# Résolution de problèmes algorithmiques

Given a grid containing some segments, find the minimum number of times segments need to be displaced such that a particular segment can escape from the grid.

## A shortest path problem in the configuration graph

We model the problem as follows. On the given 6 by 6 grid are placed n vehicles. Vehicles are numbered starting from 0, 0 being the special vehicle which has to escape the grid. Each vehicle has a width equal to 1 and length 2 or 3 grid cells. It can be placed horizontally or vertically. Its location is identified by the coordinate of its left most upper most cell. Depending on the vehicle orientation one of the coordinates is fixed, while the other one can vary. We define a configuration as the vector consisting of the variable coordinates of the vehicles.

For example the previous configuration would be encoded as follows.

# vehicle     0     1      2     3      4      5     6     7     8
horizontal = [True, False, True, False, False, True, True, True, False]
length =     [2,    2,     3,    3,     2,     2,    3,    2,    2]
fixcoor =    [2,    0,     0,    4,     2,     3,    3,    4,    5]
config =     (0,    0,     1,    0,     2,     4,    2,    3,    4)


Now in one step a single vehicle can be displaced, which consists in changing one coordinate of the configuration vector. This describes some underlying graph, where configurations are vertices and there is an edge between configurations A and B if B can be obtained from A in one step. The goal is to find a shortest path from the initial configuration to a target configuration (which is one where vehicle 0 reaches the border of the grid). This can be done by simple BFS traversal of the graph. The only new part here is that the graph is only implicitly given.

The BFS search on an implicitly given graph uses a function graph which maps a vertex to a list of neighboring vertices in the graph. It also needs a function is_target to check if with a given vertex the traversal is completed.

def bfs_implicit(graph, start, is_target):
dist = {start: 0}
to_visit = deque([start])
while to_visit:
node = to_visit.pop()
for neighbor in graph(node):
if neighbor not in dist:   # new vertex
dist[neighbor] = dist[node] + 1
to_visit.appendleft(neighbor)
if is_target(neighbor):
return dist[neighbor]
return None   # target is not reachable


So for this problem we have to start building the data structures from the given intial grid. This is done as follows by inspecting the grid on row wise order. As you can see this is the longest part of the code.

def read(grid):
""" reads the grid and builds the data structure.
returns current configuration
"""
global horizontal, fixcoor, length
row = [-1] * n                             # first coordinates seen of a vehicle
col = [-1] * n
horizontal = [None] * n                    # create data structures
length = [None] * n
for i in range(dim):                       # loop over all grid cells (i,j)
for j in range(dim):
c = grid[i][j]
if c != '.':
if c == 'x':                   # determine vehicle index from character
v = 0
else:
v = ord(c) - ord('a') + 1
if row[v] == -1:               # first time vehicle is seen
row[v] = i
col[v] = j
length[v] = 1
else:                          # rest of the vehicle
horizontal[v] = (row[v] == i)
length[v] += 1
fixcoor = []
config = []
for v in range(n):                         # set fixed coordinate
if horizontal[v]:
fixcoor.append(row[v])
config.append(col[v])
else:
fixcoor.append(col[v])
config.append(row[v])
return tuple(config)                       # current configuration


Finally to encode the graph, we just need to implement the neighbor oracle function and the target oracle test. For those we need a helper function that verifies if in a given configuration a given cell is occupied by some vehicle.

def occupied(config, r, c):
"""returns if cell (r,c) is occupied in the given configuration
"""
if not( 0 <= r < dim and 0 <= c < dim):    # cells around the grid are occupied
return True
for v in range(n):             # is vehicle v covering cell (r,c)?
if (horizontal[v] and fixcoor[v] == r and config[v] <= c < config[v] + length[v] or
not horizontal[v] and fixcoor[v] == c and config[v] <= r < config[v] + length[v]):
return True
return False

def is_target(config):
return config[0] == dim - length[0]

def graph(config):
"""iterates over all reachable configurations from the given one
"""
for v in range(n):           # loop over all vehicles
if horizontal[v]:
# right
d = 1
while not occupied(config, fixcoor[v], config[v] + d + length[v] - 1):
yield config[:v] + (config[v] + d, ) + config[v+1:]
d += 1
# left
d = 1
while not occupied(config, fixcoor[v], config[v] - d):
yield config[:v] + (config[v] - d, ) + config[v+1:]
d += 1
else:
# down
d = 1
while not occupied(config, config[v] + d + length[v] - 1, fixcoor[v]):
yield config[:v] + (config[v] + d, ) + config[v+1:]
d += 1
# up
d = 1
while not occupied(config, config[v] - d, fixcoor[v]):
yield config[:v] + (config[v] - d, ) + config[v+1:]
d += 1


We can verify in a quick and rough estimation that the overal running time is acceptable. A configuration consists of a vector of dimension at most 10 and there are 5 choices for each vector element. This gives an upper bound of $5^{10} < 2^{24}$ vertices. But the actual number must be much smaller since most of the configuration vectors would not be valid by overlapping vehicles. The number of neighbors of a configuration can also be bounded by $5\cdot 10$. Hence we can estimate the BFS to terminate within a few million operations.