计算Android屏幕解锁组合数

晚饭时和同事聊到安卓屏幕解锁时会有多少种解锁方案，觉得很有趣，吃完饭开始想办法解题，花了大概2个小时解决。思路如下：

1. 使用索引值0-9表示从左到右、从上到下的9个点，行、列号很容易从索引值得到；
2. 使用一个列表（routeList）来表示解锁路径，列表的元素为点的索引值；
3. 从任意点N出发（将index(N)放入routeList），判断与哪些点（集合NextNodes）可以构成合法路径；然后用递归的方式从NextNodes中的点出发，寻找下一个可以构成合法路径的点；直到再也无法找到可用的点供路径使用

下一步，就是在Nexus 7上捣鼓，发现Android定义合法的解锁路径必须满足下列条件：

• 构成路径的点不得少于4个
• 路径中不能有重复的点
• （这点有点绕口）构成边或对角线的两点（例如[0,2]构成边，[0,8]构成对角线），不能在路径中相邻，除非其中点已经在路径中出现

接下来就是写代码了，Let's talk with the code!

问题的解是：

 ###############################################################
  #######                   AndrUlkComb.py              #########
  ## Calculate the COMBination count for ANDroid UNLock screen ##
  ##              By grepall@gmail.com                         ##
  ###############################################################

  ###############################################################
  ## The Android unlock screen looks like below
  ## -------------------------------
  ##       o  ->  o  ->  o
  ##
  ##       o      o      o
  ##
  ##       o      o      o
  ## --------------------------------
  ##
  ## We'll use:
  ##  1. Index from 0-8 to represent all 9 nodes instead of
  ##      using row:column format.
  ##  2. An array (routeList in the code) to record the path of the unlock combination.
  ##      For example, for the unlock comb which is shown in
  ##      in the figure. There will be 3 elements in the array:
  ##      0, 1, and 2. (Although it's NOT a valid path because of
  ##      not long enough)
  ##
  ## Valid unlock path must holds true for 3 conditions:
  ##  C1: It must pass 4 different nodes at least;
  ##  C2: It cannot contains duplicate nodes;
  ##  C3: This is a little tricky. Any node cannot direct link to another node, when
  ##      any node will be past through if we connect those 2 nodes in the unlock panel. UNLESS
  ##      the crossed node already EXISTs in the path.
  ##      For example, path (0, 2, 5, 4) is not valid, because 0-->2 crosses node 1. But path
  ##      (1, 0, 2, 5) is a vliad path.
  ##      This rule holds for diagonal as well.

  #######################################################################
  # M stands for the width of the unlock panel
  # H stands for the height of the unlock panel
  # CAUTION: Modify this value will NOT generate correct result!!! That's TO BE DONE
  #
  W = 3   #Width
  H = 3   #Height

  # Unlock path
  routeList = list()

  # Combination count, which is we want to know
  routeCount = 0

  def isValidRoute(p1, route):
      # If the candidate node exists in the unlock path
      if route.count(p1) != 0:
          return False
      return True;

  def isCrossPoint(p1, p2):
      # Will the connection line (p1, p2) cross another node?
      x1 = p1%W
      y1 = p1/W
      x2 = p2%W
      y2 = p2/W
      if x1 == x2 and abs(y1-y2) > 1:
          return True # same column
      if y1 == y2 and abs(x1-x2) > 1:
          return True # same row
      if abs(y1-y2) == abs(x1-x2) and abs(y1-y2)>1:
          return True # diagonal
      return False

  def tryPoint(lastPt, route):
      # Try next valid node for the unlock path recursively
      global routeCount
      for pt in range(0, W*H):
          if isValidRoute(pt, route): #C2
              crossPt = (lastPt+pt)/2
              if isCrossPoint(lastPt, pt) and route.count(crossPt) == 0: #C3
                  continue
              route.append(pt)
              if len(route) >3: #C1
                  routeCount += 1
              tryPoint(pt, route)
              route.pop()

  # Each node may be the start node for unlock
  for i in range(0, W*H):
      routeList = [i]
      tryPoint(i, routeList)
      print "Start from point %d, combination count is %d now..." % (i, routeCount)
  print "Toatl combination counter for Android unlock screen: " + str(routeCount)