A little-endian bit vector of non-negative dimension.

Create a bit vector from a little-endian list of bits.

The following will hold:

length . takeWhile not === countLeadingZeros . fromBits
length . takeWhile not . reverse === countTrailingZeros . fromBits

Time: \(\, \mathcal{O} \left( n \right) \)

Since: 0.1.0

Examples

>>> fromBits [True, False, False]
[3]1

Create a little-endian list of bits from a bit vector.

The following will hold:

length . takeWhile not . toBits === countLeadingZeros
length . takeWhile not . reverse . toBits === countTrailingZeros

Time: \(\, \mathcal{O} \left( n \right) \)

Since: 0.1.0

Examples

>>> toBits [4]11
[True, True, False, True]

Create a bit vector of non-negative dimension from an integral value.

The integral value will be treated as an signed number and the resulting bit vector will contain the two's complement bit representation of the number.

The integral value will be interpreted as little-endian so that the least significant bit of the integral value will be the value of the 0th index of the resulting bit vector and the most significant bit of the integral value will be at index dimension − 1.

Note that if the bit representation of the integral value exceeds the supplied dimension, then the most significant bits will be truncated in the resulting bit vector.

Time: \(\, \mathcal{O} \left( 1 \right) \)

Since: 0.1.0

Examples

>>> fromNumber 8 96
[8]96

>>> fromNumber 8 -96
[8]160

>>> fromNumber 6 96
[6]32

Two's complement value of a bit vector.

Time: \(\, \mathcal{O} \left( 1 \right) \)

Since: 0.1.0

Examples

>>> toSignedNumber [4]0
0

>>> toSignedNumber [4]3
3

>>> toSignedNumber [4]7
7

>>> toSignedNumber [4]8
-8

>>> toSignedNumber [4]12
-4

>>> toSignedNumber [4]15
-1

Unsigned value of a bit vector.

Time: \(\, \mathcal{O} \left( 1 \right) \)

Since: 0.1.0

Examples

>>> toUnsignedNumber [4]0
0

>>> toUnsignedNumber [4]3
3

>>> toUnsignedNumber [4]7
7

>>> toUnsignedNumber [4]8
8

>>> toUnsignedNumber [4]12
12

>>> toUnsignedNumber [4]15
15

Get the dimension of a BitVector. Preferable to finiteBitSize as it returns a type which cannot represent a non-negative value and a BitVector must have a non-negative dimension.

Time: \(\, \mathcal{O} \left( 1 \right) \)

Since: 0.1.0

Examples

>>> dimension [2]3
2

>>> dimension [4]12
4

Determine if any bits are set in the BitVector. Faster than (0 ==) . popCount.

Time: \(\, \mathcal{O} \left( 1 \right) \)

Since: 0.1.0

Examples

>>> isZeroVector [2]3
False

>>> isZeroVector [4]0
True

Get the inclusive range of bits in BitVector as a new BitVector.

If either of the bounds of the subrange exceed the bit vector's dimension, the resulting subrange will append an infinite number of zeroes to the end of the bit vector in order to satisfy the subrange request.

Time: \(\, \mathcal{O} \left( 1 \right) \)

Since: 0.1.0

Examples

>>> subRange (0,2) [4]7
[3]7

>>> subRange (1, 3) [4]7
[3]3

>>> subRange (2, 4) [4]7
[3]1

>>> subRange (3, 5) [4]7
[3]0

>>> subRange (10, 20) [4]7
[10]0

Determine the number of set bits in the BitVector up to, but not including, index k.

To determine the number of unset bits in the BitVector, use k - rank bv k.

Uses "broadword programming." Efficient on small BitVectors (10^3).

Time: \(\, \mathcal{O} \left( \frac{n}{w} \right) \), where \(w\) is the number of bits in a Word.

Since: 1.1.0

Examples

>>> let bv = fromNumber 128 0 `setBit` 0 `setBit` 65

>>> rank bv   0  -- Count how many ones in the first 0 bits (always returns 0)
0

>>> rank bv   1  -- Count how many ones in the first 1 bits
1

>>> rank bv   2  -- Count how many ones in the first 2 bits
1

>>> rank bv  65  -- Count how many ones in the first 65 bits
1

>>> rank bv  66  -- Count how many ones in the first 66 bits
1

>>> rank bv 128  -- Count how many ones in all 128 bits
2

>>> rank bv 129  -- Out-of-bounds, fails gracefully
2

Find the index of the k-th set bit in the BitVector.

To find the index of the k-th unset bit in the BitVector, use select (complement bv) k.

Uses "broadword programming." Efficient on small BitVectors (10^3).

Time: \(\, \mathcal{O} \left( \frac{n}{w} \right) \), where \(w\) is the number of bits in a Word.

Since: 1.1.0

Examples

>>> let bv = fromNumber 128 0 `setBit` 0 `setBit` 65

>>> select bv 0  -- Find the 0-indexed position of the first one bit
Just 0

>>> select bv 1  -- Find the 0-indexed position of the second one bit
Just 65

>>> select bv 2  -- There is no 3rd set bit, `select` fails
Nothing