Part 1 (link) introduces basic bitwise operations. This is part 2 and it's mainly about (in)famous bitsets and example problems. Also, see links to very useful advanced stuff at the bottom. EDIT: here's video version of this blog (on my Youtube channel).

### Built-in functions

In C++, `__builtin_popcount(x)`

returns popcount of a number — the number of ones in the binary representation of $$$x$$$. Use `__builtin_popcountll(x)`

for long longs.

There are also `__builtin_clz`

and `__builtin_ctz`

(and their long long versions) for counting the number of leading or trailing zeros in a **positive** number. Read more here. Now, try to solve these two simple tasks in $$$O(1)$$$, then open the spoiler to check the solution:

**Compute the biggest power of 2 that is a divisor of x. Example: f(12) = 4**

**Compute the smallest power of 2 that is not smaller than x. Example: f(12) = 16**

While popcount is often needed, I rarely use the other two functions. During a contest, I would solve the two tasks above in $$$O(\log x)$$$ with simple while-loops, because it's easier and more intuitive for me. Just be aware that these can be done in $$$O(1)$$$, and use clz or ctz if you need to speed up your solution.

### Motivation behind bitsets

Consider this problem: There are $$$N \leq 5000$$$ workers. Each worker is available during some days of this month (which has 30 days). For each worker, you are given a set of numbers, each from interval $$$[1, 30]$$$, representing his/her availability. You need to assign an important project to two workers but they will be able to work on the project only when they are both available. Find two workers that are best for the job — maximize the number of days when both these workers are available.

You can compute the intersection of two workers (two sets) in $$$O(30)$$$ by using e.g. two pointers for two sorted sequences. Doing that for every pair of workers is $$$O(N^2 \cdot 30)$$$, slightly too slow.

We can think about the availability of a worker as a binary string of length $$$30$$$, which can be stored in a single `int`

. With this representation, we can count the intersection size in $$$O(1)$$$ by using `__builtin_popcount(x[i] & x[j])`

. The complexity becomes $$$O(N^2)$$$, fast enough.

What if we are given the availability for the whole year or in general for $$$D$$$ days? We can handle $$$D \leq 64$$$ in a single `unsigned long long`

, what about bigger $$$D$$$?

We can split $$$D$$$ days into convenient parts of size $$$64$$$ and store the availability of a single worker in an array of $$$\frac{D}{64}$$$ unsigned long longs. Then, the intersection can be computed in $$$O(\frac{D}{64})$$$ and the whole complexity is $$$O(N^2 \cdot \frac{D}{64})$$$.

**code**

So, we can simulate a long binary number with multiple unsigned long longs. The implementation isn't that bad but doing binary shifts would be quite ugly. Turns out all of this can be done with bitsets easily.

### Bitsets

`bitset<365>`

is a binary number with $$$365$$$ bits available, and it supports most of binary operations. The code above changes into simple:

**code**

Some functions differ, e.g. `x.count()`

instead of `__builtin_popcount(x)`

but it's only more convenient. You can read and print binary numbers, construct a bitset from int or string `bitset<100> a(17); bitset<100> b("1010");`

. You can even access particular bits with `b[i]`

. Read more in C++ reference https://en.cppreference.com/w/cpp/utility/bitset.

Note that the size of the bitset must be a constant number. You can't read $$$n$$$ and then declare `bitset<n> john;`

. If $$$n$$$ is up to $$$100$$$, just create `bitset<100>`

.

The complexity of bitwise operations is $$$O(\frac{size}{32})$$$ or $$$O(\frac{size}{64})$$$, it depends on the architecture of your computer.

### Problems

**P1. Different numbers** — You are given a sequence of $$$N \leq 10^7$$$ numbers, each from interval $$$[0, 10^9]$$$. How many different values appear in the sequence? Don't use `set`

or `unordered_set`

because they quite slow.

**solution**

**P2. Knapsack** — You are given $$$N \leq 1000$$$ items, each with some weight $$$w_i$$$. Is there a subset of items with total weight exactly $$$W \leq 10^6$$$?

**solution**

**P3. Triangles in a graph** — Given a graph with $$$n \leq 2000$$$ vertices and $$$m \leq n \cdot (n - 1) / 2$$$ edges, count triples of vertices $$$a, b, c$$$ such that there are edges $$$a-b$$$, $$$a-c$$$ and $$$b-c$$$.

**hint**

**P4. Chef and Queries** — https://www.codechef.com/problems/CHEFQUE (easy)

**P5. Odd Topic** — https://www.codechef.com/AABH2020/problems/ODTPIC (medium), thanks to Not-Afraid for the suggestion

**P6. Funny Gnomes** — https://www.codechef.com/problems/SHAIKHGN (hard)

### Bonuses

1) `m & (m-1)`

turns off the lowest bit that was set to $$$1$$$ in a positive number $$$m$$$. For example, we get $$$24$$$ for $$$m = 26$$$, as $$$11010$$$ changes into $$$11000$$$. Explanation on quora

2) A quite similar trick allows us to iterate efficiently over all submasks of a mask, article on cp-algorithms / e-maxx. This article also explains why masks-submasks iteration is $$$O(3^n)$$$.

3) DP on broken profile (grid dp) — https://cp-algorithms.com/dynamic_programming/profile-dynamics.html

4) SOS dp (sum over subset) — https://codeforces.com/blog/entry/45223 & https://www.youtube.com/watch?v=Lpvsd8WpbWc&t=5m4s

5) `_Find_next`

function and complexity notation for bitsets — https://codeforces.com/blog/entry/43718

~~I will add links to some problems in online judges, feel free to suggest some in the comments.~~ I think that bonuses 3 and 4 lack some explanation with drawings, maybe I will make some soon.

This problem from Codechef is a perfect example where bitsets comes handy.

Would you solve a couple of dp_bitmask problems in yournext post?

amazing work

It's worth noting that after adding

`#pragma GCC target("popcnt")`

`__builtin_popcount()`

is replaced to corresponding machine instruction (look at the difference). In my test this maked x2 speed up.`bitset::count()`

use`__builtin_popcount()`

call in implementation, so it's also affected by this.this problem on codechef and see comments in editorial some solved using bitset to speed up the brute force solution

Another problem where bitsets come handy.

What is the difference between $$$O(1)$$$ and $$$O(30)$$$? How is using bitsets 'true' constant time?

Good question. You can think about the number of bits in architecture (usually $$$32$$$ or $$$64$$$) as a variable and then we're talking about $$$O(n)$$$ vs. $$$O(\frac{n}{b})$$$. You come up with an algorithm and it will have speed dependent on the machine it will be run on.

Thanks for the informative tutorial. The following is a C++14 demo program on using the 32/64-bit popcount/clz/ctz bit-counting built-in functions in C++. The templates included in this demo may help beginners to use these functions without worrying about memorizing their slightly inconvenient names.

demo

Hi Errichto, I had asked you during a recent stream whether it is possible to do an iterative version of going through all bitmasks of length $$$N$$$ in $$$O(2^N)$$$ time, instead of $$$O(N*2^N)$$$. We can do it in recursion, by passing what we need to keep track of as an argument in the recursive call. You had suggested that it is possible via some DP and bit tricks. So I thought a little bit, and the trick used in Fenwick Tree came to mind. I have written a simple code, that for each mask will store all the bits that are on in it. This is the link.

I had a question regarding this, we can get the value of the last set bit using &, but we need to find the position of the bit, i.e. if last set bit value is $$$100$$$ then position is $$$2$$$. To find position, we will still have to iterate over the length of the bitmask right? Does the __builtin_ctz function take 1 operation or length of bits number of operations? What about the bitwise & and $$$|$$$ operators? Ofcourse maximum operations ( that we ever do in CP, mostly ) will only have 64 bit numbers, so you could say O(64) = O(1), but I want to know about the constants exactly.

You can use some magic like https://www.chessprogramming.org/BitScan#With_isolated_LS1B to get the position instead of the value. Don't ask me how it works, idk about that but it works. You can also do it in O(logBITS) with binary search (as mentioned in the link I gave too under divide and conquer).

Also maybe more important to know, most often optimizing these O(BITS) that you loop over bits is overkill and not necessary because you usually do some really light operations also not involving cache misses so it's actually faster than you'd expect.

Wow, the divide and conquer method was mind blowing. Another question I have always had in my mind was, when you do $$$a$$$&$$$b$$$, you must make $$$log(max(a,b))$$$ operations right? So does that make the first part of finding the last bit value itself useless? Also, does this mean Fenwick Tree has $$$log^2(N)$$$ update instead of $$$log(N)$$$?

No, a&b should be done in few cycles and should actually be cheaper than a+b because the ALU in cpus for sure have a bitwise and operation and it doesn't need to carry over the carry bit so it's fully parallel for the bits.

Oh yeah, makes sense. ( Le me: Recalls Computer Architecture course, wonders what use is university )

Update :)

I made two Youtube videos (part 1 and part2) which cover the same content as my blogs.

And added bonus (5) with a function to find the next bit 1 in a bitset,

`_Find_next(i)`

.I have a question

Can I implement find_next_bit() manually under O(log n) ?

Shift and then https://www.geeksforgeeks.org/position-of-rightmost-set-bit/

IGMaster ! I have 2 questions

Is it danger if I use signed variable ?

But if I want to know next set bit of (n) starting from (p). I still need to know the variable (n) which is from position (p) — 1 -> 0 right ?

Thanks IGMaster

IGMaster ! Can I find next unset bit without reversing it ?

I edited: Master -> IGMaster

Stop calling me master. I would negate a number first (flip every bit) but for sure something equivalent can be done. I think you're focusing too much on this. When I had your rating, I didn't know any of this.

Thanks and sorry for calling like that.

I just curious to know if there exists a simpler & effeciency bitwise implementation

Errichto: Stop calling me master.

SPyofgame: I edited: Master -> IGMaster

XDDDDDDDDDDDDDDDDDDDDDDDDDD

Edit: This should be deleted, I was just toasted and put 1LL << 31 instead of 1LL << 32.

... ... ...

Hey people, I need some help.

For chef and queries, I have the current si updating like this after each iteration:

s = ((a*s+b)% MAX_W);

where MAX_W is 1LL << 31;

the type of a, s and b is long long.

I don't get why this messes up the algorithm. If I change their types to unsigned int and remove the % operation in the update, the algorithm works fine. However, I think that the first approach should be working fine, as it is how the problem says we need to calculate si.

Can someone please guide me through the reasoning behind the issue?

Thanks!

Please check these 2 codes:

Problem:- https://codeforces.com/problemset/problem/550/B WA:- https://ideone.com/0eCeTX AC:- https://codeforces.com/contest/550/submission/11417471

The only difference is, the AC code has 0 based array, and mine has 1 based array. Is it wrong to make 1 based array for bitmasking?

Then you need a bitmask with bits on positions 1 through N, so an even number up to 2^(n+1). I don't think you understand what your code does.

Thanks. Problem fixed.

Errichto What is the complexity of

`__builtin_popcount(N)`

? (Assuming $$$N$$$ can be as arbitrarily large as possible)It will take O(32)

So in other words it's just $$$\mathcal{O}(\text{number of bits})$$$ ? I saw this comment claiming that's it's $$$\mathcal{O}(\log{(\text{number of bits})})$$$

If it is so, I don't know why it is and what exactly is working at architecture or compiler level. If anyone replies at your comment plz let me know as I also what to know how it can do in log(bits) time.

Treat int popcount as an $$$O(1)$$$ operation but it's just slightly slower than simpler operations like xor. The complexity becomes $$$O(L/32)$$$ for bitsets of length $$$L$$$.