You can watch my Youtube video (link) with the same content as this blog. Anyway, enjoy.

### Introduction

Let's learn bitwise operations that are useful in Competitive Programming. Prerequisite is knowing the binary system. For example, the following must be clear for you already.

Keep in mind that we can pad a number with leading zeros to get the length equal to the size of our type size. For example, `char`

has $$$8$$$ bits and `int`

has $$$32$$$.

### Bitwise AND, OR, XOR

You likely already know basic logical operations like AND and OR. Using `if(condition1 && condition2)`

checks if both conditions are true, while OR (`c1 || c2`

) requires at least one condition to be true.

Same can be done bit-per-bit with whole numbers, and it's called bitwise operations. You must know bitwise AND, OR and XOR, typed respectively as `& | ^`

, each with just a single character. XOR of two bits is $$$1$$$ when exactly one of those two bits is $$$1$$$ (so, XOR corresponds to `!=`

operator on bits). There's also NOT but you won't use it often. Everything is explained in Wikipedia but here's an example for bitwise AND. It shows that `53 & 28`

is equal to $$$20$$$.

```
53 = 110101
28 = 11100
110101
& 11100 // imagine padding a shorter number with leading zeros to get the same length
-------
010100 = 20
```

**C++ code for experimenting**

### Shifts

There are also bitwise shifts `<<`

and `>>`

, not having anything to do with operators used with `cin`

and `cout`

.

As the arrows suggest, the left shift `<<`

shifts bits to the left, increasing the value of the number. Here's what happens with `13 << 2`

— a number $$$13$$$ shifted by $$$2$$$ to the left.

```
LEFT SHIFT RIGHT SHIFT
13 = 1101 13 = 1101
(13 << 2) = 110100 (13 >> 2) = 11
```

If there is no overflow, an expression `x << b`

is equal to $$$x \cdot 2^b$$$, like here we had `(13 << 2) = 52`

.

Similarly, the right shift `>>`

shifts bits to the right and some bits might disappear this way, like bits `01`

in the example above. An expression `x >> b`

is equal to the floor of $$$\frac{x}{2^b}$$$. It's more complicated for negative numbers but we won't discuss it.

### So what can we do?

$$$2^k$$$ is just `1 << k`

or `1LL << k`

if you need long longs. Such a number has binary representation like `10000`

and its AND with any number $$$x$$$ can have at most one bit on (one bit equal to $$$1$$$). This way we can check if some bit is on in number $$$x$$$. The following code finds ones in the binary representation of $$$x$$$, assuming that $$$x \in [0, 10^9]$$$:

`for(int i = 0; i < 30; i++) if((x & (1 << i)) != 0) cout << i << " ";`

(we don't have to check $$$i = 30$$$ because $$$2^{30} > x$$$)

And let's do that slightly better, stopping for too big bits, and using the fact that `if(value)`

checks if `value`

is non-zero in C++.

`for(int i = 0; (1 << i) <= x; i++) if(x & (1 << i)) cout << i << " ";`

Consider this problem: You are given $$$N \leq 20$$$ numbers, each up to $$$10^9$$$. Is there a subset with sum equal to given goal $$$S$$$?

It can be solved with recursion but there's a very elegant iterative approach that iterates over every number $$$x$$$ from $$$0$$$ to $$$2^n - 1$$$ and considers $$$x$$$ to be a binary number of length $$$n$$$, where bit $$$1$$$ means taking a number and bit $$$0$$$ is not taking. Understanding this is crucial to solve any harder problems with bitwise operations. Analyze the following code and then try to write it yourself from scratch without looking at mine.

**solution code**

Two easy problems where you can practice iterating over all $$$2^N$$$ possibilities:

- https://codeforces.com/problemset/problem/1097/B

- https://codeforces.com/problemset/problem/550/B

### Speed

Time complexity of every bitwise operation is $$$O(1)$$$. These operations are very very fast (well, popcount is just fast) and doing $$$10^9$$$ of them might fit in 1 second. You will later learn about bitsets which often produce complexity like $$$O(\frac{n^2}{32})$$$, good enough to pass constraints $$$n \leq 10^5$$$.

I will welcome any feedback. Coming next: popcount, bitsets, dp with bitmasks. ~~I will also make a YT video on this.~~ YT video link is at the top.

Part 2 link

Here's a trick question: how much is

`1LL<<66`

?Nasal demons

Well, it's UB, what should I expect? My guess is $$$0$$$ on most machines, but let's hear what it is.

Or are you saying it isn't UB?

Yes, it's UB. Not a question specifically for you, just anyone who doesn't have much experience with bitwise operations, because it's so counterintuitive.

I would expect to see 4 (at least if values are runtime and not known to compiler)

what do you mean by UB

google -> what is UB in programming?

I tried this code

It computes a non-zero value for shifting by

`i`

but computes 0 for shifting by constantYeah, undefined behaviour is undefined.

It will not be UB soon, since C++20.

source? I doubt they would go with such pessimization.

according to cppreference (I know, it's not always correct ) there are changes in c++20 but still

You are right, I didn't see the restriction on the right operand.

But it seems that now

`8 << 30`

will be defined (0), however`1 << 33`

will remain undefined behaviour.Mentioned here (in

since C++20section) https://en.cppreference.com/w/cpp/language/operator_arithmetic#Bitwise_shift_operators`8<<30`

is plain signed overflow which you should avoid anyway,`8U<<30`

was defined for a long time already. C++20 just makes the left shift an exception from the rule that signed overflow is UB.When red's code is

`mask & (1 << i)`

instead of`mask >> i & 1`

....When you check the divisibility by $$$2$$$ with bitwise operations... Are you so sure that your method is simpler and easier to understand for beginners?

actually, point of my massage is about checking bits in long long mask

beginners (and not so) very often making this mistake

about understanding of methods — i think it same

Ok, that's a good point. Well, you might need a power of two in your solution, so you need to know how to use

`1LL << k`

anyway, what is mentioned in the article. Though, now I see the advantage of your method.>beginner uses it with 64-bit bitmasks

The introductory part might be a little confusing: you have

`^`

in your list of logical operators and again in your list of bitwise operators.thanks, added a note for that

EDIT: I modified that part more in response to a comment below.

It might be better to use

`^^`

to denote logical XOR and simply note that languages tend not to include it. I don't see that`^`

is analogous to`&&`

and`||`

the way you are listing things.I think some think like

`if (a ^^ b)`

can be done with`if (a ==! b)`

so they dont add the logical XOR`^^`

in C++Amazing errichto waiting for dp with bitmasks.

How soon can we expect the YT videos??

Sir thank u so much for this . i Have been looking for good researce like this

Looking forward for the next tutorial. Having trouble with those.

One quick shortcut for

`to_binary(num)`

while debugging is`bitset<8>(num)`

.You can do

`cout << bitset<8>(num) << endl;`

can you please explain me what is to_binary(num) please

expand this part in blog C++ code for experimenting

It's my function from the blog, you can ignore it. Just remember that

`bitset<8>(x)`

will cast variable $$$x$$$ into bitset.What is it's Data type ?

Is it string ? Integer ?

Rezwan.Arefin01 , Errichto

Nice informative blog

//////please someone help me with this code of 1st problem, i am getting wrong answer at test case 21

## include <bits/stdc++.h>

## define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

## define ll long long

using namespace std;

int main() { IOS; ll n,k,i,ans,x,s,mask,ch=0,r=0; cin>>n; vector v; for(i=0;i<n;i++){ cin>>x; ch+=x; v.push_back(x); } if(ch%360==0) cout<<"YES"<<endl; else if(ch%2==0){ ch/=2; for(mask=0;mask<(1LL<<n);mask++){ ans=0; for(i=0;i<n;i++){ if(mask & (1LL<<i)){ ans+=v[i]; } if(ans==ch) { cout<<"YES"<<endl; return 0; } }

}

} else{ cout<<"NO"<<endl; }

}

Updated Your solution

You can visit above link to see what wrong you're doing.

Explaination: when the input is 10 10 140 120 100 , then it can be done like this

10 + (-10) + 140 + 120 + 100 = 360 which is basically the starting point 0.

So, instead of checking if it is the half of sum, we should check are we able to rotate 360*n deg after nulliying some values. (here 10 and -10).

So the statement would be like this:

`(ch-2*ans)%360 == 0)`

I hope this helps.

Thanks for such an amazing blog. Special thanks to give additional links for problem to understand the concept clearly.

`bitset<>`

when used in conversion to binary to decimal :`cout<< bitset<8>("1001").to_ulong() << endl;`

Thanks for the tutorial. can anyone suggest some problem..so i can learn more technique- related to this topic.

$$$ O \left( \frac{n^2}{32} \right) = O(n^2)$$$, would be more correct to say $$$O \left(\frac{n^2}{\text{machine word}} \right)$$$.