EXPTREE - Editorial



Author: Aleksandr Kulkov
Primary Tester: Misha Chorniy
Editorialist: Hussain Kara Fallah




Combinatorics,Math,Number Theory


Consider an ordered tree with N vertices. Your task is to calculate the expected value of the number of vertices having exactly one child in such tree assuming that it is uniformly chosen from the set of all ordered trees of size N.

Consider the answer to be a proper fraction P/Q, where gcd(P, Q) = 1. Then your task is to output two integers PQ-1 mod 109+7 and PQ-1 mod 109+9



The number of ordered trees consisting of n nodes is equal to the (n-1)th catalan number. Imagine the Depth-First-Search order of the nodes of the trees, entering each node corresponds to an open bracket ( , and finishing each node corresponds to a closed bracket ) . So if we represented the DFS order of a tree it will form a simple balanced bracket sequence.

Two different ordered trees must have different bracket representations. So we can say that the number of ordered trees consisting of n nodes is equal to the number of balanced bracket sequences consisting of (n-1) pairs of brackets. (n-1) because our tree is connected so we must fix the root (the first and the last bracket in the expression). If the subexpression (the expression excluding the first and last bracket) is balanced, that guarantees that we will never exit the root before the last bracket and our tree is connected.

It’s well known that the number of balanced bracket sequences of n pairs of brackets is equal to the nth catalan number.

C_n = \frac{(2n)!}{(n+1)!n!}

Before reading the next observation you should be familiar with Linearity Of Expectation.


Let’s consider all different (n-1) ordered trees, and think about applying this operation. Choosing an arbitrary node of a tree and linking it to a single child (nth node) and removing all edges between our chosen node and its children, then linking them to our new node. So our new node would serve as a single child of the chosen one.

A node having one child in a tree is equivalent to a pair of brackets, with a balanced expression between them and surrounded by a pair of brackets.

Observe that each node in a tree consisting of n nodes and having only one child can be formed by applying this operation to only one tree of (n-1) nodes. It’s exactly the same as choosing an opened bracket and its corresponding closing bracket in a simple balanced bracket sequence framing the subexpression between by a pair of brackets (our new node).

Thus our nominator P would be

P = ( number of trees of n-1 nodes * (n-1) )

We multiplied by n-1 which denotes the number of ways to choose the vertex we decided to apply this operation on.

Q = ( number of trees of n nodes )

answer = \frac{C_{n-2} * (n-1)}{C_{n-1}}

According to the catalan number formula we wrote above, this can be reduced to:

answer = \frac{n * (n-1)}{4n-6}

This fraction can be reduced easily by calculating GCDs between the denominator and the nominator terms. We can calculate PQ-1 after calculating the modular inverse of Q considering each modulo.


It’s true that this solution is given without a formal proof. But the idea is quite simple and correct. Actually its correctness can be proved by intuition. In real life contests you won’t have the internet to read formal proofs and papers or search for formulas for a sequence. This solution is not hard to get from scratch and worths trying.


AUTHOR’s solution: Will be found here
TESTER’s solution: Will be found here
EDITORIALIST’s solution: Will be found here


We don’t need to cancel gcd, just taking P and Q modulo 10^9+7 and 10^9+9 is enough.


You mean gcd() of any two numbers in this case will definitely 1?

True because while applying inverse condition is that a and m should be coprime ba-1(inverse)%m and if a is not a multiple of m multiples of a also cannot be multiples of m and hence they are coprime too.

@bansal1232 No gcd in this case will not at all be 1 (in all cases may be in some) but the thing is you don’t need them to be 1 because
8/2%5=4 and 4/1%5=4.

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Observe that each node in a tree consisting of n nodes and having only one child can be formed by applying this operation to only one tree of (n-1) nodes.

Why is this true?

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I don’t know why they gave that Proper fraction part. Even if you just solve it simply by taking Q^-1 and multiply it by P we get correct ans.

"P = ( number of trees of n-1 nodes * (n-1) )

We multiplied by n-1 which denotes the number of ways to choose the vertex we decided to apply this operation on."

I don’t understand how the editorial got to this part.

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A good and comprehensive paper on the same http://www.cs.tau.ac.il/~nachumd/papers/Enumerations.pdf


i am not able to understand multiplicative inverse part

How 6/5 is 400000004 and 200000003??

5^-1 gives 400000003, multiply it by 6 and then take mod. You’ll get the desired op.

Total number of ordered trees(denominator) for n-node is (n-1)th catalan number. And the numerator is (n-2)th middle number in pascal triangle or central binomial co-efficient.
Formula for central binomial co-efficient

Now answer is n(n-1)/4n-6 (simplification).
Now for dealing large numbers like 10^18 numerator will overflow so, I calculated it in two parts. Numerator1 = n, Numerator2=(n-1).

Now final numerator =( Numerator1/gcd(num1,denominator) * Numerator2/gcd(num2,denomenator) )

Final denominator is min(denominator/gcd(num1,denominator) , denominator/gcd(num2,denomenator) )

Now comes inverse mod part.

Important point: If x is smaller than y and we want to find y mod^-1 x then it is equal to (y%x) mod^-1 x.

Now if I use Fermat’s theorem everytime it timouts. So here is link which shows how to find inverse mod much faster in linear time Inverse Mod in linear time it is in Russian translate it. Now if I find inverse mod of all numbers again timout so I find inverse mod only till 10^3 (and then i use these result to find inverse mod of all number till 10^7 or 10^9)(kind of square root decompostion it can be easily understood from formula given in link to find inverse mod).

My solution

Almost 30 submissions to get full 100 :smiley:

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you can find inverse using fermat’s theorem in logN time.

t test cases. so t*log 10^7 gives time out. O(n) compute (ony one time for n=2 to n=1000),store and then using it is more faster.

It didnt gave TLE to me when i did the same @rj25 . In fact, code passed well within the time limit.

@vijju123 can you give link to your solution ?https://www.codechef.com/viewsolution/14488182 My solution using that approach only gave me 10 points. But @vijju I think my final approach is much faster than O(log n) one.


Here @rj25

If one had found the following paper is that considered cheating? :slight_smile:


Your solution is much simpler than mine. I have complicated everything by not taking n%m right from start.

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Well getting a formal proof is something really hard, you will find yourself ending in reading papers. This solution was the easiest thing that came up to my mind. Anyway, this solution is kind of intuitive :slight_smile: