A + B without arithmetic operators, Python vs C++
Question:
I was trying to solve an old question:
Write a function that add two [integer] numbers A and B. You should not use + or any arithmetic operators.
The best solution is like this, quoted from “LintCode-A+B Problem“:
For a + b in any base, we can treat the plus as two part: 1. a + b without carry; 2. the carry generated by a +b. The a+b then equals to part 1 plus part 2. If part1+part2 generates more carry, we can then repeat this procedure, until there is no carry.
I can understand this algorithm and everything seems good, so I tested it on lintcode with code pasted below.
class Solution:
"""
@param a: The first integer
@param b: The second integer
@return: The sum of a and b
"""
def aplusb(self, a, b):
while b != 0:
carry = a & b
a = a ^ b
b = carry << 1
return a
But surprisingly, it gave me Time Limit Exceeded error in test case [100, -100]. So I ran it locally and print a, b for every loop:
(-8, 8)
(-16, 16)
(-32, 32)
(-64, 64)
(-128, 128)
(-256, 256)
(-512, 512)
(-1024, 1024)
(-2048, 2048)
(-4096, 4096)
(-8192, 8192)
(-16384, 16384)
(-32768, 32768)
(-65536, 65536)
(-131072, 131072)
...
The calculation is correct, so I think this algorithm does not work for such input but when I wrote the same algorithm in C++, it just works:
class Solution {
public:
int aplusb(int a, int b) {
while (b!=0){
int carry = a & b;
a = a^b;
b = carry << 1;
}
return a;
}
};
I don’t know what should be asked exactly, basically the questions are:
- Why does C++ give the correct output
0 while Python doesn’t?
- If I use Python, how do I modify this algorithm to make it work?
Answers:
The problem are negative numbers, or, how they are represented. In Python integer numbers have arbitrary accuracy, while C++ ints are 32bit or 64bit. So in Python, you have to handle negative numbers, e.g. subtraction, separately, or limit the number of bits by hand.
The binary, 2’s complement representation of -4 is
...11100
Yes, I really do mean infinitely many 1‘s to the left; this is a binary repeating numeral. Technically, 4 is a repeating numeral too:
...00100
it’s just repeating 0‘s to the left.
Your addition problem is
...11100
+ ...00100
--------------------
...00000
The operators ^, <<, and & have no trouble computing with infinitely many binary digits, but the problem is that there are infinitely many carries, and you are computing them one digit at a time. This will never finish.
Thus, you have to recognize when this algorithm will get stuck in this situation and do something else to account for it.
You don’t run into this problem in C/C++, because, for example, if int is 32-bits, then all of the digits except for the rightmost 31 digits get collapsed into a single bit, so it does the remaining carries all at once.
However, technically speaking, the meaning of left shifting an int is in terms of the value as an integer, rather than as a bit pattern, so you are invoking undefined behavior if the two most significant bits carry are ever different, because then carry << 1 would produce an overflow).
Following the great explanation by @Hurkyl , I stepped through the algorithm for a=4 and b=-4 using the fact that python implements an infinite two’s compliment representation:
Step 0:
a = ...(0)...000100
b = ...(1)...111100
carry = a & b = ...(0)...000100
a = a ^ b = ...(1)...111000
b = carry << 1 = ...(0)...001000
Step 1:
a = ...(1)...111000
b = ...(0)...001000
carry = a & b = ...(0)...001000
a = a ^ b = ...(1)...110000
b = carry << 1 = ...(0)...010000
Step 2:
a = ...(1)...110000
b = ...(0)...010000
carry = a & b = ...(0)...010000
a = a ^ b = ...(1)...100000
b = carry << 1 = ...(0)...100000
It is clear that there needs to be an effective cutoff to emulate something like a 32-bit signed two’s compliment integer. Once the carry bit bubbles up beyond highest bit the algorithm needs to be halted. The following appears to work:
MAX_BIT = 2**32
MAX_BIT_COMPLIMENT = -2**32
def aplusb(a, b):
while b != 0:
if b == MAX_BIT:
return a ^ MAX_BIT_COMPLIMENT
carry = a & b
a = a ^ b
b = carry << 1
return a
Results:
>>> aplusb(100,-100)
0
>>> aplusb(100,-99)
1
>>> aplusb(97,-99)
-2
>>> aplusb(1000,-500)
500
>>> aplusb(-1000,8000)
7000
If 1-bit binary math operations (^) are forbidden, go for unary!
from itertools import chain
def unary(x):
"Unary representation of x"
return ''.join(['x' for _ in range(0,x)])
def uplus(x, y):
"Unary sum of x and y"
return [c for c in chain(x,y)]
def plus(i, j):
"Return sum calculated using unary math"
return len(uplus(unary(i), unary(j)))
My solution:
def foo(a, b):
"""iterate through a and b, count iteration via a list, check len"""
x = []
for i in range(a):
x.append(a)
for i in range(b):
x.append(b)
print len(x)
As already stated, bitwise is better.
That is because python is not normally using 32 bit signed int.
See:ctypes.c_int32
Accepted solution:
class Solution:
"""
@param a: The first integer
@param b: The second integer
@return: The sum of a and b
"""
def aplusb(self, a, b):
import ctypes
a = ctypes.c_int32(a).value
a = ctypes.c_int32(a).value
while b != 0:
carry = ctypes.c_int32(a & b).value
a = ctypes.c_int32(a ^ b).value
b = ctypes.c_int32(carry << 1).value
return a
I was trying to solve an old question:
Write a function that add two [integer] numbers A and B. You should not use + or any arithmetic operators.
The best solution is like this, quoted from “LintCode-A+B Problem“:
For a + b in any base, we can treat the plus as two part: 1. a + b without carry; 2. the carry generated by a +b. The a+b then equals to part 1 plus part 2. If part1+part2 generates more carry, we can then repeat this procedure, until there is no carry.
I can understand this algorithm and everything seems good, so I tested it on lintcode with code pasted below.
class Solution:
"""
@param a: The first integer
@param b: The second integer
@return: The sum of a and b
"""
def aplusb(self, a, b):
while b != 0:
carry = a & b
a = a ^ b
b = carry << 1
return a
But surprisingly, it gave me Time Limit Exceeded error in test case [100, -100]. So I ran it locally and print a, b for every loop:
(-8, 8)
(-16, 16)
(-32, 32)
(-64, 64)
(-128, 128)
(-256, 256)
(-512, 512)
(-1024, 1024)
(-2048, 2048)
(-4096, 4096)
(-8192, 8192)
(-16384, 16384)
(-32768, 32768)
(-65536, 65536)
(-131072, 131072)
...
The calculation is correct, so I think this algorithm does not work for such input but when I wrote the same algorithm in C++, it just works:
class Solution {
public:
int aplusb(int a, int b) {
while (b!=0){
int carry = a & b;
a = a^b;
b = carry << 1;
}
return a;
}
};
I don’t know what should be asked exactly, basically the questions are:
- Why does C++ give the correct output
0while Python doesn’t? - If I use Python, how do I modify this algorithm to make it work?
The problem are negative numbers, or, how they are represented. In Python integer numbers have arbitrary accuracy, while C++ ints are 32bit or 64bit. So in Python, you have to handle negative numbers, e.g. subtraction, separately, or limit the number of bits by hand.
The binary, 2’s complement representation of -4 is
...11100
Yes, I really do mean infinitely many 1‘s to the left; this is a binary repeating numeral. Technically, 4 is a repeating numeral too:
...00100
it’s just repeating 0‘s to the left.
Your addition problem is
...11100
+ ...00100
--------------------
...00000
The operators ^, <<, and & have no trouble computing with infinitely many binary digits, but the problem is that there are infinitely many carries, and you are computing them one digit at a time. This will never finish.
Thus, you have to recognize when this algorithm will get stuck in this situation and do something else to account for it.
You don’t run into this problem in C/C++, because, for example, if int is 32-bits, then all of the digits except for the rightmost 31 digits get collapsed into a single bit, so it does the remaining carries all at once.
However, technically speaking, the meaning of left shifting an int is in terms of the value as an integer, rather than as a bit pattern, so you are invoking undefined behavior if the two most significant bits carry are ever different, because then carry << 1 would produce an overflow).
Following the great explanation by @Hurkyl , I stepped through the algorithm for a=4 and b=-4 using the fact that python implements an infinite two’s compliment representation:
Step 0:
a = ...(0)...000100
b = ...(1)...111100
carry = a & b = ...(0)...000100
a = a ^ b = ...(1)...111000
b = carry << 1 = ...(0)...001000
Step 1:
a = ...(1)...111000
b = ...(0)...001000
carry = a & b = ...(0)...001000
a = a ^ b = ...(1)...110000
b = carry << 1 = ...(0)...010000
Step 2:
a = ...(1)...110000
b = ...(0)...010000
carry = a & b = ...(0)...010000
a = a ^ b = ...(1)...100000
b = carry << 1 = ...(0)...100000
It is clear that there needs to be an effective cutoff to emulate something like a 32-bit signed two’s compliment integer. Once the carry bit bubbles up beyond highest bit the algorithm needs to be halted. The following appears to work:
MAX_BIT = 2**32
MAX_BIT_COMPLIMENT = -2**32
def aplusb(a, b):
while b != 0:
if b == MAX_BIT:
return a ^ MAX_BIT_COMPLIMENT
carry = a & b
a = a ^ b
b = carry << 1
return a
Results:
>>> aplusb(100,-100)
0
>>> aplusb(100,-99)
1
>>> aplusb(97,-99)
-2
>>> aplusb(1000,-500)
500
>>> aplusb(-1000,8000)
7000
If 1-bit binary math operations (^) are forbidden, go for unary!
from itertools import chain
def unary(x):
"Unary representation of x"
return ''.join(['x' for _ in range(0,x)])
def uplus(x, y):
"Unary sum of x and y"
return [c for c in chain(x,y)]
def plus(i, j):
"Return sum calculated using unary math"
return len(uplus(unary(i), unary(j)))
My solution:
def foo(a, b):
"""iterate through a and b, count iteration via a list, check len"""
x = []
for i in range(a):
x.append(a)
for i in range(b):
x.append(b)
print len(x)
As already stated, bitwise is better.
That is because python is not normally using 32 bit signed int.
See:ctypes.c_int32
Accepted solution:
class Solution:
"""
@param a: The first integer
@param b: The second integer
@return: The sum of a and b
"""
def aplusb(self, a, b):
import ctypes
a = ctypes.c_int32(a).value
a = ctypes.c_int32(a).value
while b != 0:
carry = ctypes.c_int32(a & b).value
a = ctypes.c_int32(a ^ b).value
b = ctypes.c_int32(carry << 1).value
return a