Reversing Digits
September 4, 2015
The answers are different in Scheme and Python, which surprised me. First the Scheme version, which uses digits and undigits from the Standard Prelude:
(define (timing rev)
(time
(let ((count 0))
(do ((n 1 (+ n 1)))
((< 2000000 n) count)
(when (= n (rev n))
(set! count (+ count 1)))))))
(define (rev1 n) ; iterative
(let loop ((n n) (z 0))
(if (zero? n) z
(loop (quotient n 10)
(+ (* z 10) (modulo n 10))))))
(define (rev2 n) ; recursive
(define (rev n r)
(if (zero? n) r
(rev (quotient n 10)
(+ (* r 10) (modulo n 10)))))
(rev n 0))
(define (rev3 n) ; numeric split
(undigits (reverse (digits n))))
(define (rev4 n) ; string split
(string->number
(list->string
(reverse
(string->list
(number->string n))))))
> (timing rev1)
(time (let ((...)) ...))
22 collections
1731 ms elapsed cpu time, including 0 ms collecting
1729 ms elapsed real time, including 1 ms collecting
192005856 bytes allocated, including 185119744 bytes reclaimed
2998
> (timing rev2)
(time (let ((...)) ...))
27 collections
1747 ms elapsed cpu time, including 0 ms collecting
1751 ms elapsed real time, including 0 ms collecting
224007136 bytes allocated, including 227317600 bytes reclaimed
2998
> (timing rev3)
(time (let ((...)) ...))
132 collections
2918 ms elapsed cpu time, including 0 ms collecting
2911 ms elapsed real time, including 1 ms collecting
1116478688 bytes allocated, including 1112394288 bytes reclaimed
2998
> (timing rev4)
(time (let ((...)) ...))
317 collections
2527 ms elapsed cpu time, including 47 ms collecting
2560 ms elapsed real time, including 18 ms collecting
2668524480 bytes allocated, including 2669892448 bytes reclaimed
2998
The iterative and recursive versions are about the same, the version that uses the convenience functions from the Standard Prelude is the worst, and the version that converts back and forth to a string is in the middle. Which is what I expected.
Now the Python version:
from time import clock
def rev(n): # iterative
r = 0
while n > 0:
r = r * 10 + n % 10
n = n // 10
return r
print "iterative:"
start = clock()
count = 0
for n in xrange(100000):
if n == rev(n):
count += 1
print count
print (clock() - start) * 1000
def rev(n, r=0): # recursive
if n == 0: return r
return rev(n // 10, r*10 + n%10)
print ""
print "recursive:"
start = clock()
count = 0
for n in xrange(100000):
if n == rev(n):
count += 1
print count
print (clock() - start) * 1000
def rev(n): # convert to string
return int("".join(reversed(str(n))))
print ""
print "convert to string:"
start = clock()
count = 0
for n in xrange(100000):
if n == rev(n):
count += 1
print count
print (clock() - start) * 1000
Running that script produces this output at ideone:
iterative: 1099 184.628 recursive: 1099 265.007 convert to string: 1099 354.793
That surprised me. The recursive version is substantially slower than the iterative version. Apparently Python has slow function-calling performance.
You can run the program at http://ideone.com/vzwkwD. You will see there that the version the converts to a string is very fast, which surprises me. Obviously the underlying cost models of Chicken Scheme, which is used by ideone, and Chez Scheme, which I run at home, are very different.
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);
}