Given:
a
(say, std::uint64_t
), which contains at least one set (1
) bit.b
which is a subset of a
(i.e. a & b == b
), and has at least one bit set.I want to select spans of contiguous 1-bits in a
which overlap with a bit in b
:
a = 0b1111001110001100;
b = 0b0000001010001000;
//c=0b0000001110001100
// XXXX YYY ZZ
The XXXX group is 0 in c
because b & XXXX
is false. The ZZ group is copied because b
has one of the Z bits set. The YYY group is also set in c
for the same reason. Notice that b
can have multiple set bits in a single group in a
.
So for every contiguous group of 1
s in a
, set all of those bits in c
if b
has a 1
in any of those positions. A more complex example:
std::uint64_t a = 0b1101110110101;
std::uint64_t b = 0b0001010010001;
// desired c == 0b0001110110001
// contiguous groups ^^^ ^^ ^ that overlap with a 1 in b
assert(a & b == b); // b is a subset of a
std::uint64_t c = some_magic_operation(a, b);
assert(c == 0b0001110110001);
Are there any bit-logic instructions/intrinsics (MMX, SSE, AVX, BMI1/BMI2), or bit manipulation tricks which allows me to calculate c
from a
and b
efficiently? (i.e. without loops)?
ADDITIONAL:
Using hint from Denis' answer I can only imagine loop-based algorithm:
std::uint64_t a = 0b0110111001001101;
std::uint64_t b = 0b0100101000001101;
assert(a & b == b); // subset
std::cout << std::bitset< 16 >(a) << std::endl;
std::cout << std::bitset< 16 >(b) << std::endl;
std::uint64_t x = (a + b) & ~a;
std::uint64_t c = 0;
while ((x = (a & (x >> 1)))) { // length of longest 1-series times
c |= x;
}
std::cout << std::bitset< 16 >(c) << std::endl;
In case of uint64_t
you may do this trick:
Let's set a = 0b11011101101
. Having at least one 0 bit is important. The bitmask has 4 separate areas, filled with 1 bits. If you do c=a+(a&b)
, then each 1-filled area will overflow if at least one bit of b
in this area is set. So you can check then, which area was overflown. For example, if you want 1-bits in 2-nd and 3-rd areas of a
, you may do so:
assert(c & 0b00100010000);
// ^^^ ^^ this segments overflows
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