Comments on: Largest Palindromic Number In Python

By: Nicholas

Nicholas — Fri, 31 Aug 2012 16:14:52 +0000

Another thing to remember: If you want a number that starts with 9, and it’s a palindrome, you should only try numbers that can come out with 9′s if you follow a multiplication table. That means only try with the following sets in the last slot: {1,9},{3,3},{7,7}
If that doesn’t work, you’ll need to do the same thing with 8s and stop incrementing as soon as the result goes over 900,000, and so on in that order until all humans are eaten.

By: vince

vince — Mon, 16 Mar 2009 19:47:23 +0000

actually, you only perform 900 * 900

By: vince

vince — Mon, 16 Mar 2009 19:46:54 +0000

Actually, you only perform 900 * 900 my mistake.

By: vince

vince — Mon, 16 Mar 2009 19:45:57 +0000

999 * 999 = 998001 iterations through every number, but I will be testing each one at least twice, so I only need to test half.

This point is an error if you consider the first one where you only go until 100.

Following your first point, means that you only do 999 * 900 iterations as opposed to 999 * 999. Why? you stop at 100. This means that you don’t consider 999 * 99, 999 * 98, and so on.

Remember: the number of digits from 100 to 999 = (999 – 100) + 1 = 900.

By: leed

leed — Sun, 26 Oct 2008 09:29:02 +0000

I think follow code is the fastest I have tried.
maxnum = 0
for i in range(990, 99, -11):
for j in range(999, i, -1):
num = i * j
if num <= maxnum:
break
else:
strnum = str(num)
if strnum == strnum[::-1]:
maxnum = num
break
print maxnum

By: i dont know

i dont know — Thu, 11 Sep 2008 15:50:14 +0000

what’s are biggest 2 3-dgits palindrome?

By: Athropos

Athropos — Sun, 16 Sep 2007 12:52:44 +0000

You should use the timeit module when you need to time something. It’s better than using the time command because you then don’t measure the startup time of the Python interpreter which can be a big part of the total time.

1
2
3
4
5
6

import timeit

...

t1 = timeit.Timer('palindrome()', 'from __main__ import palindrome')
print t1.timeit(20)

By: phoenyx

phoenyx — Sun, 16 Sep 2007 09:13:16 +0000

FYI, it’s considered bad form to post answers to problems outside of the forum.

By: Duccio

Duccio — Sun, 16 Sep 2007 07:38:38 +0000

Good post…only a doubt.
When you reach the column of the table of multiple of 11 (990,979,etc.) i think you must multiplicate this number for all the numbers smaller than this, and not only for multiple of 11 smaller than this. For example I don’t see 990*989,990*988,etc. that are multiple of 11 too. Don’t you think?

By: Aaron

Aaron — Sun, 16 Sep 2007 02:56:05 +0000

@Peter- Ahh, yes. Good call. I’ve updated my post to reflect the change.

@Seth- No problem. Let me know how far you get.

By: Seth

Seth — Sun, 16 Sep 2007 02:23:25 +0000

Never seen Project Euler before. Thanks for posting about it.

By: Peter

Peter — Sun, 16 Sep 2007 02:06:38 +0000

A few thoughts:

- 5 digit palindromes are not necessarily divisible by 11, though that doesn’t matter too much here as we’re looking for 6 digit ones
abcba = 10001a + 1010b + 100c

- quick one liner (takes a few seconds here)

1	max([xy for x in range(100,1000) for y in range(100,1000) if str(xy)==str(x*y)[::-1]])

By: Justin

Justin — Sat, 15 Sep 2007 23:19:13 +0000

1
2
3

def is_palin(num):
s = str(num)
return s == s[::-1]