"Bit Mash Techniques" Code Answer's

Bit Mash Techniques

By Bright Batfish on May 25, 2021

1. REPRESENTATION: A 32 (or 64)-bit signed integer for up to 32 (or 64) items. ( To avoid issues with the 
    two’s complement representation, use a 32-bit/64-bit signed integer to represent bitmasks of up to
    30/62 items only, respectively ).

    For example:                          5| 4  | 3 | 2 | 1 | 0   <- 0-based indexing from right
                                         32| 16 | 8 | 4 | 2 | 1   <- power of 2
                     A= 34 (base 10) =   1 | 0  | 0 | 0 | 1 | 0   <- in binary
                                         F | E  | D | C | B | A   <- alternative alphabet label
   In the example above,the integer A = 34 or 100010 in binary also represents a small set {1, 5} with a
   0-based indexing scheme in increasing digit significance ( or {B, F} using the alternative alphabet
   label )because the second and the sixth bits (counting from the right) of A are on ( 1 ).

 2. To multiply/divide an integer by 2: 
                                    We only need to shift the bits in the integer left/right, respectively.
    Notice that the truncation in the shift right operation automatically rounds the division-by-2 down,
    e.g. 17/2  = 8.

    For example:         A = 34 (base 10)                  = 100010 (base 2)
                         A = A << 1 = A * 2 = 68 (base 10) = 1000100 (base 2)
                         A = A >> 2 = A / 4 = 17 (base 10) = 10001 (base 2)
                         A = A >> 1 = A / 2 = 8 (base 10) = 1000 (base 2) <- LSB( Least Significant Bit )is gone

 3. Add the jth object to the subset (set the jth bit from 0 to 1):
     use the bitwise OR operation A |= (1 << j).

     For example:     A = 34 (base 10) = 100010 (base 2)
                      j = 3, 1 << j    = 001000 <- bit ‘1’ is shifted to the left 3 times
                                        -------- OR (true if either of the bits is true)
                      A = 42 (base 10) = 101010 (base 2) // update A to this new value 42

4. Remove the jth object from the subset (set the jth bit from 1 to 0):
     use the bitwise AND operation A &= ∼(1 << j).

     For example:         A = 42 (base 10) = 101010 (base 2)
                          j = 1, ~(1 << j) = 111101 <- ‘~’ is the bitwise NOT operation
                                             -------- AND
                          A = 40 (base 10) = 101000 (base 2) // update A to this new value 40

5. Check whether the jth object is in the subset (check whether jth bit is 1):
   use the bitwise AND operation T = A & (1 << j).
   If T = 0, then the j-th item of the set is off.
   If T != 0 (to be precise, T = (1 << j)), then the j-th item of the set is on.

   For example:    A = 42 (base 10) = 101010 (base 2)
                   j = 3, 1 << j    = 001000 <- bit ‘1’ is shifted to the left 3 times
                                     -------- AND (only true if both bits are true)
                   T = 8 (base 10)  = 001000 (base 2) -> not zero, the 3rd item is on

6. To toggle (flip the status of) the j-th item of the set:
   use the bitwise XOR operation A ∧ = (1 << j).

   For example:       A = 40 (base 10) = 101000 (base 2)
                      j = 2, (1 << j)  = 000100 <- bit ‘1’ is shifted to the left 2 times
                                        -------- XOR <- true if both bits are different
                      A = 44 (base 10) = 101100 (base 2) // update A to this new value 44

7. To get the value of the least significant bit that is on (first from the right):
   use T = (A & (-A)).

   For example:     A =  40 (base 10) = 000...000101000 (32 bits, base 2)
                   -A = -40 (base 10) = 111...111011000 (two’s complement)
                                       ----------------- AND
                    T =   8 (base 10) = 000...000001000 (3rd bit from right is on)

8. To turn on all bits in a set of size n: (be careful with overflows)
   use A = (1 << n) - 1 ;

9. Iterate through all subsets of a set of size n:
           `for ( x = 0; x < (1 << n); ++x )`  

10. Iterate through all subsets of a subset y (not including empty set):
           `for ( x = y; x > 0; x = ( y & (x-1) ) )`

Source: codeforces.com

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