. Sep 3, 2018 in Python 146 views

## 1 answer to this question.

Could use itertools:

```>>> for comb in itertools.combinations_with_replacement(range(9, -1, -1), 3):
print comb

(9, 9, 9)
(9, 9, 8)
(9, 9, 7)
(9, 9, 6)
...
(4, 0, 0)
(3, 3, 3)
(3, 3, 2)
(3, 3, 1)
(3, 3, 0)
(3, 2, 2)
(3, 2, 1)
(3, 2, 0)
(3, 1, 1)
(3, 1, 0)
(3, 0, 0)
(2, 2, 2)
(2, 2, 1)
(2, 2, 0)
(2, 1, 1)
(2, 1, 0)
(2, 0, 0)
(1, 1, 1)
(1, 1, 0)
(1, 0, 0)
(0, 0, 0)```

Or recursively, appending more and more digits until enough, which can more directly produce intobjects instead of digit tuples (not sure whether that's what you actually need):

```def build(enough, prefix=0):
if prefix >= enough:
print(prefix)
return
for digit in range(prefix % 10 + 1) if prefix else range(1, 10):
build(enough, prefix * 10 + digit)```

Demo (note it leaves out "000", not sure whether you'd want that anyway):

```>>> n = 3
>>> build(10**(n-1))
100
110
111
200
210
211
220
221
222
300
310
311
320
321
322
330
331
332
333
400
410
411
420``` answered Sep 3, 2018 by
• 57,530 points

+1 vote

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