Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
- Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
- base 2
- base 3
- Display 0 (zero) as the first number found, even though some other definitions ignore it.
- Optionally, show the decimal number found in its binary and ternary form.
- Show all output here.
It's permissible to assume the first two numbers and simply list them.
- See also
- Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
11l
V digits = ‘0123456789abcdefghijklmnopqrstuvwxyz’
F baseN(=num, b)
I num == 0
R ‘0’
V result = ‘’
L num != 0
(num, V d) = divmod(num, b)
result ‘’= :digits[Int(d)]
R reversed(result)
F pal2(num)
I num == 0 | num == 1
R 1B
V based = bin(num)
R based == reversed(based)
F pal_23(limit)
V r = [Int64(0), 1]
V n = 1
L
n++
V b = baseN(n, 3)
V revb = reversed(b)
L(trial) (b‘’revb, b‘0’revb, b‘1’revb, b‘2’revb)
V t = Int64(trial, radix' 3)
I pal2(t)
r.append(t)
I r.len == limit
R r
L(pal23) pal_23(6)
print(pal23‘ ’baseN(pal23, 3)‘ ’baseN(pal23, 2))- Output:
0 0 0 1 1 1 6643 100010001 1100111110011 1422773 2200021200022 101011011010110110101 5415589 101012010210101 10100101010001010100101 90396755477 22122022220102222022122 1010100001100000100010000011000010101
Ada
Simple Technique (Brute Force)
with Ada.Text_IO, Base_Conversion;
procedure Brute is
type Long is range 0 .. 2**63-1;
package BC is new Base_Conversion(Long);
function Palindrome (S : String) return Boolean is
(if S'Length < 2 then True
elsif S(S'First) /= S(S'Last) then False
else Palindrome(S(S'First+1 .. S'Last-1)));
function Palindrome(N: Long; Base: Natural) return Boolean is
(Palindrome(BC.Image(N, Base =>Base)));
package IIO is new Ada.Text_IO.Integer_IO(Long);
begin
for I in Long(1) .. 10**8 loop
if Palindrome(I, 3) and then Palindrome(I, 2) then
IIO.Put(I, Width => 12); -- prints I (Base 10)
Ada.Text_IO.Put_Line(": " & BC.Image(I, Base => 2) & "(2)" &
", " & BC.Image(I, Base => 3) & "(3)");
-- prints I (Base 2 and Base 3)
end if;
end loop;
end Brute;
- Output:
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)
For larger numbers, this is a bit slow.
Advanced Technique
To speed this up, we directly generate palindromes to the base 3 and then check if these are also palindromes to the base 2. We use the fact that these palindromes (to the base 3) must have an odd number of digits (to the base 2 and 3) and that the number in the middle is a 1 (to the base 3). We use unsigned 64 bits integers that we consider as array of 64 bits thanks to a function Int_To_Bits (an instantiation of a generic conversion function) that goes from one type to the other immediately. We can now access the i'th bit of an Int in a very efficient way (since it is chosen by the compiler itself). This trick gives us a very efficient function to test if a number is a base-2 palindrome.
The code is then very fast and also very much readable than if we had done the bit manipulations by hand.
with Ada.Text_IO, Ada.Unchecked_Conversion;use Ada.Text_IO;
procedure Palindromic is
type Int is mod 2**64; -- the size of the unsigned values we will test doesn't exceed 64 bits
type Bits is array (0..63) of Boolean;
for Bits'Component_Size use 1;
-- This function allows us to get the i'th bit of an Int k by writing Int_To_Bits(k)(i)
function Int_To_Bits is new Ada.Unchecked_Conversion(Int,Bits);
-- an inline function to test if k is palindromic in a very efficient way since we leave the loop
-- as soon as two bits are not symmetric). Number_Of_Digits is the number of digits (in base 2) of k minus 1
function Is_Pal2 (k : Int;Number_Of_Digits : Natural) return Boolean is
(for all i in 0..Number_Of_Digits=>Int_To_Bits(k)(i)=Int_To_Bits(k)(Number_Of_Digits-i));
function Reverse_Number (k : Int) return Int is --returns the symmetric representation of k (base-3)
n : Int := 0;
p : Int := k;
begin
while 0<p loop
n := n * 3 + p mod 3;
p := p / 3;
end loop;
return n;
end reverse_number;
procedure Print (n : Int) is
package BC is new Ada.Text_IO.Modular_IO (Int); use BC; -- allows us to express a variable of modular type in a given base
begin
Put (n, Base=>2, Width=>65); Put (n, Base=>3, Width=>45); put_line (" " & n'Img);
end Print;
p3, n, bound, count_pal: Int := 1;
begin
Print (0); -- because 0 is the special case asked to be treated, that is why count_pal=1
Process_Each_Power_Of_4 : for p in 0..31 loop -- because 4^p < 2^64
-- adjust the 3-power of the number to test so that the palindrome built with it has an odd number of digits in base-2
while (3*p3+1)*p3 < 2**(2*p) loop p3 := 3*p3;end loop;
bound := 2**(2*p)/(3*p3);
for k in Int range Int'Max(p3/3, bound) .. Int'Min (2*bound,p3-1) loop
n := (3*k+1)*p3 + Reverse_Number (k); -- n is a 2p+1 digits number in base 2 and is a palindrome in base 3.
if Is_Pal2 (n, 2*p) then
Print (n);
count_pal := count_pal + 1;
exit Process_Each_Power_Of_4 when count_pal = 7;
end if;
end loop;
end loop Process_Each_Power_Of_4;
end Palindromic;
- Output:
On a modern machine, (core i5 for example), this code, compiled with the -O3 and -gnatp options, takes less than 5 seconds to give the seven first palindromes smaller than 2^64.
2#0# 3#0# 0
2#1# 3#1# 1
2#1100111110011# 3#100010001# 6643
2#101011011010110110101# 3#2200021200022# 1422773
2#10100101010001010100101# 3#101012010210101# 5415589
2#1010100001100000100010000011000010101# 3#22122022220102222022122# 90396755477
2#10101001100110110110001110011011001110001101101100110010101# 3#2112200222001222121212221002220022112# 381920985378904469
AppleScript
On my machine, this gets the first five results practically instantaneously and the sixth about eight seconds later.
on intToText(int, base) -- Simple version for brevity.
script o
property digits : {int mod base as integer}
end script
set int to int div base
repeat until (int = 0)
set beginning of o's digits to int mod base as integer
set int to int div base
end repeat
return join(o's digits, "")
end intToText
on join(lst, delim)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to delim
set txt to lst as text
set AppleScript's text item delimiters to astid
return txt
end join
on task()
set output to {"0 0 0", "1 1 1"} -- Results for 0 and 1 taken as read.
set b3Hi to 0 -- Integer value of a palindromic ternary number, minus its low end.
set msdCol to 1 -- Column number (0-based) of its leftmost ternary digit.
set lsdCol to 1 -- Column number of the high end's rightmost ternary digit.
set lsdUnit to 3 ^ lsdCol as integer -- Value of 1 in this column.
set lrdCol to 1 -- Column number of the rightmost hi end digit to mirror in the low end.
set lrdUnit to 3 ^ lrdCol as integer -- Value of 1 in this column.
set columnTrigger to 3 ^ (msdCol + 1) as integer -- Next value that will need an additional column.
repeat until ((count output) = 6)
-- Notionally increment the high end's rightmost column.
set b3Hi to b3Hi + lsdUnit
-- If this carries through to start an additional column, adjust the presets.
if (b3Hi = columnTrigger) then
set lrdCol to lrdCol + msdCol mod 2
set lrdUnit to 3 ^ lrdCol as integer
set msdCol to msdCol + 1
set lsdCol to (msdCol + 1) div 2
set lsdUnit to 3 ^ lsdCol as integer
set columnTrigger to columnTrigger * 3
end if
-- Work out a mirroring low end value and add it in.
set b3Lo to b3Hi div lrdUnit mod 3
repeat with p from (lrdCol + 1) to msdCol
set b3Lo to b3Lo * 3 + b3Hi div (3 ^ p) mod 3
end repeat
set n to b3Hi + b3Lo
-- See if the result's palindromic in base 2 too. It must be an odd number and the value of the
-- reverse of its low end (including any overlap) must match that of its truncated high end.
set oL2b to n mod 2
if (oL2b = 1) then
set b2Hi to n
repeat while (b2Hi > oL2b)
set b2Hi to b2Hi div 2
if (b2Hi > oL2b) then set oL2b to oL2b * 2 + b2Hi mod 2
end repeat
-- If it is, append its decimal, binary, and ternary representations to the output.
if (b2Hi = oL2b) then ¬
set end of output to join({intToText(n, 10), intToText(n, 2), intToText(n, 3)}, " ")
end if
end repeat
set beginning of output to "decimal binary ternary:"
return join(output, linefeed)
end task
task()
- Output:
"decimal binary ternary:
0 0 0
1 1 1
6643 1100111110011 100010001
1422773 101011011010110110101 2200021200022
5415589 10100101010001010100101 101012010210101
90396755477 1010100001100000100010000011000010101 22122022220102222022122"
Arturo
pal2?: function [n][
digs2: digits.base:2 n
return digs2 = reverse digs2
]
revNumber: function [z][
u: z
result: 0
while [u > 0][
result: result + (2*result) + u%3
u: u/3
]
return result
]
pal23: function [][
p3: 1
cnt: 1
print [
pad (to :string 0)++" :" 14
pad.right join to [:string] digits.base:2 0 37 "->"
join to [:string] digits.base:3 0
]
loop 0..31 'p [
while [(p3*(1+3*p3)) < shl 1 2*p]-> p3: p3*3
bound: (shl 1 2*p)/3*p3
limDown: max @[p3/3, bound]
limUp: min @[2*bound, p3-1]
if limUp >= limDown [
loop limDown..limUp 'k [
n: (revNumber k) + (1+3*k)*p3
if pal2? n [
print [
pad (to :string n)++" :" 14
pad.right join to [:string] digits.base:2 n 37 "->"
join to [:string] digits.base:3 n
]
cnt: cnt + 1
if cnt=6 -> return null
]
]
]
]
]
pal23
- Output:
0 : 0 -> 0
1 : 1 -> 1
6643 : 1100111110011 -> 100010001
1422773 : 101011011010110110101 -> 2200021200022
5415589 : 10100101010001010100101 -> 101012010210101
90396755477 : 1010100001100000100010000011000010101 -> 22122022220102222022122
C
Per the observations made by the Ruby code (which are correct), the numbers must have odd number of digits in base 3 with a 1 at the middle, and must have odd number of digits in base 2.
#include <stdio.h>
typedef unsigned long long xint;
int is_palin2(xint n)
{
xint x = 0;
if (!(n&1)) return !n;
while (x < n) x = x<<1 | (n&1), n >>= 1;
return n == x || n == x>>1;
}
xint reverse3(xint n)
{
xint x = 0;
while (n) x = x*3 + (n%3), n /= 3;
return x;
}
void print(xint n, xint base)
{
putchar(' ');
// printing digits backwards, but hey, it's a palindrome
do { putchar('0' + (n%base)), n /= base; } while(n);
printf("(%lld)", base);
}
void show(xint n)
{
printf("%llu", n);
print(n, 2);
print(n, 3);
putchar('\n');
}
xint min(xint a, xint b) { return a < b ? a : b; }
xint max(xint a, xint b) { return a > b ? a : b; }
int main(void)
{
xint lo, hi, lo2, hi2, lo3, hi3, pow2, pow3, i, n;
int cnt;
show(0);
cnt = 1;
lo = 0;
hi = pow2 = pow3 = 1;
while (1) {
for (i = lo; i < hi; i++) {
n = (i * 3 + 1) * pow3 + reverse3(i);
if (!is_palin2(n)) continue;
show(n);
if (++cnt >= 7) return 0;
}
if (i == pow3)
pow3 *= 3;
else
pow2 *= 4;
while (1) {
while (pow2 <= pow3) pow2 *= 4;
lo2 = (pow2 / pow3 - 1) / 3;
hi2 = (pow2 * 2 / pow3 - 1) / 3 + 1;
lo3 = pow3 / 3;
hi3 = pow3;
if (lo2 >= hi3)
pow3 *= 3;
else if (lo3 >= hi2)
pow2 *= 4;
else {
lo = max(lo2, lo3);
hi = min(hi2, hi3);
break;
}
}
}
return 0;
}
- Output:
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) 381920985378904469 10101001100110110110001110011011001110001101101100110010101(2) 2112200222001222121212221002220022112(3)
C#
No strings involved. Ternary numbers (only of odd length and with a 1 in the middle) are generated by permutating powers of 3
and then checked to see if they are palindromic in binary.
The first 6 numbers take about 1/10th of a second. The 7th number takes about 3 and a half minutes.
using System;
using System.Collections.Generic;
using System.Linq;
public class FindPalindromicNumbers
{
static void Main(string[] args)
{
var query =
PalindromicTernaries()
.Where(IsPalindromicBinary)
.Take(6);
foreach (var x in query) {
Console.WriteLine("Decimal: " + x);
Console.WriteLine("Ternary: " + ToTernary(x));
Console.WriteLine("Binary: " + Convert.ToString(x, 2));
Console.WriteLine();
}
}
public static IEnumerable<long> PalindromicTernaries() {
yield return 0;
yield return 1;
yield return 13;
yield return 23;
var f = new List<long> {0};
long fMiddle = 9;
while (true) {
for (long edge = 1; edge < 3; edge++) {
int i;
do {
//construct the result
long result = fMiddle;
long fLeft = fMiddle * 3;
long fRight = fMiddle / 3;
for (int j = f.Count - 1; j >= 0; j--) {
result += (fLeft + fRight) * f[j];
fLeft *= 3;
fRight /= 3;
}
result += (fLeft + fRight) * edge;
yield return result;
//next permutation
for (i = f.Count - 1; i >= 0; i--) {
if (f[i] == 2) {
f[i] = 0;
} else {
f[i]++;
break;
}
}
} while (i >= 0);
}
f.Add(0);
fMiddle *= 3;
}
}
public static bool IsPalindromicBinary(long number) {
long n = number;
long reverse = 0;
while (n != 0) {
reverse <<= 1;
if ((n & 1) == 1) reverse++;
n >>= 1;
}
return reverse == number;
}
public static string ToTernary(long n)
{
if (n == 0) return "0";
string result = "";
while (n > 0) { {
result = (n % 3) + result;
n /= 3;
}
return result;
}
}
- Output:
Decimal: 0 Ternary: 0 Binary: 0 Decimal: 1 Ternary: 1 Binary: 1 Decimal: 6643 Ternary: 100010001 Binary: 1100111110011 Decimal: 1422773 Ternary: 2200021200022 Binary: 101011011010110110101 Decimal: 5415589 Ternary: 101012010210101 Binary: 10100101010001010100101 Decimal: 90396755477 Ternary: 22122022220102222022122 Binary: 1010100001100000100010000011000010101
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 );
}
- Output:
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
Common Lisp
Unoptimized version
(defun palindromep (str)
(string-equal str (reverse str)) )
(loop
for i from 0
with results = 0
until (>= results 6)
do
(when (and (palindromep (format nil "~B" i))
(palindromep (format nil "~3R" i)) )
(format t "n:~a~:* [2]:~B~:* [3]:~3R~%" i)
(incf results) ))
- Output:
n:0 [2]:0 [3]:0 n:1 [2]:1 [3]:1 n:6643 [2]:1100111110011 [3]:100010001 n:1422773 [2]:101011011010110110101 [3]:2200021200022 n:5415589 [2]:10100101010001010100101 [3]:101012010210101 n:90396755477 [2]:1010100001100000100010000011000010101 [3]:22122022220102222022122 n:381920985378904469 [2]:10101001100110110110001110011011001110001101101100110010101 [3]:2112200222001222121212221002220022112
D
import core.stdc.stdio, std.ascii;
bool isPalindrome2(ulong n) pure nothrow @nogc @safe {
ulong x = 0;
if (!(n & 1))
return !n;
while (x < n) {
x = (x << 1) | (n & 1);
n >>= 1;
}
return n == x || n == (x >> 1);
}
ulong reverse3(ulong n) pure nothrow @nogc @safe {
ulong x = 0;
while (n) {
x = x * 3 + (n % 3);
n /= 3;
}
return x;
}
void printReversed(ubyte base)(ulong n) nothrow @nogc {
' '.putchar;
do {
digits[n % base].putchar;
n /= base;
} while(n);
printf("(%d)", base);
}
void main() nothrow @nogc {
ulong top = 1, mul = 1, even = 0;
uint count = 0;
for (ulong i = 0; true; i++) {
if (i == top) {
if (even ^= 1)
top *= 3;
else {
i = mul;
mul = top;
}
}
immutable n = i * mul + reverse3(even ? i / 3 : i);
if (isPalindrome2(n)) {
printf("%llu", n);
printReversed!3(n);
printReversed!2(n);
'\n'.putchar;
if (++count >= 6) // Print first 6.
break;
}
}
}
- Output:
0 0(3) 0(2) 1 1(3) 1(2) 6643 100010001(3) 1100111110011(2) 1422773 2200021200022(3) 101011011010110110101(2) 5415589 101012010210101(3) 10100101010001010100101(2) 90396755477 22122022220102222022122(3) 1010100001100000100010000011000010101(2)
EasyLang
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
- Output:
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)
Elixir
defmodule Palindromic do
import Integer, only: [is_odd: 1]
def number23 do
Stream.concat([0,1], Stream.unfold(1, &number23/1))
end
def number23(i) do
n3 = Integer.to_string(i,3)
n = (n3 <> "1" <> String.reverse(n3)) |> String.to_integer(3)
n2 = Integer.to_string(n,2)
if is_odd(String.length(n2)) and n2 == String.reverse(n2),
do: {n, i+1},
else: number23(i+1)
end
def task do
IO.puts " decimal ternary binary"
number23()
|> Enum.take(6)
|> Enum.each(fn n ->
n3 = Integer.to_charlist(n,3) |> :string.centre(25)
n2 = Integer.to_charlist(n,2) |> :string.centre(39)
:io.format "~12w ~s ~s~n", [n, n3, n2]
end)
end
end
Palindromic.task
- Output:
decimal ternary binary
0 0 0
1 1 1
6643 100010001 1100111110011
1422773 2200021200022 101011011010110110101
5415589 101012010210101 10100101010001010100101
90396755477 22122022220102222022122 1010100001100000100010000011000010101
Emacs Lisp
Thanks to notes in Ruby code that made this solution possible:
(require 'calc-bin)
(defun is-palindrome-p (str)
"Test if STR is a palindrome."
(equal str (reverse str)))
(defun ends-in-zero-p (str)
"Test if STR ends in zero (0)."
(string-match-p "0\\'" str))
(defun base-3-to-base-2 (str)
"Convert STR from base 3 to base 2.
STR must be a string containing only digits 0-2."
(let ((calc-number-radix 2)
(base-10-number (string-to-number str 3)))
;; return string of base 2 number
(format "%s" (math-format-radix base-10-number))))
(defun base-3-to-base-10 (str)
"Convert STR from base 3 to base 10.
STR must be a string containing only digits 0-2."
(let ((calc-number-radix 10)
(base-10-number (string-to-number str 3)))
;; return string of base 10 number
(format "%s" (number-to-string base-10-number))))
(defun find-palindromes (number)
"Find NUMBER of pairs of palindrome numbers.
The pair consists of one number which is a palindrome in
base 3 and a base 2 number of the same value that is
also a palindrome in base 2."
(let ((calc-number-radix 3)
(base-3-partial-string)
(base-3-partial-number 0)
(base-3-string)
(base-2-string)
(base-10-string)
(count 2))
(insert (format "\n%12s %23s %37s" "Base_10" "Base_3" "Base_2"))
(insert "\n----------------------------------------------------------------------------")
(insert (format "\n%12s %23s %37s" "0" "0" "0"))
(insert (format "\n%12s %23s %37s" "1" "1" "1"))
(while (< count number)
(setq base-3-partial-number (1+ base-3-partial-number))
(setq base-3-partial-string (format "%s" (math-format-radix base-3-partial-number)))
(setq base-3-string (concat base-3-partial-string "1" (reverse base-3-partial-string)))
(unless (ends-in-zero-p base-3-string)
(setq base-2-string (base-3-to-base-2 base-3-string))
(when (is-palindrome-p base-2-string)
(setq count (1+ count))
(setq base-10-string (base-3-to-base-10 base-3-string))
(insert (format "\n%12s %23s %37s" base-10-string base-3-string base-2-string)))))))
- Output:
(find-palindromes 6)
Base_10 Base_3 Base_2
----------------------------------------------------------------------------
0 0 0
1 1 1
6643 100010001 1100111110011
1422773 2200021200022 101011011010110110101
5415589 101012010210101 10100101010001010100101
90396755477 22122022220102222022122 1010100001100000100010000011000010101
F#
// Find palindromic numbers in both binary and ternary bases. December 19th., 2018
let fG(n,g)=(Seq.unfold(fun(g,e)->if e<1L then None else Some((g%3L)*e,(g/3L,e/3L)))(n,g/3L)|>Seq.sum)+g+n*g*3L
Seq.concat[seq[0L;1L;2L];Seq.unfold(fun(i,e)->Some (fG(i,e),(i+1L,if i=e-1L then e*3L else e)))(1L,3L)]
|>Seq.filter(fun n->let n=System.Convert.ToString(n,2).ToCharArray() in n=Array.rev n)|>Seq.take 6|>Seq.iter (printfn "%d")
- Output:
Finding 6 takes no time.
0 1 6643 1422773 5415589 90396755477 Real: 00:00:00.482, CPU: 00:00:00.490, GC gen0: 77, gen1: 0
Finding 7 takes a little longer.
0 1 6643 1422773 5415589 90396755477 381920985378904469 Real: 00:23:09.114, CPU: 00:23:26.430, GC gen0: 209577, gen1: 1
Factor
This implementation uses the methods for reducing the search space discussed in the Ruby example.
USING: combinators.short-circuit formatting io kernel lists
lists.lazy literals math math.parser sequences tools.time ;
IN: rosetta-code.2-3-palindromes
CONSTANT: info $[
"The first 6 numbers which are palindromic in both binary "
"and ternary:" append
]
: expand ( n -- m ) 3 >base dup <reversed> "1" glue 3 base> ;
: 2-3-pal? ( n -- ? )
expand >bin
{ [ length odd? ] [ dup <reversed> sequence= ] } 1&& ;
: first6 ( -- seq )
4 0 lfrom [ 2-3-pal? ] lfilter ltake list>array
[ expand ] map { 0 1 } prepend ;
: main ( -- )
info print nl first6 [
dup [ >bin ] [ 3 >base ] bi
"Decimal : %d\nBinary : %s\nTernary : %s\n\n" printf
] each ;
[ main ] time
- Output:
The first 6 numbers which are palindromic in both binary and ternary: 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 Running time: 0.555949118 seconds
FreeBASIC
As using a brute force approach will be too slow for this task we instead create ternary palindromes and check if they are also binary palindromes using the optimizations which have been noted in some of the other language solutions :
' FB 1.05.0 Win64
'converts decimal "n" to its ternary equivalent
Function Ter(n As UInteger) As String
If n = 0 Then Return "0"
Dim result As String = ""
While n > 0
result = (n Mod 3) & result
n \= 3
Wend
Return result
End Function
' check if a binary or ternary numeric string "s" is palindromic
Function isPalindromic(s As String) As Boolean
' we can assume "s" will have an odd number of digits, so can ignore the middle digit
Dim As UInteger length = Len(s)
For i As UInteger = 0 To length \ 2 - 1
If s[i] <> s[length - 1 - i] Then Return False
Next
Return True
End Function
' print a number which is both a binary and ternary palindrome in all three bases
Sub printPalindrome(n As UInteger)
Print "Decimal : "; Str(n)
Print "Binary : "; bin(n)
Print "Ternary : "; ter(n)
Print
End Sub
' create a ternary palindrome whose left part is the ternary equivalent of "n" and return its decimal equivalent
Function createPalindrome3(n As UInteger) As UInteger
Dim As String ternary = Ter(n)
Dim As UInteger power3 = 1, sum = 0, length = Len(ternary)
For i As Integer = 0 To Length - 1 ''right part of palindrome is mirror image of left part
If ternary[i] > 48 Then '' i.e. non-zero
sum += (ternary[i] - 48) * power3
End If
power3 *= 3
Next
sum += power3 '' middle digit must be 1
power3 *= 3
sum += n * power3 '' value of left part is simply "n" multiplied by appropriate power of 3
Return sum
End Function
Dim t As Double = timer
Dim As UInteger i = 1, p3, count = 2
Dim As String binStr
Print "The first 6 numbers which are palindromic in both binary and ternary are :"
Print
' we can assume the first two palindromic numbers as per the task description
printPalindrome(0) '' 0 is a palindrome in all 3 bases
printPalindrome(1) '' 1 is a palindrome in all 3 bases
Do
p3 = createPalindrome3(i)
If p3 Mod 2 > 0 Then ' cannot be even as binary equivalent would end in zero
binStr = Bin(p3) '' Bin function is built into FB
If Len(binStr) Mod 2 = 1 Then '' binary palindrome must have an odd number of digits
If isPalindromic(binStr) Then
printPalindrome(p3)
count += 1
End If
End If
End If
i += 1
Loop Until count = 6
Print "Took ";
Print Using "#.###"; timer - t;
Print " seconds on i3 @ 2.13 GHz"
Print "Press any key to quit"
Sleep
- Output:
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 Took 0.761 seconds on i3 @ 2.13 GHz
Go
On my modest machine (Intel Celeron @1.6ghz) this takes about 30 seconds to produce the 7th palindrome. Curiously, the C version (GCC 5.4.0, -O3) takes about 55 seconds on the same machine. As it's a faithful translation, I have no idea why.
package main
import (
"fmt"
"strconv"
"time"
)
func isPalindrome2(n uint64) bool {
x := uint64(0)
if (n & 1) == 0 {
return n == 0
}
for x < n {
x = (x << 1) | (n & 1)
n >>= 1
}
return n == x || n == (x>>1)
}
func reverse3(n uint64) uint64 {
x := uint64(0)
for n != 0 {
x = x*3 + (n % 3)
n /= 3
}
return x
}
func show(n uint64) {
fmt.Println("Decimal :", n)
fmt.Println("Binary :", strconv.FormatUint(n, 2))
fmt.Println("Ternary :", strconv.FormatUint(n, 3))
fmt.Println("Time :", time.Since(start))
fmt.Println()
}
func min(a, b uint64) uint64 {
if a < b {
return a
}
return b
}
func max(a, b uint64) uint64 {
if a > b {
return a
}
return b
}
var start time.Time
func main() {
start = time.Now()
fmt.Println("The first 7 numbers which are palindromic in both binary and ternary are :\n")
show(0)
cnt := 1
var lo, hi, pow2, pow3 uint64 = 0, 1, 1, 1
for {
i := lo
for ; i < hi; i++ {
n := (i*3+1)*pow3 + reverse3(i)
if !isPalindrome2(n) {
continue
}
show(n)
cnt++
if cnt >= 7 {
return
}
}
if i == pow3 {
pow3 *= 3
} else {
pow2 *= 4
}
for {
for pow2 <= pow3 {
pow2 *= 4
}
lo2 := (pow2/pow3 - 1) / 3
hi2 := (pow2*2/pow3-1)/3 + 1
lo3 := pow3 / 3
hi3 := pow3
if lo2 >= hi3 {
pow3 *= 3
} else if lo3 >= hi2 {
pow2 *= 4
} else {
lo = max(lo2, lo3)
hi = min(hi2, hi3)
break
}
}
}
}
- Output:
Sample run:
The first 7 numbers which are palindromic in both binary and ternary are : Decimal : 0 Binary : 0 Ternary : 0 Time : 1.626245ms Decimal : 1 Binary : 1 Ternary : 1 Time : 3.076839ms Decimal : 6643 Binary : 1100111110011 Ternary : 100010001 Time : 4.026575ms Decimal : 1422773 Binary : 101011011010110110101 Ternary : 2200021200022 Time : 5.014413ms Decimal : 5415589 Binary : 10100101010001010100101 Ternary : 101012010210101 Time : 5.949399ms Decimal : 90396755477 Binary : 1010100001100000100010000011000010101 Ternary : 22122022220102222022122 Time : 24.878073ms Decimal : 381920985378904469 Binary : 10101001100110110110001110011011001110001101101100110010101 Ternary : 2112200222001222121212221002220022112 Time : 30.090048188s
Haskell
import Data.Char (digitToInt, intToDigit, isDigit)
import Data.List (transpose, unwords)
import Numeric (readInt, showIntAtBase)
---------- PALINDROMIC IN BOTH BINARY AND TERNARY --------
dualPalindromics :: [Integer]
dualPalindromics =
0 :
1 :
take
4
( filter
isBinPal
(readBase3 . base3Palindrome <$> [1 ..])
)
base3Palindrome :: Integer -> String
base3Palindrome =
((<>) <*> (('1' :) . reverse)) . showBase 3
isBinPal :: Integer -> Bool
isBinPal n =
( \s ->
( \(q, r) ->
(1 == r)
&& drop (succ q) s == reverse (take q s)
)
$ quotRem (length s) 2
)
$ showBase 2 n
--------------------------- TEST -------------------------
main :: IO ()
main =
mapM_
putStrLn
( unwords
<$> transpose
( ( fmap
=<< flip justifyLeft ' '
. succ
. maximum
. fmap length
)
<$> transpose
( ["Decimal", "Ternary", "Binary"] :
fmap
( (<*>) [show, showBase 3, showBase 2]
. return
)
dualPalindromics
)
)
)
where
justifyLeft n c s = take n (s <> replicate n c)
-------------------------- BASES -------------------------
readBase3 :: String -> Integer
readBase3 = fst . head . readInt 3 isDigit digitToInt
showBase :: Integer -> Integer -> String
showBase base n = showIntAtBase base intToDigit n []
- Output:
Decimal Ternary Binary 0 0 0 1 1 1 6643 100010001 1100111110011 1422773 2200021200022 101011011010110110101 5415589 101012010210101 10100101010001010100101 90396755477 22122022220102222022122 1010100001100000100010000011000010101
J
Solution:
isPalin=: -: |. NB. check if palindrome
toBase=: #.inv"0 NB. convert to base(s) in left arg
filterPalinBase=: ] #~ isPalin@toBase/ NB. palindromes for base(s)
find23Palindromes=: 3 filterPalinBase 2 filterPalinBase ] NB. palindromes in both base 2 and base 3
showBases=: [: ;:inv@|: <@({&'0123456789ABCDEFGH')@toBase/ NB. display numbers in bases
NB.*getfirst a Adverb to get first y items returned by verb u
getfirst=: adverb define
100000x u getfirst y
:
res=. 0$0
start=. 0
blk=. i.x
whilst. y > #res do.
tmp=. u start + blk
start=. start + x
res=. res, tmp
end.
y{.res
)
Usage:
find23Palindromes i. 2e6 NB. binary & ternary palindromes less than 2,000,000
0 1 6643 1422773
10 2 3 showBases find23Palindromes getfirst 6 NB. first 6 binary & ternary palindomes
0 0 0
1 1 1
6643 1100111110011 100010001
1422773 101011011010110110101 2200021200022
5415589 10100101010001010100101 101012010210101
90396755477 1010100001100000100010000011000010101 22122022220102222022122
Java
This takes a while to get to the 6th one (I didn't time it precisely, but it was less than 2 hours on an i7)
public class Pali23 {
public static boolean isPali(String x){
return x.equals(new StringBuilder(x).reverse().toString());
}
public static void main(String[] args){
for(long i = 0, count = 0; count < 6;i++){
if((i & 1) == 0 && (i != 0)) continue; //skip non-zero evens, nothing that ends in 0 in binary can be in this sequence
//maybe speed things up through short-circuit evaluation by putting toString in the if
//testing up to 10M, base 2 has slightly fewer palindromes so do that one first
if(isPali(Long.toBinaryString(i)) && isPali(Long.toString(i, 3))){
System.out.println(i + ", " + Long.toBinaryString(i) + ", " + Long.toString(i, 3));
count++;
}
}
}
}
- Output:
0, 0, 0 1, 1, 1 6643, 1100111110011, 100010001 1422773, 101011011010110110101, 2200021200022 5415589, 10100101010001010100101, 101012010210101 90396755477, 1010100001100000100010000011000010101, 22122022220102222022122
Alternatively, using a simple and efficient algorithm, the first six number are found in less than a second.
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();
}
}
- Output:
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
JavaScript
ES6
(() => {
'use strict';
// GENERIC FUNCTIONS
// range :: Int -> Int -> [Int]
const range = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
// compose :: (b -> c) -> (a -> b) -> (a -> c)
const compose = (f, g) => x => f(g(x));
// listApply :: [(a -> b)] -> [a] -> [b]
const listApply = (fs, xs) =>
[].concat.apply([], fs.map(f =>
[].concat.apply([], xs.map(x => [f(x)]))));
// pure :: a -> [a]
const pure = x => [x];
// curry :: Function -> Function
const curry = (f, ...args) => {
const go = xs => xs.length >= f.length ? (f.apply(null, xs)) :
function () {
return go(xs.concat([].slice.apply(arguments)));
};
return go([].slice.call(args, 1));
};
// transpose :: [[a]] -> [[a]]
const transpose = xs =>
xs[0].map((_, iCol) => xs.map(row => row[iCol]));
// reverse :: [a] -> [a]
const reverse = xs =>
typeof xs === 'string' ? (
xs.split('')
.reverse()
.join('')
) : xs.slice(0)
.reverse();
// take :: Int -> [a] -> [a]
const take = (n, xs) => xs.slice(0, n);
// drop :: Int -> [a] -> [a]
const drop = (n, xs) => xs.slice(n);
// maximum :: [a] -> a
const maximum = xs =>
xs.reduce((a, x) => (x > a || a === undefined ? x : a), undefined);
// quotRem :: Integral a => a -> a -> (a, a)
const quotRem = (m, n) => [Math.floor(m / n), m % n];
// length :: [a] -> Int
const length = xs => xs.length;
// justifyLeft :: Int -> Char -> Text -> Text
const justifyLeft = (n, cFiller, strText) =>
n > strText.length ? (
(strText + cFiller.repeat(n))
.substr(0, n)
) : strText;
// unwords :: [String] -> String
const unwords = xs => xs.join(' ');
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// BASES AND PALINDROMES
// show, showBinary, showTernary :: Int -> String
const show = n => n.toString(10);
const showBinary = n => n.toString(2);
const showTernary = n => n.toString(3);
// readBase3 :: String -> Int
const readBase3 = s => parseInt(s, 3);
// base3Palindrome :: Int -> String
const base3Palindrome = n => {
const s = showTernary(n);
return s + '1' + reverse(s);
};
// isBinPal :: Int -> Bool
const isBinPal = n => {
const
s = showBinary(n),
[q, r] = quotRem(s.length, 2);
return (r !== 0) && drop(q + 1, s) === reverse(take(q, s));
};
// solutions :: [Int]
const solutions = [0, 1].concat(range(1, 10E5)
.map(compose(readBase3, base3Palindrome))
.filter(isBinPal));
// TABULATION
// cols :: [[Int]]
const cols = transpose(
[
['Decimal', 'Ternary', 'Binary']
].concat(
solutions.map(
compose(
xs => listApply([show, showTernary, showBinary], xs),
pure
)
)
)
);
return unlines(
transpose(cols.map(col => col.map(
curry(justifyLeft)(maximum(col.map(length)) + 1, ' ')
)))
.map(unwords));
})();
- Output:
Decimal Ternary Binary 0 0 0 1 1 1 6643 100010001 1100111110011 1422773 2200021200022 101011011010110110101 5415589 101012010210101 10100101010001010100101 90396755477 22122022220102222022122 1010100001100000100010000011000010101
jq
Adapted from Wren
The C (jq) and Go (gojq) implementations of jq produce correct results for the first 6 numbers, as shown below. jq's "number" type lacks the integer arithmetic precision required to compute the next number in the sequence, and gojq's memory management is not up to the task even on a generously endowed machine.
Generic Utilities
# Convert the input integer to a string in the specified base (2 to 36 inclusive)
def convert(base):
def stream:
recurse(if . >= base then ./base|floor else empty end) | . % base ;
[stream] | reverse
| if base < 10 then map(tostring) | join("")
elif base <= 36 then map(if . < 10 then 48 + . else . + 87 end) | implode
else error("base too large")
end;
# integer division using integer operations only
def idivide($i; $j):
($i % $j) as $mod
| ($i - $mod) / $j ;
def idivide($j):
idivide(.; $j);
# If cond then show the result of update before recursing
def iterate(cond; update):
def i: select(cond) | update | (., i);
i;The Task
def isPalindrome2:
if (. % 2 == 0) then . == 0
else {x:0, n: .}
| until(.x >= .n;
.x = .x*2 + (.n % 2)
| .n |= idivide(2) )
| .n == .x or .n == (.x|idivide(2))
end;
def reverse3:
{n: ., x: 0}
| until (.n == 0;
.x = .x*3 + (.n % 3)
| .n |= idivide(3) )
| .x;
def show:
"Decimal : \(.)",
"Binary : \(convert(2))",
"Ternary : \(convert(3))",
"";
def task($count):
"The first \($count) numbers which are palindromic in both binary and ternary are:",
(0|show),
({cnt:1, lo:0, hi:1, pow2:1, pow3:1}
| iterate( .cnt < $count;
.emit = null
| .i = .lo
| until (.i >= .hi or .emit;
((.i*3+1)*.pow3 + (.i|reverse3)) as $n
| if $n|isPalindrome2
then .emit = [$n|show]
| .cnt += 1
else .
end
| .i += 1 )
| if .cnt == $count then . # all done
else if .i == .pow3
then .pow3 *= 3
else .pow2 *= 4
end
| .break = false
| until( .break;
until(.pow2 > .pow3; .pow2 *= 4)
| .lo2 = idivide( idivide(.pow2;.pow3) - 1; 3)
| .hi2 = (idivide(idivide(.pow2*2;.pow3)-1;3) + 1)
| .lo3 = (.pow3|idivide(3))
| .hi3 = .pow3
| if .lo2 >= .hi3 then .pow3 *= 3
elif .lo3 >= .hi2 then .pow2 *= 4
else .lo = ([.lo2, .lo3]|max)
| .hi = ([.hi2, .hi3]|min)
| .break = true
end )
end)
| select(.emit).emit[] );
task(6)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
Julia
ispalindrome(n, bas) = (s = string(n, base=bas); s == reverse(s))
prin3online(n) = println(lpad(n, 15), lpad(string(n, base=2), 40), lpad(string(n, base=3), 30))
reversebase3(n) = (x = 0; while n != 0 x = 3x + (n %3); n = div(n, 3); end; x)
function printpalindromes(N)
lo, hi, pow2, pow3, count, i = 0, 1, 1, 1, 1, 0
println(lpad("Number", 15), lpad("Base 2", 40), lpad("Base 3", 30))
prin3online(0)
while true
for j in lo:hi-1
i = j
n = (3 * j + 1) * pow3 + reversebase3(j)
if ispalindrome(n, 2)
prin3online(n)
count += 1
if count >= N
return
end
end
end
if i == pow3
pow3 *= 3
else
pow2 *= 4
end
while true
while pow2 <= pow3
pow2 *= 4
end
lo2 = div(div(pow2, pow3) - 1, 3)
hi2 = div(div(pow2 * 2, pow3), 3) + 1
lo3 = div(pow3, 3)
hi3 = pow3
if lo2 >= hi3
pow3 *= 3
elseif lo3 >= hi2
pow2 *= 4
else
lo = max(lo2, lo3)
hi = min(hi2, hi3)
break
end
end
end
end
printpalindromes(6)
- Output:
Number Base 2 Base 3
0 0 0
1 1 1
6643 1100111110011 100010001
1422773 101011011010110110101 2200021200022
5415589 10100101010001010100101 101012010210101
90396755477 1010100001100000100010000011000010101 22122022220102222022122
Kotlin
// version 1.0.5-2
/** converts decimal 'n' to its ternary equivalent */
fun Long.toTernaryString(): String = when {
this < 0L -> throw IllegalArgumentException("negative numbers not allowed")
this == 0L -> "0"
else -> {
var result = ""
var n = this
while (n > 0) {
result += n % 3
n /= 3
}
result.reversed()
}
}
/** wraps java.lang.Long.toBinaryString in a Kotlin extension function */
fun Long.toBinaryString(): String = java.lang.Long.toBinaryString(this)
/** check if a binary or ternary numeric string 's' is palindromic */
fun isPalindromic(s: String): Boolean = (s == s.reversed())
/** print a number which is both a binary and ternary palindrome in all three bases */
fun printPalindrome(n: Long) {
println("Decimal : $n")
println("Binary : ${n.toBinaryString()}")
println("Ternary : ${n.toTernaryString()}")
println()
}
/** create a ternary palindrome whose left part is the ternary equivalent of 'n' and return its decimal equivalent */
fun createPalindrome3(n: Long): Long {
val ternary = n.toTernaryString()
var power3 = 1L
var sum = 0L
val length = ternary.length
for (i in 0 until length) { // right part of palindrome is mirror image of left part
if (ternary[i] > '0') sum += (ternary[i].toInt() - 48) * power3
power3 *= 3L
}
sum += power3 // middle digit must be 1
power3 *= 3L
sum += n * power3 // value of left part is simply 'n' multiplied by appropriate power of 3
return sum
}
fun main(args: Array<String>) {
var i = 1L
var p3: Long
var count = 2
var binStr: String
println("The first 6 numbers which are palindromic in both binary and ternary are:\n")
// we can assume the first two palindromic numbers as per the task description
printPalindrome(0L) // 0 is a palindrome in all 3 bases
printPalindrome(1L) // 1 is a palindrome in all 3 bases
do {
p3 = createPalindrome3(i)
if (p3 % 2 > 0L) { // cannot be even as binary equivalent would end in zero
binStr = p3.toBinaryString()
if (binStr.length % 2 == 1) { // binary palindrome must have an odd number of digits
if (isPalindromic(binStr)) {
printPalindrome(p3)
count++
}
}
}
i++
}
while (count < 6)
}
- Output:
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
Mathematica /Wolfram Language
palindromify3[n_] :=
Block[{digits},
If[Divisible[n, 3], {},
digits = IntegerDigits[n, 3];
FromDigits[#, 3] & /@
{Join[Reverse[digits], digits], Join[Reverse[Rest[digits]], {First[digits]}, Rest[digits]]}
]
];
base2PalindromeQ[n_] := IntegerDigits[n, 2] === Reverse[IntegerDigits[n, 2]];
Select[Flatten[palindromify3 /@ Range[1000000]], base2PalindromeQ]
- Output:
{1, 6643, 1422773, 5415589, 90396755477}
Nim
import bitops, strformat, times
#---------------------------------------------------------------------------------------------------
func isPal2(k: uint64; digitCount: Natural): bool =
## Return true if the "digitCount" + 1 bits of "k" form a palindromic number.
for i in 0..digitCount:
if k.testBit(i) != k.testBit(digitCount - i):
return false
result = true
#---------------------------------------------------------------------------------------------------
func reverseNumber(k: uint64): uint64 =
## Return the reverse number of "n".
var p = k
while p > 0:
result += 2 * result + p mod 3
p = p div 3
#---------------------------------------------------------------------------------------------------
func toBase2(n: uint64): string =
## Return the string representation of "n" in base 2.
var n = n
while true:
result.add(chr(ord('0') + (n and 1)))
n = n shr 1
if n == 0: break
#---------------------------------------------------------------------------------------------------
func toBase3(n: uint64): string =
## Return the string representation of "n" in base 3.
var n = n
while true:
result.add(chr(ord('0') + n mod 3))
n = n div 3
if n == 0: break
#---------------------------------------------------------------------------------------------------
proc print(n: uint64) =
## Print the value in bases 10, 2 and 3.
echo &"{n:>18} {n.toBase2():^59} {n.toBase3():^41}"
#---------------------------------------------------------------------------------------------------
proc findPal23() =
## Find the seven first palindromic numbers in binary and ternary bases.
var p3 = 1u64
var countPal = 1
print(0)
for p in 0..31:
while (3 * p3 + 1) * p3 < 1u64 shl (2 * p):
p3 *= 3
let bound = 1u64 shl (2 * p) div (3 * p3)
for k in max(p3 div 3, bound) .. min(2 * bound, p3 - 1):
let n = (3 * k + 1) * p3 + reverseNumber(k)
if isPal2(n, 2 * p):
print(n)
inc countPal
if countPal == 7:
return
#———————————————————————————————————————————————————————————————————————————————————————————————————
let t0 = cpuTime()
findPal23()
echo fmt"\nTime: {cpuTime() - t0:.2f}s"
- Output:
0 0 0
1 1 1
6643 1100111110011 100010001
1422773 101011011010110110101 2200021200022
5415589 10100101010001010100101 101012010210101
90396755477 1010100001100000100010000011000010101 22122022220102222022122
381920985378904469 10101001100110110110001110011011001110001101101100110010101 2112200222001222121212221002220022112
Time: 4.89s
PARI/GP
check(n)={ \\ Check for 2n+1-digit palindromes in base 3
my(N=3^n);
forstep(i=N+1,2*N,[1,2],
my(base2,base3=digits(i,3),k);
base3=concat(Vecrev(base3[2..n+1]), base3);
k=subst(Pol(base3),'x,3);
base2=binary(k);
if(base2==Vecrev(base2), print1(", "k))
)
};
print1("0, 1"); for(i=1,11,check(i))- Output:
0, 1, 6643, 1422773, 5415589, 90396755477
PascalABC.NET
// https://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases#Perl#PascalABC.NET
uses School;
function ToBase(n: int64; base: integer): string;
begin
var sb := new StringBuilder;
while n>0 do
begin
sb.Insert(0,n mod base);
n := n div base;
end;
Result := sb.ToString;
end;
function IsPalindrome(s: string): boolean := s = s.Inverse;
begin
Println(0,0,0);
for var i: int64 := 0 to 1000000 do
begin
var d3 := ToBase(i,3);
d3 := d3 + '1' + d3.Inverse;
var d := Dec(d3,3);
var d2 := ToBase(d,2);
if IsPalindrome(d2) then
Println(d,d3,d2)
end;
end.
- Output:
0 0 0 1 1 1 6643 100010001 1100111110011 1422773 2200021200022 101011011010110110101 5415589 101012010210101 10100101010001010100101 90396755477 22122022220102222022122 1010100001100000100010000011000010101
Perl
use ntheory qw/fromdigits todigitstring/;
print "0 0 0\n"; # Hard code the 0 result
for (0..2e5) {
# Generate middle-1-palindrome in base 3.
my $pal = todigitstring($_, 3);
my $b3 = $pal . "1" . reverse($pal);
# Convert base 3 number to base 2
my $b2 = todigitstring(fromdigits($b3, 3), 2);
# Print results (including base 10) if base-2 palindrome
print fromdigits($b2,2)," $b3 $b2\n" if $b2 eq reverse($b2);
}
- Output:
0 0 0 1 1 1 6643 100010001 1100111110011 1422773 2200021200022 101011011010110110101 5415589 101012010210101 10100101010001010100101 90396755477 22122022220102222022122 1010100001100000100010000011000010101
Phix
Alternative approach. Works by finding the next palindrome, in either base, in an attempt to skip fairly large chunks of the search space. Prints the first 6 palindromes (the limit on 32 bit) in about a second, but the 7th (on 64 bit only) takes just over half an hour.
Theoretically it could be made a fair bit faster by replacing the string handling (which I hope you will find very easy to follow) with maths/bit-fiddling, however my attempts at that turned out noticeably slower.
with javascript_semantics -- widths and limits for 32/64 bit running (see output below): constant {dsize,w3,w2,limit} = iff(machine_bits()=32?{12,23,37,6} :{18,37,59,7}), -- [atoms on 32-bit have only 53 bits of precision, but 7th ^^^^ requires 59] dfmt = sprintf("%%%dd",dsize), -- ie "%12d" or "%18d" esc = #1B function center(string s, integer l) l = max(0,floor((l-length(s))/2)) string space = repeat(' ',l) s = space & s & space return s end function integer count = 1 procedure show(atom n, string p2, p3) if count=1 then printf(1," %s %s %s\n",{pad_head("decimal",dsize),center("ternary",w3),center(" binary",w2)}) end if string ns = sprintf(dfmt,n) printf(1,"%2d: %s %s %s\n",{count, ns, center(p3,w3), center(p2,w2)}) count += 1 end procedure procedure progress64(string e, p2, p3) e = pad_head(e,dsize) printf(1,"--: %s %s %s\r",{e, center(p3,w3), center(p2,w2)}) end procedure function to_base(atom i, integer base) string s = "" while i>0 do s = append(s,remainder(i,base)+'0') i = floor(i/base) end while s = reverse(s) if s="" then s = "0" end if return s end function function from_base(string s, integer base) atom res = 0 for i=1 to length(s) do res = res*base+s[i]-'0' end for return res end function function sn(string s, integer f, base) -- helper function, return s mirrored (if f!=0) -- and as a (decimal) number (if base!=0) -- all returns from next_palindrome() get fed through here. if f then s[f+2..$] = reverse(s[1..f]) end if atom n = iff(base?from_base(s,base):0) return {s,n} end function function next_palindrome(integer base, object s) -- -- base is 2 or 3 -- s is not usually a palindrome, but derived from one in <5-base> -- -- all done with very obvious string manipulations, plus a few -- less obvious optimisations (odd length, middle 1 in base 3). -- -- example: next_palindrome(2,"10001000100") -> "10001010001" -- if not string(s) then s = to_base(s,base) end if integer l = length(s), f = floor(l/2), m = f+1, c if mod(l,2) then -- optimisation: palindromes must be odd-length -- 1) is a plain mirror greater? (as in the example just given) {string r} = sn(s,f,0) -- optimisation: base 3 palindromes have '1' in the middle if base=3 and r[m]!='1' then r[m] = '1' end if if r>s then return sn(r,0,base) end if -- 2) can we (just) increment the middle digit? c = s[m]-'0'+1 if base=2 or c=1 then if c<base then s[m] = c+'0' return sn(s,f,base) end if s[m] = '0' elsif base=3 then s[m] = '1' end if -- 3) can we increment left half (or is it all <base-1>s?) for i=f to 1 by -1 do if s[i]<base-1+'0' then s[i] += 1 return sn(s,f,base) else s[i] = '0' end if end for l += 2 -- (stay odd) else l += 1 -- (even->odd) end if -- 4) well then, next palindrome is longer, 1000..0001-style s = sprintf("1%s1",{repeat('0',l-2)}) -- optimisation: base 3 palindromes have '1' in the middle if base=3 then m = (l+1)/2 s[m] = '1' end if return sn(s,0,base) end function string p2 = "0", p3 = "0" -- palindromes as strings in base 2 and 3 atom n2 = 0, n3 = 0, -- decimal equivalents of the above. t0 = time(), t1 = time()+1 while count<=limit do if n2=n3 then show(n2,p2,p3) {p2,n2} = next_palindrome(2,p2) {p3,n3} = next_palindrome(3,p3) elsif n2<n3 then {p2,n2} = next_palindrome(2,n3-1) elsif n2>n3 then {p3,n3} = next_palindrome(3,n2-1) end if if time()>t1 and platform()!=JS then progress64(elapsed_short(time()-t0),p2,p3) t1 = time()+1 if find(get_key(),{'q','Q',esc}) then exit end if end if end while ?elapsed(time()-t0)
- Output:
32 bit:
decimal ternary binary 1: 0 0 0 2: 1 1 1 3: 6643 100010001 1100111110011 4: 1422773 2200021200022 101011011010110110101 5: 5415589 101012010210101 10100101010001010100101 6: 90396755477 22122022220102222022122 1010100001100000100010000011000010101 "1.0s"
64-bit:
decimal ternary binary 1: 0 0 0 2: 1 1 1 3: 6643 100010001 1100111110011 4: 1422773 2200021200022 101011011010110110101 5: 5415589 101012010210101 10100101010001010100101 6: 90396755477 22122022220102222022122 1010100001100000100010000011000010101 --: 56s 2001210110001221221000110121002 1100000101111010101010010100101010101111010000011 7: 381920985378904469 2112200222001222121212221002220022112 10101001100110110110001110011011001110001101101100110010101 "33 minutes and 08s"
between 6 and 7 I have shown progress() in action, which is constantly overwritten, and mesmerising to watch.
much simpler version
(slightly but not alot faster)
with javascript_semantics function to_base(atom n, integer base) string result = "" while true do result &= remainder(n,base) n = floor(n/base) if n=0 then exit end if end while return result end function procedure show(integer count, atom n) string n2 = sq_add('0',to_base(n,2)), n3 = sq_add('0',to_base(n,3)), p2 = repeat(' ',(37-length(n2))/2), p3 = repeat(' ',(23-length(n3))/2) printf(1,"%2d: %12d %s%s%s %s%s\n",{count, n, p3,n3,p3, p2,n2}) end procedure function createpalindrome3(integer n) atom tot = 0, power3 = 1 string ternary = to_base(n,3) for i=length(ternary) to 1 by -1 do tot += ternary[i] * power3 power3 *= 3 end for return tot + power3 + n*power3*3 end function atom t0 = time() printf(1,"%16s %15s %30s\n",{"decimal","ternary","binary"}) show(0,0) show(1,1) integer count = 2, n = 1 while count<6 do atom n3 = createpalindrome3(n) if remainder(n3,2) then string n2 = to_base(n3,2) if n2[$]=1 and n2=reverse(n2) then show(count,n3) count += 1 end if end if n += 1 end while ?elapsed(time()-t0)
- Output:
decimal ternary binary 0: 0 0 0 1: 1 1 1 2: 6643 100010001 1100111110011 3: 1422773 2200021200022 101011011010110110101 4: 5415589 101012010210101 10100101010001010100101 5: 90396755477 22122022220102222022122 1010100001100000100010000011000010101 "0.6s"
much faster version
Inspired by Scala 😏
with javascript_semantics function to_base(string s, integer base) -- convert decimal string s to specified base assert(base>=2 and base<=9) -- (>9 as below) string res = "" while length(s) do integer q, r = 0 for i=1 to length(s) do q = r*10+s[i]-'0' s[i] = floor(q/base)+'0' r = mod(q,base) end for res &= r+'0' -- +(r>9)*('A'-'9'-1) s = trim_head(s,'0') end while -- res = reverse(res) -- (if not palindromic!) return res end function constant A = {"0","1","6643","1422773","5415589","90396755477", "381920985378904469","1922624336133018996235", "2004595370006815987563563", "8022581057533823761829436662099", "392629621582222667733213907054116073", "32456836304775204439912231201966254787", "428027336071597254024922793107218595973", "1597863243206403857787246920544522912361", "30412638162199251273509758127730300026189", "32345684491703244980406880704479906642045", "24014998963383302600955162866787153652444049"} for i=1 to length(A) do printf(1,"%=145s\n",to_base(A[i],2)) printf(1,"%=145s\n",to_base(A[i],3)) end for
- Output:
0
0
1
1
1100111110011
100010001
101011011010110110101
2200021200022
10100101010001010100101
101012010210101
1010100001100000100010000011000010101
22122022220102222022122
10101001100110110110001110011011001110001101101100110010101
2112200222001222121212221002220022112
11010000011100111000101110001110011011001110001110100011100111000001011
122120102102011212112010211212110201201021221
110101000011111010101010100101111011110111011110111101001010101010111110000101011
221010112100202002120002212200021200202001211010122
1100101010000100101101110000011011011111111011000011100001101111111101101100000111011010010000101010011
21000020210011222122220212010000100001021202222122211001202000012
10010111001111000100010100010100000011011011000101011011100000111011010100011011011000000101000101000100011110011101001
122102120011102000101101000002010021111120010200000101101000201110021201221
11000011010101111010110010100010010011011010101001101000001000100010000010110010101011011001001000101001101011110101011000011
1222100201002211120110022121002012121101011212102001212200110211122001020012221
101000010000000110001000011111100101011110011100001110100011100010001110001011100001110011110101001111110000100011000000010000101
222001200110022102121001000200200202022111220202002002000100121201220011002100222
10010110010000110010100010001000111010010000111000010010100010111011101000101001000011100001001011100010001000101001100001001101001
10121021220121202021201220210001211020122122102011210001202210212020212102212012101
101100101011111111011010000110101101100101010100101101010000001000000000100000010101101001010101001101101011000010110111111110101001101
2112120210211212121000000011202022210210101010120122202021100000001212121120120212112
101111100001110001100000011101111000001111001000110100111001010101101101010100111001011000100111100000111101110000001100011100001111101
2200221111210202000010122020002221112212101012122111222000202210100002020121111220022
1000100111010110110111101001100011100110100000011011001010100101011001100011001101010010101001101100000010110011100011001011110110110101110010001
2202021211210100110100002202101000110000220121210220000110001012022000010110010121121202022
Picat
import sat.
to_num(List, Base, Num) =>
Len = length(List),
Num #= sum([List[I] * Base**(Len-I) : I in 1..Len]).
palindrom(S) =>
N = len(S),
Start :: 1..N, % start at the first non-zero position:
foreach(I in 1..N)
I1 #= max(1, min(N, N-(I-Start))), % I1 is the symmetry index partner of I (if relevant)
element(I1, S, S1), % S1 is the respective digit
I #< Start #=> S[I] #= 0, % skip leading 0´s
I #= Start #=> S[I] #> 0, % Start points to the first non-zero digit
I #>= Start #=> S[I] #= S1 % palindromic symmetry
end.
constrain(Max, B, X) =>
Len = floor(log(Max) / log(B)) + 1, % length of Max in Base B representation
Digits = new_list(Len), Digits :: 0..B-1,
to_num(Digits, B, X), % Digits show the Base B representation of X
palindrom(Digits).
main =>
N = 11, % maximum number of decimal digits for search, can be set freely
Max = 10**N - 1, % maximum number
X :: 2..Max,
constrain(Max, 2, X),
constrain(Max, 3, X),
Pnumbers = solve_all([X]),
foreach([Y] in [[0], [1]] ++ Pnumbers.sort()) % start with 0 and 1, then show solutions > 1
printf("%w %s %s%n", Y, to_radix_string(Y,2), to_radix_string(Y,3))
end.Output:
0 0 0 1 1 1 6643 1100111110011 100010001 1422773 101011011010110110101 2200021200022 5415589 10100101010001010100101 101012010210101 90396755477 1010100001100000100010000011000010101 22122022220102222022122
PicoLisp
(de ternary (N)
(if (=0 N)
(cons N)
(make
(while (gt0 N)
(yoke (% (swap 'N (/ N 3)) 3)) ) ) ) )
(de p? (L1 L2)
(and
(= L1 (reverse L1))
(= L2 (reverse L2)) ) )
(zero N)
(for (I 0 (> 6 I))
(let (B2 (chop (bin N)) B3 (ternary N))
(when (p? B2 B3)
(println N (pack B2) (pack B3))
(inc 'I) )
(inc 'N) ) )- Output:
0 "0" "0" 1 "1" "1" 6643 "1100111110011" "100010001" 1422773 "101011011010110110101" "2200021200022" 5415589 "10100101010001010100101" "101012010210101" 90396755477 "1010100001100000100010000011000010101" "22122022220102222022122"
Python
Imperative
from itertools import islice
digits = "0123456789abcdefghijklmnopqrstuvwxyz"
def baseN(num,b):
if num == 0: return "0"
result = ""
while num != 0:
num, d = divmod(num, b)
result += digits[d]
return result[::-1] # reverse
def pal2(num):
if num == 0 or num == 1: return True
based = bin(num)[2:]
return based == based[::-1]
def pal_23():
yield 0
yield 1
n = 1
while True:
n += 1
b = baseN(n, 3)
revb = b[::-1]
#if len(b) > 12: break
for trial in ('{0}{1}'.format(b, revb), '{0}0{1}'.format(b, revb),
'{0}1{1}'.format(b, revb), '{0}2{1}'.format(b, revb)):
t = int(trial, 3)
if pal2(t):
yield t
for pal23 in islice(pal_23(), 6):
print(pal23, baseN(pal23, 3), baseN(pal23, 2))
- Output:
0 0 0 1 1 1 6643 100010001 1100111110011 1422773 2200021200022 101011011010110110101 5415589 101012010210101 10100101010001010100101 90396755477 22122022220102222022122 1010100001100000100010000011000010101
Functional
'''Numbers with palindromic digit strings in both binary and ternary'''
from itertools import (islice)
# palinBoth :: Generator [Int]
def palinBoth():
'''Non finite stream of dually palindromic integers.'''
yield 0, '0', '0'
ibt = 1, '1', '1'
yield ibt
while True:
ibt = until(isBoth)(psucc)(psucc(ibt))
yield int(ibt[2], 3), ibt[1], ibt[2]
# isBoth :: (Int, String, String) -> Bool
def isBoth(ibt):
'''True if the binary string is palindromic (as
the ternary string is already known to be).
'''
b = ibt[1]
return b == b[::-1]
# psucc :: (Int, String, String) -> (Int, String, String)
def psucc(ibt):
'''The next triple of index, binary
and (palindromic) ternary string
'''
d = 1 + ibt[0]
s = showBase3(d)
pal = s + '1' + s[::-1]
return d, bin(int(pal, 3))[2:], pal
# showBase3 :: Int -> String
def showBase3(n):
'''Ternary digit string for integer n.'''
return showIntAtBase(3)(
lambda i: '012'[i]
)(n)('')
# ------------------------- TEST -------------------------
def main():
'''Integers with palindromic digits in
both binary and ternary bases.
'''
xs = take(6)(palinBoth())
d, b, t = xs[-1]
bw = len(b)
tw = len(t)
print(
fTable(
label('rjust')(('Decimal', len(str(d)))) +
''.join(map(
label('center'),
[('Binary', bw), ('Ternary', tw)]
)) + '\n'
)(compose(str)(fst))(
lambda p: p[1].center(bw, ' ') +
' ' + p[2].center(tw, ' ')
)(identity)(xs)
)
# ----------------------- GENERIC ------------------------
# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
'''Right to left function composition.'''
return lambda f: lambda x: g(f(x))
# fst :: (a, b) -> a
def fst(tpl):
'''First member of a pair.'''
return tpl[0]
# identity :: a -> a
def identity(x):
'''The identity function.'''
return x
# showIntAtBase :: Int -> (Int -> String) -> Int -> String -> String
def showIntAtBase(base):
'''String representation of an integer in a given base,
using a supplied function for the string representation
of digits.
'''
def wrap(toChr, n, rs):
def go(nd, r):
n, d = nd
r_ = toChr(d) + r
return go(divmod(n, base), r_) if 0 != n else r_
return 'unsupported base' if 1 >= base else (
'negative number' if 0 > n else (
go(divmod(n, base), rs))
)
return lambda toChr: lambda n: lambda rs: (
wrap(toChr, n, rs)
)
# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.
'''
def go(xs):
return (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)
return go
# until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
'''The result of repeatedly applying f until p holds.
The initial seed value is x.
'''
def go(f):
def g(x):
v = x
while not p(v):
v = f(v)
return v
return g
return go
# ---------------------- FORMATTING ----------------------
# label :: Method String -> (String, Int)
def label(k):
'''Stringification, using the named justification
method (ljust|centre|rjust) of the label,
and the specified amount of white space.
'''
def go(sw):
s, w = sw
return getattr(s, k)(w, ' ') + ' '
return go
# fTable :: String -> (a -> String) ->
# (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
# MAIN ---
if __name__ == '__main__':
main()
- Output:
Decimal Binary Ternary
0 -> 0 0
1 -> 1 1
6643 -> 1100111110011 100010001
1422773 -> 101011011010110110101 2200021200022
5415589 -> 10100101010001010100101 101012010210101
90396755477 -> 1010100001100000100010000011000010101 22122022220102222022122
Racket
#lang racket
(require racket/generator)
(define (digital-reverse/base base N)
(define (inr n r)
(if (zero? n) r (inr (quotient n base) (+ (* r base) (modulo n base)))))
(inr N 0))
(define (palindrome?/base base N)
(define (inr? n m)
(if (= n 0)
(= m N)
(inr? (quotient n base) (+ (* m base) (modulo n base)))))
(inr? N 0))
(define (palindrome?/3 n)
(palindrome?/base 3 n))
(define (b-palindromes-generator b)
(generator
()
;; it's a bit involved getting the initial palindroms, so we do them manually
(for ((p (in-range b))) (yield p))
(let loop ((rhs 1) (mx-rhs b) (mid #f) (mx-rhs*b (* b b)))
(cond
[(= rhs mx-rhs)
(cond
[(not mid) (loop (quotient mx-rhs b) mx-rhs 0 mx-rhs*b)]
[(zero? mid) (loop mx-rhs mx-rhs*b #f (* mx-rhs*b b))])]
[else
(define shr (digital-reverse/base b rhs))
(cond
[(not mid)
(yield (+ (* rhs mx-rhs) shr))
(loop (add1 rhs) mx-rhs #f mx-rhs*b)]
[(= mid (- b 1))
(yield (+ (* rhs mx-rhs*b) (* mid mx-rhs) shr))
(loop (+ 1 rhs) mx-rhs 0 mx-rhs*b)]
[else
(yield (+ (* rhs mx-rhs*b) (* mid mx-rhs) shr))
(loop rhs mx-rhs (add1 mid) mx-rhs*b)])]))))
(define (number->string/base n b)
(define (inr acc n)
(if (zero? n) acc
(let-values (((q r) (quotient/remainder n b)))
(inr (cons (number->string r) acc) q))))
(if (zero? n) "0" (apply string-append (inr null n))))
(module+ main
(for ((n (sequence-filter palindrome?/3 (in-producer (b-palindromes-generator 2))))
(i (in-naturals))
#:final (= i 5))
(printf "~a: ~a_10 ~a_3 ~a_2~%"
(~a #:align 'right #:min-width 3 (add1 i))
(~a #:align 'right #:min-width 11 n)
(~a #:align 'right #:min-width 23 (number->string/base n 3))
(~a #:align 'right #:min-width 37 (number->string/base n 2)))))
(module+ test
(require rackunit)
(check-true (palindrome?/base 2 #b0))
(check-true (palindrome?/base 2 #b10101))
(check-false (palindrome?/base 2 #b1010))
(define from-oeis:A060792
(list 0 1 6643 1422773 5415589 90396755477 381920985378904469
1922624336133018996235 2004595370006815987563563
8022581057533823761829436662099))
(check-match from-oeis:A060792
(list (? (curry palindrome?/base 2)
(? (curry palindrome?/base 3))) ...))
(check-eq? (digital-reverse/base 2 #b0) #b0)
(check-eq? (digital-reverse/base 2 #b1) #b1)
(check-eq? (digital-reverse/base 2 #b10) #b01)
(check-eq? (digital-reverse/base 2 #b1010) #b0101)
(check-eq? (digital-reverse/base 10 #d0) #d0)
(check-eq? (digital-reverse/base 10 #d1) #d1)
(check-eq? (digital-reverse/base 10 #d10) #d01)
(check-eq? (digital-reverse/base 10 #d1010) #d0101)
(define pg ((b-palindromes-generator 2)))
(check-match
(map (curryr number->string 2) (for/list ((i 16) (p (in-producer (b-palindromes-generator 2)))) p))
(list "0" "1" "11" "101" "111" "1001" "1111" "10001" "10101" "11011"
"11111" "100001" "101101" "110011" "111111" "1000001")))
- Output:
1: 0_10 0_3 0_2 2: 1_10 1_3 1_2 3: 6643_10 100010001_3 1100111110011_2 4: 1422773_10 2200021200022_3 101011011010110110101_2 5: 5415589_10 101012010210101_3 10100101010001010100101_2 6: 90396755477_10 22122022220102222022122_3 1010100001100000100010000011000010101_2
Raku
(formerly Perl 6) Instead of searching for numbers that are palindromes in one base then checking the other, generate palindromic trinary numbers directly, then check to see if they are also binary palindromes (with additional simplifying constraints as noted in other entries). Outputs the list in decimal, binary and trinary.
constant palindromes = 0, 1, |gather for 1 .. * -> $p {
my $pal = $p.base(3);
my $n = :3($pal ~ '1' ~ $pal.flip);
next if $n %% 2;
my $b2 = $n.base(2);
next if $b2.chars %% 2;
next unless $b2 eq $b2.flip;
take $n;
}
printf "%d, %s, %s\n", $_, .base(2), .base(3) for palindromes[^6];
- Output:
0, 0, 0 1, 1, 1 6643, 1100111110011, 100010001 1422773, 101011011010110110101, 2200021200022 5415589, 10100101010001010100101, 101012010210101 90396755477, 1010100001100000100010000011000010101, 22122022220102222022122
REXX
version 1
Programming note: This version is quite a bit faster than the previous REXX program that was entered.
For this REXX program, a few deterministic assumptions were made:
- for the requirement of binary palindromes, the number of binary digits have to be odd.
- for the requirement of ternary palindromes, the numbers can't end in zero (in base 3).
The method used is to (not find, but) construct a binary palindrome by:
- using the binary version of a number (abcdef), which may end in binary zeroes,
- flipping the binary digits (fedcba) [note that a is always 1 (one)],
- constructing two binary palindromes:
- abcdef || 0 || fedcba and
- abcdef || 1 || fedcba
- (the above two concatenation ( || ) steps ensures an odd number of binary digits),
- ensure the decimal versions are not evenly divisible by 3,
- convert the decimal numbers to base 3,
- ensure that the numbers in base 3 are palindromic.
/*REXX program finds numbers that are palindromic in both binary and ternary. */
digs=50; numeric digits digs /*biggest known B2B3 palindrome: 44 dig*/
parse arg maxHits .; if maxHits=='' then maxHits=6 /*use six as a limit.*/
hits=0; #= 'fiat' /*the number of palindromes (so far). */
call show 0,0,0; call show 1,1,1 /*show the first two palindromes (fiat)*/
!.= /* [↓] build list of powers of three. */
do i=1 until !.i>10**digs; !.i=3**i; end /*compute powers of three for radix3.*/
p=1 /* [↓] primary search: bin palindromes*/
do #=digs /*use all numbers, however, DEC is odd.*/
binH=x2b( d2x(#) ) + 0 /*convert some decimal number to binary*/
binL=reverse(binH) /*reverse the binary digits (or bits).*/
dec=x2d( b2x( binH'0'binL) ); if dec//3\==0 then call radix3
dec=x2d( b2x( binH'1'binL) ); if dec//3\==0 then call radix3
end /*#*/ /* [↑] crunch 'til found 'nuff numbers*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
radix3: parse var dec x 1 $,q /* [↓] convert decimal # ──► ternary.*/
do j=p while !.j<=x; end /*find upper limit of power of three. */
p=j-1 /*use this power of three for next time*/
do k=p by -1 for p; _=!.k; d=x%_; q=q || d; x=x//_; end /*k*/
t=q || x /*handle residual of ternary conversion*/
if t\==reverse(t) then return /*is T ternary number not palindromic? */
call show $, t, strip(x2b(d2x($)), , 0) /*show number: decimal, ternary, binary*/
return /* [↑] RADIX3 subroutine is sluggish.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
show: hits=hits+1; say /*bump the number of palindromes found.*/
say right('['hits"]", 5) right( arg(1), digs) '(decimal), ternary=' arg(2)
say right('', 5+1+ digs) ' binary =' arg(3)
if hits>2 then if hits//2 then #=#'0'
if hits<maxHits then return /*Not enough palindromes? Keep looking*/
exit /*stick a fork in it, we're all done. */
- output when using the default input of: 7
[1] 0 (decimal), ternary= 0
binary = 0
[2] 1 (decimal), ternary= 1
binary = 1
[3] 6643 (decimal), ternary= 100010001
binary = 1100111110011
[4] 1422773 (decimal), ternary= 2200021200022
binary = 101011011010110110101
[5] 5415589 (decimal), ternary= 101012010210101
binary = 10100101010001010100101
[6] 90396755477 (decimal), ternary= 22122022220102222022122
binary = 1010100001100000100010000011000010101
[7] 381920985378904469 (decimal), ternary= 2112200222001222121212221002220022112
binary = 10101001100110110110001110011011001110001101101100110010101
[Output note: the 6th number (above) took a couple of seconds to compute.]
version 2
This REXX version takes advantage that the palindromic numbers (in both binary and ternary bases) seem to only have a modulus nine residue of 1, 5, 7, or 8. With this assumption, the following REXX program is about 25% faster.
/*REXX program finds numbers that are palindromic in both binary and ternary. */
digs=50; numeric digits digs /*biggest known B2B3 palindrome: 44 dig*/
parse arg maxHits .; if maxHits=='' then maxHits=6 /*use six as a limit.*/
hits=0; #= 'fiat' /*the number of palindromes (so far). */
call show 0,0,0; call show 1,1,1 /*show the first two palindromes (fiat)*/
#.=0; #.1=1; #.5=1; #.7=1; #.8=1 /*modulus nine results that are OK. */
!.= /* [↓] build list of powers of three. */
do i=1 until !.i>10**digs; !.i=3**i; end /*compute powers of three for radix3.*/
p=1 /* [↓] primary search: bin palindromes*/
do #=digs /*use all numbers, however, DEC is odd.*/
binH=x2b( d2x(#) ) + 0 /*convert some decimal number to binary*/
binL=reverse(binH) /*reverse the binary digits (or bits).*/
dec=x2d( b2x( binH'0'binL) ); _=dec//9; if #._ then if dec//3\==0 then call radix3
dec=x2d( b2x( binH'1'binL) ); _=dec//9; if #._ then if dec//3\==0 then call radix3
end /*#*/ /* [↑] crunch 'til found 'nuff numbers*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
radix3: parse var dec x 1 $,q /* [↓] convert decimal # ──► ternary.*/
do j=p while !.j<=x; end /*find upper limit of power of three. */
p=j-1 /*use this power of three for next time*/
do k=p by -1 for p; _=!.k; d=x%_; q=q || d; x=x//_; end /*k*/
t=q || x /*handle residual of ternary conversion*/
if t\==reverse(t) then return /*is T ternary number not palindromic? */
call show $, t, strip(x2b(d2x($)), , 0) /*show number: decimal, ternary, binary*/
return /* [↑] RADIX3 subroutine is sluggish.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
show: hits=hits+1; say /*bump the number of palindromes found.*/
say right('['hits"]", 5) right( arg(1), digs) '(decimal), ternary=' arg(2)
say right('', 5+1+ digs) ' binary =' arg(3)
if hits>2 then if hits//2 then #=#'0'
if hits<maxHits then return /*Not enough palindromes? Keep looking*/
exit /*stick a fork in it, we're all done. */
- output is identical to the 1st REXX version.
Ring
# Project: Find palindromic numbers in both binary and ternary bases
max = 6
nr = 0
pal = 0
see "working..." + nl
see "wait for done..." + nl
while true
binpal = basedigits(nr,2)
terpal = basedigits(nr,3)
bool1 = ispalindrome(binpal)
bool2 = ispalindrome(terpal)
if bool1 = 1 and bool2 = 1
pal = pal + 1
see string(nr) + " " + binpal + "(2) " + terpal + "(3)" + nl
if pal = max
exit
ok
ok
nr = nr + 1
end
see "done..." + nl
func basedigits(n,base)
if n = 0
return "0"
ok
result = ""
while n > 0
result = string(n % base) + result
n = floor(n/base)
end
return result
func ispalindrome(astring)
if astring = "0"
return 1
ok
bString = ""
for i=len(aString) to 1 step -1
bString = bString + aString[i]
next
if aString = bString
return 1
else
return 0
ok- Output:
working... wait for done... 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) done...
Ruby
This program is based on the fact that the double palindromic numbers in base 3 all have a "1" right in the middle. Also, both base 2 and base 3 representations have an odd number of digits.
- 1 digit under the number of the palindromic doesn't become zero.
- As for the N numbering-system, at the time of the multiple of N, 1 digit below becomes zero.
- Palindromic by the even-number digit binary system is 3 multiples.
- Palindromic by the even-number digit ternary-system is 4 multiples.
- In palindromic by the ternary-system of the odd digit, the value of the center position is an even number in case of "0" or "2".
This program constructs base 3 palindromes using the above "rules" and checks if they happen to be binary palindromes.
pal23 = Enumerator.new do |y|
y << 0
y << 1
for i in 1 .. 1.0/0.0 # 1.step do |i| (Ruby 2.1+)
n3 = i.to_s(3)
n = (n3 + "1" + n3.reverse).to_i(3)
n2 = n.to_s(2)
y << n if n2.size.odd? and n2 == n2.reverse
end
end
puts " decimal ternary binary"
6.times do |i|
n = pal23.next
puts "%2d: %12d %s %s" % [i, n, n.to_s(3).center(25), n.to_s(2).center(39)]
end
- Output:
decimal ternary binary 0: 0 0 0 1: 1 1 1 2: 6643 100010001 1100111110011 3: 1422773 2200021200022 101011011010110110101 4: 5415589 101012010210101 10100101010001010100101 5: 90396755477 22122022220102222022122 1010100001100000100010000011000010101
Scala
Functional programmed, (tail) recursive
- Output:
Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
import scala.annotation.tailrec
import scala.compat.Platform.currentTime
object Palindrome23 extends App {
private val executionStartTime = currentTime
private val st: Stream[(Int, Long)] = (0, 1L) #:: st.map(xs => nextPalin3(xs._1))
@tailrec
private def nextPalin3(n: Int): (Int, Long) = {
@inline
def isPali2(i: BigInt): Boolean = {
val s = i.toString(2)
if ((s.length & 1) == 0) false else s == s.reverse
}
def palin3(i: BigInt): Long = {
val n3 = i.toString(3)
java.lang.Long.parseLong(n3 + "1" + n3.reverse, 3)
}
val actual: Long = palin3(n)
if (isPali2(actual)) (n + 1, actual) else nextPalin3(n + 1)
}
println(f"${"Decimal"}%18s${"Binary"}%35s${"Ternary"}%51s")
(Stream(0L) ++ st.map(_._2)).take(6).foreach(n => {
val bigN = BigInt(n)
val (bin, ter) = (bigN.toString(2), bigN.toString(3))
println(f"${n}%18d, ${
bin + " " * ((60 - bin.length) / 2)}%60s, ${
ter + " " * ((37 - ter.length) / 2)}%37s")
})
println(s"Successfully completed without errors. [total ${currentTime - executionStartTime} ms]")
}
Fastest and high yields (17) solution 😏
- Output:
Best seen running in your browser either by Scastie (remote JVM).
import scala.io.Source
object FastPalindrome23 extends App {
val rawText = Source.fromURL("http://oeis.org/A060792/b060792.txt")
var count = 0
rawText.getLines().map(_.split(" "))
.foreach(s => {
val n = BigInt(s(1))
val (bin, ter) = (n.toString(2), n.toString(3))
count += 1
println(
f"Decimal : ${n}%-44d , Central binary digit: ${bin(bin.length / 2)}")
println(f"Binary : ${bin}")
println(f"Ternary : ${ter + " " * ((91 - ter.length) / 2)}%91s")
println(f"Central : ${"^"}%46s%n---%n")
})
println(s"${count} palindromes found.")
}
- Output:
Decimal : 0 , Central binary digit: 0 Binary : 0 Ternary : 0 Central : ^ --- Decimal : 1 , Central binary digit: 1 Binary : 1 Ternary : 1 Central : ^ --- Decimal : 6643 , Central binary digit: 1 Binary : 1100111110011 Ternary : 100010001 Central : ^ --- Decimal : 1422773 , Central binary digit: 1 Binary : 101011011010110110101 Ternary : 2200021200022 Central : ^ --- Decimal : 5415589 , Central binary digit: 0 Binary : 10100101010001010100101 Ternary : 101012010210101 Central : ^ --- Decimal : 90396755477 , Central binary digit: 0 Binary : 1010100001100000100010000011000010101 Ternary : 22122022220102222022122 Central : ^ --- Decimal : 381920985378904469 , Central binary digit: 0 Binary : 10101001100110110110001110011011001110001101101100110010101 Ternary : 2112200222001222121212221002220022112 Central : ^ --- Decimal : 1922624336133018996235 , Central binary digit: 0 Binary : 11010000011100111000101110001110011011001110001110100011100111000001011 Ternary : 122120102102011212112010211212110201201021221 Central : ^ --- Decimal : 2004595370006815987563563 , Central binary digit: 1 Binary : 110101000011111010101010100101111011110111011110111101001010101010111110000101011 Ternary : 221010112100202002120002212200021200202001211010122 Central : ^ --- Decimal : 8022581057533823761829436662099 , Central binary digit: 1 Binary : 1100101010000100101101110000011011011111111011000011100001101111111101101100000111011010010000101010011 Ternary : 21000020210011222122220212010000100001021202222122211001202000012 Central : ^ --- Decimal : 392629621582222667733213907054116073 , Central binary digit: 0 Binary : 10010111001111000100010100010100000011011011000101011011100000111011010100011011011000000101000101000100011110011101001 Ternary : 122102120011102000101101000002010021111120010200000101101000201110021201221 Central : ^ --- Decimal : 32456836304775204439912231201966254787 , Central binary digit: 1 Binary : 11000011010101111010110010100010010011011010101001101000001000100010000010110010101011011001001000101001101011110101011000011 Ternary : 1222100201002211120110022121002012121101011212102001212200110211122001020012221 Central : ^ --- Decimal : 428027336071597254024922793107218595973 , Central binary digit: 1 Binary : 101000010000000110001000011111100101011110011100001110100011100010001110001011100001110011110101001111110000100011000000010000101 Ternary : 222001200110022102121001000200200202022111220202002002000100121201220011002100222 Central : ^ --- Decimal : 1597863243206403857787246920544522912361 , Central binary digit: 0 Binary : 10010110010000110010100010001000111010010000111000010010100010111011101000101001000011100001001011100010001000101001100001001101001 Ternary : 10121021220121202021201220210001211020122122102011210001202210212020212102212012101 Central : ^ --- Decimal : 30412638162199251273509758127730300026189 , Central binary digit: 0 Binary : 101100101011111111011010000110101101100101010100101101010000001000000000100000010101101001010101001101101011000010110111111110101001101 Ternary : 2112120210211212121000000011202022210210101010120122202021100000001212121120120212112 Central : ^ --- Decimal : 32345684491703244980406880704479906642045 , Central binary digit: 0 Binary : 101111100001110001100000011101111000001111001000110100111001010101101101010100111001011000100111100000111101110000001100011100001111101 Ternary : 2200221111210202000010122020002221112212101012122111222000202210100002020121111220022 Central : ^ --- Decimal : 24014998963383302600955162866787153652444049 , Central binary digit: 0 Binary : 1000100111010110110111101001100011100110100000011011001010100101011001100011001101010010101001101100000010110011100011001011110110110101110010001 Ternary : 2202021211210100110100002202101000110000220121210220000110001012022000010110010121121202022 Central : ^ --- 17 palindromes found.
Scheme
(import (scheme base)
(scheme write)
(srfi 1 lists)) ; use 'fold' from SRFI 1
;; convert number to a list of digits, in desired base
(define (r-number->list n base)
(let loop ((res '())
(num n))
(if (< num base)
(cons num res)
(loop (cons (remainder num base) res)
(quotient num base)))))
;; convert number to string, in desired base
(define (r-number->string n base)
(apply string-append
(map number->string
(r-number->list n base))))
;; test if a list of digits is a palindrome
(define (palindrome? lst)
(equal? lst (reverse lst)))
;; based on Perl/Ruby's insight
;; -- construct the ternary palindromes in order
;; using fact that their central number is always a 1
;; -- convert into binary, and test if result is a palindrome too
(define (get-series size)
(let loop ((results '(1 0))
(i 1))
(if (= size (length results))
(reverse results)
(let* ((n3 (r-number->list i 3))
(n3-list (append n3 (list 1) (reverse n3)))
(n10 (fold (lambda (d t) (+ d (* 3 t))) 0 n3-list))
(n2 (r-number->list n10 2)))
(loop (if (palindrome? n2)
(cons n10 results)
results)
(+ 1 i))))))
;; display final results, in bases 10, 2 and 3.
(for-each
(lambda (n)
(display
(string-append (number->string n)
" in base 2: "
(r-number->string n 2)
" in base 3: "
(r-number->string n 3)))
(newline))
(get-series 6))
- Output:
0 in base 2: 0 in base 3: 0 1 in base 2: 1 in base 3: 1 6643 in base 2: 1100111110011 in base 3: 100010001 1422773 in base 2: 101011011010110110101 in base 3: 2200021200022 5415589 in base 2: 10100101010001010100101 in base 3: 101012010210101 90396755477 in base 2: 1010100001100000100010000011000010101 in base 3: 22122022220102222022122
Sidef
var format = "%11s %24s %38s\n"
format.printf("decimal", "ternary", "binary")
format.printf(0, 0, 0)
for n in (0 .. 2e5) {
var pal = n.base(3)||''
var b3 = (pal + '1' + pal.flip)
var b2 = Num(b3, 3).base(2)
if (b2 == b2.flip) {
format.printf(Num(b2, 2), b3, b2)
}
}
- Output:
decimal ternary binary
0 0 0
1 1 1
6643 100010001 1100111110011
1422773 2200021200022 101011011010110110101
5415589 101012010210101 10100101010001010100101
90396755477 22122022220102222022122 1010100001100000100010000011000010101
Swift
import Foundation
func isPalin2(n: Int) -> Bool {
var x = 0
var n = n
guard n & 1 != 0 else {
return n == 0
}
while x < n {
x = x << 1 | n & 1
n >>= 1
}
return n == x || n == x >> 1
}
func reverse3(n: Int) -> Int {
var x = 0
var n = n
while n > 0 {
x = x * 3 + (n % 3)
n /= 3
}
return x
}
func printN(_ n: Int, base: Int) {
var n = n
print(" ", terminator: "")
repeat {
print("\(n % base)", terminator: "")
n /= base
} while n > 0
print("(\(base))", terminator: "")
}
func show(n: Int) {
print(n, terminator: "")
printN(n, base: 2)
printN(n, base: 3)
print()
}
private var count = 0
private var lo = 0
private var (hi, pow2, pow3) = (1, 1, 1)
show(n: 0)
while true {
var n: Int
for i in lo..<hi {
n = (i * 3 + 1) * pow3 + reverse3(n: i)
guard isPalin2(n: n) else {
continue
}
show(n: n)
count += 1
guard count < 7 else {
exit(0)
}
}
if hi == pow3 {
pow3 *= 3
} else {
pow2 *= 4
}
while true {
while pow2 <= pow3 {
pow2 *= 4
}
let lo2 = (pow2 / pow3 - 1) / 3
let hi2 = (pow2 * 2 / pow3 - 1) / 3 + 1
let lo3 = pow3 / 3
let hi3 = pow3
if lo2 >= hi3 {
pow3 *= 3
} else if lo3 >= hi2 {
pow2 *= 4
} else {
lo = max(lo2, lo3)
hi = min(hi2, hi3)
break
}
}
}
- Output:
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) 381920985378904469 10101001100110110110001110011011001110001101101100110010101(2) 2112200222001222121212221002220022112(3)
Tcl
We can use [format %b] to format a number as binary, but ternary requires a custom proc:
proc format_%t {n} {
while {$n} {
append r [expr {$n % 3}]
set n [expr {$n / 3}]
}
if {![info exists r]} {set r 0}
string reverse $r
}
Identifying palindromes is simple. This form is O(n) with a large constant factor, but good enough:
proc pal? {s} {expr {$s eq [string reverse $s]}}
The naive approach turns out to be very slow:
proc task {{find 6}} {
for {set i 0} {$find} {incr i} {
set b [format %b $i]
set t [format_%t $i]
if {[pal? $b] && [pal? $t]} {
puts "Palindrome: $i ($b) ($t)"
incr find -1
}
}
}
puts [time {task 4}]
- Output:
Palindrome: 0 (0) (0) Palindrome: 1 (1) (1) Palindrome: 6643 (1100111110011) (100010001) Palindrome: 1422773 (101011011010110110101) (2200021200022) 21944474 microseconds per iteration
22 seconds for only the first four elements .. not good enough! We can do much better than that by naively iterating the binary palindromes. This is nice to do in a coroutine:
package require Tcl 8.5 ;# for coroutines
proc 2pals {} {
yield 0
yield 1
while 1 {
incr i
set a [format %b $i]
set b [string reverse $a]
yield ${a}$b
yield ${a}0$b
yield ${a}1$b
}
}
The binary strings emitted by this generator are not in increasing order, but for this particular task, that turns out to be unimportant.
Our main loop needs only minor changes:
proc task {{find 6}} {
coroutine gen apply {{} {yield; 2pals}}
while {$find} {
set b [gen]
set i [scan $b %b]
set t [format_%t $i]
if {[pal? $t]} {
puts "Palindrome: $i ($b) ($t)"
incr find -1
}
}
rename gen {}
}
puts [time task]
This version finds the first 6 in under 4 seconds, which is good enough for the task at hand:
- Output:
Palindrome: 0 (0) (0) Palindrome: 1 (1) (1) Palindrome: 6643 (1100111110011) (100010001) Palindrome: 1422773 (101011011010110110101) (2200021200022) Palindrome: 5415589 (10100101010001010100101) (101012010210101) Palindrome: 90396755477 (1010100001100000100010000011000010101) (22122022220102222022122) 3643152 microseconds per iteration
Plenty more optimisations are possible! Exploiting the observations in Ruby's implementation should make the 7th element reachable in reasonable time ...
VBA
Public Declare Function GetTickCount Lib "kernel32.dll" () As Long
'palindromes both in base3 and base2
'using Decimal data type to find number 6 and 7, although slowly
Private Function DecimalToBinary(DecimalNum As Long) As String
Dim tmp As String
Dim n As Long
n = DecimalNum
tmp = Trim(CStr(n Mod 2))
n = n \ 2
Do While n <> 0
tmp = Trim(CStr(n Mod 2)) & tmp
n = n \ 2
Loop
DecimalToBinary = tmp
End Function
Function Dec2Bin(ByVal DecimalIn As Variant, _
Optional NumberOfBits As Variant) As String
Dec2Bin = ""
DecimalIn = Int(CDec(DecimalIn))
Do While DecimalIn <> 0
Dec2Bin = Format$(DecimalIn - 2 * Int(DecimalIn / 2)) & Dec2Bin
DecimalIn = Int(DecimalIn / 2)
Loop
If Not IsMissing(NumberOfBits) Then
If Len(Dec2Bin) > NumberOfBits Then
Dec2Bin = "Error - Number exceeds specified bit size"
Else
Dec2Bin = Right$(String$(NumberOfBits, _
"0") & Dec2Bin, NumberOfBits)
End If
End If
End Function
Public Sub base()
'count integer n from 0 upwards
'display representation in base 3
Time1 = GetTickCount
Dim n As Long
Dim three(19) As Integer
Dim pow3(19) As Variant
Dim full3 As Variant
Dim trail As Variant
Dim check As Long
Dim len3 As Integer
Dim carry As Boolean
Dim i As Integer, j As Integer
Dim s As String
Dim t As String
pow3(0) = CDec(1)
For i = 1 To 19
pow3(i) = 3 * pow3(i - 1)
Next i
Debug.Print String$(5, " "); "iter"; String$(7, " "); "decimal"; String$(18, " "); "binary";
Debug.Print String$(30, " "); "ternary"
n = 0: full3 = 0: t = "0": s = "0"
Debug.Print String$(8 - Len(CStr(n)), " "); n; String$(12 - Len(CStr(full3)), " ");
Debug.Print full3; String$((41 - Len(t)) / 2, " "); t; String$((41 - Len(t)) / 2, " ");
Debug.Print String$((31 - Len(s)) / 2, " "); s
n = 0: full3 = 1: t = "1": s = "1"
Debug.Print String$(8 - Len(CStr(n)), " "); n; String$(12 - Len(CStr(full3)), " ");
Debug.Print full3; String$((41 - Len(t)) / 2, " "); t; String$((41 - Len(t)) / 2, " ");
Debug.Print String$((31 - Len(s)) / 2, " "); s
number = 0
n = 1
len3 = 0
full3 = 3
Do 'For n = 1 To 200000 '20000000 takes 1000 seconds and number 7 not found yet
three(0) = three(0) + 1
carry = False
If three(0) = 3 Then
three(0) = 0
carry = True
j = 1
Do While carry
three(j) = three(j) + 1
If three(j) = 3 Then
three(j) = 0
j = j + 1
Else
carry = False
End If
Loop
If len3 < j Then
trail = full3 - (n - 1) * pow3(len3 + 2) - pow3(len3 + 1)
len3 = j
full3 = n * pow3(len3 + 2) + pow3(len3 + 1) + 3 * trail
For i = 0 To j - 1
full3 = full3 - 2 * pow3(len3 - i)
Next i
full3 = full3 + 1 'as j=len3 now and 1=pow3(len3 - j)
Else
full3 = full3 + pow3(len3 + 2)
For i = 0 To j - 1
full3 = full3 - 2 * pow3(len3 - i)
Next i
full3 = full3 + pow3(len3 - j)
End If
Else
full3 = full3 + pow3(len3 + 2) + pow3(len3)
End If
s = ""
For i = 0 To len3
s = s & CStr(three(i))
Next i
'do we have a hit?
t = Dec2Bin(full3) 'CStr(DecimalToBinary(full3))
If t = StrReverse(t) Then
'we have a hit
number = number + 1
s = StrReverse(s) & "1" & s
If n < 200000 Then
Debug.Print String$(8 - Len(CStr(n)), " "); n; String$(12 - Len(CStr(full3)), " ");
Debug.Print full3; String$((41 - Len(t)) / 2, " "); t; String$((41 - Len(t)) / 2, " ");
Debug.Print String$((31 - Len(s)) / 2, " "); s
If number = 4 Then
Debug.Print "Completed in"; (GetTickCount - Time1) / 1000; "seconds"
Time2 = GetTickCount
Application.ScreenUpdating = False
End If
Else
Debug.Print n, full3, Len(t), t, Len(s), s
Debug.Print "Completed in"; (Time2 - Time1) / 1000; "seconds";
Time3 = GetTickCount
End If
End If
n = n + 1
Loop Until number = 5 'Next n
Debug.Print "Completed in"; (Time3 - Time1) / 1000; "seconds"
Application.ScreenUpdating = True
End Sub
- Output:
' iter decimal binary ternary' 0 0 0 0 ' 0 1 1 1 ' 27 6643 1100111110011 100010001 ' 650 1422773 101011011010110110101 2200021200022 ' 825 5415589 10100101010001010100101 101012010210101 ' 170097 90396755477 1010100001100000100010000011000010101 22122022220102222022122 'Completed in 5,14 seconds ' 328601606 381920985378904469 59 10101001100110110110001110011011001110001101101100110010101 37 2112200222001222121212221002220022112 Completed in 5,14 secondsCompleted in 16394,64 seconds
Wren
Just the first 6 palindromes as the 7th is too large for Wren to process without resorting to BigInts.
import "./fmt" for Fmt
var isPalindrome2 = Fn.new { |n|
var x = 0
if (n % 2 == 0) return n == 0
while (x < n) {
x = x*2 + (n%2)
n = (n/2).floor
}
return n == x || n == (x/2).floor
}
var reverse3 = Fn.new { |n|
var x = 0
while (n != 0) {
x = x*3 + (n%3)
n = (n/3).floor
}
return x
}
var start
var show = Fn.new { |n|
Fmt.print("Decimal : $d", n)
Fmt.print("Binary : $b", n)
Fmt.print("Ternary : $t", n)
Fmt.print("Time : $0.3f ms\n", (System.clock - start)*1000)
}
var min = Fn.new { |a, b| (a < b) ? a : b }
var max = Fn.new { |a, b| (a > b) ? a : b }
start = System.clock
System.print("The first 6 numbers which are palindromic in both binary and ternary are :\n")
show.call(0)
var cnt = 1
var lo = 0
var hi = 1
var pow2 = 1
var pow3 = 1
while (true) {
var i = lo
while (i < hi) {
var n = (i*3+1)*pow3 + reverse3.call(i)
if (isPalindrome2.call(n)) {
show.call(n)
cnt = cnt + 1
if (cnt >= 6) return
}
i = i + 1
}
if (i == pow3) {
pow3 = pow3 * 3
} else {
pow2 = pow2 * 4
}
while (true) {
while (pow2 <= pow3) pow2 = pow2 * 4
var lo2 = (((pow2/pow3).floor - 1)/3).floor
var hi2 = (((pow2*2/pow3).floor-1)/3).floor + 1
var lo3 = (pow3/3).floor
var hi3 = pow3
if (lo2 >= hi3) {
pow3 = pow3 * 3
} else if (lo3 >= hi2) {
pow2 = pow2 * 4
} else {
lo = max.call(lo2, lo3)
hi = min.call(hi2, hi3)
break
}
}
}
- Output:
The first 6 numbers which are palindromic in both binary and ternary are : Decimal : 0 Binary : 0 Ternary : 0 Time : 0.054 ms Decimal : 1 Binary : 1 Ternary : 1 Time : 0.147 ms Decimal : 6643 Binary : 1100111110011 Ternary : 100010001 Time : 0.290 ms Decimal : 1422773 Binary : 101011011010110110101 Ternary : 2200021200022 Time : 0.919 ms Decimal : 5415589 Binary : 10100101010001010100101 Ternary : 101012010210101 Time : 1.408 ms Decimal : 90396755477 Binary : 1010100001100000100010000011000010101 Ternary : 22122022220102222022122 Time : 184.314 ms
zkl
VERY slow after six but does find it.
fcn pal23W{ //--> iterator returning (index,palindromic number)
Walker.tweak(fcn(ri,r){ // references to loop start and count of palindromes
foreach i in ([ri.value..*]){
n3:=i.toString(3);
n:=String(n3,"1",n3.reverse()).toInt(3); // create base 3 palindrome
n2:= n.toString(2);
if(n2.len().isOdd and n2==n2.reverse()){ // stop here, return answer
ri.set(i+1); // continue loop from this value at next iteration
return(r.inc(),n);
}
}
}.fp(Ref(3),Ref(3))).push(T(1,0),T(2,1)) // seed with first two results
}foreach idx,n in (pal23W().walk(6)){
println("%2d: %,d == %.3B(3) == %.2B(2)".fmt(idx,n,n,n))
}- Output:
1: 0 == 0(3) == 0(2) 2: 1 == 1(3) == 1(2) 3: 6,643 == 100010001(3) == 1100111110011(2) 4: 1,422,773 == 2200021200022(3) == 101011011010110110101(2) 5: 5,415,589 == 101012010210101(3) == 10100101010001010100101(2) 6: 90,396,755,477 == 22122022220102222022122(3) == 1010100001100000100010000011000010101(2)
Categories:
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