Hello Codeforces. Today I'm writing about a Math topic that is really simple, but resources are limited.

SLAE stands for system of linear algebraic equations. Basically, consider we have a set of equations of the form :

*a* _{0}·*x* _{0} + *a* _{1}·*x* _{1} + *a* _{2}·*x* _{2} + ... + *a* _{ n - 1}·*x* _{ n - 1} = *val* _{0}

*b* _{0}·*x* _{0} + *b* _{1}·*x* _{1} + *b* _{2}·*x* _{2} + .... + *b* _{ n - 1}·*x* _{ n - 1} = *val* _{1}

*c* _{0}·*x* _{0} + *c* _{1}·*x* _{1} + *c* _{2}·*x* _{2} + .... + *c* _{ n - 1}·*x* _{ n - 1} = *val* _{2}

.....

Note that all *a*, *b*, *c*... are real-valued arrays and all *val* _{ i}, *x* _{ i} are arbitrary reals. Realize how *x* _{0}, *x* _{1}, ...*x* _{ n - 1} appear in each of the equations. In the post below, it is assumed we are dealing with reals, and not only integers.

Now, we want to find values of [*x* _{0}, *x* _{1}...*x* _{ n - 1}] that satisfy each of the given equations listed, **given** all *a*, *b*, *c*... and *val* _{ i}. The simplest method to find such solutions is to use **Gaussian Elimination**, that solves the problem in *O*(*N* ^{3}), where *N* = number of equations = number of variables .

To Learn about Gaussian Elimination, click here. Today, we shall learn about 2 special class of problems that can be solved using Gaussian Elimination.

**Problem 1** : Markov Chains and Cyclic Expected Values :

**Pre-requisite** : Gaussian-Elimination, Expectation of a random variable.

Many a times as a part of expected value problems, you are expected to sum up infinite series that hold as limits, as probabilities lie in the closed interval [0, 1]. For example,

, as

However, not always can we expect the variables whose Expected value we need to calculate to be **independent**. Consider you have *N* random variables , where , there are cyclic dependencies among the variables for their expected values, i.e consider depends on , on and depends on . So, there exists an infinite loop for calculating the Expected values of the random variables.

For example, consider the following problem :

You are given Tree *T* consisting of *N* nodes. Initially there is a player in node *S*. In a single move, he moves to one of the adjacent nodes of the node he is currently at, each with equal probability. What is the expected number of moves before he reaches node *T* ?.

Here, we need to understand that the expected values are infinite sums as well as cyclic. Creating a simple formula for the answer is quite hard. The Expected value starting from node *S* depends on some neighbor of node *S*, however, the Expected value of some neighbor of node *S* depends on Expected value of node *S*. Notice that whenever we reach a particular node, the probability of moving to any other node regardless of the number of steps performed always remains the same. So, this is a **Markov Chain**. Let's consider the transition matrix of this chain.

Create a matrix *P*, where *P*[*i*][*j*] = probability of moving from node *i* to *j* in a single move. Now, Let denote the expected number of steps needed to reach node *T* from node *i*.

.

Try and take a moment and think about why this formula is correct.

**Spoiler**

Surprise Surprise, this can be modeled as SLAE. Rewrite equations as :

. So the system is :

\begin{equation} \begin{pmatrix} 1-P[0][0] & -P[0][1] & ... & -P[0][N-1] \newline -P[1][0] & 1-P[1][1] & ... & -P[1][N-1] \newline .... \newline -P[N-1][0] & -P[N-1][1] & ... & 1-P[N-1][N-1] \end{pmatrix} \cdot \begin{pmatrix} \mathbb E(0) \newline \mathbb E(1) \newline .. \newline .. \newline \mathbb E(n-1) \end{pmatrix} = \begin{pmatrix} 1 \newline 1 \newline .. \newline 1 \end{pmatrix} \end{equation}

This is the equation (*I* _{ N} - *P*)·*E* = 1, We need to find the column vector *E*. Note that for node *T*, we need to have *P*[*t*][*i*] = 0, *t* ≠ *i* and *P*[*t*][*t*] = 1, as we won't move from node *T*, it is an absorbent state of the Markov chain. So, the *T* ^{ th} row of matrix *I* _{ N} - *P* will be all zeros. Also, the equation does not hold true for node *T*. Also, we know . So, the part does not affect any of the equations. So, just remove the *T* ^{ th} row and column from both sides of the equation.

The matrix is now a square (*N* - 1)·(*N* - 1) matrix, that is invertible. Invert the matrix using Gaussian Elimination augmenting with the RHS, to obtain *E*, i.e.

We can use this generic technique in all cases where the expected values are cyclic in nature , i.e expected value of state *A* depends on state B, and expected value of state *B* depends on state *A*. We can use any prime mod too, to obtain expected value in Modulo. Just remember : dependent random variables, use this. Practice Problems :

One (Same problem as above) My Code

**Problem 2** : Xor's using SLAE

**Pr-requisite** : Vector Space properties, Linear Algebra.

mod 2

mod 2

mod 2

So, xor is just bit-wise addition mod 2. We can represent the xor of two integer's *x*, *y* as vector addition in . For example ,

i.e. ,

\begin{equation} \begin{pmatrix} 0 \newline 1 \newline 0 \end{pmatrix} + \begin{pmatrix} 1 \newline 1 \newline 1 \end{pmatrix} \equiv \begin{pmatrix} 1 \newline 0 \newline 1 \end{pmatrix}

Mod \space 2

\end{equation}

So, we can use this addition to replace xor.The main advantage of this scheme is that we have converted the subset xor problem to solving a linear system instead. Consider we want to find *a*, *b*, *c* such that:

*a*·*v*1 + *b*·*v*2 + *c*·*v*3 ≡ *x* **Mod** 2, given *v*1, *v*2, *v*3 and *x*. Here *v*1, *v*2, *v*3, *x* are arbitrary binary column vectors. This is equivalent to solving the linear system :

\begin{equation} \begin{pmatrix} v1 & v2 & v3 \end{pmatrix} \cdot \begin{pmatrix} a \newline b \newline c \end{pmatrix} \equiv x \hspace{0.2cm} Mod \hspace{0.2cm} 2 \end{equation}.

Since *a*, *b*, *c* can only belong to {0, 1}, this is precisely finding a solution to subset xor.

Note that the **span** of any given set of size *N* is a vector space. There is a concept called as Basis of a vector space , i.e a smallest size subset of a given set that spans the entire vector space spanned by the original set given. We can solve the same problem over smaller sized basis rather than using all the elements of the set.

Via the Basis, we can solve useful xor based problems such as :

1> Given a set *S* of size *N*, find the number of distinct integers that can be represented using xor over the set of the given elements.

**Solution**

2> How many sub sequences of a given set *S* of size *N* have xor equal to *X*. (Do it yourself).

**Hint**

3> What is the maximum possible xor you can have using a subset of a given set :

**Solution**

All of these problems can be modeled using SLAE in Mod 2. We can do operations faster in Mod 2 using bitset having complexity .

Problems :

This is my first time wiring a tutorial blog, so please excuse any minor mistakes. Please give you feedback, it is always useful. Thank You.

Great tutorial, but

`O(N^3/64)`

is`O(N^3)`

, you can use`~N^3/64 operations`

or`O(N^3) with very small constant`

. Now we can solve this problem?It seems to be usual jargon (although technically incorrect). If you really want you can talk about O(N^3 / WORD_SIZE) (because this is actually what it is).

For the problem in the link, I have a solution for strings of length up to 500. How can we solve it for such large strings ? Also, what I meant by

Nin the last line isN= number of bits. I didn't understand your queryI mean that you are not using big-O notation correctly from a mathematical point of view. "O(N^3 / WORD_SIZE)" or "O(N^3) with very small constant" or ≈

N^{3}/ 64 — correctI dont know how to solve it, but 168 users have successfully solved it.

"Here, we need to understand that the probabilities are infinite...". As far as I know probabilities are finite (always in [0, 1]). What do you mean with these sentence?

I have modified the sentence, thanx for pointing it out

I understand how to derive formula $$$E[i] = \sum_{j=0}^{n-1} P(i,j) . E[j]$$$ by considering the next node from node $$$i$$$, however, from where did an additional $$$1$$$ appear in your formula ?

AJ_Coder

Edit, I got it: $$$1$$$ is added since we have already performed a move to visit node $$$j$$$.