Trait malachite_base::num::arithmetic::traits::CeilingSqrt

source · [−]

pub trait CeilingSqrt {
    type Output;

    fn ceiling_sqrt(self) -> Self::Output;
}

Expand description

Finds the ceiling of the square root of a number.

Required Associated Types

source

Required Methods

source

fn ceiling_sqrt(self) -> Self::Output

Implementations on Foreign Types

source

impl CeilingSqrt for u8

source

fn ceiling_sqrt(self) -> u8

Returns the ceiling of the square root of a u8.

$f(x) = \lceil\sqrt{x}\rceil$.

Notes

The u8 implementation uses a lookup table.

impl CeilingSqrt for u32

source

fn ceiling_sqrt(self) -> u32

Returns the ceiling of the square root of an integer.

$f(x) = \lceil\sqrt{x}\rceil$.

Notes

For u32 and u64, the square root is computed using Newton’s method.

impl CeilingSqrt for u64

source

fn ceiling_sqrt(self) -> u64

Returns the ceiling of the square root of an integer.

$f(x) = \lceil\sqrt{x}\rceil$.

Notes

For u32 and u64, the square root is computed using Newton’s method.

impl CeilingSqrt for u16

source

fn ceiling_sqrt(self) -> u16

Returns the ceiling of the square root of a u16.

$f(x) = \lceil\sqrt{x}\rceil$.

Notes

The u16 implementation calls the implementation for u32s.

impl CeilingSqrt for usize

source

fn ceiling_sqrt(self) -> usize

Returns the ceiling of the square root of a usize.

$f(x) = \lceil\sqrt{x}\rceil$.

Notes

The usize implementation calls the u32 or u64 implementations.

impl CeilingSqrt for u128

source

fn ceiling_sqrt(self) -> u128

Returns the ceiling of the square root of a u128.

$f(x) = \lceil\sqrt{x}\rceil$.

Worst-case complexity

$T(n) = O(n)$

$M(n) = O(1)$

where $T$ is time, $M$ is additional memory, and $n$ is self.significant_bits().

For u128, using a floating-point approximation and refining the result works, but the number of necessary adjustments becomes large for large u128s. To overcome this, large u128s switch to a binary search algorithm. To get decent starting bounds, the following fact is used:

If $x$ is nonzero and has $b$ significant bits, then

$2^{b-1} \leq x \leq 2^b-1$,

$2^{b-1} \leq x \leq 2^b$,

$2^{2\lfloor (b-1)/2 \rfloor} \leq x \leq 2^{2\lceil b/2 \rceil}$,

$2^{2(\lceil b/2 \rceil-1)} \leq x \leq 2^{2\lceil b/2 \rceil}$,

$\lfloor\sqrt{2^{2(\lceil b/2 \rceil-1)}}\rfloor \leq \lfloor\sqrt{x}\rfloor \leq \lfloor\sqrt{2^{2\lceil b/2 \rceil}}\rfloor$, since $x \mapsto \lfloor\sqrt{x}\rfloor$ is weakly increasing,

$2^{\lceil b/2 \rceil-1} \leq \lfloor\sqrt{x}\rfloor \leq 2^{\lceil b/2 \rceil}$.

For example, since $10^9$ has 30 significant bits, we know that $2^{14} \leq \lfloor\sqrt{10^9}\rfloor \leq 2^{15}$.