Update fibonacci-numbers.md

src/algebra/fibonacci-numbers.md

@@ -74,6 +74,7 @@ The encoding of an integer $n$ can be done with a simple greedy algorithm:
 
 To decode a code word, first remove the final $1$. Then, if the $i$-th bit is set (indexing from 0 from the leftmost to the rightmost bit), sum $F_{i+2}$ to the number.
 
+
 ## Formulas for the $n^{\text{th}}$ Fibonacci number { data-toc-label="Formulas for the <script type='math/tex'>n</script>-th Fibonacci number" }
 
 The $n$-th Fibonacci number can be easily found in $O(n)$ by computing the numbers one by one up to $n$. However, there are also faster ways, as we will see.
@@ -94,17 +95,71 @@ where the square brackets denote rounding to the nearest integer.
 
 As these two formulas would require very high accuracy when working with fractional numbers, they are of little use in practical calculations.
 
+### Fibonacci fast method
+
+The $n$-th Fibonacci number can be easily found in $O(n)$ by computing the numbers one by one up to $n$. However, there are also faster ways, as we will see.
+
+We can start from an iterative approach, to take advantage of the use of the formula $F_n = F_{n-1} + F_{n-2}$, therefore, we will simply precalculate those values in an array. Taking into account the base cases for $F_0$ and $F_1$.
+
+```cpp
+int fib(int n) {
+ int a = 0;
+ int b = 1;
+ for (int i = 0; i < n; i++) {
+ int tmp = a + b;
+ a = b;
+ b = tmp;
+ }
+ return b;
+}
+```
+
+In this way, we obtain a linear solution, $O(n)$ time, saving all the values prior to $n$ in the sequence.
+
 ### Matrix form
 
 It is easy to prove the following relation:
 
-$$\begin{pmatrix}F_{n-1} & F_{n} \cr\end{pmatrix} = \begin{pmatrix}F_{n-2} & F_{n-1} \cr\end{pmatrix} \cdot \begin{pmatrix}0 & 1 \cr 1 & 1 \cr\end{pmatrix}$$
-
-Denoting $P \equiv \begin{pmatrix}0 & 1 \cr 1 & 1 \cr\end{pmatrix}$, we have:
+$$\begin{pmatrix} 1 & 1 \cr 1 & 0 \cr\end{pmatrix} ^ n = \begin{pmatrix} F_{n+1} & F_{n} \cr F_{n} & F_{n-1} \cr\end{pmatrix}$$
+
+Thus, in order to find $F_n$ in $O(log n)$ time, we must raise the matrix to n. (See [Binary exponentiation](https://github.com/cp-algorithms/cp-algorithms/blob/master/src/algebra/binary-exp.md))
+
+``` cpp
+typedef long long ll;
+
+struct matrix{
+ ll mat[2][2];
+ matrix friend operator *(const matrix &a, const matrix &b){
+ matrix c;
+ for(int i = 0; i < 2; i++)
+ for(int j = 0; j < 2; j++){
+ c.mat[i][j] = 0;
+ for(int k = 0; k < 2; k++){
+ c.mat[i][j] += (a.mat[i][k]*b.mat[k][j]);
+ }
+ }
+ return c;
+ }
+};
+
+matrix matpow(matrix base, ll n){
+ matrix ans {{{1, 0}, {0, 1}}};
+ while(n){
+ if(n&1)
+ ans = ans*base;
+ base = base*base;
+ n >>= 1;
+ }
+ return ans;
+}
 
-$$\begin{pmatrix}F_n & F_{n+1} \cr\end{pmatrix} = \begin{pmatrix}F_0 & F_1 \cr\end{pmatrix} \cdot P^n$$
+long long fib(int n) { 
+ matrix base {{{1, 1}, {1, 0}}};
+ base = matpow(base, n);
+ return base.mat[0][1];
+} 
 
-Thus, in order to find $F_n$, we must raise the matrix $P$ to $n$. This can be done in $O(\log n)$ (see [Binary exponentiation](binary-exp.md)).
+```
 
 ### Fast Doubling Method
 
@@ -153,4 +208,7 @@ We now choose two pairs of identical remainders with the smallest indices in the
 * [Project Euler - Even Fibonacci numbers](https://www.hackerrank.com/contests/projecteuler/challenges/euler002/problem)
 * [DMOJ - Fibonacci Sequence](https://dmoj.ca/problem/fibonacci)
 * [DMOJ - Fibonacci Sequence (Harder)](https://dmoj.ca/problem/fibonacci2)
+* [DMOJ UCLV - Numbered sequence of pencils](https://dmoj.uclv.edu.cu/problem/secnum)
+* [DMOJ UCLV - Fibonacci 2D](https://dmoj.uclv.edu.cu/problem/fibonacci)
+* [DMOJ UCLV - fibonacci calculation](https://dmoj.uclv.edu.cu/problem/fibonaccicalculatio)