cln/tests/exam_I_sqrtp.cc

#include "test_I.h"
#define floor(a,b)  ((a) / (b))

int test_sqrtp (void)
{
        int error = 0;
        // As our algorithm for sqrtp does pre-filtering mod (64*63*65*11),
        // we check that no square is thrown away by this pre-filtering.
        {
                uintL a, b, c, d;
                for (a = 0; a <= floor(64,2); a++)
                  for (b = 0; b <= floor(63,2); b++)
                    for (c = 0; c <= floor(65,2); c++)
                      for (d = 0; d <= floor(11,2); d++) {
                        // Produce a number x == a mod 64, x == b mod 63,
                        // x == c mod 65, x == d mod 11, and square it.
                        uintL x = 1306305*a + 1967680*b + 133056*c + 2358720*d;
                        x = x % 2882880; // not strictly necessary
                        cl_I w;
                        ASSERT4(sqrtp(expt_pos(x,2),&w) && w == x, a,b,c,d);
                }
        }
        return error;
}
1	kreckel	1.1	#include "test_I.h"
2			#define floor(a,b) ((a) / (b))
3
4			int test_sqrtp (void)
5			{
6			int error = 0;
7			// As our algorithm for sqrtp does pre-filtering mod (646365*11),
8			// we check that no square is thrown away by this pre-filtering.
9			{
10			uintL a, b, c, d;
11			for (a = 0; a <= floor(64,2); a++)
12			for (b = 0; b <= floor(63,2); b++)
13			for (c = 0; c <= floor(65,2); c++)
14			for (d = 0; d <= floor(11,2); d++) {
15			// Produce a number x == a mod 64, x == b mod 63,
16			// x == c mod 65, x == d mod 11, and square it.
17			uintL x = 1306305a + 1967680b + 133056c + 2358720d;
18			x = x % 2882880; // not strictly necessary
19			cl_I w;
20			ASSERT4(sqrtp(expt_pos(x,2),&w) && w == x, a,b,c,d);
21			}
22			}
23			return error;
24			}
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