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Neckbearded Basement Dweller 15/05/27(Wed)17:08 No. 4737

File 143273931665.png - (12.44KB , 1025x176 , 145.png )

I'm trying to figure out this problem. Here's my code in python:

def reversible(limit):
num_list = ()
for num in range(1,limit):
num = str(num)
if num[-1] == 0:
sum = int(num) + int(num[::-1])
checker = 0
for i in str(sum):

if int(i) % 2 == 0:
checker += 1
if checker == len(str(sum)):
if int(num) not in num_list:
if int(num[-1]) != 0:
num_list = num_list + (num,)

checker = 0
count = 0
for x in num_list:
count += 1
return count

def main():


it works, but the only problem is it takes extremely long to solve. I've only managed to solve up to 20 million, and that took 10 -20 minutes. I used a tuple instead of a list, and that helped a bit but it's still slow. Is there an alternative, like more conditions to break or not use range?

Neckbearded Basement Dweller 15/05/28(Thu)17:33 No. 4738

I dont think you need to keep a listing, just print them out because you should be able to predict the duplicates
I dont know the answer but for the numbers 11 thru 99 if you arrange them as a 2D matrix (ie 11..19 is the 1st row, 21..29 is the 2nd row, etc) you should see the pattern of where the duplicates are. And I believe (I am not sure) as you go thru each row you can cancel out a column in the matrix.

Neckbearded Basement Dweller 15/05/28(Thu)17:38 No. 4739

Say you have a number 13739, is there an algorithm to break that down to its two components x and reverse(x)?

Neckbearded Basement Dweller 15/06/15(Mon)21:45 No. 4761

Most of the cause of the slow speed is because python loops are slow. You can look up numba, which is a jit compiler that can speed up numerical computations like this very well.

If you really want to reverse a number without converting to string, you could do mod 10, subtract result and divide 10. Numba should speed that up nicely since it's a numerical loop.

Neckbearded Basement Dweller 15/07/13(Mon)20:52 No. 4768

my solution (quick n dirty)

n=0for i in xrange(10,1000000000): if i%10==0: pass else: c=False j=int(str(i)[::-1]) for k in str(i+j): if int(k)%2==0: c=True if c==False: n+=1print n

Solution still unknown coz the program is still running (on an ARM v6 lol), I expect it to finish by tomorrow morning.

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