Codesignal Solution: isTournament

Original Problem

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 the following final form:

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

Using an adjacency matrix \(M\) we built before, we can solve the problem more elegantly by using an environment like Julia or Matlab:

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

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