我无法理解方程式的解，该解说找到最小的数，使得x加y等于x按位或y

Question

The problem is to find kth smallest number(y) which satisfies the following equations x + y = x | y .

eg: x = 5 kthSmallest = 1 gives y = 2 as 5+2 = 5 | 2

I coded it using the basic approach

long kthPlusOrSolutionMine(int x, int k){
int kthSolution = 1;
for(int i=0;i>=0 ; i++ ){
    int rhs = x | i;
    int lhs = x + i;
    if(rhs == lhs){
        kthSolution++;
        if(kthSolution == k)
            return i;
    }
    return 0;
}
}

However I found another solution of the same problem but I can't understand it completely.

long kthPlusOrSolution(int _x, int k) {
long x = _x;
long y=0,t;
for(t=1; k; t=t<<1, k=k>>1){
    //cout << "t is " << t << " x is " << x << " k is " << k <<endl;
    //cout << (t&x) << endl;
    while(t&x)
        t=t<<1;
    if(k&1)
        y = y|t;
}
return y;
}

Can somebody please help me in understanding that how does it works.

Answer 1

Bitwise, you can compare the truth table of addition and OR (C is the addition carry) :

A B + C | A+B==A|B
0 0 0 0 0   yep
1 0 1 0 1   yep
0 1 1 0 1   yep
1 1 0 1 1   nope

So you can say that

• for each bit of X set to 1, the corresponding bit of Y must be zero.
• for each bit of X set to 0, the corresponding bit of Y can be either 0 or 1.

Let's number the bits of y that are allowed to vary from 0 to I.

x 0 0 1 0 0 1 1
i 3 2 - 1 0 - -

Note that if we call Y _i a number with the i _th bit set to 1, Y _i is the solution number 2 ⁱ +1 of the equation.

for instance:

x  = 010011
y    0-00--
y1 = 0-01-- 2nd
y2 = 0-10-- 3rd
     0-11-- 
y3 = 1-00-- 5th
     1-01--
     1-10--
     1-11-- 
y4 =10-00-- 9th

Also note that the rank difference between a solution with the i _th bit set to zero and the same bit set to 1 is 2 ⁱ .

example: the bit of rank 2 is varying

0-00-- 1st
1-00-- 5th

difference in rank: 2 ² = 4

This means that setting the i _th bit of y to 1 skips 2 ⁱ solutions .
That is the magic property that fuels the algorithm.

during each loop, t is set to the i _th bit of x (eg if x = 11, successive values of t will be 4, 8, 16 etc)

k is decomposed in powers of 2. During each loop, the n _th bit of k in base 2 is examined, and each time y is updated, 2 ⁿ solutions are eliminated using our magic property.

exemple: if k = 5 (1001b)

1st loop k = 5 -> eliminate 2^0 = 1 solution
2nd loop k = 2 -> does nothing
3rd look k = 1 -> eliminate 2^2 = 4 solutions

at exit, y is set to the 5th solution.
Or more precisely, the 5th solution different from 0,
which is the 6th if you count zero as a (trivial) solution.