Find palindromic numbers in both binary and ternary bases: Difference between revisions

</pre>

=={{header|C++}}==

<syntaxhighlight lang="c++">

#include <algorithm>

#include <cstdint>

#include <iostream>

// Convert the given decimal number to the given number base

// and return it converted to a string

std::string to_base_string(const uint64_t& number, const uint32_t& base) {

uint64_t n = number;

if ( n == 0 ) {

return "0";

}

std::string result;

while ( n > 0 ) {

result += std::to_string(n % base);

n /= base;

}

std::reverse(result.begin(), result.end());

return result;

}

void display(const uint64_t& number) {

std::cout << "Decimal: " << number << std::endl;

std::cout << "Binary : " << to_base_string(number, 2) << std::endl;

std::cout << "Ternary: " << to_base_string(number, 3) << std::endl << std::endl;

}

bool is_palindromic(const std::string& number) {

std::string copy = number;

std::reverse(copy.begin(), copy.end());

return number == copy;

}

// Create a ternary palindrome whose left part is the ternary equivalent of the given number

// and return it converted to a decimal

uint64_t create_ternary_palindrome(const uint64_t number) {

std::string ternary = to_base_string(number, 3);

uint64_t power_of_3 = 1;

uint64_t result = 0;

for ( uint64_t i = 0; i < ternary.length(); ++i ) { // Right part of palindrome is the mirror image of left part

if ( ternary[i] > '0' ) {

result += ( ternary[i] - '0' ) * power_of_3;

}

power_of_3 *= 3;

}

result += power_of_3; // Middle digit must be 1

power_of_3 *= 3;

result += number * power_of_3; // Left part is the given number multiplied by the appropriate power of 3

return result;

}

int main() {

std::cout << "The first 6 numbers which are palindromic in both binary and ternary are:" << std::endl;

display(0); // 0 is a palindrome in all 3 bases

display(1); // 1 is a palindrome in all 3 bases

uint64_t number = 1;

uint32_t count = 2;

do {

uint64_t ternary = create_ternary_palindrome(number);

if ( ternary % 2 == 1 ) { // Cannot be an even number since its binary equivalent would end in zero

std::string binary = to_base_string(ternary, 2);

if ( binary.length() % 2 == 1 ) { // Binary palindrome must have an odd number of digits

if ( is_palindromic(binary) ) {

display(ternary);

count++;

}

}

}

number++;

}

while ( count < 6 );

}

</syntaxhighlight>

{{ out }}

<pre>

The first 6 numbers which are palindromic in both binary and ternary are:

Decimal: 0

Binary : 0

Ternary: 0

Decimal: 1

Binary : 1

Ternary: 1

Decimal: 6643

Binary : 1100111110011

Ternary: 100010001

Decimal: 1422773

Binary : 101011011010110110101

Ternary: 2200021200022

Decimal: 5415589

Binary : 10100101010001010100101

Ternary: 101012010210101

Decimal: 90396755477

Binary : 1010100001100000100010000011000010101

Ternary: 22122022220102222022122