# Codesignal Solution: isTournament

Determine if the given directed graph is a tournament. A **tournament** is a directed graph in which every pair of distinct vertices is connected by a single directed edge.

**Example**

For`n = 5`

,`fromV = [2, 1, 3, 4, 4, 4, 1, 2, 3, 4]`

and`toV = [3, 2, 1, 3, 2, 1, 5, 5, 5, 5]`

,

the output should be`isTournament(n, fromV, toV) = true`

.

Here's how the given graph looks like:

**Input/Output**

**[time limit] 4000ms (js)**

**[input] integer n**A positive integer

`n`

representing the number of vertices in the given graph.*Guaranteed constraints:*`1 ≤ n ≤ 10`

.**[input] array.integer fromV**An array of integers containing integers less than or equal to

`n`

.*Guaranteed constraints:*`0 ≤ fromV.length ≤ 50`

,`1 ≤ fromV[i] ≤ n`

.**[input] array.integer toV**For each

`i`

in range`[0, fromV.length)`

there is an edge from the vertex number`fromV[i]`

to the vertex`toV[i]`

in the given directed graph.*Guaranteed constraints:*`toV.length = fromV.length`

,`1 ≤ toV[i] ≤ n`

.**[output] boolean**`true`

if the given graph is a tournament,`false`

otherwise.

## Solution

If we list all possible pairs of nodes for \(n=5\) and mark the ones for the given graph it may look like this:

(1, 1),(1, 2), (1, 3), (1, 4),(1, 5)

(2, 1), (2, 2),(2, 3), (2, 4),(2, 5)(3, 1), (3, 2), (3, 3), (3, 4),(3, 5)(4, 1),(4, 2),(4, 3), (4, 4),(4, 5)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5)

We now mirror the connections and see this nice pattern, where only the diagonal remains unmarked:

(1, 1),(1, 2),(1, 3),(1, 4),(1, 5)(2, 1), (2, 2),(2, 3),(2, 4),(2, 5)(3, 1),(3, 2), (3, 3),(3, 4),(3, 5)(4, 1),(4, 2),(4, 3), (4, 4),(4, 5)(5, 1),(5, 2),(5, 3),(5, 4), (5, 5)

Creating a set of unique connection tuples, is pretty straightforward using python3, by concatenating the two arrays \(f\) and \(t\). Duplicates get kicked out by the set creation: `{*zip(f + t, f + t)}`

. Since the given constraints \(1 \leq f[i] \leq n\) and \(1 \leq t[i] \leq n\) all numbers are valid and we only need to work with the count. How many distinct tuples do we have? n-squared minus the diagonal. This allows us to implement the following routine:

isTournament = lambda n, f, t: len({*zip(f + t, t + f)}) == n * n - n

The routine would be perfectly small, but fails if the number is correct, but a duplicate connection gets removed. That's why we also need to check if the original length is okay to form the actual matrix:

isTournament = lambda n, f, t: len({*zip(f + t, t + f)}) == n * n - n and n * n - n == 2 * len(f)

Merging the two conditions to form a smaller solution yields:

isTournament = lambda n, f, t: n * n - n == 4 * len(f) + len({*zip(f + t, t + f)})

Using the Adjacency matrix \(M\), we can solve the problem more elegant by using an environment like Julia or Matlab:

isequal(M + M', 1 - eye(size(M)...))

All images are copyright to Codesignal