How to tell who wins in rock paper scissors

RockPaperScissors.cs

enum RockPaperScissorsChoice
{
	Rock = 1,
	Paper = 2,
	Scissors = 4
}

static string DescribeRockPaperScissorsResult(RockPaperScissorsChoice first, RockPaperScissorsChoice second)
{
	if (first == second)
			return "Tie.";
	if ((((int)first >> 1) | ((int)first << 2) & 5) == (int)second)
			return $"{first} beats {second}.";
	return $"{second} beats {first}.";
}

static void Test() {
	Assert(RockPaperScissorsChoice.Rock, RockPaperScissorsChoice.Rock, "Tie.");
	Assert(RockPaperScissorsChoice.Rock, RockPaperScissorsChoice.Paper, "Paper beats Rock.");
	Assert(RockPaperScissorsChoice.Rock, RockPaperScissorsChoice.Scissors, "Rock beats Scissors.");
	Assert(RockPaperScissorsChoice.Paper, RockPaperScissorsChoice.Rock, "Paper beats Rock.");
	Assert(RockPaperScissorsChoice.Paper, RockPaperScissorsChoice.Paper, "Tie.");
	Assert(RockPaperScissorsChoice.Paper, RockPaperScissorsChoice.Scissors, "Scissors beats Paper.");
	Assert(RockPaperScissorsChoice.Scissors, RockPaperScissorsChoice.Rock, "Rock beats Scissors.");
	Assert(RockPaperScissorsChoice.Scissors, RockPaperScissorsChoice.Paper, "Scissors beats Paper.");
	Assert(RockPaperScissorsChoice.Scissors, RockPaperScissorsChoice.Scissors, "Tie.");
}

static void Assert(RockPaperScissorsChoice first, RockPaperScissorsChoice second, string expected)
{
	var actual = DescribeRockPaperScissorsResult(first, second);
	if (actual == expected)
			Debug.WriteLine($"PASS Expected: {expected} \tActual: {actual}");
	else
			Debug.Fail($"FAIL Expected: {expected} \tActual: {actual}");
}

Theory of operation

As any schoolchild will tell you, the possible outcomes of a rock-paper-scissors game are described by a directed cyclic graph on three vertices, where each player choice is a vertex and the edges connect each choice to what it beats, or alternatively, to what beats it. Starting with such a graph, determining the winner between choices X and Y can then be done by following the edge from X and checking whether it takes you to Y. If it does, X wins (or X loses, depending on how you defined the graph). If X = Y, the outcome is a tie. If following the edge from X did not lead to Y and there was no tie, then Y must beat X (or X wins, depending on how you defined the graph). The funny idea that led to this code was to realize this traversal of the cycle using a bitwise rotation of a 3-bit integer. I submit it as a simultaneously obfuscated and succinct solution to this well-known toy problem.