macocci7/php-math-integer

PHP 数论数学库（仅限自然数）。

维护者

详细信息

github.com/macocci7/PHP-Math-Integer

源代码

问题

安装: 12

依赖关系: 0

建议者: 0

安全: 0

星级: 1

关注者: 1

分支: 0

开放问题: 0

1.1.3 2024-04-18 01:58 UTC

Requires

  • php: >=8.1

Requires (Dev)

MIT 238f8d42a30e9e56431499fc30d6960e9710385e

  • macocci7 <macocci7.woop@yahoo.co.jp>

This package is auto-updated.

Last update: 2024-09-18 02:54:10 UTC

README

1. 功能

PHP-Math-Integer 是一个 PHP 库，用于处理数论主题（仅限自然数）。

可用主题

  • 数的初步知识
  • 质数的初步知识
  • 约数的初步知识
  • 倍数的初步知识
  • 欧几里得算法的初步知识
  • 普通分数的初步知识
  • 贝祖定理的初步知识

2. 内容

3. 要求

4. 安装

composer require macocci7/php-math-integer

5. 使用方法

5.1. Macocci7\PhpMathInteger\Number

此类处理数的初步知识。

  • PHP

    <?php

require_once __DIR__ . '/../vendor/autoload.php';

use Macocci7\PhpMathInteger\Number;

$n = new Number();

echo "Is 1 int? - " . ($n->isInt(1) ? 'Yes' : 'No') . ".\n";
echo "Are [ 1, 2, 3, ] all int? - " . ($n->isIntAll([ 1, 2, 3, ]) ? 'Yes' : 'No') . ".\n";
echo "Is 1 natural? - " . ($n->isNatural(1) ? 'Yes' : 'No') . ".\n";
echo "Are [ 1, 2, 3, ] all natural? - " . ($n->isNaturalAll([ 1, 2, 3, ]) ? 'Yes' : 'No') . ".\n";
echo "Is 1.2 float? - " . ($n->isFloat(1.2) ? 'Yes' : 'No') . ".\n";
echo "Are [ -2.1, 0.0, 1.2, ] all float? - " . ($n->isFloatAll([ -2.1, 0.0, 1.2, ]) ? 'Yes' : 'No') . ".\n";
echo "Is 1.2 number? - " . ($n->isNumber(1.2) ? 'Yes' : 'No') . ".\n";
echo "Are [ -2, 0.1, 3, ] all number? - " . ($n->isNumberAll([ -2, 0.1, 3, ]) ? 'Yes' : 'No') . ".\n";
echo "Is 0.1 fraction? - " . ($n->isFraction(0.1) ? 'Yes' : 'No') . ".\n";
echo "Are [ -0.99, 0.1, 0.99, ] all fraction? - " . ($n->isFractionAll([ -0.99, 0.1, 0.99, ]) ? 'Yes' : 'No') . ".\n";
echo "Sign of -2.5 is " . $n->sign(-2.5) . ".\n";
echo "Integer part of 3.14 is " . $n->int(3.14) . ".\n";
echo "Fractional part of 3.14 is " . $n->fraction(3.14) . ".\n";
echo "-3th digit of 123.4567 is " . $n->nthDigit(-3, 123.4567) . ".\n";
echo "Number of digits -123.4567 is " . $n->numberOfDigits(-123.4567) . ".\n";
echo "Number of fractional digits -12.3456 is " . $n->numberOfFractionalDigits(-12.3456) . ".\n";

  • 结果

    Is 1 int? - Yes.
Are [ 1, 2, 3, ] all int? - Yes.
Is 1 natural? - Yes.
Are [ 1, 2, 3, ] all natural? - Yes.
Is 1.2 float? - Yes.
Are [ -2.1, 0.0, 1.2, ] all float? - Yes.
Is 1.2 number? - Yes.
Are [ -2, 0.1, 3, ] all number? - Yes.
Is 0.1 fraction? - Yes.
Are [ -0.99, 0.1, 0.99, ] all fraction? - Yes.
Sign of -2.5 is -1.
Integer part of 3.14 is 3.
Fractional part of 3.14 is 0.14.
-3th digit of 123.4567 is 6.
Number of digits -123.4567 is 3.
Number of fractional digits -12.3456 is 4.

  • 方法

5.2. Macocci7\PhpMathInteger\Prime

此类处理质数的初步知识。

  • PHP

    <?php

require_once __DIR__ . '/../vendor/autoload.php';

use Macocci7\PhpMathInteger\Prime;

$p = new Prime();

// judge if $n is prime or not
$n = 3;
echo sprintf("Is %d prime? - %s.\n", $n, $p->isPrime($n) ? 'Yes' : 'No');

// judge if all of $n are prime or not
$n = [ 2, 3, 5, ];
echo sprintf(
    "Are all of [%s] prime? - %s.\n\n",
    implode(', ', $n),
    $p->isPrimeAll($n) ? 'Yes' : 'No'
);

// a prime previous to $n
$n = 5;
echo sprintf("A prime previous to %d is %d.\n", $n, $p->previous($n));

// a prime next to $n
$n = 5;
echo sprintf("A prime next to %d is %d.\n\n", $n, $p->next($n));

// primes between $a and $b
$a = 6;
$b = 14;
echo sprintf(
    "Primes between %d and %d are [%s].\n\n",
    $a,
    $b,
    implode(', ', $p->between($a, $b))
);

// factorize
$n = 1234567890;
echo sprintf("Factorize %d:\n\n", $n);

$r = $p->factorize($n);
$l1 = $p->numberOfDigits(max(array_column($r, 0)));
$l2 = $p->numberOfDigits(max(array_column($r, 1)));
$s = str_repeat(' ', $l1 + 1);
$b = $s . str_repeat('-', $l2);

foreach ($r as $f) {
    echo (
        $f[0]
        ? sprintf("%" . $l1 . "d)%" . $l2 . "d\n%s\n", $f[0], $f[1], $b)
        : sprintf("%s%" . $l2 . "d\n", $s, $f[1])
    );
}
echo "\n";

// Factorized formula
echo $n . " = " . $p->factorizedFormula($n)['formula'] . "\n";

  • 结果

    Is 3 prime? - Yes.
Are all of [2, 3, 5] prime? - Yes.

A prime previous to 5 is 3.
A prime next to 5 is 7.

Primes between 6 and 14 are [7, 11, 13].

Factorize 1234567890:

  2)1234567890
    ----------
  3) 617283945
    ----------
  3) 205761315
    ----------
  5)  68587105
    ----------
3607)  13717421
    ----------
          3803

1234567890 = 2 * 3 ^ 2 * 5 * 3607 * 3803

  • 方法

5.3. Macocci7\PhpMathInteger\Divisor

此类处理约数的初步知识。

  • PHP

    <?php

require_once __DIR__ . '/../vendor/autoload.php';

use Macocci7\PhpMathInteger\Divisor;

$d = new Divisor();
$a = 12;
$b = 18;

// Number of divisors
echo sprintf("%d has %d divisors.\n", $a, $d->count($a));

// List of all divisors
echo sprintf("[%s]\n", implode(', ', $d->list($a)));

// Common factors
echo sprintf("%d = %s\n", $a, $d->formula($a));
echo sprintf("%d = %s\n", $b, $d->formula($b));
echo sprintf(
    "common factors : %s\n",
    $d->formula($d->value($d->commonFactors($a, $b)))
);
echo sprintf(
    "common divisors : [%s]\n",
    implode(', ', $d->commonDivisors($a, $b))
);

// greatest common factor (divisor)
echo sprintf(
    "greatest common factor (divisor) : %s\n",
    $d->greatestCommonFactor($a, $b)
);

// Reducing fraction
$r = $d->reduceFraction($a, $b);
$ra = $d->value($r[0]);
$rb = $d->value($r[1]);
echo sprintf("%d/%d reduces to %d/%d\n", $a, $b, $ra, $rb);

  • 结果

    12 has 6 divisors.
[1, 2, 3, 4, 6, 12]
12 = 2 ^ 2 * 3
18 = 2 * 3 ^ 2
common factors : 2 * 3
common divisors : [1, 2, 3, 6]
greatest common factor (divisor) : 6
12/18 reduces to 2/3

  • 方法

5.4. Macocci7\PhpMathInteger\Multiple

此类处理倍数的初步知识。

  • PHP

    <?php

require_once __DIR__ . '/../vendor/autoload.php';

use Macocci7\PhpMathInteger\Multiple;

$m = new Multiple();
$a = 12;
$b = 18;

// least common multiple
echo sprintf(
    "least common multiple of %d and %d is %d\n",
    $a,
    $b,
    $m->leastCommonMultiple($a, $b)
);

  • 结果

    least common multiple of 12 and 18 is 36

  • 方法

5.5. Macocci7\PhpMathInteger\Euclid

此类处理欧几里得算法的初步知识。

  • PHP

    <?php

require_once __DIR__ . '/../vendor/autoload.php';

use Macocci7\PhpMathInteger\Euclid;

// Presets values
$e = new Euclid();
$a = 390;
$b = 273;
$c = 39;

// Judges if $c is GCD($a, $b) or not
echo sprintf(
    "Is %d GCD(%d, %d)? - %s.\n",
    $c,
    $a,
    $b,
    $e->isGcdOf($c, $a, $b) ? 'Yes' : 'No'
);

// Euclidean Algorithm
$r = $e->run($a, $b);
echo "Euclidean Algorithm:\n";
foreach ($r['processText'] as $t) {
    echo $t . "\n";
}

// Formula of remainders
echo "Remainders can be expressed as:\n";
foreach ($r['processData'] as $d) {
    echo sprintf("%d = %d - %d * %d\n", $d['r'], $d['a'], $d['b'], $d['c']);
}

// Judges if $a and $b are coprime or not
echo sprintf(
    "Are %d and %d coprime? - %s.\n",
    $a,
    $b,
    $e->isCoprime($a, $b) ? 'Yes' : 'No'
);

// GCD($a, $b)
echo sprintf(
    "Because the Greatest Common Divisor of %d and %d is %d.\n",
    $a,
    $b,
    $e->gcd($a, $b)
);

  • 结果

    Is 39 GCD(390, 273)? - Yes.
Euclidean Algorithm:
390 = 273 * 1 + 117
273 = 117 * 2 + 39
117 = 39 * 3 + 0
Remainders can be expressed as:
117 = 390 - 273 * 1
39 = 273 - 117 * 2
0 = 117 - 39 * 3
Are 390 and 273 coprime? - No.
Because the Greatest Common Divisor of 390 and 273 is 39.

  • 方法

5.6. Macocci7\PhpMathInteger\Fraction

此类处理普通分数的初步知识。

  • PHP

    <?php

require_once __DIR__ . '/../vendor/autoload.php';

use Macocci7\PhpMathInteger\Fraction;

// prest: 1 and 2/4
$f = new Fraction('1 2/4');

// is reduced or not?
echo $f->text() . " is " . ($f->isReduced() ? '' : 'not ') . "reduced.\n";

// is proper or not?
echo $f->text() . " is " . ($f->isProper() ? '' : 'not ') . "a proper fraction.\n";

// is improper or not?
echo $f->text() . " is " . ($f->isImproper() ? '' : 'not ') . "a improper fraction.\n";

// is mixed or not?
echo $f->text() . " is " . ($f->isMixed() ? '' : 'not ') . "a mixed fraction.\n";

// convert $f into a improper fraction
echo $f->text() . " = " . $f->improper()->text() . "\n";

// convert $f into a mixed fraction
echo $f->text() . " = " . $f->mixed()->text() . "\n";

// reduce fraction
echo $f->text() . " reduces to " . $f->reduce()->text() . "\n";

// integral part
echo "integral part of " . $f->text() . " is " . $f->int() . "\n";

// change into a float
echo $f->text() . " = " . $f->float() . "\n";

// four arithmetic operations
$f1 = new Fraction('1/3');
$f2 = new Fraction('1/6');
echo $f1->text() . ' + ' . $f2->text() . ' = ' . $f1->add($f2)->text() . "\n";
$f1->set('2/3');
$f2->set('1/6');
echo $f1->text() . ' - ' . $f2->text() . ' = ' . $f1->substract($f2)->text() . "\n";
$f1->set('2/3');
$f2->set('1/6');
echo $f1->text() . ' * ' . $f2->text() . ' = ' . $f1->multiply($f2)->text() . "\n";
$f1->set('2/3');
$f2->set('1/6');
echo $f1->text() . ' / ' . $f2->text() . ' = ' . $f1->divide($f2)->text() . "\n";

// reduce fractions to a common denominator
$f1->set('1/3');
$f2->set('2/5');
echo "reduce the fractions of " . $f1->text() . " and " . $f2->text()
    . " to a common denominator:\n";
$f1->toCommonDenominator($f2);
echo $f1->text() . " and " . $f2->text() . "\n";

  • 结果

    1 2/4 is not reduced.
1 2/4 is not a proper fraction.
1 2/4 is not a improper fraction.
1 2/4 is a mixed fraction.
1 2/4 = 6/4
6/4 = 1 2/4
1 2/4 reduces to 1 1/2
integral part of 1 1/2 is 1
1 1/2 = 1.5
1/3 + 1/6 = 1/2
2/3 - 1/6 = 1/2
2/3 * 1/6 = 1/9
2/3 / 1/6 = 4/1
reduce the fractions of 1/3 and 2/5 to a common denominator:
5/15 and 6/15

  • 方法

5.7. Macocci7\PhpMathInteger\Bezout

此类处理贝祖定理的初步知识。

  • PHP

    <?php

require_once __DIR__ . '/../vendor/autoload.php';

use Macocci7\PhpMathInteger\Bezout;

// Bezout's Identity: 3x + 4y = 1
$b = new Bezout([3, 4, 1, ]);
echo sprintf("Bezout's Identity: %s\n", $b->identity());

// Solvable or not
echo sprintf("Is it solvable? - %s.\n", ($b->isSolvable() ? 'Yes' : 'No'));

// A solution set
$s = $b->solution()['solution'];
echo sprintf("A solutionset: (x, y) = (%d, %d)\n", $s['x'], $s['y']);

// General solution
$g = $b->generalSolution()['generalSolution']['formula'];
echo sprintf("General solution:\n\t%s\n\t%s\n", $g['x'], $g['y']);

  • 结果

    Bezout's Identity: 3x + 4y = 1
Is it solvable? - Yes.
A solutionset: (x, y) = (-1, 1)
General solution:
  x = 4k - 1
  y = 3k + 1

  • 方法

6. 示例

7. 许可证

MIT

文档创建日期：2023/10/19

文档更新日期：2024/04/18

版权所有 2023 - 2024 macocci7