Trait num_integer::roots::Roots
source · pub trait Roots: Integer {
// Required method
fn nth_root(&self, n: u32) -> Self;
// Provided methods
fn sqrt(&self) -> Self { ... }
fn cbrt(&self) -> Self { ... }
}
Expand description
Provides methods to compute an integer’s square root, cube root,
and arbitrary n
th root.
Required Methods§
sourcefn nth_root(&self, n: u32) -> Self
fn nth_root(&self, n: u32) -> Self
Returns the truncated principal n
th root of an integer
– if x >= 0 { ⌊ⁿ√x⌋ } else { ⌈ⁿ√x⌉ }
This is solving for r
in rⁿ = x
, rounding toward zero.
If x
is positive, the result will satisfy rⁿ ≤ x < (r+1)ⁿ
.
If x
is negative and n
is odd, then (r-1)ⁿ < x ≤ rⁿ
.
§Panics
Panics if n
is zero:
println!("can't compute ⁰√x : {}", 123.nth_root(0));
or if n
is even and self
is negative:
println!("no imaginary numbers... {}", (-1).nth_root(10));
§Examples
use num_integer::Roots;
let x: i32 = 12345;
assert_eq!(x.nth_root(1), x);
assert_eq!(x.nth_root(2), x.sqrt());
assert_eq!(x.nth_root(3), x.cbrt());
assert_eq!(x.nth_root(4), 10);
assert_eq!(x.nth_root(13), 2);
assert_eq!(x.nth_root(14), 1);
assert_eq!(x.nth_root(std::u32::MAX), 1);
assert_eq!(std::i32::MAX.nth_root(30), 2);
assert_eq!(std::i32::MAX.nth_root(31), 1);
assert_eq!(std::i32::MIN.nth_root(31), -2);
assert_eq!((std::i32::MIN + 1).nth_root(31), -1);
assert_eq!(std::u32::MAX.nth_root(31), 2);
assert_eq!(std::u32::MAX.nth_root(32), 1);
Provided Methods§
sourcefn sqrt(&self) -> Self
fn sqrt(&self) -> Self
Returns the truncated principal square root of an integer – ⌊√x⌋
This is solving for r
in r² = x
, rounding toward zero.
The result will satisfy r² ≤ x < (r+1)²
.
§Panics
Panics if self
is less than zero:
println!("no imaginary numbers... {}", (-1).sqrt());
§Examples
use num_integer::Roots;
let x: i32 = 12345;
assert_eq!((x * x).sqrt(), x);
assert_eq!((x * x + 1).sqrt(), x);
assert_eq!((x * x - 1).sqrt(), x - 1);
sourcefn cbrt(&self) -> Self
fn cbrt(&self) -> Self
Returns the truncated principal cube root of an integer –
if x >= 0 { ⌊∛x⌋ } else { ⌈∛x⌉ }
This is solving for r
in r³ = x
, rounding toward zero.
If x
is positive, the result will satisfy r³ ≤ x < (r+1)³
.
If x
is negative, then (r-1)³ < x ≤ r³
.
§Examples
use num_integer::Roots;
let x: i32 = 1234;
assert_eq!((x * x * x).cbrt(), x);
assert_eq!((x * x * x + 1).cbrt(), x);
assert_eq!((x * x * x - 1).cbrt(), x - 1);
assert_eq!((-(x * x * x)).cbrt(), -x);
assert_eq!((-(x * x * x + 1)).cbrt(), -x);
assert_eq!((-(x * x * x - 1)).cbrt(), -(x - 1));