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Bit Manipulation Techniques

the_unconventional_coder profile image ABHINAVA GHOSH (he/him) ・4 min read

LIKE MYSELF I HAVE SEEN MANY PEOPLE SPECIALLY BEGINNERS STRUGGLING WITH BIT MANIPULATIONS IN THE BEGINNING!
SO I THOUGHT I WOULD WRITE A POST SO THAT PEOPLE CAN BENEFIT FROM IT!

THIS IS JUST THE NOTES I MADE A WHILE BACK AS A LOOKUP NOTES WHENEVER I WAS STUCK AT ANY BIT MANIPULATION QUESTIONS!

BIT HACKS:

🎯 binary numbers are 0 indexed
REMEMBER left shifting a number by i multiplies that number with 2^i
and right shifting by i divides it by 2^i


🎯 N-1 INVERTS EVERY BIT STARTING FROM RIGHTMOST ONE till INDEX β€˜0’
IF N=0b00000110
N-1=0b00000101


🎯 1<< left shifts 1 by i positions
1>>i
right shifts 1 by i positions


🎯 whenever we want to toggle ,set,unset ith bit of a number we always use 1 as our helper with different operators

🎯 toggle ith bit:toggling means 1->0 and 0->1
1ST STEP: left shift 1 to the position where we want to toggle
2ND STEP: we have taken the 1 to the ith bit(remember all rest are 0s) then we need to make it a 1 if its a zero and make it 0 if its a one
3RD STEP: so we have the 1 fixed ,using this we need to make 0->1 or 1->0
if we do '&'
0&1=0(incorrect)
1&1=1(incorrect)

'|'
0|1=1(correct)
1|1=1(incorrect)

'^'
0^1=1(correct)
1^1=0(correct)

hence we could only do xor(^)


🎯 set ith bit(means make the ith bit 0->1)

we again use 1<<i to reach the ith position
then we do '|' with 1<<i
therefore if bit was 0 it becomes 1
if it was already 1 then it still remains 1
NOTE-USING XOR IS NOT RECOMMENDED ,BECAUSE IF THE ith BIT IS ALREADY SET THEN XOR WILL GIVE WRONG RESULT!


🎯 unset/clear ith bit (make 1->0)

we need again the 1< now we do a negation ~ to this 1 we took so that the 1 in our 1 initially becomes 0 and every other bit becomes 1
suppose i=3
we do 1<<3 to get
0b00000001->0b00001000
then ~ gives
0b00001000->0b11110111
so we have ith bit if our 1 as 0 and rest 1
now we just & it with the number whose ith bit is to be unset!


🎯 Checking if bit at nth position is set or unset:

Left shift β€˜1’ to given position and then β€˜AND'(β€˜&’).


Some more quick hacks:

🎯 Inverting every bit of a number/1’s complement:
If we want to invert every bit of a number i.e change bit β€˜0’ to β€˜1’ and bit β€˜1’ to β€˜0’.We can do this with the help of β€˜~’ operator. For example : if number is num=00101100 (binary representation) so β€˜~num’ will be β€˜11010011’.

This is also the β€˜1s complement of number’.


🎯 Two’s complement of the number: 2’s complement of a number is 1’s complement + 1.
So formally we can have 2’s complement by finding 1s complement and adding 1 to the result i.e (~num+1)


🎯 *Stripping off the lowest set bit *:
In many situations we want to strip off the lowest set bit for example in Binary Indexed tree data structure, counting number of set bit in a number.

We do like this:

X = X & (X-1)

Let us see this by taking an example, let X = 1100.

(X-1) inverts all the bits till it encounter lowest set β€˜1’ and it also invert that lowest set β€˜1’.

X-1 becomes 1011. After β€˜ANDing’ X with X-1 we get lowest set bit stripped.


🎯 Getting lowest set bit of a number:
This is done by using expression β€˜X &(-X)’Let us see this by taking an example:Let X = 00101100.
So ~X(1’s complement) will be β€˜11010011’ and 2’s complement will be (~X+1 or -X) i.e β€˜11010100’.
So if we β€˜AND’ original number β€˜X’ with its two’s complement which is β€˜-X’, we get lowest set bit.


🎯 SOME MORE IMPORTANT HACKS:

x&1 gives the lowest bit(helps in finding whether number is even or odd i.e if last bit is 0 then it is even otherwise odd)
x & (x-1) will clear the lowest set bit of x
x & ~(x-1) extracts the lowest set bit of x (all others are clear). Pretty patterns when applied to a linear sequence.
x & (x + (1 0<< n)) = x, with the run of set bits (possibly length 0) starting at bit n cleared.
x & ~(x + (1 << n)) = the run of set bits (possibly length 0) in x, starting at bit n.
x | (x + 1) = x with the lowest cleared bit set.
x | ~(x + 1) = extracts the lowest cleared bit of x (all others are set).
x | (x - (1 << n)) = x, with the run of cleared bits (possibly length 0) starting at bit n set.
x | ~(x - (1 << n)) = the lowest run of cleared bits (possibly length 0) in x, starting at bit n are the only clear bits.


HOPE IT HELPS!

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Have a Good Day! πŸ˜ƒ

Some of my other posts:

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Java Access Modifiers goto Article
Java Generics goto Article
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