Find palindromic numbers in both binary and ternary bases: Difference between revisions
Find palindromic numbers in both binary and ternary bases (view source)
Revision as of 21:14, 15 April 2024
, 1 month ago→{{header|EasyLang}}
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// 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;
Line 717:
5415589 101012010210101(3) 10100101010001010100101(2)
90396755477 22122022220102222022122(3) 1010100001100000100010000011000010101(2)</pre>
=={{header|EasyLang}}==
<syntaxhighlight>
fastfunc ispalin2 n .
m = n
while m > 0
x = x * 2 + m mod 2
m = m div 2
.
if n = x
return 1
.
.
fastfunc reverse3 n .
while n > 0
r = r * 3 + n mod 3
n = n div 3
.
return r
.
func$ itoa n b .
if n > 0
return itoa (n div b) b & n mod b
.
.
proc main . .
print "0 0(2) 0(3)"
print "1 1(2) 1(3)"
pow3 = 3
while 1 = 1
for i = pow3 / 3 to pow3 - 1
# assumption that the middle digit must be 1
n = (i * 3 + 1) * pow3 + reverse3 i
if ispalin2 n = 1
print n & " " & itoa n 2 & "(2) " & itoa n 3 & "(3)"
cnt += 1
if cnt = 6 - 2
return
.
.
.
pow3 *= 3
.
.
main
</syntaxhighlight>
{{out}}
<pre>
0 0(2) 0(3)
1 1(2) 1(3)
6643 1100111110011(2) 100010001(3)
1422773 101011011010110110101(2) 2200021200022(3)
5415589 10100101010001010100101(2) 101012010210101(3)
90396755477 1010100001100000100010000011000010101(2) 22122022220102222022122(3)
</pre>
=={{header|Elixir}}==
Line 1,231 ⟶ 1,286:
5415589, 10100101010001010100101, 101012010210101
90396755477, 1010100001100000100010000011000010101, 22122022220102222022122</pre>
Alternatively, using a simple and efficient algorithm, the first six number are found in less than a second.
<syntaxhighlight lang="java">
public final class FindPalindromicNumbersBases23 {
public static void main(String[] aArgs) {
System.out.println("The first 7 numbers which are palindromic in both binary and ternary are:");
display(0); // 0 is a palindrome in all 3 bases
display(1); // 1 is a palindrome in all 3 bases
long number = 1;
int count = 2;
do {
long ternary = createTernaryPalindrome(number);
if ( ternary % 2 == 1 ) { // Cannot be an even number since its binary equivalent would end in zero
String binary = toBinaryString(ternary);
if ( binary.length() % 2 == 1 ) { // Binary palindrome must have an odd number of digits
if ( isPalindromic(binary) ) {
display(ternary);
count++;
}
}
}
number++;
}
while ( count < 7 );
}
// Create a ternary palindrome whose left part is the ternary equivalent of the given number
// and return its decimal equivalent
private static long createTernaryPalindrome(long aNumber) {
String ternary = toTernaryString(aNumber);
long powerOf3 = 1;
long sum = 0;
for ( int i = 0; i < ternary.length(); i++ ) { // Right part of a palindrome is the mirror image of left part
if ( ternary.charAt(i) > '0' ) {
sum += ( ternary.charAt(i) - '0' ) * powerOf3;
}
powerOf3 *= 3;
}
sum += powerOf3; // Middle digit must be 1
powerOf3 *= 3;
sum += aNumber * powerOf3; // Left part is the given number multiplied by the appropriate power of 3
return sum;
}
private static boolean isPalindromic(String aNumber) {
return aNumber.equals( new StringBuilder(aNumber).reverse().toString() );
}
private static String toTernaryString(long aNumber) {
if ( aNumber == 0 ) {
return "0";
}
StringBuilder result = new StringBuilder();
while ( aNumber > 0 ) {
result.append(aNumber % 3);
aNumber /= 3;
}
return result.reverse().toString();
}
private static String toBinaryString(long aNumber) {
return Long.toBinaryString(aNumber);
}
private static void display(long aNumber) {
System.out.println("Decimal: " + aNumber);
System.out.println("Binary : " + toBinaryString(aNumber));
System.out.println("Ternary: " + toTernaryString(aNumber));
System.out.println();
}
}
</syntaxhighlight>
{{ out }}
<pre>
The first 7 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
Decimal: 381920985378904469
Binary : 10101001100110110110001110011011001110001101101100110010101
Ternary: 2112200222001222121212221002220022112
</pre>
=={{header|JavaScript}}==
Line 3,327 ⟶ 3,489:
{{libheader|Wren-fmt}}
Just the first 6 palindromes as the 7th is too large for Wren to process without resorting to BigInts.
<syntaxhighlight lang="
var isPalindrome2 = Fn.new { |n|
| |||