Emirps
November 2, 2010
We give three solutions. The first uses a function emirp? to identify emirps and enumerates them by testing each n in range:
(define (rev n) (undigits (reverse (digits n))))
(define (palin? n) (= n (rev n)))
(define (emirp? n) (and (prime? n) (prime? (rev n)) (not (palin? n))))
(define (emirps1 n) (filter emirp? (range 2 n)))
There are 36 emirps less than a thousand:
> (emirps1 1000)
(13 17 31 37 71 73 79 97 107 113 149 157 167 179 199 311 337
347 359 389 701 709 733 739 743 751 761 769 907 937 941 953
967 971 983 991)
Emirps1 operates in linear time. But prime? is fairly slow, and sieving is fast, so it makes sense to identify the emirps using a sieve:
(define (emirps2 n)
(let ((n10 (next-power-of-ten n)))
(let loop ((ps (filter (complement palin?) (primes n10))) (es '()))
(cond ((null? ps) (take-while (right-section < n) (sort < es)))
((member (rev (car ps)) (cdr ps))
(loop (cdr ps) (cons (car ps) (cons (rev (car ps)) es))))
(else (loop (cdr ps) es))))))
It is tempting, but wrong, to sieve up to n instead of 10⌈log10n⌉. (You can probably guess that I succumbed to the temptation in my first version of the function.) The problem is that primes whose reversal is bigger than n will mistakenly disappear from the result. The calculation next-power-of-ten was originally written as (inexact->exact (expt 10 (ceiling (log10 n)))))), but was changed due to a nagging insecurity that somewhere, someday, (log10 1000) might be 3.0000000000000004, leading to an incorrect answer:
(define (next-power-of-ten n)
(let loop ((i 1) (ten^i 10))
(if (<= n ten^i) (expt 10 i)
(loop (+ i 1) (* ten^i 10)))))
Testing shows that emirps2 is faster than emirps1 for n=1000:
> (time (length (emirps1 1000)))
(time (length (emirps1 1000)))
no collections
14 ms elapsed cpu time
22 ms elapsed real time
76344 bytes allocated
36
> (time (length (emirps2 1000)))
(time (length (emirps2 1000)))
no collections
1 ms elapsed cpu time
1 ms elapsed real time
29032 bytes allocated
36
But consider what happens when we increase n from a thousand to a million:
> (time (length (emirps1 1000000)))
(time (length (emirps1 1000000)))
111 collections
16760 ms elapsed cpu time, including 289 ms collecting
17245 ms elapsed real time, including 312 ms collecting
465986352 bytes allocated, including 469509920 bytes reclaimed
11184
> (time (length (emirps2 1000000)))
(time (length (emirps2 1000000)))
5 collections
24402 ms elapsed cpu time, including 33 ms collecting
26126 ms elapsed real time, including 34 ms collecting
22172848 bytes allocated, including 19376512 bytes reclaimed
11184
Oops! The problem is that member performs linear search, making emirps2 quadratic rather than linear. But the idea of sieving for primes instead of testing for them is good. Here is a third version of emirps:
(define (emirps3 n)
(let* ((n10 (next-power-of-ten n))
(ps (filter (complement palin?) (primes n10)))
(qs (sort < (append ps (map rev ps)))))
(let loop ((qs qs) (es '()))
(cond ((or (< n (car qs)) (null? (cdr qs))) (reverse es))
((= (car qs) (cadr qs))
(loop (cddr qs) (cons (car qs) es)))
(else (loop (cdr qs) es))))))
Here ps is the non-palindromic primes, and qs combines those primes with their reverses, in order; for n=1000, the beginning of the qs list is (13 13 14 16 17 17 19 23 …). Then a simple scan through qs, stopping at n, identifies twins, which indicate that both a prime and its reverse are in the list. This is a linear algorithm, like emirps1, but much faster, since it uses a sieve rather than testing each number for primality, and since the loop is only over the primes rather than the complete range. Here are some timings, in milliseconds, for (length (emirps n)):
n emirps1 emirps2 emirps3 length
------- ------- ------- ------- ------
1000 14 0 36
10000 163 15 6 240
50000 1020 724 60 980
150000 2529 23734 664 2954
500000 8427 24415 671 6700
1000000 16760 24402 690 11184
By comparison, (length (primes 1000000)) takes 123ms and finds 78498 primes.
We used take-while, filter, range, sort, digits, undigits, log10, complement and right-section from the Standard Prelude, and primes and prime? from two previous exercises. You can run the program at http://programmingpraxis.codepad.org/WCYfLGjL.
Programming Praxis 
[…] Praxis – Emirps By Remco Niemeijer In today’s Programming Praxis, our task is to enumerate all the non-palindrome prime numbers that are still […]
My Haskell solution (see http://bonsaicode.wordpress.com/2010/11/02/programming-praxis-emirps/ for a version with comments):
import Data.Numbers.Primes emirps :: [Integer] emirps = [p | p <- primes, let rev = reverse (show p) , isPrime (read rev), show p /= rev] main :: IO () main = print $ takeWhile (< 1000000) emirpsHere is my Haskell solution: http://www.gleocadie.net/?p=178&lang=en
import Data.List isPrime l = isPrimeHelper l primes isPrimeHelper a (p:ps) | p*p > a = True | a `mod` p == 0 = False | otherwise = isPrimeHelper a ps primes = 2 : filter isPrime [3,5..] permirps = drop 4 (takeWhile (<1000000) primes) isEmirps x = let sx = show x in let rev = reverse sx in sx /= rev && isPrimeHelper (read rev) primes emirps = filter isEmirps permirps main = print $ emirpsComments, advices are welcomed.
I’m loving this website! Thanks a lot for the problems.
import Data.Numbers.Primes emirps :: [Integer] emirps = filter (\a -> isEmirp a) . filter (\a -> not . isPalindrome $ a) . takeWhile (< 1000000) $ primes where isPalindrome :: Integer -> Bool isPalindrome x = (reverse . show $ x) == (show x) isEmirp :: Integer -> Bool isEmirp x = isPrime . read . reverse . show $ x main :: IO () main = print . show $ emirpsI’m loving this website, thanks a lot!
Haskell:
import Data.Numbers.Primes emirps :: [Integer] emirps = filter (\a -> isEmirp a) . filter (\a -> not . isPalindrome $ a) . takeWhile (< 1000000) $ primes where isPalindrome :: Integer -> Bool isPalindrome x = (reverse . show $ x) == (show x) isEmirp :: Integer -> Bool isEmirp x = isPrime . read . reverse . show $ x main :: IO () main = print . show $ emirpsLove the site.
Sorry, line 14 should read
if prime(int(str(n)[::-1])) and str(n) != str(n)[::-1]:
@Joe: your prime() function isn’t quite correct. For instance, prime(10) returns True.
Wrote it in python (though for speed, a different language may be better). In the interest of speed, I wrote a function called reverse(n) that writes n backwards using only numerical operations (avoiding strings). Interestingly, trying to memoize my is_prime() function made everything slower (perhaps due to memory usage?). Running ./emirps.py 1000000 took just under 15 seconds on my aging laptop. For more speed I used pypy (Python with a just in time compiler); it finished in just under 3 seconds.
#!/usr/bin/env python2.6 import math def reverse(n): """ Given integer n, returns n written backwards. Note: uses only numerical operations (not string ones) for speed. """ d, r = 0, 0 while n > 0: l = int(math.log10(n)) r += (10 ** d) * (n // (10 ** (l))) d += 1 n %= (10**l) return r def is_prime(n): """ Simple check for primality, testing n mod k for odd k up to 1 + sqrt(n). """ if n == 2: return True elif n % 2 == 0: return False else: for k in xrange(3, 1 + int(math.sqrt(n)), 2): if n % k == 0: return False return True def main(n): """ Prints out all emirps less than n. """ for p in xrange(2, n): r = reverse(p) if (is_prime(p)) and (r != p) and (is_prime(r)): print p if __name__ == '__main__': import sys main(int(sys.argv[1]))Changing line 36 in main(n) from
to
yields a decent speed up; normal execution finishes in just under 11 seconds, while pypy finishes it in under 2.
Probably if you avoid using math.log10(n), you’ll end up with a faster execution.
My C Implementation
http://codepad.org/xVdzeVs6
This one was easy one. ;)
My C Implementation
http://codepad.org/xVdzeVs6
This one was easy one. ;)
We can build on the sieve of Erastosthenes exercise to create a list of primes < 1,000,000. Since the output array of the sieve is sorted, we can use binary search on the table for O(log n) search of the sieve, when checking for whether each reversal is prime. The Factor code for sieve is already posted on this blog and won't be reproduced here.
USING: kernel sequences vectors math math.parser locals
binary-search sieve ;
IN: emirp
: emirp? ( n vec — ? )
swap
number>string dup reverse 2dup =
[ 3drop f ]
[ nip string>number swap sorted-member? ] if ;
:: emirp-filter ( primes — semirp )
V{ } clone
primes
[ dup primes emirp?
[ suffix ]
[ drop ] if
] each ;
: Semirp ( n — vec )
primes emirp-filter ;
Factor session:
( scratchpad ) 100000 Semirp
— Data stack:
V{ 13 17 31 37 71 73 79 97 107 113 149 157 167 179 199…
( scratchpad ) length .
1646
Sorry, somehow the code block didn’t work on last post, probably got a tag wrong somewhere…
We can build on the sieve of Erastosthenes exercise to create a list of primes < 1,000,000. Since the output array of the sieve is sorted, we can use binary search on the table for O(log n) search of the sieve, when checking for whether each reversal is prime. The Factor code for sieve is already posted on this blog and won't be reproduced here.
USING: kernel sequences vectors math math.parser locals
binary-search sieve ;
IN: emirp
: emirp? ( n vec — ? )
swap
number>string dup reverse 2dup =
[ 3drop f ]
[ nip string>number swap sorted-member? ] if ;
:: emirp-filter ( primes — semirp )
V{ } clone
primes
[ dup primes emirp?
[ suffix ]
[ drop ] if
] each ;
: Semirp ( n — vec )
primes emirp-filter ;
Factor session:
( scratchpad ) 100000 Semirp
— Data stack:
V{ 13 17 31 37 71 73 79 97 107 113 149 157 167 179 199…
( scratchpad ) length .
1646
ruby solution : http://codepad.org/0ZlE1j4F
November 2nd, 2010.c:
#ifndef LEONARDO #error "This program requires Leonardo IDE to run." #endif #include "seal_bool.h" /* <http://GitHub.com/sealfin/C-and-C-Plus-Plus/blob/master/seal_bool.h> */ #include <stdlib.h> #include <stdio.h> #include <string.h> #include <time.h> #define N 1000000L #if N <= 0 #error "N ≤ 0." #endif long f_ReverseDigits_A( long p ) { long result = 0; while( p != 0 ) { result *= 10; result += ( p % 10 ); p /= 10; } return result; } size_t f_NumberOfDigits( const long p ) { long i = 10; size_t number_of_digits = 1; while( p / i != 0 ) { i *= 10; number_of_digits ++; } return number_of_digits; } char *g_s; long f_ReverseDigits_B( const long p ) { static char *s = NULL; size_t i = 0, k; if( s == NULL ) { s = ( char* )malloc( sizeof( char ) * ( f_NumberOfDigits( N ) + 1 )); g_s = s; } sprintf( s, "%ld", p ); k = strlen( s ) - 1; while( i < k ) { const char c = s[ i ]; s[ i ] = s[ k ]; s[ k ] = c; i ++; k --; } return atol( s ); } bool g_isPrime[ N ]; long g_numberOfPrimes = 0, *g_primes; bool f_IsPrime( const long p ) { if( p % 2 == 0 ) return p == 2; else return g_isPrime[ p ]; } void p_EnumerateEmirps( long ( *p_reverseDigits )( long ), const char * const p_fileName, long *p_secondsTaken ) { const time_t beginning_time = time( NULL ); long i = 0; FILE *f = fopen( p_fileName, "w" ); for( ; i < g_numberOfPrimes; i ++ ) { const long prime = g_primes[ i ]; const long reversed_prime = p_reverseDigits( prime ); if( f_IsPrime( reversed_prime ) && ( reversed_prime != prime )) fprintf( f, "%ld\n", prime ); } fclose( f ); *p_secondsTaken = time( NULL ) - beginning_time; } char *f_AllocateMemoryForAndSetMessage( const long p_secondsTaken, const char p_variant, const char p_terminator ) { size_t message_length = strlen( "• " ); char *message; message_length += f_NumberOfDigits( p_secondsTaken ); message_length += strlen( " second" ); if( p_secondsTaken != 1 ) message_length ++; message_length += strlen( " using the f_ReverseDigits_ function " ); message = ( char* )malloc( sizeof( char ) * ( message_length + 1 )); sprintf( message, "• %ld second%s using the f_ReverseDigits_%c function%c", p_secondsTaken, ( p_secondsTaken != 1 )?"s":"", p_variant, p_terminator ); return message; } void main( void ) { long i = 0, k, seconds_taken_A, seconds_taken_B; /* Firstly, let's determine the prime numbers in the range [ 0, N ) so that later we can test if a reversed prime number is still a prime number. */ for( ; i < N; i ++ ) g_isPrime[ i ] = true; if( N >= 1 ) g_isPrime[ 0 ] = false; if( N >= 2 ) g_isPrime[ 1 ] = false; if( N >= 3 ) g_numberOfPrimes ++; for( i = 3; i < N; i += 2 ) if( g_isPrime[ i ] ) { g_numberOfPrimes ++; for( k = i + i; k < N; k += i ) g_isPrime[ k ] = false; } /* Secondly, let's create a list of the prime numbers in the range [ 0, N ) so that later we can iterate through that list of prime numbers, testing if each prime number is also an emirp. */ g_primes = ( long* )malloc( sizeof( long ) * g_numberOfPrimes ); k = 0; if( N >= 3 ) g_primes[ k ++ ] = 2; for( i = 3; k < g_numberOfPrimes; i += 2 ) if( f_IsPrime( i )) g_primes[ k ++ ] = i; p_EnumerateEmirps( f_ReverseDigits_A, "November 2nd, 2010 (A).out", &seconds_taken_A ); p_EnumerateEmirps( f_ReverseDigits_B, "November 2nd, 2010 (B).out", &seconds_taken_B ); free( g_primes ); free( g_s ); { float height_A, height_B; char *message = ( char* )malloc( sizeof( char ) * ( strlen( "To enumerate the emirps less than " ) + f_NumberOfDigits( N ) + strlen( " took:" ) + 1 )), *message_A, *message_B; sprintf( message, "To enumerate the emirps less than %ld took:", N ); if( seconds_taken_A > seconds_taken_B ) { height_A = 100; height_B = 100 * (( float )seconds_taken_B / seconds_taken_A ); } else { height_A = 100 * (( float )seconds_taken_A / seconds_taken_B ); height_B = 100; } // Set-up the view. /** View( Out 0 ); ViewOrigin( Out 100, Out 112, 0 ); SmallText( Out 0, Out 0, Out 11, Out String, Out L, Out Left, 0 ) Assign L = message; **/ // Set-up the bar chart for the f_ReverseDigits_A function. message_A = f_AllocateMemoryForAndSetMessage( seconds_taken_A, 'A', ';' ); /** RectangleFrameColor( 1, Out Cyan, 0 ); Rectangle( Out 1, Out -36, Out Y, Out 30, Out H, 0 ) Assign Y = -1 * height_A H = height_A; LineColor( 1, Out Cyan, 0 ); Line( Out 1, Out -30, Out Y1, Out 0, Out Y2, 0 ) Assign Y1, Y2 = -1 * height_A - 6; Line( Out 1, Out 0, Out Y, Out 0, Out -6, 0 ) Assign Y = -1 * height_A - 6; Line( Out 1, Out -36, Out Y1, Out -30, Out Y2, 0 ) Assign Y1 = -1 * height_A Y2 = -1 * height_A - 6; Line( Out 1, Out -6, Out Y1, Out 0, Out Y2, 0 ) Assign Y1 = -1 * height_A Y2 = -1 * height_A - 6; Line( Out 1, Out -6, Out -1, Out 0, Out -6, 0 ); SmallTextColor( 1, Out Cyan, 0 ); SmallText( Out 1, Out 0, Out 22, Out String, Out L, Out Left, 0 ) Assign L = message_A; **/ // Set-up the bar chart for the f_ReverseDigits_B function. message_B = f_AllocateMemoryForAndSetMessage( seconds_taken_B, 'B', '.' ); /** RectangleFrameColor( 2, Out Magenta, 0 ); Rectangle( Out 2, Out 6, Out Y, Out 30, Out H, 0 ) Assign Y = -1 * height_B H = height_B; LineColor( 2, Out Magenta, 0 ); Line( Out 2, Out 12, Out Y1, Out 42, Out Y2, 0 ) Assign Y1, Y2 = -1 * height_B - 6; Line( Out 2, Out 42, Out Y, Out 42, Out -6, 0 ) Assign Y = -1 * height_B - 6; Line( Out 2, Out 6, Out Y1, Out 12, Out Y2, 0 ) Assign Y1 = -1 * height_B Y2 = -1 * height_B - 6; Line( Out 2, Out 36, Out Y1, Out 42, Out Y2, 0 ) Assign Y1 = -1 * height_B Y2 = -1 * height_B - 6; Line( Out 2, Out 36, Out -1, Out 42, Out -6, 0 ); SmallTextColor( 2, Out Magenta, 0 ); SmallText( Out 2, Out 0, Out 33, Out String, Out L, Out Left, 0 ) Assign L = message_B; **/ free( message ); free( message_A ); free( message_B ); } }The solution interpreted using Leonardo IDE 3.4.1 on an Apple Power Mac G4 (AGP Graphics) (450MHz processor, 1GB memory) running Mac OS 9.2.2 (International English).
“November 2nd, 2010 (A).out” & “November 2nd, 2010 (B).out”:
I interpreted the instruction that I “should strive for maximum speed” as an invitation to compare two functions for reversing the digits of an integer. The results were surprising: the function
f_ReverseDigits_B, which reverses the digits of an integer by converting that integer into a string, reversing the characters of that string, and converting that string back into an integer, was as fast – and on occasion faster and on no occasion slower – than the functionf_ReverseDigits_A, which reverses the digits of an integer arithmetically.(I’m just trying to solve the problems posed by this ‘site whilst I try to get a job; I’m well aware that my solutions are far from the best – but, in my defence, I don’t have any traditional qualifications in computer science :/ )