Sum of Two Integers without using "+" operator in python

Question

Sum of Two Integers without using "+" operator in python

Need some help understanding python solutions of leetcode 371. "Sum of Two Integers". I found https://discuss.leetcode.com/topic/49900/python-solution/2 is the most voted python solution, but I am having problem understand it.

How to understand the usage of "% MASK" and why "MASK = 0x100000000"?
How to understand "~((a % MIN_INT) ^ MAX_INT)"?
When sum beyond MAX_INT, the functions yells negative value (for example getSum(2147483647,2) = -2147483647), isn't that incorrect?

class Solution(object):

    def getSum(self, a, b):
        """
        :type a: int
        :type b: int
        :rtype: int
        """
        MAX_INT = 0x7FFFFFFF
        MIN_INT = 0x80000000
        MASK = 0x100000000
        while b:
            a, b = (a ^ b) % MASK, ((a & b) << 1) % MASK
        return a if a <= MAX_INT else ~((a % MIN_INT) ^ MAX_INT)

python

asked on Stack Overflow Jul 24, 2016 by

user1269298 • edited Jul 24, 2016 by

user1269298

2 Answers

Let's disregard the MASK, MAX_INT and MIN_INT for a second.

Why does this black magic bitwise stuff work?

The reason why the calculation works is because (a ^ b) is "summing" the bits of a and b. Recall that bitwise xor is 1 when the bits differ, and 0 when the bits are the same. For example (where D is decimal and B is binary), 20D == 10100B, and 9D = 1001B:

and 11101B == 29D.

But, if you have a case with a carry, it doesn't work so well. For example, consider adding (bitwise xor) 20D and 20D.

10100
10100
-----
00000

Oops. 20 + 20 certainly doesn't equal 0. Enter the (a & b) << 1 term. This term represents the "carry" for each position. On the next iteration of the while loop, we add in the carry from the previous loop. So, if we go with the example we had before, we get:

# First iteration (a is 20, b is 20)
10100 ^ 10100 == 00000 # makes a 0
(10100 & 10100) << 1 == 101000 # makes b 40

# Second iteration:
000000 ^ 101000 == 101000 # Makes a 40
(000000 & 101000) << 1 == 0000000 # Makes b 0

Now b is 0, we are done, so return a. This algorithm works in general, not just for the specific cases I've outlined. Proof of correctness is left to the reader as an exercise ;)

What do the masks do?

All the masks are doing is ensuring that the value is an integer, because your code even has comments stating that a, b, and the return type are of type int. Thus, since the maximum possible int (32 bits) is 2147483647. So, if you add 2 to this value, like you did in your example, the int overflows and you get a negative value. You have to force this in Python, because it doesn't respect this int boundary that other strongly typed languages like Java and C++ have defined. Consider the following:

def get_sum(a, b):
    while b:
        a, b = (a ^ b), (a & b) << 1
    return a

This is the version of getSum without the masks.

print get_sum(2147483647, 2)

outputs

2147483649

while

 print Solution().getSum(2147483647, 2)

outputs

-2147483647

due to the overflow.

The moral of the story is the implementation is correct if you define the int type to only represent 32 bits.

answered on Stack Overflow Jul 25, 2016 by

Matt Messersmith • edited Nov 7, 2016 by

Elsa Lin

Here is solution works in every case

cases
- -
- +
+ -
+ +

solution
python default int size is not 32bit, it is very large number, so to prevent overflow and stop running into infinite loop, we use 32bit mask to limit int size to 32bit (0xffffffff)

a,b=-1,-1
mask=0xffffffff 
while (b & mask):
    carry=a & b
    a=a^b
    b=carray <<1
print( (a&Mask) if b>0 else a)

answered on Stack Overflow Jan 21, 2021 by

user13966865

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