Question

I’ve been working on Euler project problem 4, my code works fine but it takes too much time (0.41 seconds).How can I optimize it so that it would take less time.Is there a trick I’m missing, or special functions that I’m not aware of?
This is the code:

#Note: tpal is a function to test if number is palindrome

pal =0

for i in range(999,100,-1):
    if pal >= i*999:    #A way to get out of loop and to not test on all numbers
        break 
    for j in range(999,100,-1):
        if pal >= i*999:
            break 
        if j > i:                         #numbers would already have been tested so I skip them 
            continue
        pot=i*j
        if ((tpal(pot)==1)and(pot> pal)):
            pal=pot
            i1=i
            j1=j

print(i1,j1,pal)

def tpal(num):
    num=list(str(num))
    Le=len(num)
    if Le == 1: # if number is of one digit than palindrome
        return 1

    le=len(num)

    if le%2 !=0: #4 example 10101even nbr
        le-=1
    le/2    

    for i in range(0,le):
       if num[i]!=num[Le-i-1]:
           return 0

    return 1                  

Answer 1

Now that it turns out the code has < 1 s runtime, it’s not really that interesting any more. You could modify the code to test fewer numbers and give up sooner. But there’s one obvious optimization which is kind of cute. This line:

        if ((tpal(pot)==1)and(pot> pal)):

checks whether something is a palindrome each time, even if pot <= pal. The palindrome test is expensive. If you simply swap the order: (note you don’t need the ==1):

        if (pot > pal) and tpal(pot):

then you could save a lot of time:

In [24]: timeit orig()
1 loops, best of 3: 201 ms per loop

In [25]: timeit orig_swapped()
10 loops, best of 3: 30.1 ms per loop

because A and B doesn’t evaluate B if A is already false and so it knows that A and B has to be false. (This is called ‘short-circuiting’; the same thing happens with ‘A or B’ if A is true.)

Incidentally, the last line here:

if le%2 !=0: #4 example 10101even nbr
    le-=1
le/2    
^^^^

doesn’t change le. I think these three lines are meant to amount to le //= 2.