text: # Copyright 2019 The Google Research Authors.
  #
  # Licensed under the Apache License, Version 2.0 (the "License");
  # you may not use this file except in compliance with the License.
  # You may obtain a copy of the License at
  #
  #     http://www.apache.org/licenses/LICENSE-2.0
  #
  # Unless required by applicable law or agreed to in writing, software
  # distributed under the License is distributed on an "AS IS" BASIS,
  # WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
  # See the License for the specific language governing permissions and
  # limitations under the License.
  
  """Utility functions used by generate_graph.py."""
  from __future__ import absolute_import
  from __future__ import division
  from __future__ import print_function
  
  import hashlib
  import itertools
  
  import numpy as np
  
  
  def gen_is_edge_fn(bits):
      """Generate a boolean function for the edge connectivity.
  
      Given a bitstring FEDCBA and a 4x4 matrix, the generated matrix is
        [[0, A, B, D],
         [0, 0, C, E],
         [0, 0, 0, F],
         [0, 0, 0, 0]]
  
      Note that this function is agnostic to the actual matrix dimension due to
      order in which elements are filled out (column-major, starting from least
      significant bit). For example, the same FEDCBA bitstring (0-padded) on a 5x5
      matrix is
        [[0, A, B, D, 0],
         [0, 0, C, E, 0],
         [0, 0, 0, F, 0],
         [0, 0, 0, 0, 0],
         [0, 0, 0, 0, 0]]
  
      Args:
        bits: integer which will be interpreted as a bit mask.
  
      Returns:
        vectorized function that returns True when an edge is present.
      """
      def is_edge(x, y):
          """Is there an edge from x to y (0-indexed)?"""
          if x >= y:
              return 0
          # Map x, y to index into bit string
          index = x + (y * (y - 1) // 2)
          return (bits >> index) % 2 == 1
  
      return np.vectorize(is_edge)
  
  
  def is_full_dag(matrix):
      """Full DAG == all vertices on a path from vert 0 to (V-1).
  
      i.e. no disconnected or "hanging" vertices.
  
      It is sufficient to check for:
        1) no rows of 0 except for row V-1 (only output vertex has no out-edges)
        2) no cols of 0 except for col 0 (only input vertex has no in-edges)
  
      Args:
        matrix: V x V upper-triangular adjacency matrix
  
      Returns:
        True if the there are no dangling vertices.
      """
      shape = np.shape(matrix)
  
      rows = matrix[:shape[0]-1, :] == 0
      rows = np.all(rows, axis=1)     # Any row with all 0 will be True
      rows_bad = np.any(rows)
  
      cols = matrix[:, 1:] == 0
      cols = np.all(cols, axis=0)     # Any col with all 0 will be True
      cols_bad = np.any(cols)
  
      return (not rows_bad) and (not cols_bad)
  
  
  def num_edges(matrix):
      """Computes number of edges in adjacency matrix."""
      return np.sum(matrix)
  
  
  def hash_module(matrix, labeling, dense_label=False):
      """Computes a graph-invariance MD5 hash of the matrix and label pair.
  
      Args:
        matrix: np.ndarray square upper-triangular adjacency matrix.
        labeling: list of int labels of length equal to both dimensions of
          matrix.
  
      Returns:
        MD5 hash of the matrix and labeling.
      """
  
      vertices = np.shape(matrix)[0]
      in_edges = np.sum(matrix, axis=0).tolist()
      out_edges = np.sum(matrix, axis=1).tolist()
      assert len(in_edges) == len(out_edges) == len(labeling)
  
      if dense_label:
          # May not be necessary!!
          dense_labeling = []
          for i in range(vertices):
              _lbl = ''
              for j in range(vertices):
                  _lbl += labeling[j, i] + ","
              dense_labeling.append(_lbl)
          print("processing labeling into ", dense_labeling)
          labeling = dense_labeling
  
      hashes = list(zip(out_edges, in_edges, labeling))
      hashes = [hashlib.md5(str(h).encode('utf-8')).hexdigest() for h in hashes]
      # Computing this up to the diameter is probably sufficient but since the
      # operation is fast, it is okay to repeat more times.
      for _ in range(vertices):
          new_hashes = []
          for v in range(vertices):
              in_neighbors = [hashes[w] for w in range(vertices) if matrix[w, v]]
              out_neighbors = [hashes[w] for w in range(vertices) if matrix[v, w]]
              new_hashes.append(hashlib.md5(
                  (''.join(sorted(in_neighbors)) + '|' +
                   ''.join(sorted(out_neighbors)) + '|' +
                   hashes[v]).encode('utf-8')).hexdigest())
          hashes = new_hashes
      fingerprint = hashlib.md5(str(sorted(hashes)).encode('utf-8')).hexdigest()
      return fingerprint
  
  
  def permute_graph(graph, label, permutation):
      """Permutes the graph and labels based on permutation.
  
      Args:
        graph: np.ndarray adjacency matrix.
        label: list of labels of same length as graph dimensions.
        permutation: a permutation list of ints of same length as graph dimensions.
  
      Returns:
        np.ndarray where vertex permutation[v] is vertex v from the original graph
      """
      # vertex permutation[v] in new graph is vertex v in the old graph
      forward_perm = zip(permutation, list(range(len(permutation))))
      inverse_perm = [x[1] for x in sorted(forward_perm)]
      edge_fn = lambda x, y: graph[inverse_perm[x], inverse_perm[y]] == 1
      new_matrix = np.fromfunction(np.vectorize(edge_fn),
                                   (len(label), len(label)),
                                   dtype=np.int8)
      new_label = [label[inverse_perm[i]] for i in range(len(label))]
      return new_matrix, new_label
  
  
  def is_isomorphic(graph1, graph2):
      """Exhaustively checks if 2 graphs are isomorphic."""
      matrix1, label1 = np.array(graph1[0]), graph1[1]
      matrix2, label2 = np.array(graph2[0]), graph2[1]
      assert np.shape(matrix1) == np.shape(matrix2)
      assert len(label1) == len(label2)
  
      vertices = np.shape(matrix1)[0]
      # Note: input and output in our constrained graphs always map to themselves
      # but this script does not enforce that.
      for perm in itertools.permutations(range(0, vertices)):
          pmatrix1, plabel1 = permute_graph(matrix1, label1, perm)
          if np.array_equal(pmatrix1, matrix2) and plabel1 == label2:
              return True
  
      return False
  