## SUM And XOR

### April 1, 2016

The naive solution iterates over a and b up to s, which is quadratic. But a little algebra makes an easy linear solution. Given

```s = a + b
x = a ^ b```

then

`x = (s - b) ^ b`

Since s and x are known, check all integers b from 0 to s and report those that work:

```(define (sum-xor s x)
(list-of (list (- s b) b)
(b range 1 s)
(= x (bitwise-xor (- s b) b))))```
```> (sum-xor 9 5)
((7 2) (6 3) (3 6) (2 7))```

A simple optimization works half-way through the range, generating both a pair and its conjugate whenever it finds a suitable b, but we won’t bother.

We used list comprehensions from the Standard Prelude. You can run the program at http://ideone.com/eDfLxn.

Pages: 1 2

### 8 Responses to “SUM And XOR”

1. Marcos said
```def sums_target(target):
a = target - 1

for i in range(1, target):
yield i, a
a -= 1

def sum_and_xor(sum, xor):
for pair in sums_target(sum):
a, b = pair
if a ^ b == xor:
yield a, b

for solution in sum_and_xor(9, 5):
print(solution)
```
2. Ernie said

Actually you only have to loop through the range 0 to s/2 since if (a,b) is a solution, then (b,a) is a solution

3. matthew said

Hard to resist a nice bit-twiddling challenge: s and x differ just at positions where there is a carry in from the right, and x tells us where the two numbers differ: if the number are the same at a particular bit position, then there is a carry just if they are both 1, if both 0, there can be no carry. If the digits are different at a given position, then because addition and xor are symmetric, we can have 0 and 1 either way round in the original numbers, regardless of the carries. So, the solutions have a certain number of bits fixed, and we can have all possible combinations of the other bits, so we can just count through those combinations: if a is a possible value, we can find the next possible value by setting all fixed bits to 1, incrementing, then setting the fixed bits back to their correct values. Maybe some code is clearer:

```#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
typedef unsigned int T;
int main(int argc, char *argv[]) {
T a0 = strtoul(argv,NULL,0);
T b0 = strtoul(argv,NULL,0);
T s = a0 + b0;
T x = a0 ^ b0;            // a and b are different at these positions
T carries = (s ^ x) >> 1; // Positions where there is a carry out
T fixed = ~x & carries;   // a & ~x == b & ~x == fixed, for all solutions
T a = fixed;
printf("s = %u x = %u\n",s,x);
while(true) {
T b = a ^ x;
if (b < a) break;
assert(a+b == s);
printf("%u %u\n",a,b);
a = (((a | ~x) + 1) & x) | fixed;
}
}
```
4. J.Fossy said

For each x=2^k :
i) a b = a + 2^k
s = a + a + 2^k = 2a + 2^k
a = (s – 2^k)/2

ii) a >= 2^k => b = a – 2^k
s = a + a – 2^k = 2a – 2^k
a = (s + 2^k)/2

This is true for each bit of x => so a = (s +/- all combinations of 2^k of x)/2 or there is no result

//————————————————————————–
#include

//————————————————————————–
int main(int argn, char *argv[]) {

if ( argn != 3 ) {
printf(“usage: %s \n\n”, argv);
return 1; //==========================================================>
}

int s = atoi(argv);
int x = atoi(argv);

printf(“s=%d x=%d\n:\n”, s, x);

int i, j;
int n;
int m;
int b;

for ( i = 0, m = 1, n = 0;
i m;
i++, m <<= 1
) {
if ( x & m ) {
b[n] = m;
n++;
}
}

for ( i = 0;
i < (1 << n);
i++
) {

int a = s;

for ( j = 0, m = 1;
j < n;
j++, m <>= 1;

int b = s – a;

if ( (a ^ b) != x ) {
//printf(“a=%d b=%d s=%d x=%d FALSE\n”, a, b, a + b, a ^ b);
return 1; //========================================================>
}

return 0;
}

5. matthew said

@J.Fossy: That looks interesting but it’s all got a bit garbled, you might want to have a look at https://en.support.wordpress.com/code/posting-source-code/

6. Chris Judge said

In c#:

```        static List<Tuple<int, int>> getSumXOR (int s, int x) {
List<Tuple<int, int>> t = new List<Tuple<int, int>>();

for (int i = 1; i <= s / 2; i++) {
int j = i ^ x;
if (s == i + j) {
}
}

return t;
}
```
7. jay said

JS:

```var coupleSumXOR = function(sum, cant) {
var couples = [];
var j;
var halfway = Math.floor(sum / 2);
cant = cant <= halfway ? cant : sum - cant;
for (var i = 1; i <= halfway; i++) {
if (i !== cant) {
j = sum - i;
i === j ? couples.push([i,j]) : couples.push([i, j], [j,i]);
};
};
return couples;
};

console.log(coupleSumXOR(10, 3));
```

thanks for posting problems, good exercise