How to resolve the algorithm Unprimeable numbers step by step in the Pike programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Unprimeable numbers step by step in the Pike programming language

Table of Contents

Problem Statement

As used here, all unprimeable numbers   (positive integers)   are always expressed in base ten.

───── Definition from OEIS ─────: Unprimeable numbers are composite numbers that always remain composite when a single decimal digit of the number is changed.

───── Definition from Wiktionary   (referenced from Adam Spencer's book) ─────: (arithmetic)   that cannot be turned into a prime number by changing just one of its digits to any other digit.   (sic)

Unprimeable numbers are also spelled:   unprimable. All one─ and two─digit numbers can be turned into primes by changing a single decimal digit.

190   isn't unprimeable,   because by changing the zero digit into a three yields   193,   which is a prime.

The number   200   is unprimeable,   since none of the numbers   201, 202, 203, ··· 209   are prime, and all the other numbers obtained by changing a single digit to produce   100, 300, 400, ··· 900,   or   210, 220, 230, ··· 290   which are all even.

It is valid to change   189   into   089   by changing the   1   (one)   into a   0   (zero),   which then the leading zero can be removed,   and then treated as if the   "new"   number is   89.

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Let's start with the solution:

Step by Step solution about How to resolve the algorithm Unprimeable numbers step by step in the Pike programming language

Source code in the pike programming language

bool is_unprimeable(int i)
{
    string s = i->digits();
    for(int offset; offset < sizeof(s); offset++) {
	foreach("0123456789"/1, string repl) {
	    array chars = s/1;
	    chars[offset] = repl;
	    int testme = (int)(chars*"");
	    if( testme->probably_prime_p() )
		return false;
	}
    }
    return true;
}

void main()
{
    int i, count;
    array unprimes = ({});
    mapping first_enders = ([]); // first unprimeable ending with each digit 
    while(sizeof(first_enders) != 10) {
	i++;
	if( is_unprimeable(i) ) {
	    count++;
	    unprimes += ({ i });
	    string last_digit = i->digits()[<0..];
	    if( !first_enders[last_digit] )
		first_enders[last_digit] = i;
	}
	werror("%d\r", i); // Progress output
    }
    
    write("First 35 unprimeables: %s\n\n", (array(string))unprimes[0..34]*" ");
    write("The 600th unprimeable is %d\n\n", unprimes[599]);
    write("The first unprimeable number that ends in\n");
    foreach(sort(indices(first_enders)), string e) {
	write("  %s is: %9d\n", e, first_enders[e]);
    }
}


  

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