### Roundgod's blog

By Roundgod, history, 5 months ago, ,

Hello, Codeforces! The reason why I am writing this blog is that my ACM/ICPC teammate calabash_boy_love_15 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 distinct, 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.

• +323

 » 5 months ago, # |   +33 Auto comment: topic has been updated by Roundgod (previous revision, new revision, compare).
 » 5 months ago, # |   +8 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)?
•  » » 5 months ago, # ^ |   +40 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).
•  » » » 5 months ago, # ^ |   0 Thanks, got it.
 » 3 months ago, # | ← Rev. 2 →   +9 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
•  » » 3 months ago, # ^ |   +3 I don't think it can be solved using inclusion-exclusion. You can use digit dp to solve it. Here's my solution.
 » 3 months ago, # |   +10 This is really a nice tutorial, liked it. Hope to get more nice tutorials like this in future.
 » 3 months ago, # |   +3 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}}.
•  » » 3 months ago, # ^ |   +2 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 months ago, # ^ | ← Rev. 2 →   +3 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 months ago, # ^ | ← Rev. 4 →   +18 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.
•  » » » » » 3 months ago, # ^ |   0 Thanks for clearing things up. Waiting for part 2.
•  » » » » » 3 months ago, # ^ |   +8 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 months ago, # ^ | ← Rev. 3 →   +8 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: $g(i,j)=g(i-j,j)+g(i-j,j-1)-g(i-(n+1),j-1)$
 » 3 months ago, # |   +10 Auto comment: topic has been updated by Roundgod (previous revision, new revision, compare).
 » 3 months ago, # |   -28 The maths is beyond my level used in it . i couldnt undersatand anything
 » 3 months ago, # | ← Rev. 3 →   0 You can talk about the Inclusion-Exclusion principle with the Mobius function?