# My 3rd Problem for HackerRank // Towers

## Problem statement

You're given infinitely many bricks of at most 15 different heights (other dimensions are not relevant here). You're also given a big integer $n$. The task is to count the time needed to build every possible tower of height N from the given bricks. Bricks are stacked vertically only and stand in an upright position. Two towers are different if their brick height sequences are different and building one tower takes exactly 2 time units regardless of its height.

Here is the full problem statement.

## Solution

In the editorial I omit the fact that we have to count the time needed to build all towers - we will count here just the number of different towers and to get the result, you have to multiply this number by 2.

### Simpler problem first

There is a restriction, that every bricks is of height at most 15 and $N$ is at most $10^{18}$. Lets first try to solve a simpler problem and assume that there are only bricks of height 1 or 2 and $N \leq 10^6$.

Let $f_n$ be the number of different towers of height $n$ which can be build of given bricks. In our simpler problem we have:

$f_1 = 1$, because there is only one tower of height 1 consisting of one bricks of height 1

$f_2 = 2$, because there are exactly two towers of height 2 - one consisting of two bricks of height 1 and the other consisting of one brick of height 2

We can now give a recursive definition for $f_n$ when $n > 2$:

$f_n = f_{n - 1} + f_{n - 2}$

because we can get one unique tower of height $n$ by placing a brick of height 1 on any tower of height $n - 1$ and one unique tower of height also $n$ by placing a brick of height 2 on any tower of height $n - 2$.

If we compute $f_n$ bottom-up (like in dynamic programming) rather than using recursion explicitly, we can do it in $O(n)$ time which is fine if $n$ is not as big as in the problem statement.

### Removing heights restriction

Lets remove the first restriction, so we have to deal now with bricks of at most 15 different heights (from 1 to 15).

Let $h_i$ indicates whether we have a brick of height $i$:

then the general recursion equation is:

$f_n = 0$ for $n < 0$

$f_0 = 1$

$f_n = h_1 \cdot f_{n - 1} + h_2 \cdot f_{n - 2} + h_3 \cdot f_{n - 3} + \ldots + h_15 \cdot f_{n - 15}$ for $n > 0$

### Solving the original problem

The last thing that we have do deal with is the really big $n$ - up to $10^{18}$. We will do it defining the recursive equation as a matrix multiplication problem and solve it using fast matrix exponentiation.

Lets assume that we've computed $f_n$ for $n \leq 15$ using dynamic programming and let $M$ be a $15 \times 15$ matrix defined as follows:

$M = \begin{bmatrix}0&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\0&0&1&0&0&0&0&0&0&0&0&0&0&0&0\\0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&1&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&0&0&0&0&0&1&0&0\\0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\h_{15}&h_{14}&h_{13}&h_{12}&h_{11}&h_{10}&h_9&h_8&h_7&h_6&h_5&h_4&h_3&h_2&h_1\end{bmatrix}$

Let $V$ be the following vector:

$V={\begin{bmatrix}f_1&f_2&f_3&f_4&f_5&f_6&f_7&f_8&f_9&f_{10}&f_{11}&f_{12}&f_{13}&f_{14}&f_{15}\end{bmatrix}}^T$

if we multiply $M \cdot V$ we get:

$R={\begin{bmatrix}f_2&f_3&f_4&f_5&f_6&f_7&f_8&f_9&f_10&f_{11}&f_{12}&f_{13}&f_{14}&f_{15}&f_{16}\end{bmatrix}}^T$

and we can return $f_{16}$ from here, so in order to compute $f_n$ for $n > 15$ we can compute:

R = \underbrace{(M \cdot (M \cdot \dots \cdot (M}_{n - 15} \cdot V)))

which can be written in the following form, because matrix multiplication is associative:

R = M^{(n-15)} \cdot V

and we can compute $n$-th power of a matrix of size $m \times m$ it $O(m^3 \cdot \log_2 {n})$ using fast exponentiation by squaring and in our problem we have $m = 15$ and $n \leq 10^{18}$ which is perfectly fine in terms of time limits. One speedup here can be to do exponentiation iteratively rather than recursively

# Codechef Long // July Challenge - GERALD09 the tiebreaker problem

The June challenge is over. I am very happy, because I finished 19th overall, which is my personal best:

While editorials for all problems are available on Codechef forum, almost every solution to the tie breaker problem is unique, so I want to share mine, because it is completely different from the one posted in the official editorial.

## Problem statement

The full problem statement is here. You are given a matrix $T$ with $n \leq 15$ rows and $m \leq 15$ colums. In addition, you are given an integer $k$, where $1 \leq k \leq n^2 \cdot m^2$. Your task is to fill the matrix with letters from 4-letters alphabet $\{A, C, G, T\}$ in such a way, that the number of different submatrices of matrix $T$ is as close to $k$ as possible - the less the difference is the greater score you have.

## Solution

I used the concept of 2 dimensional variant of polynomial hashing in order to count the number of different submatrices in time $O(n^2 \cdot m^2)$. The concept allows to compare any two submatrices in $O(1)$ time.

Lets make some observations:

1. It is impossible to create a matrix with less than $n \cdot m$ different submatrices, because every 2 submatrices of different size are different. It follows that,if $k \leq n \cdot m$, the best you can do is to fill the matrix with only one letter, lets cay $A$.
2. It is impossible to create a matrix with more than $M = n \cdot (n + 1) / 2 \cdot m \cdot (m + 1) / 2$ different submatrices, because this is the number of all possible submatrices. It follows that if $k \geq M$, the best you can do is to return a matrix with $n$ rows and $m$ colums with the maximal number of different submatrices.
My approach is to precompute for every of 225 pairs $(n, m)$ the matrix with the maximum number of different submatrices. I did that by repeatedly exchanging some random letter in a matrix and if the resulting matrix has a greater number of different submatrices than it had before, then it becomes the resulting matrix. Of course it might not return the best matrix, but if you repeat the exchange many times it should be at least quite close to the optimal matrix. It is worth to mention, that such a a process can stuck in a local maximum, so it is better to choose the result from many different runs.
3. Since the maximum allowed source file limit on Codechef is 50kb, you can store a lot precomputed information in it. Let $R := \{s : n \cdot m < s < M\}$. Based on two above cases, we know that $k \in R$. For every pair of $(n, m)$, I assigned to $mem[n][m][s]$ a matrix of size $n \times m$ with exactly $s$ different submatrices, where $s$ is equally distributed over $R$. For example, let $n, m = 10$, then $100 < s < 3025$ and we chose to memorize matrices which differ in number of different submatrices by 500, then we compute $mem[10][10][s]$ for $s \in \{600, 1100, 1600, 2100, 2600\}$. It helps us to chose the initial matrix for the beginning of the on-line computation. For example, if $n, m = 10$, $k = 1220$ and we computed $mem[10][10]$ as above, we select $mem[10][10][1100]$ as the initial matrix. I computed $mem$ using the same random method as in Step 2.
4. Lets fill the matrix with the letter $A$ only and consider, how many different submatrices we can generate by placing letter $C$ in the $i$-th row and $j$-th column. If we do that, then every submatrix with the upper left corner in $(i_1, j_1)$ where $i_1 \leq i$ and $j_2 \leq j$ and the lower right corner in $(i_2, j_2)$, where $i_2 \geq i$ and $j_2 \geq j$ is unique, because any other matrix of the same size has either no letter $C$ or has letter $C$ at different relative position. It follows that using that method, we can create a matrix with up to about $n^2 / 4 \cdot m^2 / 4 + n \cdot m$ different submatrices. The natural extension is to insert letters $G$ and $T$ at some positions different than $(i, j)$ in our matrix, which helps us to create a matrix with up to about $3/16 \cdot n^2 \cdot m^2 + n \cdot m$ different submatrices.

We are ready now for the on-line computation. We choose as the initial matrix the one, which can be generated in any of the above steps with the closest number of different submatrices to $k$.

My online computation consists of only one technique. For any pair of $(n, m)$, I set the number of computation steps to $c \cdot 225 / (n^2 \cdot m^2)$ for the largest possible $c$ which doesn't exceed the time limit, in which I try to exchange some random elements in the matrix and hope to generate a better one.

At the beginning, I count the number of different submatrices in my initial matrix. Then, for every computations step, I exchange exactly $\sqrt{n \cdot m}$ random elements, again count the number of different submatrices and if it generated a better matrix, I store it as a current matrix.

## Conclusion

While the obvious method is to try to exchange on-line some elements in order to achieve a better solution, the crucial step is to set up the initial matrix in the best possible way. I am pretty satisfied with my solution. It results in 20th best score in that problem and gave me 19th place in the contest.