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!
🎯 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 ‘0’
left shifts 1 by i positions
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 '&'
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
we do 1<<3 to get
then ~ gives
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 ‘1’ to given position and then ‘AND'(‘&’).
🎯 Inverting every bit of a number/1’s complement:
If we want to invert every bit of a number i.e change bit ‘0’ to ‘1’ and bit ‘1’ to ‘0’.We can do this with the help of ‘~’ operator. For example : if number is num=00101100 (binary representation) so ‘~num’ will be ‘11010011’.
This is also the ‘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 ‘1’ and it also invert that lowest set ‘1’.
X-1 becomes 1011. After ‘ANDing’ X with X-1 we get lowest set bit stripped.
🎯 Getting lowest set bit of a number:
This is done by using expression ‘X &(-X)’Let us see this by taking an example:Let X = 00101100.
So ~X(1’s complement) will be ‘11010011’ and 2’s complement will be (~X+1 or -X) i.e ‘11010100’.
So if we ‘AND’ original number ‘X’ with its two’s complement which is ‘-X’, we get lowest set bit.
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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