🔒 count set bits in succession

[violet ice]

does anyone know how i can count the number of bits in succession in an 8 bit number?

for example, there are lines of 3 bits and lines of 2 bits:

0b 1110 0011

something with binary instructions would be really helpful. thank you

[candid yew]

@violet ice

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[violet ice]

would a lookup table work?

[solar snow]

there are only 256 possible values for a byte so yes

[crystal jewel]

I would rather calculate on demand

But, it's possible

[violet ice]

that's what i would say

calculate on the fly would be better

[solar snow]

but also, int.bit_count exists, and you can extract selected bits with &, so

(x & 0b1111).bit_count()```
gets the amount of lower 4 bits set

[violet ice]

not that

how many bits are set IN SUCCESSION

so in a "line"

[crystal jewel]

That would be 3 because 000 and 111 correct?

[violet ice]

0b11100110 has 5 bits set, but has 1 line of 3 and 1 line of 2

[crystal jewel]

Either happen -> 3

[violet ice]

set bits

[solar snow]

ah so the spaces in your original example are just for readability

[violet ice]

sorry if i didn't mention that in the message

[violet ice]

yes

[crystal jewel]

Oh ok

Even easier

Is that a homework?

[violet ice]

no

personal project

[solar snow]

if you have these tools in your toolbox:

• you know how to get some limit for the amount of bits (you do, it is at most 8)
• you know how to loop and get each bit
• you know how to keep track of state across loop iterations
  then you have everything you need to create an algorithm that deals with spans of set bits

[olive rapids]

num&1 to see current last bit
num>>1 to move the bits right 1 space (thus removing last bit, and second last becomes new last). Save the result of that to use

You'd want to repeat those 2 operations in a loop - until the num becomes 0 (no more bits set to check) and manually keep count of groups

That's the low-memory way.

Memory-intensive way would be to treat it as sequence of characters (so each bit becomes one byte...) and throw it into something like intertools.groupby

[violet ice]

alright

makes sense

i'm using this to score connect 4 bitboard positions

i was lead to this SO post

didn't really help since i had zero idea what they meant

[crystal jewel]

Do you want me to show the solution to you?

[violet ice]

sure

that works

[crystal jewel]

def solver(v: int) -> int:
    max_c = c = 0
    while v:
        if v & 1:
            c+=1
        else:
            if max_c < c:
                max_c = c
            c = 0
        v >>= 1
    return c

You could do v %= 256 or v &= 255 to make sure it only work with 8 bit number tho (the last 8 bit number if exceed)

[cedar widget]

was groupby suggested?

>>> s = '11100011'
>>> from itertools import groupby
>>> for k, g in groupby(s):
...     print(k, len(list(g)))
...
1 3
0 3
1 2

[crystal jewel]

Yes

Memory-intensive way would be to treat it as sequence of characters (so each bit becomes one byte...) and throw it into something like intertools.groupby

[violet ice]

where's the source code for itertools.groupby?

[cedar widget]

docs are here

[candid yew]

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