Largest palindrome product: Difference between revisions

{{trans|Wren}}

 

V r = Int64(0)

L n > 0

k -= 2

I nextN

 

{{out}}

=={{header|ALGOL 68}}==

{{Trans|Wren}}

# returns n with the digits reversed #

PROC reverse = ( LONG INT v )LONG INT:

OD

OD

 

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{{Trans|Ring}}

Also showing the maximum for 2 and 4 .. 7 digit numbers. Tests for a better product before testing for palindromicity.

PROC is pal = ( LONG INT n )BOOL:

BEGIN

limit start *:= 10

OD

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<pre>

 

=={{header|AWK}}==

# syntax: GAWK -f LARGEST_PALINDROME_PRODUCT.AWK

BEGIN {

return(rts)

}

{{out}}

<pre>

 

=={{header|Ksh}}==

#!/bin/ksh

 

 

print "Largest palindrome product of two 3-digit factors = ${MAXPPROD}"

{{out}}<pre>

Largest palindrome product of two 3-digit factors = 906609</pre>

===Main Task===

{{trans|Paper & Pencil}}

class Program {

 

Console.Write("{0} {1} μs", bs, sw.Elapsed.TotalMilliseconds * 1000.0);

}

{{out|Output @ Tio.run}}

<pre>913 x 993 = 906609 245.2 μs</pre>

 

===Stretch===

 

class Program {

Console.Write("{0} sec", sw.Elapsed.TotalSeconds);

}

{{out|Output @ Tio.run}}

Showing results for 2 through 10 digit factors.

 

=={{header|F_Sharp|F#}}==

// Largest palindrome product. Nigel Galloway: November 3rd., 2021

let fN g=let rec fN g=[yield g%10; if g>=10 then yield! fN(g/10)] in let n=fN g in n=List.rev n

printfn "%d" ([for n in 100..999 do for g in n..999->n*g]|>List.filter fN|>List.max)

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<pre>

=={{header|FreeBASIC}}==

===Version 1===

'turn a number into a palindrom with twice as many digits

dim as string ns, ret

next n

nextd: 'yes, I used a goto. sue me.

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<pre>99 * 91 = 9009

===Version 2===

This version is based on Version 1 with only a few changes and some extra code based on the fact that one divisor can be divided by 11, this speeds it even more up and a option for using goto, exit or continue. highest n is 9, (highest possible for unsigned 64bit integers).

' compile with: fbc -s console

 

Print : Print "hit any key to end program"

Sleep

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<pre>99 * 91 = 9009

{{trans|Wren}}

18 digit integers are within the range of Go's uint64 type though finding the result for 9-digit number products takes a while - around 15 seconds on my machine.

 

import "fmt"

}

}

 

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{{works with|jq}}

'''Works with gojq, the Go implementation of jq'''

tostring|explode|reverse|implode|tonumber;

.emit ) ;

 

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<pre>

 

=={{header|Julia}}==

 

function twoprodpal(factorlength)

twoprodpal(i)

end

<pre>

For factor length 2, 91 * 99 = 9009

=== Faster version ===

{{trans|Python}}

 

const T = [Set([(0, 0)])]

largestpalindrome(k)

end

<pre>

1 9 (9, 1) 9

 

=={{header|Mathematica}} / {{header|Wolfram Language}}==

Block[{digits = IntegerDigits@n}, digits == Reverse[digits]]

 

 

Grid[Join[{{"factors", "largest palindrome"}}, {#, Times @@ #} & /@

 

{{out}}<pre>

 

=={{header|Paper & Pencil}}==

 

lowest possible 6 digit palindrome starting with 9 is 900009

979 x 931 = 911449‬, not a palindrome, so continue

979 x 921 = 901659‬, not a palindrome, and less than the best found so far, so stop

 

=={{header|Perl}}==

{{libheader|ntheory}}

use warnings;

use feature 'say';

say "Largest palindromic product of two @{[$l]}-digit integers: $f[1] × $f[0] = $p" and last LOOP;

}

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<pre>Largest palindromic product of two 2-digit integers: 91 × 99 = 9009

and further optimised as per the C# comments (on this very rosettacode page).

You can run this online [http://phix.x10.mx/p2js/Largest_palindrome_product.htm here].

<span style="color: #000080;font-style:italic;">-- demo\rosetta\Largest_palindrome_product.exw</span>

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">fmt</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">sp</span><span style="color: #0000FF;">,</span><span style="color: #000000;">sx</span><span style="color: #0000FF;">,</span><span style="color: #000000;">sy</span><span style="color: #0000FF;">,</span><span style="color: #000000;">e</span><span style="color: #0000FF;">})</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>

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<pre>

=={{header|Python}}==

Original author credit to user peijunz at Leetcode.

 

T=[set([(0, 0)])]

for k in range(1, 14):

largestPalindrome(k)

<pre>

1 9 (9, 1) 9

=={{header|Raku}}==

 

my $p5 = Inline::Perl5.new();

$p5.use: 'ntheory';

}

$p

<pre>Largest palindromic product of two 2-digit integers: 91 × 99 = 9009

Largest palindromic product of two 3-digit integers: 913 × 993 = 906609

 

=={{header|Ring}}==

 

prod = 1

ok

end

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<pre>working...

 

=={{header|Sidef}}==

 

for k in ((10**n - 1) `downto` 10**(n-1)) {

var (a,b) = largest_palindrome_product(n)

say "Largest palindromic product of two #{n}-digit integers: #{a} * #{b} = #{a*b}"

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<pre>

=={{header|Wren}}==

The approach here is to manufacture palindromic numbers of length 2n in decreasing order and then see if they're products of two n-digit numbers.

var r = 0

while (n > 0) {

if (nextN) break

}

 

{{out}}

 

=={{header|XPL0}}==

int A, B;

[B:= 0;

];

IntOut(0, Max);

 

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