Here is my solution to the Programming Praxis exercise from Dec. 21, 2012. The solution is written in python.

The task:

It is sometimes possible, starting with a prime, to add a digit to the front of the number that forms a new prime. For instance, starting from prime 7, adding 1 forms prime 17, adding 3 forms prime 317, adding 6 forms prime 6317, adding 2 forms prime 26317, and so on.

Your task is to write a program to find the largest prime that can be formed in this way.

My solution:

# Author: Will Clausen # # Date: Jan. 13, 2013 # # This program will solve the problem from Programming Praxis # on Dec. 21, 2013 import math import random # The answer is generated in these first two functions, # but the meat of the computation is in checking primality def buildPrime(): maxPrime = 1 # Implement a depth-first search approach by generating # successively large primes starting from the initial # one-digit primes, 2, 3, 5, and 7 for x in range(3, 10): if probIsPrime(x, 30): nextPrime = buildPrimes(x, 1) if nextPrime > maxPrime: maxPrime = nextPrime return maxPrime # A helper function that ensures new digits are added to # front of the old prime to check for a new prime def buildPrimes(num, power): maxPrime = num for x in range(1, 10): check = num + x*pow(10, power) if probIsPrime(int(check), 30): nextPrime = buildPrimes(check, power+1) if nextPrime > maxPrime: maxPrime = nextPrime return maxPrime def probIsPrime(num, numChecks): # Some optimizations if num == 2: return True if num % 2 == 0: return False # Here I use a list of the first 100 primes to use # as bases when checking primality. This will allow for # accurate calculations for primes of any reasonable size # (or even an unreasonable size, like 10^100) LoP = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541] i = 0 # Unfortunately, using numbers greater than num for checking # causes errors, so some checking is necessary here to ensure # accuracy if (num < LoP[numChecks]): LoP = range(1, num) # The meat of the computation while numChecks > 0: check = LoP[i] i += 1 # Be careful not to index out of the list of primes... if (i == len(LoP)): i = 0 if not strongPseudoprimeTest(num, check): break numChecks -= 1 return numChecks == 0 """ Function to check primality. It effectively employs Fermat's Little Theorem in a creative way to deterimine if a number, (the check) demonstrates the compositeness of a number (the num). """ def strongPseudoprimeTest(num, check): # Initialize some helpful variables, nice names are unnecessary. d = num - 1 s = 0 numLessOne = num - 1 # If d is even divide it by two and increment s while d % 2 == 0: d /= 2 s += 1 t = pow(check, d, num) if ((t == 1) or (t == numLessOne)): return True s -= 1 while s > 0: s -= 1 t = pow(t, 2, num) if ((t == 1) or (t == numLessOne)): return True return False # This code produced the correct solution of 357686312646216567629137L