Blog Archive

Saturday, October 4, 2014

Perseverance Quotes and Proverbs

Perseverance Quotes and Proverbs: "I hold a doctrine, to which I owe not much, indeed, but all the little I ever had, namely, that with ordinary talent and extraordinary perseverance, all things are attainable.
Sir T. F. Buxton"



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Thursday, October 2, 2014

maze

http://www.cs.carleton.edu/faculty/jgoldfea/cs201/spring11/assignments/Maze.py



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Find Shortest Path With Obstacle MAZE

Find Shortest Path With Obstacle:



"""
2D array of characters
 
1 1 1 1 1
S 1 X 1 1
1 1 1 1 1
X 1 1 E 1
1 1 1 1 X
 
 
S is the starting point
E is the ending point
X means you cannot traverse to that point
 
1. Find if there is a path from S to E
2. Find the length of shortest path from S to E given the above matrix
3. Find the shortest path from S to E given the above matrix
 
 
Restriction: Move to 8 positions
"""
 
def find_path(map):
M = len(map)
N = len(map[0])
S = None
for i in range(M):
if S:
break
for j in range(N):
if map[i][j] == 'S':
S = (i, j)
break
 
queue = [ ( S, [] ) ]
 
while len(queue) > 0:
current, path = queue.pop()
x, y = current
if x < 0 or y < 0 or x >= M or y >= N or map[x][y] == 'X':
continue
 
path.append(current)
if map[x][y] == 'E':
return path # Or len(path) for question 2
 
map[x][y] = 'X' # Mark as visited
 
queue.insert(0, ((x-1, y-1), path[:]))
queue.insert(0, ((x-1, y ), path[:]))
queue.insert(0, ((x-1, y+1), path[:]))
queue.insert(0, ((x , y-1), path[:]))
queue.insert(0, ((x , y+1), path[:]))
queue.insert(0, ((x+1, y-1), path[:]))
queue.insert(0, ((x+1, y ), path[:]))
queue.insert(0, ((x+1, y+1), path[:]))
 
return None
 
map = [ ['1', '1', '1', '1', '1',],
['S', '1', 'X', '1', '1',],
['1', '1', '1', '1', '1',],
['X', '1', '1', 'E', '1',],
['1', '1', '1', '1', 'X',],
]
 
print find_path(map)
 
# 讨论
# 1. 无论用何种方法,最好的复杂度应该都是O(m*n)因为最坏情况都是要遍历完整个图
# 2. 但是单就最好情况来说,想讨论下有没有更好的方法,主要考虑如下:
# 因为实际已知 S 和 E 的坐标,这里用到的BFS实际是和不知道终点位置是一样的,都是BFS最先的情况就是答案
# 但我觉得方法应该是可以向着终点方向走,遇到障碍让开走,如果能够实现这种方法,肯定是最优的解法,
# 希望大家能给我点思路