"""Unit tests for the chain decomposition functions."""
from itertools import cycle
from itertools import islice

import networkx as nx


def cycles(seq):
    """Yields cyclic permutations of the given sequence.

    For example::

        >>> list(cycles("abc"))
        [('a', 'b', 'c'), ('b', 'c', 'a'), ('c', 'a', 'b')]

    """
    n = len(seq)
    cycled_seq = cycle(seq)
    for x in seq:
        yield tuple(islice(cycled_seq, n))
        next(cycled_seq)


def cyclic_equals(seq1, seq2):
    """Decide whether two sequences are equal up to cyclic permutations.

    For example::

        >>> cyclic_equals("xyz", "zxy")
        True
        >>> cyclic_equals("xyz", "zyx")
        False

    """
    # Cast seq2 to a tuple since `cycles()` yields tuples.
    seq2 = tuple(seq2)
    return any(x == tuple(seq2) for x in cycles(seq1))


class TestChainDecomposition:
    """Unit tests for the chain decomposition function."""

    def assertContainsChain(self, chain, expected):
        # A cycle could be expressed in two different orientations, one
        # forward and one backward, so we need to check for cyclic
        # equality in both orientations.
        reversed_chain = list(reversed([tuple(reversed(e)) for e in chain]))
        for candidate in expected:
            if cyclic_equals(chain, candidate):
                break
            if cyclic_equals(reversed_chain, candidate):
                break
        else:
            self.fail("chain not found")

    def test_decomposition(self):
        edges = [
            # DFS tree edges.
            (1, 2),
            (2, 3),
            (3, 4),
            (3, 5),
            (5, 6),
            (6, 7),
            (7, 8),
            (5, 9),
            (9, 10),
            # Nontree edges.
            (1, 3),
            (1, 4),
            (2, 5),
            (5, 10),
            (6, 8),
        ]
        G = nx.Graph(edges)
        expected = [
            [(1, 3), (3, 2), (2, 1)],
            [(1, 4), (4, 3)],
            [(2, 5), (5, 3)],
            [(5, 10), (10, 9), (9, 5)],
            [(6, 8), (8, 7), (7, 6)],
        ]
        chains = list(nx.chain_decomposition(G, root=1))
        assert len(chains) == len(expected)

    # This chain decomposition isn't unique
    #        for chain in chains:
    #            print(chain)
    #            self.assertContainsChain(chain, expected)

    def test_barbell_graph(self):
        # The (3, 0) barbell graph has two triangles joined by a single edge.
        G = nx.barbell_graph(3, 0)
        chains = list(nx.chain_decomposition(G, root=0))
        expected = [[(0, 1), (1, 2), (2, 0)], [(3, 4), (4, 5), (5, 3)]]
        assert len(chains) == len(expected)
        for chain in chains:
            self.assertContainsChain(chain, expected)

    def test_disconnected_graph(self):
        """Test for a graph with multiple connected components."""
        G = nx.barbell_graph(3, 0)
        H = nx.barbell_graph(3, 0)
        mapping = dict(zip(range(6), "abcdef"))
        nx.relabel_nodes(H, mapping, copy=False)
        G = nx.union(G, H)
        chains = list(nx.chain_decomposition(G))
        expected = [
            [(0, 1), (1, 2), (2, 0)],
            [(3, 4), (4, 5), (5, 3)],
            [("a", "b"), ("b", "c"), ("c", "a")],
            [("d", "e"), ("e", "f"), ("f", "d")],
        ]
        assert len(chains) == len(expected)
        for chain in chains:
            self.assertContainsChain(chain, expected)

    def test_disconnected_graph_root_node(self):
        """Test for a single component of a disconnected graph."""
        G = nx.barbell_graph(3, 0)
        H = nx.barbell_graph(3, 0)
        mapping = dict(zip(range(6), "abcdef"))
        nx.relabel_nodes(H, mapping, copy=False)
        G = nx.union(G, H)
        chains = list(nx.chain_decomposition(G, root="a"))
        expected = [
            [("a", "b"), ("b", "c"), ("c", "a")],
            [("d", "e"), ("e", "f"), ("f", "d")],
        ]
        assert len(chains) == len(expected)
        for chain in chains:
            self.assertContainsChain(chain, expected)