gmp_gcdext
(PHP 4 >= 4.0.4, PHP 5)
gmp_gcdext — Calculate GCD and multipliers
Описание
array gmp_gcdext
( resource $a
, resource $b
)
Calculates g, s, and t, such that a*s + b*t = g = gcd(a,b), where gcd is the greatest common divisor. Returns an array with respective elements g, s and t.
This function can be used to solve linear Diophantine equations in two variables. These are equations that allow only integer solutions and have the form: a*x + b*y = c. For more information, go to the » "Diophantine Equation" page at MathWorld
Список параметров
Возвращаемые значения
An array of GMP numbers.
Примеры
Пример #1 Solving a linear Diophantine equation
<?php
// Solve the equation a*s + b*t = g
// where a = 12, b = 21, g = gcd(12, 21) = 3
$a = gmp_init(12);
$b = gmp_init(21);
$g = gmp_gcd($a, $b);
$r = gmp_gcdext($a, $b);
$check_gcd = (gmp_strval($g) == gmp_strval($r['g']));
$eq_res = gmp_add(gmp_mul($a, $r['s']), gmp_mul($b, $r['t']));
$check_res = (gmp_strval($g) == gmp_strval($eq_res));
if ($check_gcd && $check_res) {
$fmt = "Solution: %d*%d + %d*%d = %d\n";
printf($fmt, gmp_strval($a), gmp_strval($r['s']), gmp_strval($b),
gmp_strval($r['t']), gmp_strval($r['g']));
} else {
echo "Error while solving the equation\n";
}
// output: Solution: 12*2 + 21*-1 = 3
?>
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- GNU Вычисления с увеличенной точностью
- gmp_abs
- gmp_add
- gmp_and
- gmp_clrbit
- gmp_cmp
- gmp_com
- gmp_div_q
- gmp_div_qr
- gmp_div_r
- gmp_div
- gmp_divexact
- gmp_export
- gmp_fact
- gmp_gcd
- gmp_gcdext
- gmp_hamdist
- gmp_import
- gmp_init
- gmp_intval
- gmp_invert
- gmp_jacobi
- gmp_legendre
- gmp_mod
- gmp_mul
- gmp_neg
- gmp_nextprime
- gmp_or
- gmp_perfect_square
- gmp_popcount
- gmp_pow
- gmp_powm
- gmp_prob_prime
- gmp_random_bits
- gmp_random_range
- gmp_random_seed
- gmp_random
- gmp_root
- gmp_rootrem
- gmp_scan0
- gmp_scan1
- gmp_setbit
- gmp_sign
- gmp_sqrt
- gmp_sqrtrem
- gmp_strval
- gmp_sub
- gmp_testbit
- gmp_xor
Коментарии
The extended GCD can be used to calculate mutual modular inverses of two
coprime numbers. Internally gmp_invert uses this extended GCD routine,
but effectively throws away one of the inverses.
If gcd(a,b)=1, then r.a+s.b=1
Therefore r.a == 1 (mod s) and s.b == 1 (mod r)
Note that one of r and s will be negative, and so you'll want to
canonicalise it.