# [Tutor] Is there a better way to do this?

jb jb at riseup.net
Thu Jul 22 03:45:15 CEST 2004

```On Wed, Jul 21, 2004 at 09:22:26PM -0400, Hee-Seng Kye wrote:
> I'm trying to write a program that computes six-digit numbers, in which
> the left digit is always smaller than its following digit (i.e., it's
> always ascending).  The output of the program starts with "0 1 2 3 4 5"
> and ends on "6 7 8 9 A B."  The best I could do was to have many
> embedded 'for' statements:
>
> c = 1
> for p0 in range(0, 7):
>   for p1 in range(1, 12):
>     for p2 in range(2, 12):
>       for p3 in range(3, 12):
>         for p4 in range(4, 12):
>           for p5 in range(5, 12):
>             if p0 < p1 < p2 < p3 < p4 < p5:
>               print repr(c).rjust(3), "\t",
>               print "%X %X %X %X %X %X" % (p0, p1, p2, p3, p4, p5)
>               c += 1
> print "...Done"
>
> This works, except that it's very slow.  I need to get it up to
> nine-digit numbers, in which case it's significantly slower.  I was
> wondering if there is a more efficient way to do this.
>

this version is a bit quicker:

c = 1
for p0 in range(0, 7):
for p1 in range(p0+1, 12):
for p2 in range(p1+1, 12):
for p3 in range(p2+1, 12):
for p4 in range(p3+1, 12):
for p5 in range(p4+1, 12):
print repr(c).rjust(3), "\t",
print "%X %X %X %X %X %X" % (p0, p1, p2, p3, p4, p5)
c += 1
print "...Done"

```