 # nathanchere's solution

## to Dominoes in the C# Track

Published at Sep 17 2019 · 0 comments
Instructions
Test suite
Solution

Make a chain of dominoes.

Compute a way to order a given set of dominoes in such a way that they form a correct domino chain (the dots on one half of a stone match the dots on the neighbouring half of an adjacent stone) and that dots on the halves of the stones which don't have a neighbour (the first and last stone) match each other.

For example given the stones `[2|1]`, `[2|3]` and `[1|3]` you should compute something like `[1|2] [2|3] [3|1]` or `[3|2] [2|1] [1|3]` or `[1|3] [3|2] [2|1]` etc, where the first and last numbers are the same.

For stones `[1|2]`, `[4|1]` and `[2|3]` the resulting chain is not valid: `[4|1] [1|2] [2|3]`'s first and last numbers are not the same. 4 != 3

Some test cases may use duplicate stones in a chain solution, assume that multiple Domino sets are being used.

## Running the tests

To run the tests, run the command `dotnet test` from within the exercise directory.

Initially, only the first test will be enabled. This is to encourage you to solve the exercise one step at a time. Once you get the first test passing, remove the `Skip` property from the next test and work on getting that test passing. Once none of the tests are skipped and they are all passing, you can submit your solution using `exercism submit Dominoes.cs`

## Further information

For more detailed information about the C# track, including how to get help if you're having trouble, please visit the exercism.io C# language page.

### DominoesTest.cs

``````// This file was auto-generated based on version 2.1.0 of the canonical data.

using System;
using Xunit;

public class DominoesTest
{
[Fact]
public void Empty_input_empty_output()
{
var dominoes = Array.Empty<(int, int)>();
Assert.True(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Singleton_input_singleton_output()
{
var dominoes = new[] { (1, 1) };
Assert.True(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Singleton_that_cant_be_chained()
{
var dominoes = new[] { (1, 2) };
Assert.False(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Three_elements()
{
var dominoes = new[] { (1, 2), (3, 1), (2, 3) };
Assert.True(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Can_reverse_dominoes()
{
var dominoes = new[] { (1, 2), (1, 3), (2, 3) };
Assert.True(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Cant_be_chained()
{
var dominoes = new[] { (1, 2), (4, 1), (2, 3) };
Assert.False(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Disconnected_simple()
{
var dominoes = new[] { (1, 1), (2, 2) };
Assert.False(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Disconnected_double_loop()
{
var dominoes = new[] { (1, 2), (2, 1), (3, 4), (4, 3) };
Assert.False(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Disconnected_single_isolated()
{
var dominoes = new[] { (1, 2), (2, 3), (3, 1), (4, 4) };
Assert.False(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Need_backtrack()
{
var dominoes = new[] { (1, 2), (2, 3), (3, 1), (2, 4), (2, 4) };
Assert.True(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Separate_loops()
{
var dominoes = new[] { (1, 2), (2, 3), (3, 1), (1, 1), (2, 2), (3, 3) };
Assert.True(Dominoes.CanChain(dominoes));
}

[Fact(Skip = "Remove to run test")]
public void Nine_elements()
{
var dominoes = new[] { (1, 2), (5, 3), (3, 1), (1, 2), (2, 4), (1, 6), (2, 3), (3, 4), (5, 6) };
Assert.True(Dominoes.CanChain(dominoes));
}
}``````
``````﻿using System.Collections.Generic;
using System.Linq;

public static class Dominoes
{
public static bool CanChain(IEnumerable<(int, int)> input)
{
var dominoes = input.ToList();
if (!dominoes.Any()) return true;

// If any of the domino values appears an odd number of times, a full circular chain is not possible
var values = dominoes.FlattenTuples();
if (values.IsUnbalancedSet()) return false;

// From this point we only care about unique values to speed up procesing
dominoes = dominoes.Distinct().ToList();
values = values.Distinct().ToList();

// Starting at the first unique value, we'll track all other values which have a domino connecting it
var connectedValues = new HashSet<int> { values.First() };
values.RemoveAt(0);

foreach (var value in values)
{
if (connectedValues.Contains(value)) continue;

// If there are no dominoes joining the current value to any of the already connected values, a chain is not possible
if (!dominoes.Any(d => d.ContainsAny(value, connectedValues))) return false;

// Record any other values we can chain to from this value
var matches = dominoes.Where(d => d.Contains(value)).ToList();
foreach (var domino in matches)
{
dominoes.Remove(domino);
}
}

// If we reach here, we've proven it is possible to create a circular chain from all of the dominoes provided
return true;
}

// Indicates if a domino consists of one specific number AND one of various possible second numbers
private static bool ContainsAny(this (int, int) domino, int expected1, HashSet<int> expected2) =>
domino.Item1 == expected1 && expected2.Contains(domino.Item2) ||
expected2.Contains(domino.Item1) && domino.Item2 == expected1;

// Indicates if a domino contains the expected face value on either side
private static bool Contains(this (int, int) domino, int expected) =>
domino.Item1 == expected || domino.Item2 == expected;

private static List<int> FlattenTuples(this IEnumerable<(int, int)> dominoes) =>
dominoes.SelectMany(d => new[] { d.Item1, d.Item2 }).ToList();

// If any of the face values appear an uneven number of times, a full chain is not possible
private static bool IsUnbalancedSet(this IEnumerable<int> dominoValues) =>
dominoValues
.GroupBy(i => i)
.Select(grp => grp.Count())
.Any(count => count % 2 != 0);
}``````