仅利用30行Python代码来展示X算法

X = {1, 2, 3, 4, 5, 6, 7}
Y = {
  'A': [1, 4, 7],
  'B': [1, 4],
  'C': [4, 5, 7],
  'D': [3, 5, 6],
  'E': [2, 3, 6, 7],
  'F': [2, 7]}

X = {
  1: {'A', 'B'},
  2: {'E', 'F'},
  3: {'D', 'E'},
  4: {'A', 'B', 'C'},
  5: {'C', 'D'},
  6: {'D', 'E'},
  7: {'A', 'C', 'E', 'F'}}

def solve(X, Y, solution=[]):
  if not X:
    yield list(solution)
  else:
    c = min(X, key=lambda c: len(X[c]))
    for r in list(X[c]):
      solution.append(r)
      cols = select(X, Y, r)
      for s in solve(X, Y, solution):
        yield s
      deselect(X, Y, r, cols)
      solution.pop()
 
def select(X, Y, r):
  cols = []
  for j in Y[r]:
    for i in X[j]:
      for k in Y[i]:
        if k != j:
          X[k].remove(i)
    cols.append(X.pop(j))
  return cols
 
def deselect(X, Y, r, cols):
  for j in reversed(Y[r]):
    X[j] = cols.pop()
    for i in X[j]:
      for k in Y[i]:
        if k != j:
          X[k].add(i)

X = {j: set(filter(lambda i: j in Y[i], Y)) for j in X}

X = {j: set() for j in X}
for i in Y:
  for j in Y[i]:
    X[j].add(i)