CHEFWORD - Editorial


Author: Dmytro Berezin

Tester: Sergey Kulik

Editorialist: Florin Chirica




matrix multiplication; basic probability theory


You’re given an initial word S. You do K claps: each clap a character from S turns into another character with some given probability. You’re also given a list of final words. What’s the probability the word S will be one from the final list after those K claps?


A word S can transform in other word s[i] if and only if their length are equal. Probability the word S to transform in s[i] is prob(S[1], s[1]) * prob(S[2], s[2]) * …, where prob(a, b) is probability the character a will transform in the character b after K claps.

We can calculate dp[k][a][b] = probability character a will turn into character b after k claps. We find the recurrence dp[k][a][b] = sum of dp[1][a][a’] * dp[k - 1][a’][b]. This recurrence can be improved by using matrix multiplication, since it is a linear recurrence.


We need to compute probability that word S will become one of the words from s list. When is it possible the initial word S to become one of words s[i]? If the words S and s[i] have the same length, there’s some probability (possibly 0) that word S will become word s[i]. If word S and word s[i] have different lengths, then we know for sure it’s not possible, since we can’t replace a letter by SPACE character.

Let’s consider all words s[i] from s list. We’ll see what’s the probability word S will transform in word s[i] using exactly K claps. The answer of the problem is the sum of those probabilities.

So suppose we want to transform word A in word B using exactly K claps. This means A[1] needs to become B[1], A[2] needs to become B[2] and so on. Denote by P(char1, char2) = probability that character char1 will become char2 after exactly K claps. All the events P(A[1], B[1]) P(A[2], B[2]) and so on need to happen. We know from probability theory that the probability for some independent events to all happen (each of event needs to happen) is the product of each probability. So, in this case, probability word A will become word B is P(A[1], B[1]) * P(A[2], B[2]) * … * P(A[n], B[n]) where by n I noted length of A array.

The problem is reduced to: calculate P(a, b) = probability that character a will transform in character b after K claps. Calculating this in the current form seems difficult, so let’s try to add one more parameter.

Let P(a, b, x) = probability that character a will transform to character b after exactly x steps. For x = 1, the answers are given in the input.

We’ll next try to do dynamic programming. Suppose we calculated P(something1, something2, x - 1) for each something1 and something2, let’s try to calculate P(a, b, x) for each a and b. We focus now to calculate it for two fixed values of a and b. Then, we can iterate each two values a and b and whole P will be calculated for given parameter x.

Suppose we do one step. Let’s imagine character a becomes a’ after 1 clap. We need to do x - 1 claps which must transform character a’ into b. Suppose we fixed a’ now. What’s the probability a will turn into b, given that after one clap it will become a’? It is P(a, a’, 1) * P(a’, b, x - 1). Why? First, character a transforms in a’ using a clap with probabilty P(a, a’, 1). Then, a’ transfors in b after x - 1 steps with probability P(a’, b, x-1). The probability that both independent events to happen is P(a, a’, 1) * P(a’, b, x - 1).

Let’s solve for general case calculation of P(a, b, x). After one step, a character a will turn into some character a’. If we iterate all characters a’ possible, then all cases are covered. We need to calculate the sum of P(1, a, a’) * P(x-1, a’, b). Why do we need the sum? All events are independent, and exactly one event from the list needs to happen (since the character after 1st clap is always different between any two events, they’re independent). Probability that one event will happen is sum of probabilities of all events.

So we get that P(x, a, b) = sum of all P(1, a, a’) * P(x-1, a’, b).

This is how we get an O(K * 26^3) solution. This should time out. However, using some more math theory we can reduce the complexity.

Let’s recall how matrix multiplication is done: suppose you’re given two matrixes A and B and want to calculate C = A * B. We know that C[a][b] = sum of all A[a][a’] * B[a’][b]. This recurrence matches with the one from above to calculate P(x, a, b).

If we note C[a][b] = P(x, a, b), A[a][b] = P(1, a, b) and B[a][b] = P(x-1, a, b), we’re reduced our task to matrix multiplication.

The final observation to solve the task is: suppose A[a][b] is probability character a will turn to character b after exactly 1 clap. Let’s determine probability that character a will turn into character b after x claps. It’s enough to raise matrix A to power x-1, then print element from (A^(x-1))[a][b].

Since matrix multiplication is associative, it can be done by exponentiation by squaring. Hence, our total runtime is reduced to O(26^3 * (log(K)).


To be updated soon
To be updated soon


That problem almost made me mad - I knew that it is matrix multiplication, but I was getting WA again and again without an idea what’s wrong. And then suddenly, I had a question - “are all strings s distinct?”.

Applying one of the golden rules: “Do not assume something not written in problem statement” and modifying my solution got AC finally…

So the answer for same input as in problem statement with string


is still 0.125


I used the fact that the matrix becomes almost constant after being raised to the power >=40. Since we are allowed error <= 10^-6, I computed using dp for k = min(given_k,40). This ensures O(k * 26^2 * strlen(string)) to pass in 0.71s. :slight_smile:


Question to problem setter: Why |S| <= 3 ? Why not 100 for example?


I was getting a lot of wrong answers in this. I just saw @betlista’s answer. It is my personal opinion that such there shouldn’t be such trickery in the problem statement. It was an excellent question, but if I were a problem setter, I personally wouldn’t want people to have a headache figuring out twists in problem statement.

I really hope I don’t spend 10-15 hours for this sort of trickery in future.


Another way to think about the problem can be using graph theory. Let us say the given probability matrix represents adjancy matrix for a weighted directed graph, where a character represent a node. Now Our task is to find how many different paths exist from a given node to another node.

Now this is a well know theorem that in a graph the adjancy matrix raised to power K gives all possible paths from one node to another. For case of probabilities it will directly give probability(why ?).

Hence the answer is directly the given matrix raised to power K and [i][j] element will give probability of getting to character j from i in exactly K claps.

As stated the matrix exponentiation can be done easily using linear recurrence in 26^3*logK.


Just for the record this branch of probability is known as
markov chain.


Are there anyone can give me some example?I always get WA ,but i get the same answer with arsh anand’ AC solution for test case that i generated randomly

my solution:
arsh anand‘s solution:

This problem is basically this one

Hey Guys ,can anybody tell me what is wrong with my solution:

@khitish The max value of ‘hash_string’ would be 26 * 26 * 26 + 26 * 26 + 26, not 26 * 26 * 26

Can You explain why “Let’s determine probability that character a will turn into character b after x claps. It’s enough to raise matrix A to power x-1, then print element from (A^(x-1))[a][b].” but not (A^(x)[a][b])

@zentropy …still getting WA …

I used the matrix exponentiation in the contest as that was quite easy compared to a graph solution. I had a graph solution in mind as well. Here is my way of approaching the problem through graph theory -

There are maximum three components in this graph - depending on which alphabet you start from in the original word.
For each component, I have to find the probability of an alphabet occurring at the i th position in our string after k claps separately. So,

Find all the different paths from the root node of the selected component to the required alphabet.
Maintain probability of each path vs distance/claps on each path in say set 1.

In set 1, we reached the required alphabet from the root via all the paths. Now to reach this vertex again further down the graph, it has to be a part of a cycle.

Find all the cycles that the required alphabet lies on.
Maintain probability decrease per cycle vs no of vertices/claps in each cycle in say set 2.

find if given k can be obtained using
k = a+b.n
where a is any element in set 1, b is any element in set 2 and n is 0,1,2,…
find all valid (a,b) and add probabilities for each pair (a,b)
for a pair prob = p(a).p(b)^n

Once we have the three (max 3 letters) tables ready - i.e probability of each alphabet of the starting string converting into some other, the problem is almost solved.

I think this should be right but didnt have the time to implement it due to the end sems. Would be glad if someone could point out any mistakes / correctness of the approach.

I am getting AC on all case Task 2 & 3. I am getting TLE on 2 & 3 and I can’t figure out why.
Can anyone please look at my code and tell me what more optimization can be done?


Same has happened to me in many problems before. And nowadays after getting WA i do the same :stuck_out_tongue: Remind myself to never assume anything not given in the problem…


Nooooooooooooo :’( :’( Didn’t realize it during the contest. Oh well, another lesson learned :slight_smile:


Same with me…after 2 or 3 submissions of WA…i reread the problem statement…saw the constraints…where the number of query strings was greater than the total possible combinations…and then it struck “DUPLICATES”…good problem btw!!!

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Same was the algo used by me…the only variation being instead of an upper limit on ‘k’…i let the loop compute till the difference in probabilities of the current iteration and the previous iteration were less than an ‘eps’ value!!!


good catch @kunal361, I checked that |S| is at most 3 and I was not using it later, because it was not a limiting factor for me…

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