# [Numpy-discussion] Is there a better way to do this?

Hee-Seng Kye kyeser at earthlink.net
Wed Jul 21 18:19:07 CDT 2004

```My question is not directly related to NumPy, but since many people
here deal with numbers, I was wondering if I could get some help; it
would be even better if there is a NumPy (or Numarray) function that
takes care of what I want!

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 best I could do was to have many embedded 'for'
statement:

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 slow.  I was
wondering if there is a more efficient way to do this.

I would highly appreciate it if anyone could help.

Many thanks.

-Kye

```