Comments on: The Sum Of Two Squares http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/ A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer Sat, 11 Feb 2012 09:48:16 +0000 hourly 1 http://wordpress.com/ By: Drake http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-4126 Sat, 31 Dec 2011 00:40:35 +0000 http://programmingpraxis.com/?p=1856#comment-4126 At the end of the chapter Dijkstra includes this: “Note. Obvious improvements, such as testing whether r mod 4 =3, and exploiting
the recurrence relation (x+1)^2 = x^2 + (2x +1) are left as exercises.”

]]> By: Stephen Walcher http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-2285 Sat, 08 Jan 2011 20:31:49 +0000 http://programmingpraxis.com/?p=1856#comment-2285 Hey guys. Was looking at this code for an example. Ended up moving on but not before porting one of the functions over to PHP. Hope someone can use eventually.

<?php
function squares($n) {
	$pairs = array();

	$x = sqrt($n);

	while ( $x-- > sqrt($n / 2) ) {
		$y = sqrt($n - $x * $x);

		while ($x * $x + $y * $y <= $n) {
			if ($y > $x)
				break;
			if ($x * $x + $y * $y == $n)
				array_push($pairs, array($x, $y));
			++$y;
		}
	}

	echo sprintf("%d -- Pairs: %d\r\n", $n, count($pairs));
}

squares(50);
squares(48612265);
?>
]]> By: Mike http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-1142 Fri, 09 Apr 2010 02:45:55 +0000 http://programmingpraxis.com/?p=1856#comment-1142 Because x >= y, decrementing x always causes a bigger change in the sum than incrementing y. So the loop in the my routine above can be shortened to:

def sumofsquares(n):
”’Generate (x,y) pairs, where x*x + y*y == n and x >= y >= 0.

Find max x, such that x*x <= n. Then loop, comparing y*y + x*x to n.
If the sum is too small, increment y; if too big, decrement x.

"sos" is the sum of the squares.
'''

x, y, sos = 0, 0, 0
while sos = y:
while sos < n:
sos += 2*y + 1
y += 1

if sos == n:
yield x,y

sos -= 2*x – 1
x -= 1

]]> By: Mike http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-1141 Fri, 09 Apr 2010 01:07:05 +0000 http://programmingpraxis.com/?p=1856#comment-1141 Here is a solution that only uses adds and multiply by 2 (could use a shift).

def sumofsquares(n):
    '''Generate (x,y) pairs, where x*x + y*y == n and x >= y >= 0.

    Find max x, such that x*x <= n.  Then loop, comparing y*y + x*x to n.
    If the sum is too small, increment y; if too big, decrement x.
    '''
    x, y, sos = 0, 0, 0
    while sos < n:
        sos += 2*x + 1
        x += 1

    while x >= y:
        if sos < n:
            sos += 2*y + 1
            y += 1

        elif sos > n:
            sos -= 2*x - 1
            x -= 1

        else:
            yield x,y
            sos += 2*( y -x + 1 )
            y += 1
            x -= 1
]]> By: Jebb http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-957 Sun, 24 Jan 2010 17:11:33 +0000 http://programmingpraxis.com/?p=1856#comment-957 I’ve tried to pay extra attention to the boundaries of both while loops, to avoid any unnecessary iteration. I’m not convinced it’s optimal, though: I don’t particularly like the “if (y > x) break;”.

#include <stdio.h>
#include <math.h>

int main()
{
    unsigned int n, x, y;
    printf("Enter the integer n:\n");
    scanf("%d", &n);
    x = sqrt(n);
    while (x-- > sqrt(n / 2)) {
        y = sqrt(n - x * x);
        while (x * x + y * y <= n) {
            if (y > x)
                break;
            if (x * x + y * y == n)
                printf("%d %d\n", x, y);
            ++y;
        }
    }
    return 0;
}
]]> By: Jamie Hope http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-889 Fri, 08 Jan 2010 16:40:31 +0000 http://programmingpraxis.com/?p=1856#comment-889 Oh, I see. It should be “language” and not “lang”. But the HOWTO page says “lang”:
http://programmingpraxis.com/contents/howto-posting-source-code/

Please fix, as I’ll probably forget this by the time I go to post source code again. Thanks!

]]> By: Kevin http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-886 Fri, 08 Jan 2010 07:21:44 +0000 http://programmingpraxis.com/?p=1856#comment-886 OH CRAP I’m so sorry for being a noob. I JUST read the “Posting Source Code” page. So sorry!!

#include<iostream>
#include<cmath>
using namespace std;
int main(void)
{
	int n, i=0;
	cout << "Integer? ";
	cin >> n;
	double temp_frac, a, b;
	do
	{
		a = pow(i,2);
		b = sqrt(n-a);
		if( a > n )
			break;
		else if( i > b )
			break;
		else if( modf(b, &temp_frac) == 0 )
			cout << i << " , " << b << endl;
		++i;
	}
	while(1);
	return 0;
}
]]> By: Kevin http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-885 Fri, 08 Jan 2010 07:17:48 +0000 http://programmingpraxis.com/?p=1856#comment-885 Here’s mine in C++…it’s somewhat bruteforce. If there are no solutions it wont give any output. Please, gimme some feedback! I’m a student that’s trying to practice/learn and get a better feel of C++. I wanna make it my goto language :D

#include
#include
using namespace std;
int main(void)
{
int n, i=0;
cout <> n;
double temp_frac, a, b;
do
{
a = pow(i,2);
b = sqrt(n-a);
if( a > n )
break;
else if( i > b )
break;
else if( modf(b, &temp_frac) == 0 )
cout << i << " , " << b << endl;
++i;
}
while(1);
return 0;
}

]]> By: programmingpraxis http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-883 Thu, 07 Jan 2010 23:42:00 +0000 http://programmingpraxis.com/?p=1856#comment-883 Andrew, Jaime: I fixed your comments so the source code is properly formatted.

]]> By: David Humphreys http://programmingpraxis.com/2010/01/05/the-sum-of-two-squares/#comment-882 Thu, 07 Jan 2010 23:40:12 +0000 http://programmingpraxis.com/?p=1856#comment-882 Andrew: see http://en.support.wordpress.com/code/posting-source-code/
Jamie: I think it should be language, not lang.

A solution in Clojure. It works fine (but slowly) for large values. I’m sure there are more elegant ways of doing it.

(ns Sum-of-Squares)
(defmacro add-squares [x y] `(+ (* ~x ~x) (* ~y ~y)))
(defn test-single-square [x y n result]
 (if (< x y)
  result
  (let [sum (add-squares x y)]
   (cond
    (< sum n)	#(test-single-square x (inc y) n result)
    (> sum n)	#(test-single-square (dec x) y n result)
    :else	#(test-single-square (dec x) (inc y) n (conj result [x y]))))))
(defn #^{:test (fn []
 (let [check-values (fn [test-value]
  (assert (every?
   #(= test-value
    (add-squares (first %) (last %)))
    (find-squares test-value))))]
 (map check-values [1790119876545 10000798002 48612265 999 50])))}
 find-squares [n]
  (let [x (int (java.lang.Math/sqrt n)) y 0]
 (trampoline (test-single-square x y n []))))
(test #'find-squares)
]]>