Reversing a converted string in python

Question

I would like to find the largest palindromic product of 2 3-digit numbers. This is my code:

for i in range(100,1000):
    for x in range(100,1000):
        number = i*x 
        number = str(number)
    if number==number[::-1]:
        print(number)
print("DONE")

It is doing nothing and just printing Done. I have converted the number to a string and done the slice thing but not working... just printing Done . How do I fix this?

EDIT

I wanted to make the palindromes go in a list and sort them in order of how big they are.

mylist=[]
for i in range(999,99,-1):
    for x in range(999, i-1, -1,):
        number=i*x
        number=str(number)
        if number==number[::-1]:
            print(number)
            mylist.append(number)
mylist=sorted(mylist)
print("DONE")
for i in mylist:
    for x in mylist:
        if int(i) <= int(x):
            mylist.pop(int(i))
            if x<= i:
                mylist.pop(int(x))

But still its not working... How to fix?

take care for your indentation. Your if palindrome check occurs after the for x in ... loop has finished. — Adam Smith
– Adam Smith, Commented Jan 31, 2017 at 20:14
@Adam Smith thx , didn't see that! Then i assume i have to also indent the print statement? — S.L
– S.L, Commented Feb 1, 2017 at 7:06

Michael · Accepted Answer · 2017-02-01 04:58:21Z

3

About code

You shoud check EVERY number, so your if statement should be inside a loop.

That's all about code.

Performance

For faster checking you may check from LARGEST number, and then go downwards.

Example: for i in range(999, 99, -1).

U used 999 here because left side is INCLUSIVE, but right threshold is NOT.

Next step is to optimize inner loop. You don't need to check each value. Just loop from 999 to i for optimal checking.

With this alghoritm your code will find largest palindromic number without checking iversive x and i order.

In your code it may check 30/70, then 70/30 for example

Fixed code

for i in range(999, 99, -1):
  for x in range(999, i - 1, -1):
    number = str(x * i)
    if number == number[::-1]:
      print(number)
print('Done!')

-1 in range is step. So negative step it's going downwards to 2nd number

edited Feb 1, 2017 at 4:58

answered Jan 31, 2017 at 20:16

Michael

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4 Comments

Michael Over a year ago

Pardon, i, not y. It is difficult to type on phone

Aaron Over a year ago

This algorithm will not find the largest palindrome at first occurrence. It would find (for instance) 924 * 962 = 888888 before 913 * 993 = 906609 because when checking 993 it wouldn't loop down to 913

Michael Over a year ago

Yeah, Let's subtract some more... We know math, so it will be nit so hard... But ok, it is faster. That's only plus for now(Checking only unique values)

S.L Over a year ago

@MauriceMeyer i doesnt need to be defined because we are just using it to iterate through

dawg · Accepted Answer · 2017-01-31 23:04:00Z

I assume you are working on Problem 4 of Project Euler

You can use max:

>>> max((i*j,i,j) for i in range(999,99,-1) for j in range(999,i-1,-1) if str(i*j)==str(i*j)[::-1])
(906609, 913, 993)

That works in this case because the sheer number of elements is not that great for a modern computer.

But a cooler way to do this is to generate these numbers one by one from largest to smallest. Then the max is just the first one:

def factors(n):    
    ''' function to return the factors of a number '''
    return set(reduce(list.__add__, 
            ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))

def pals(n, stop_n=0, length=3):
    ''' Generator to produce palindromes from largest to smallest '''
    while n>stop_n:
        if str(n)==str(n)[::-1]:
            for f1 in factors(n):
                f2=n//f1
                if len(str(int(f1)))==length and len(str(int(f2)))==length:
                    yield n, f1, f2     
        n-=1        


>>> next(pals(999*999))
(906609, 993, 913)

And can be used for larger numbers easily (knowing that larger numbers can take a long time but monumentally faster than generating all of them):

>>> next(pals(99999*99999,length=5))
(9966006699, 99979, 99681)
>>> next(pals(999999*999999,length=6))
(999000000999, 999999, 999001)

greedy52 · Accepted Answer · 2017-01-31 20:15:22Z

0

The if statement should be inside the inner for loop.

for i in range(100,1000):
    for x in range(100,1000):
        number = i*x
        number = str(number)
        if number==number[::-1]:
            print(number)
print("DONE")

answered Jan 31, 2017 at 20:15

greedy52

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