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 (denoted by 0). You need to get the board to the position presented below by repeatedly moving one of the tiles to the free space:

$$\begin{matrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 0 \end{matrix}$$

The game “15 Puzzle” was created by Noyes Chapman in 1880.

## 1. Existence Of The Solution

Let’s consider this problem: given a position on the board, determine whether a sequence of moves which leads to a solution exists.

Suppose we have some position on the board:

$$\begin{matrix} a_1 & a_2 & a_3 & a_4 \\ a_5 & a_6 & a_7 & a_8 \\ a_9 & a_{10} & a_{11} & a_{12} \\ a_{13} & a_{14} & a_{15} & a_{16} \end{matrix}$$

where one of the elements equals zero and indicates an empty cell $a_z = 0$

Let’s consider the permutation:

$$a_1 a_2 … a_{z-1} a_{z+1} … a_{15} a_{16}$$

i.e. the permutation of numbers corresponding to the position on the board without a zero element

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. using our convention, $K = (z – 1) \div \ 4 + 1$).

Then, **the solution exists iff $N + K$ is even**.

## 2. Implementation

The algorithm above can be illustrated with the following program code:

int a[16]; for (int i=0; i<16; ++i) cin >> a[i]; int inv = 0; for (int i=0; i<16; ++i) if (a[i]) for (int j=0; j<i; ++j) if (a[j] > a[i]) ++inv; for (int i=0; i<16; ++i) if (a[i] == 0) inv += 1 + i / 4; puts ((inv & 1) ? "No Solution" : "Solution Exists");

## 3. 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 a much simpler proof. You can see below: