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: , .
If we can compute the number of integers in the interval such that is coprime with , denoted as , then the answer is . How we're gonna compute ? Instead of counting the numbers that are coprime with , we can count the numbers that aren't coprime with , that is, sharing at least one prime factor with . To do this, we can first sieve all primes not exceeding and then find all prime factors of . Let be the set of numbers that are divisible by , then the answer is the , which may be hard to compute directly. However, using the restricted inclusion-exclusion principle, we can convert the problem into computing the size of the intersection of sets, which is trivial. Time complexity is , with equals to the number of distinct prime factors of .
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 .
The only thing we need to handle is to get rid of that annoying constraint . To do that, we apply the inclusion-exclusion principle. Let , then is our desired answer. Clearly, this set of properties is homogeneous. Take , then is the number of solutions with . Setting , and it's the same as the number of solutions of the system
, thus the answer is therefore
The complexity of the solution is , due to precomputing factorials and the modular inverses of factorials.
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: .
An idea one should come up with instantly is to let as the number of permutations of length with inversions. The recurrence is also trivial: . This is , and can be optimized to using prefix sums, which is still not enough due to the given constraints.
Consider the dp formula, for the th element, we add a number to the number of inversions, so the answer is equal to the number of solutions to satisfying that .
Like the last example, we can apply the inclusion-exclusion principle. Let , then is our desired answer. Applying the similar method we've done solving the last example, we can notice that if the sum of elements of equals to , then the number of solutions to the equation is . Therefore, we can group those sets together. By inclusion-exclusion principle,
So the problem becomes calculating all the . We can use the technique as we can computing partition numbers. Partition into two cases where there exist an in the set or not, and then we get the recurrence . Another important observation is that there are at most valid values for . Therefore, the problem is solved in .
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: .
Clearly we need to factorize . One may try to use Pollard-Rho algorithm under such constraints. But there's a simpler method: one can observe here only the exponents of each prime matters, no the prime itself. So we can first try to divide it using all primes up to , and then what remains, is a prime , a square of a prime , or a product of two distinct primes pq. We can check if it's the first case using Miller-Rabin algorithm, can iterate over to check if it's the second case, and otherwise the last case.
After factorizing, we have . We need to avoid counting the cases with and , thus we should apply inclusion-exclusion principle. Let and . Then the answer is . Directly computing this would lead to the complexity of ,which is too much for in te worst case. However, noticing that two of the options of for each prime divisor lead to same computations, the complexity can be reduced to .
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.