I'm working on modernizing Rosetta Code's infrastructure. Starting with communications. Please accept this time-limited open invite to RC's Slack.. --Michael Mol (talk) 20:59, 30 May 2020 (UTC)

Find square difference

From Rosetta Code
Find square difference is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task


Find and show on this page the least positive integer number n, where difference of n*n and (n-1)*(n-1) greater than 1000.
The result is 501 because 501*501 - 500*500 = 251001 - 250000 = 1001 > 1000.



ALGOL 68[edit]

Using the same school maths ( or math for those in the US ) in the Wren version but using a calculation.
Also showng the least positive integer where the difference between n^2 and (n-1)^2 is greater than 32 000 and 2 000 000 000 - using only 32 bit integers.

BEGIN # find the lowest positive n where the difference between n^2 and (n-1)^2 is > 1000 #
INT required diff = 1 000;
INT medium diff = 32 000;
INT larger diff = 2 000 000 000;
# return s space-separated into groups of 3 digits #
PROC space separate = ( STRING unformatted )STRING:
BEGIN
STRING result := "";
INT ch count := 0;
FOR c FROM UPB unformatted BY -1 TO LWB unformatted DO
IF ch count <= 2 THEN ch count +:= 1
ELSE ch count := 1; " " +=: result
FI;
unformatted[ c ] +=: result
OD;
result
END # space separate # ;
# n^2 - ( n - 1 )^2 is n^2 - n^2 + 2n - 1, i.e. 2n - 1 #
# so 2n - 1 > 1000 or n > 1001/2 #
print( ( "Smallest n where n^2 - (n-1)^2 is > ", space separate( whole( required diff, -12 ) ), ": "
, space separate( whole( ( ( required diff + 1 ) OVER 2 ) + 1, -12 ) )
, newline
)
);
print( ( "Smallest n where n^2 - (n-1)^2 is > ", space separate( whole( medium diff, -12 ) ), ": "
, space separate( whole( ( ( medium diff + 1 ) OVER 2 ) + 1, -12 ) )
, newline
)
);
print( ( "Smallest n where n^2 - (n-1)^2 is > ", space separate( whole( larger diff, -12 ) ), ": "
, space separate( whole( ( ( larger diff + 1 ) OVER 2 ) + 1, -12 ) )
, newline
)
)
END
Output:
Smallest n where n^2 - (n-1)^2 is >           1 000:             501
Smallest n where n^2 - (n-1)^2 is >          32 000:          16 001
Smallest n where n^2 - (n-1)^2 is >   2 000 000 000:   1 000 000 001

AWK[edit]

 
# syntax: GAWK -f FIND_SQUARE_DIFFERENCE.AWK
BEGIN {
n = 1001
while (i^2-(i-1)^2 < n) {
i++
}
printf("%d\n",i)
exit(0)
}
 
Output:
501

C[edit]

#include<stdio.h>
#include<stdlib.h>
 
int f(int n) {
int i, i1;
for(i=1;i*i-i1*i1<n;i1=i, i++);
return i;
}
 
int main(void) {
printf( "%d\n", f(1000) );
return 0;
}
Output:
501

C++[edit]

The C solution is also idomatic in C++. An alterate approach is to use Ranges from C++20.

#include <iostream>
#include <ranges>
 
int main()
{
const int maxSquareDiff = 1000;
auto squareCheck = [maxSquareDiff](int i){return 2 * i - 1 > maxSquareDiff;};
auto answer = std::views::iota(1) | // {1, 2, 3, 4, 5, ....}
std::views::filter(squareCheck) | // filter out the ones that are below 1000
std::views::take(1); // take the first one
std::cout << answer.front() << '\n';
}
 
Output:
501

Dart[edit]

import 'dart:math';
 
int leastSquare(int gap) {
for (int n = 1;; n++) {
if (pow(n, 2) - pow((n - 1), 2) > gap) {
return n;
}
}
}
 
void main() {
print(leastSquare(1000));
}
Output:
501

F#[edit]

 
let n=1000 in printfn $"%d{((n+1)/2)+1}"
 
Output:
501

Factor[edit]

The difference between squares is the odd numbers, so ls(n)=⌈n/2+1⌉.

Works with: Factor version 0.99 2021-06-02
USING: math math.functions prettyprint ;
 
: least-sq ( m -- n ) 2 / 1 + ceiling ;
 
1000 least-sq .
Output:
501

Fermat[edit]

Func F(n) =
i:=0;
while i^2-(i-1)^2<n do i:=i+1 od; i.;
 
!!F(1000);
Output:
501

FreeBASIC[edit]

function fpow(n as uinteger) as uinteger
dim as uinteger i
while i^2-(i-1)^2 < n
i+=1
wend
return i
end function
 
print fpow(1001)
Output:
501

Julia[edit]

julia> findfirst(n -> n*n - (n-1)*(n-1) > 1000, 1:1_000_000)
501
 

Pari/GP[edit]

F(n)=i=0;while(i^2-(i-1)^2<n,i=i+1);return(i);
print(F(1000))
Output:
501

Pencil and Paper[edit]

Find the smallest positive integer number n, where the difference of n2 and (n - 1)2 is greater than 1000.

r: roots of squares
s: successive squares
d: differences between successive squares,
   (a.k.a, the list of positive odd integers)

r: 0, 1, 2, 3,  4,  5,  6,  7,  8,  9,  10, ...
s: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ...
d: 1, 3, 5, 7,  9, 11, 13, 15, 17, 19, ...

r: n
s: n * n
d: n * 2 + 1

solve for d > 1,000
the first odd integer greater than 1,000 is 1,001
(1,001 + 1) / 2 = 501 ( = n)

Perl[edit]

#!/usr/bin/perl
 
use strict; # https://rosettacode.org/wiki/Least_square
use warnings;
 
my $n = 1;
$n++ until $n ** 2 - ($n-1) ** 2 > 1000;
print "$n\n";
Output:
501

PL/M[edit]

This can be compiled with the original 8080 PL/M compiler and run under CP/M or an emulator.
Note that the original 8080 PL/M compiler only supports 8 and 16 bit unisgned numbers.

100H: /* FIND THE LEAST +VE N WHERE N SQUARED - (N-1) SQUARED > 1000 */
 
BDOS: PROCEDURE( FN, ARG ); /* CP/M BDOS SYSTEM CALL */
DECLARE FN BYTE, ARG ADDRESS;
GOTO 5;
END BDOS;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N );
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR( 6 ) BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
 
PRINT$LEAST: PROCEDURE( DIFF );
DECLARE DIFF ADDRESS;
CALL PR$STRING( . 'THE LOWEST N WHERE THE SQUARES OF N AND N-1 $' );
CALL PR$STRING( . 'DIFFER BY MORE THAN $' );
CALL PR$NUMBER( DIFF );
CALL PR$STRING( .' IS: $' );
CALL PR$NUMBER( ( ( DIFF + 1 ) / 2 ) + 1 );
CALL PR$NL;
END PRINT$LEAST ;
CALL PRINT$LEAST( 1000 );
CALL PRINT$LEAST( 32000 );
CALL PRINT$LEAST( 65000 );
 
EOF
Output:
THE LOWEST N WHERE THE SQUARES OF N AND N-1 DIFFER BY MORE THAN 1000 IS: 501
THE LOWEST N WHERE THE SQUARES OF N AND N-1 DIFFER BY MORE THAN 32000 IS: 16001
THE LOWEST N WHERE THE SQUARES OF N AND N-1 DIFFER BY MORE THAN 65000 IS: 32501

Phix[edit]

Essentially Wren equivalent, but explained in excruciating detail especially for enyone that evidently needs elp, said Eeyore.

with javascript_semantics
printf(1,"""
n*n - (n - 1)*(n - 1) > 1000
n*n - (n*n - 2*n + 1) > 1000
n*n - n*n + 2*n - 1 > 1000
2*n - 1 > 1000
2*n > 1001
n > 500.5
n = %d
""",ceil(500.5))
Output:
n*n - (n - 1)*(n - 1) > 1000
n*n - (n*n - 2*n + 1) > 1000
n*n - n*n + 2*n - 1 > 1000
2*n - 1 > 1000
2*n > 1001
n > 500.5
n = 501

Python[edit]

 
import math
print("working...")
limit1 = 6000
limit2 = 1000
oldSquare = 0
newSquare = 0
 
for n in range(limit1):
newSquare = n*n
if (newSquare - oldSquare > limit2):
print("Least number is = ", end = "");
print(int(math.sqrt(newSquare)))
break
oldSquare = n*n
 
print("done...")
print()
 
Output:
working...
Least number is = 501
done...

Raku[edit]

say first { $_² - ($_-1)² > 1000 }, ^Inf;
Output:
501

Ring[edit]

 
load "stdlib.ring"
see "working..." + nl
limit1 = 6000
limit2 = 1000
oldPrime = 0
newPrime = 0
 
for n = 1 to limit1
newPrime = n*n
if newPrime - oldPrime > limit2
see "Latest number is = " + sqrt(newPrime) + nl
exit
ok
oldPrime = n*n
next
 
see "done..." + nl
 
Output:
working...
Latest number is = 501
done...

Wren[edit]

The solution n for some non-negative integer k needs to be such that: n² - (n² - 2n + 1) > k which reduces to: n > (k + 1)/2.

var squareDiff = Fn.new { |k| ((k + 1) * 0.5).ceil }
System.print(squareDiff.call(1000))
Output:
501

XPL0[edit]

n^2 - (n - 1)^2 > 1000
n^2 - (n^2 - 2n + 1) > 1000
2n - 1 > 1000
2n > 1001
n > 500.5
n = 501