Intro to Bitwise Operators

At this point, you should already be familiar with the three main bitwise operators (AND, OR, and XOR). Let's take a look at some examples to better understand them.

Solution - Take a Guess

In fact, we can figure out the sum of two numbers using just their AND, OR and XOR values! Suppose we know their XOR values, we can use the following property:

$a + b = 2 \cdot (a \& b) + a \oplus b$

The proof is as follows:

$a \oplus b$ is essentially just $a + b$ in base $2$ but we never carry over to the next bit. Recall a bit in $a \oplus b$ is $1$ only if the bit in $a$ is different from the bit in $b$, thus one of them must be a $1$. However, when we add two $1$ bits, we yield a $0$, but we do not carry that $1$ to the next bit. This is where $a \& b$ comes in.

$a \& b$ is just the carry bits themselves, since a bit is $1$ only if it's a $1$ in both $a$ and $b$, which is exactly what we need. We multiply this by $2$ to shift all the bits to the left by one, so every value carries over to the next bit.

To acquire the XOR values of the two numbers, we can use the following:

$a \oplus b = \lnot(a \& b) \& (a | b)$

The proof is as follows:

Recall a bit in $a \oplus b$ is $1$ only if the bit in $a$ is different from the bit in $b$. By negating $a \& b$, the bits that are left on are in the following format:

• If it's $1$ in $a$ and $0$ in $b$
• If it's $0$ in $a$ and $1$ in $b$
• If it's $0$ in $a$ and $0$ in $b$

Now this looks pretty great, but we need to get rid of the third case. By taking the bitwise AND with $a | b$, the ones that are left on is only if there is a $1$ in either $a$ or $b$. Obviously, the third case isn't included in $a | b$ since both bits are off, and we successfully eliminate that case.

Now that we can acquire the sum of any two numbers in two queries, we can easily solve the problem now. We can find the values of the first three numbers of the array using a system of equations involving their sum (note $n \geq 3$). Once we have acquired their independent values, we can loop through the rest of the array.

Copy
#include <bits/stdc++.h>
using namespace std;

int ask(string s, int a, int b) {
	cout << s << ' ' << a << ' ' << b << endl;
	int res;
	cin >> res;
	return res;
}

Copy
import java.io.*;
import java.util.*;

public class TakeAGuess {
	private static BufferedReader br =
	    new BufferedReader(new InputStreamReader(System.in));

	public static void main(String[] args) throws IOException {
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());

Copy
def ask(s: str, a: int, b: int) -> int:
	print(f"{s} {a} {b}", flush=True)
	return int(input())


def sum(a: int, b: int) -> int:
	""":return: the sum of the elements at a and b (0-indexed)"""
	a += 1
	b += 1
	and_ = ask("and", a, b)

Example - Addition

Now let's take a look at implementing addition and multiplication using only bitwise operators. Before we do so, though, try implementing addition using bitwise operators on your own! You can test your implementation here.

Solution - Addition

If we perform addition without carrying, then we are simply performing the XOR (^) operator. Then, the bits that we carry over are those equivalent to $1$ in both numbers: $a\&b$.

Copy
int add(int a, int b) {
	while (b > 0) {
		int carry = a & b;
		a ^= b;
		b = carry << 1;
	}
	return a;
}

Copy
public static int add(int a, int b) {
	while (b > 0) {
		int carry = a & b;
		a ^= b;
		b = carry << 1;
	}
	return a;
}

Copy
def add(a: int, b: int) -> int:
	while b > 0:
		carry = a & b
		a ^= b
		b = carry << 1
	return a

Example - Multiplication

Now try implementing multiplication using bitwise operators! If you want to test your implementation of multiplication, you can do so here.

Solution - Multiplication

For simplicity, we will use the sum functions defined above. If we divide up $b$ into $2^{b_1}+2^{b_2}+\dots+2^{b_n}$, we get the following:

(This same idea is used in binary exponentiation!)

Copy
int prod(int a, int b) {
	int c = 0;
	while (b > 0) {
		if ((b & 1) == 1) {
			c = add(c, a);  // Use the addition function we coded previously
		}
		a <<= 1;
		b >>= 1;
	}
	return c;
}

Copy
public static int prod(int a, int b) {
	int c = 0;
	while (b > 0) {
		if ((b & 1) == 1) {
			c = add(c, a);  // Use the addition function we coded previously
		}
		a <<= 1;
		b >>= 1;
	}
	return c;
}

Copy
def prod(a: int, b: int) -> int:
	c = 0
	while b > 0:
		if b & 1:
			c = add(c, a)  # Use the addition function we coded previously
		a <<= 1
		b >>= 1
	return c

XOR Operation

Perhaps one of the most common binary operations in practice is bitwise xor. The special property that differentiates it from the other bitwise operations is that xor is its own inverse, i.e. $x \oplus x = 0$.

Explanation

Let's analyze the bits of each xor-sum independently. In this case, we'll check for every bit position, i.e. for every power of two, whether or not it's set in the xor-sum of every contiguous subsequence. This way, we transformed the problem into counting the number of subsequences with non-zero xor-sum; this will be applied on the array form by the bits on the same position in all the values.

Implementation

Time Complexity: $\mathcal{O}(N)$

Copy
#include <iostream>
#include <numeric>
#include <vector>

using namespace std;

int main() {
	int n;
	cin >> n;
	vector<int> v(n);

Copy
n = int(input())
v = list(map(int, input().split()))

res = 0
"""
For every bit position
check if it's set on in the xor-sum of every subsequence
"""
for i in range(30):
	s = 0

Quiz