Reversing Digits
September 4, 2015
After Tuesday’s exercise on Lychrel numbers, I answered a question about Lychrel numbers over at Stack Overflow, using a recursive function to extract and reverse the digits one-by-one. I was challenged by a reader who claimed “using int("".join(reversed(str(n)))) has got to be far more efficient than using your recursive function. Your function is clever, but clever isn’t always best.” He was wrong, of course: arithmetic on numbers is better than converting a number to a string, creating a generator to reverse the individual characters of the string, appending them one-by-one to a new string, and converting the new string back to a number, which I demonstrated to him in a timing experiment.
Your task is to find the best way to reverse the digits of a positive integer, in the language of your choice. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
In Python. On my machine, the clear winner is “reverse”. The “arithmetic” version is the slowest.
def reverse(n): return int(str(n)[::-1]) def reverse2(n): return int("".join(reversed(str(n)))) def reverse3(n, r=0): if n == 0: return r return reverse3(n // 10, r * 10 + n % 10) from util.timers3 import Timer from time import clock def bm(method, N, n): t0 = clock() for _ in range(N): x = method(n) print(method.__name__, clock() - t0) n = 123451234512345 print("number:", n) for m in (reverse, reverse2, reverse3): bm(m, 10000000, 12345) """ number: 12345 reverse 10.065856912671778 reverse2 17.162799552531382 reverse3 29.920382914088076 number: 123451234512345 reverse 10.152792060510894 reverse2 17.137077804119585 reverse3 29.87959194145961 """Oeps. In my last post the number was hardcoded. Anyway, thease timings are OK:
number: 12345123451234512345123451234512345
reverse 0.17244116711310897
reverse2 0.32812997116085263
reverse3 2.555884828145366
The function below is some 7% faster with repeated use than “reverse”.
Reader is right for large numbers.
E.g. n = 2 ** 2048 is a number with 616 digits. Using Paul’s reverse algorithms, we see that the mathematical version is ~40 times slower than the string reverse int cast version.
On the other hand, for example for n = 32 the math version is ~2 times faster.
The answer to this question is therefor:
if n < threshold:
return reverse3(n)
else:
return reverse0(n)
with an environment dependent threshold value (threshold = 1000 here).
Here a few more timings.
rs = slice(None, None, -1) def reverse0(n): return int(str(n)[rs]) def reverse1(n): return int(str(n)[::-1]) def reverse2(n): return int("".join(reversed(str(n)))) def reverse4(n): # iterative math version r = 0 while n: r = r * 10 + n % 10 n //= 10 return r def reverse(n): # iterative math combined with reverse0 if n > 1000: return int(str(n)[rs]) r = 0 while n: r = r * 10 + n % 10 n //= 10 return r import timeit N = 1000000 def bm(m, n, N): name = m.__name__ stmt = "{name}({n})".format(**locals()) setup = "from __main__ import {name}; n = {n}".format(**locals()) print ("{:12.2f}".format( min(timeit.repeat(stmt, setup, number=N, repeat=5))), end="") print("\nN={}".format(N)) nums = (2, 23, 123, 1234, 12345, 123456789) methods = (reverse0, reverse1, reverse2, reverse4, reverse) print(" n ", end="") for m in methods: print("{:>12s}".format(m.__name__), end = "") print() for n in nums: print("{:10d}".format(n), end="") for m in methods: bm(m, n, N) print() """ n reverse0 reverse1 reverse2 reverse4 reverse 2 0.87 0.92 1.55 0.52 0.61 23 0.87 0.93 1.50 0.89 0.97 123 0.89 0.94 1.54 1.32 1.41 1234 0.90 0.95 1.59 1.73 0.98 12345 0.91 0.96 1.63 2.13 1.00 123456789 0.96 1.01 1.80 4.05 1.05 """Presumably the string reverse function in Python mainly involves a couple of calls to no doubt highly optimized library functions for reading and writing bignums, rather than rattling around the byte code interpreter loop a bazillion times for the arithmetic solution. In a properly compiled language like Haskell though, the string solution will still win out, but only for much larger inputs, we need 1000 digits or so before rev2 beats rev1:
rev1 :: Integer -> Integer rev1 = aux 0 where aux a 0 = a aux a n = aux (10*a + d) n' where (n',d) = quotRem n 10 rev2 :: Integer -> Integer rev2 = read . reverse . showOn the other hand, the string implementation only wins because the library has better algorithms for radix conversion, various divide and conquer strategies are possible, for example, but we can do the same thing arithmetically.
Here, we determine how many digits are needed with integer log 10, split the number in half, recurse and then rejoin the reversed halves. For the base case we just use the iterative solution, taking care to produce a reverse of the right width (and we can do this with efficient fixed width Ints rather than variable width Integers). With a nice integer log implementation by Remco Niemeijer taken from an earlier Programming Praxis problem, it takes about 3 seconds to reverse a million digit number, versus a minute for the string solution:
-- Remco Niemeijer's integer log code from Programming Praxis, 2010 ilog10 :: Integer -> Int ilog10 n = f (div ubound 2) ubound where ubound = until (\e -> 10 ^ e > n) (* 2) 1 f lo hi | mid == lo = if 10 ^ hi == n then hi else lo | 10 ^ mid < n = f mid hi | 10 ^ mid > n = f lo mid | otherwise = mid where mid = div (lo + hi) 2 rev3 :: Integer -> Integer rev3 n = aux n (1+ilog10 n) where aux :: Integer -> Int -> Integer aux n k -- Base case, use the linear algorithm | k <= 6 = toInteger (aux2 0 (fromInteger n) k) | otherwise = let k1 = k `div` 2 k2 = k - k1 -- Split in to roughly equal halves (n2,n1) = quotRem n (10^k1) -- Recurse n1' = aux n1 k1 n2' = aux n2 k2 -- And rejoin in n1'*(10^k2)+n2' -- aux2 is like rev1 above, but always runs k times. aux2 :: Int -> Int -> Int -> Int aux2 a n 0 = a aux2 a n k = aux2 (10*a + d) n' (k-1) where (n',d) = quotRem n 10Reblogged this on A Modern Prometheus and commented:
The following is the METAL BASIC 1.0ß source code output by a pre-compiler I wrote as a potential A level Computing project.
cls k_LIMIT = 100000 print "Method #1:" t = timer number_reversed = 0 for i = 1 to k_LIMIT if i mod 10 <> 0 then number_reversed = number_reversed + 1 multiplier = 1 l_MULTIPLIER_LOOP: multiplier = multiplier * 10 if int( i / multiplier ) <> 0 then goto l_MULTIPLIER_LOOP multiplier = multiplier / 10 i_copy = i result = 0 l_LOOP: digit = i_copy mod 10 i_copy = int( i_copy / 10 ) result = result + digit * multiplier multiplier = multiplier / 10 if i_copy <> 0 then goto l_LOOP end if next i print number_reversed; " integers were reversed in "; timer - t; " seconds." print print "Method #2:" t = timer number_reversed = 0 for i = 1 to k_LIMIT if i mod 10 <> 0 then number_reversed = number_reversed + 1 s$ = str$( i ) s_reversed$ = "" for k = len( s$ ) to 0 step -1 s_reversed$ = s_reversed$ + mid$( s$, k, 1 ) next k result = val( s_reversed$ ) end if next i print number_reversed; " integers were reversed in "; timer - t; " seconds."#include
int main()
{
int num=12345,l=0,val=0;
for(;num;val=val*10+(num%10),num/=10);
printf(“number = %d\n”,val);
}