### Roundgod's blog

By Roundgod, history, 3 years ago, Hello, Codeforces! The reason why I am writing this blog is that my ACM/ICPC teammate calabash_boy is learning this technique recently(he is a master in string algorithms,btw), and he wanted me to provide some useful resources on this topic. I found that although many claim that they do know this topic well, problems concerning inclusion-exclusion principle are sometimes quite tricky and not that easy to deal with. Also, after some few investigations, the so-called "Inclusion-Exclusion principle" some people claim that they know wasn't the generalized one, and has little use when solving problems. So, what I am going to pose here, is somewhat the "Generalized Inclusion-Exclusion Principle". Most of the describing text are from the graduate text book Graduate Text in Mathematics 238, A Course in Enumeration, and the problems are those that I encountered in real problem set, so if possible, I'll add a link to the real problem so that you can solve it by yourself. I'll start with the basic formula, one can choose to skip some of the text depending on your grasp with the topic.

Consider a finite set and three subsets , To obtain , we take the sum + + . Unless are pairwise disjoint, we have an overcount, since the elements of has been counted twice. So we subtract . Now the count is correct except for the elements in which have been added three times, but also subtracted three times. The answer is therefore , or equivalently, The following formula addresses the case applied to more sets.

The Restricted Inclusion-Exclusion Principle. Let be subsets of . Then This is a formula which looks familiar to many people, I'll call it The Restricted Inclusion-Exclusion Principle, it can convert the problem of calculating the size of the union of some sets into calculating the size of the intersection of some sets. It's not hard to prove the correctness of this formula, we can just check how often an element is counted in both sides. If , then it's counted once on either side. Suppose , and more precisely, that is in exactly of the sets . The count on the left-hand side is , and on the right hand side, we have for , thus the equality holds.

Example 1. Let's see an example problem Co-prime where this principle could be applied: Given , you need to compute the number of integers in the interval such that is coprime with , that is, . There are testcases. Constraints: , .

Solution

The standard interpretation leads to the principle of inclusion-exclusion. Suppose we are given a set , called the universe, and a set of properties that the elements of may or may not process. Here we can define the properties as we like, such as , , or even . Let be the subset of elements that enjoy property (and possibly others). Then is the number of elements that process none of the properties. Clearly, is the set of elements that possess the properties (and maybe others). Using the notation  we arrive at the inclusion-exclusion principle.

Inclusion-Exclusion Principle. Let be a set, and a set of properties. Then The formula becomes even simpler when depends only on the size . We can then write for , and call a homogeneous set of properties, and in this case also depends only on the cardinality of . Hence for homogeneous properties, we have This is the very essence of Inclusion-Exclusion Principle . Please make sure you understand every notation before you proceed. One can figure out, by letting , we arrive at the restricted inclusion-exclusion principle.

Example 2. This problem Character Encoding requires you to compute the number of solutions to the equation , satisfying that , modulo . Constraints: . Hint: the number of non-negative integer solutions to is given by .

Solution

Example 3. Well, this one is a well-known problem. K-Inversion Permutations. The statement is neat and simple. Given N, K, you need to output the number of permutations of length N with K inversions, taken modulo . Constraint: .

Solution

Example 4. This problem comes from XVIII Open Cup named after E.V. Pankratiev. Grand Prix of Gomel.Problem K,(Yes, created by tourist:) ) which gives a integer , and requires one to find out the number of non-empty sets of positive integers, such that their greatest common divisor is , and their least common multiple is , taken modulo .Constraint: .

Solution

I guess that's the end of this tutorial. IMO, understanding all the solutions to the example problems above is fairly enough to solve most of the problems that require the inclusion-exclusion principle(but only for the IEP part XD). This is my first time of writing an tutorial. Please feel free to ask any questions in the comments below or point out any mistakes in the blog. Comments (24)
 » Auto comment: topic has been updated by Roundgod (previous revision, new revision, compare).
 » I think in the formula of example 3, it should be f(s) instead of f(i). Furthermore, can you explain how did you calculate f(n)?
•  » » Fixed, thank you! Since each possible combination of j distinct numbers in the range [1, N] that sum to n contributes to f(n) according to the inclusion-exclusion princple, and the inclusion-exclusion coefficient(that is, the ( - 1)i term) is determined by the parity of j. So if we can find out the number of ways to choose j distinct numbers in the range [1, N] that sum to n, which is g(n, j), efficiently, then we can then calculate f(n).
•  » » » Thanks, got it.
 » 3 years ago, # | ← Rev. 2 →   Can Someone Please tell me how to solve this Problem. I think it can also be done using inclusion exclusion but dont know how.UPD: Image of the problem: link
•  » » I don't think it can be solved using inclusion-exclusion. You can use digit dp to solve it. Here's my solution.
 » This is really a nice tutorial, liked it. Hope to get more nice tutorials like this in future.
 » In the formula for f(n), the condition should be i>=0, so that f(0) = 1.Also, the formula for g(i,j) is incomplete. For i = 12, j = 3, N = 5, it gives answer 3, whereas correct answer is 1 i.e, {{3, 4, 5}}.
•  » » Oops. I wrote it wrong. It should be $g(i,j)=g(i-j,j)+g(i-1,j-1)$. Now fixed, thank you!
•  » » » 3 years ago, # ^ | ← Rev. 2 →   Sorry, but the formula still gives wrong answer for the example I mentioned.According to the updated formula for N = 5, $g(12, 3) = g(9, 3) + g(11, 2)$.Now, $g(9, 3) = 2$ i.e, {{1,3,5},{2,3,4}} and $g(11, 2) = 0$ i.e, {}, which implies $g(12, 3) = 2 + 0 = 2$.This is wrong as the only way to do this is {3,4,5}, so $g(12,3)=1$.Also, would you please explain how are you getting the formula?
•  » » » » 3 years ago, # ^ | ← Rev. 4 →   Oh, I forgot the constraint of $N$. So basically what I wanted to represent is to split the cases into two: whether there is a one in the set of numbers, that is, $g(i,j)=g(i-j,j)+g(i-1,j-1)$, but we may have an $N$ present in the set of $g(i-j,j)$, which is forbidden. Thus the final formula should be $g(i,j)=g(i-j,j)+g(i-1,j-1)-g(i-j-N,j-1)$. Thanks for pointing out! UPD: for getting the formula, we check if there is an one in the set, if there is, we erase it, otherwise we let all numbers in the set minus one, which is basically the recurrence one may reach when calculating some partition numbers.
•  » » » » » Thanks for clearing things up. Waiting for part 2.
•  » » » » » Can't it now happen that $1$ is already part of $g(i-1,j-1)$?I.e. shouldn't the formula instead be more like $g(i,j) = g(i-j,j) - g(i-j-N,j-1) + g(i-j,j-1) - g(i-j-N,j-2)$? But aren't we then in turn running the risk of subtracting sets with $N$ too many times as $g(i-j-N,j-1)$ already may contain $N$?(Thumbs up on a great blog post btw.)
•  » » » » » » 3 years ago, # ^ | ← Rev. 3 →   Ahh...My bad. Another(hope also is the last) edit: for this problem, we may use another way of constructing sets, we check if there is an one in the set, if there isn't, we let all numbers in the set minus one, otherwise we erase the one, and then let all remaining numbers in the set minus one. Also we need to subtract the case where the biggest number was $n$ in the reduced set.Let's restate the formula: $g(i,j)=g(i-j,j)+g(i-j,j-1)-g(i-(n+1),j-1)$
 » Auto comment: topic has been updated by Roundgod (previous revision, new revision, compare).
 » The maths is beyond my level used in it . i couldnt undersatand anything
 » 3 years ago, # | ← Rev. 3 →   You can talk about the Inclusion-Exclusion principle with the Mobius function?
 » Sorry for asking these silly questions but what knowledge or definitions I need to learn in order to understand what is $N_{\supseteq T}$, $N_{= T}$, $N_{= \emptyset}$, ... and what really $N$ and $T$ represents for in this definition: (Because it just pops up from nowhere and I could not understand).  Thanks for reading and helping me!!.
 » i have query related to generalisation of example 3. x1+x2+x3.....xn=k and xi< rand(1,n) . can it be solved using inclusion exclusion.
 » I am interested in example 3, can I optimize furthur, about ($O(m \times polylog(m))$ time complexity) ?
 » I don't understand the part from "The formula becomes even simpler when ...", specifically and what does homogeneous mean ?
 » There is a straightforward $O((N+K)logK)$ solution (but with bad constant factor) to the 3rd problem K-inversion Permutations using generating functions and Fast Fourier Transform.Required answer is the coefficient of $x^K$ in the polynomial $P(x) = (1) * (1 + x) * (1 + x + x^2) * ... * (1 + x + ... + x^{(N-1)})$It can be rewritten as $P(x) = \prod _{i=1}^N \dfrac{1 - x ^ i}{1 - x}$It is tough to calculate this polynomial, but easy to calculate the log of this, which will be:$Q(x) = logP(x) = (\sum _{i=1}^N log(1 - x ^ i)) - N \cdot log(1 - x)$We know that: $log(1-x^i) = -\sum _{j = 1} ^\infty \dfrac{x^{i \cdot j}}{j}$. We only need to calculate this uptill $x^K$. It can be done naively in $O((N+K)logK)$Now all we need is the coefficient of $x^K$ in $e^{Q(x)}$. This can be done using FFT in $O(KlogK)$. It is a bit complicated but it is described here — https://cp-algorithms.com/algebra/polynomial.html
•  » » Wow, amazing! How did you think of that generating function? If I am not wrong, you are choosing, for every index i, how many elements to the left of ith index are bigger than the element at the ith index. And these things, together will determine the permutation, somehow... Wow!