Update 15-puzzle.md

tcNickolas · web-flow · commit 34df18a1cfbc · 2016-10-31T09:01:51.000-07:00
diff --git a/src/others/15-puzzle.md b/src/others/15-puzzle.md
@@ -2,47 +2,39 @@
 
 # 15 Puzzle Game: Existence Of The Solution
 
-This game is played on board, which size is 4×4 cells. On this board there are 15 playing tiles numbered from 1 to 15. One cell is left empty. It is required to come to position presented below by moving some tiles to the free space:
+This game is played on a $4 \times 4$ board. On this board there are 15 playing tiles numbered from 1 to 15. One cell is left empty. You need to get the board to the position presented below by repeatedly moving one of the tiles to the free space:
 
-![some image description](http://e-maxx.ru/tex2png/cache/a20a65d11dd5382b08465279d9bef0ca.png)
+![15 puzzle goal position](http://e-maxx.ru/tex2png/cache/a20a65d11dd5382b08465279d9bef0ca.png)
 
 The game "15 Puzzle” was created by Noyes Chapman in 1880.
 
-
-
 ## Existence Of The Solution
 
+Let's consider this problem: given position on the board, determine whether a sequence of moves which leads to a solution exists.
 
+Suppose we have some position on the board:
 
-Let's consider this problem: given position on the board to determine whether exists such a sequence of moves, which leads to a solution.
-
-Suppose, we have some position on the board:
-
-![some image description](http://e-maxx.ru/tex2png/cache/a20a65d11dd5382b08465279d9bef0ca.png)
+![15 puzzle board position](http://e-maxx.ru/tex2png/cache/4bce648a6f89c9fc25289fa27fd7a107.png)
 
 where one of the elements equals zero and indicates an empty cell $a_z  = 0$
 
 Let’s consider the permutation:
 
-![some image description](http://e-maxx.ru/tex2png/cache/37db640bfba4e1e8372e3098c8e0b1f1.png)
+$$a_1 a_2 ... a_{z-1} a_{z+1} ... a_{15} a_{16}$$
 
-(i.e. the permutation of numbers corresponding to the position on the board, without a zeroth element)
+(i.e. the permutation of numbers corresponding to the position on the board without a zero element)
 
-Let $N$ be a quantity of inversions in this permutation (i.e. the quantity of such elements $a_i$  and $a_j$  that $i < j$, but $a_i  > a_j$).
+Let $N$ be the number of inversions in this permutation (i.e. the number of such elements $a_i$  and $a_j$  that $i < j$, but $a_i  > a_j$).
 
 Suppose $K$ is an index of a row where the empty element is located (i.e. in our indications $K = (z - 1) \ div \ 4 + 1$).
 
 Then, **the solution exists iff $N + K$ is even**.
 
-
-
 ## Implementation
 
-
-
 The algorithm above can be illustrated with the following program code:
 
-````cpp
+```cpp
 int a[16];
 for (int i=0; i<16; ++i)
     cin >> a[i];
@@ -58,14 +50,16 @@ for (int i=0; i<16; ++i)
         inv += 1 + i / 4;
 
 puts ((inv & 1) ? "No Solution" : "Solution Exists");
-````
-
+```
 
 ## Proof
 
-
 In 1879 Johnson proved that if $N + K$ is odd, then the solution doesn’t exist, and in the same year Story proved that all positions when $N + K$ is even have a solution.
 
 However, all these proofs were quite complex.
 
-In 1999 Archer proposed much more simple proof (you can download his article [here](http://www.cs.cmu.edu/afs/cs/academic/class/15859-f01/www/notes/15-puzzle.pdf)).
+In 1999 Archer proposed a much simpler proof (you can download his article [here](http://www.cs.cmu.edu/afs/cs/academic/class/15859-f01/www/notes/15-puzzle.pdf)).
+
+## Practice Problems
+
+* [Hackerrank - N-puzzle](https://www.hackerrank.com/challenges/n-puzzle)
