Reputation: 1773
Pathfinding in a grid system
I'm currently building a small grid-based game in C++ using SDL. I've made a tile class, which represents each individual tile on a map. This tile class is used in a 2 dimensional vector, one dimension representing the X axis and the other one the Y axis.
I'm having an algorithm issue, I do not even know where to start with this, let's say I have this map:
0 0 1 1 0 E
0 0 0 1 0 1
0 C 0 1 0 1
0 1 0 1 0 1
0 1 1 1 1 1
C is my character, E is an exit, and 1 is a floor tile.
I would like to figure out the most optimal way to figure out if there's a way for the character to reach the exit. I know I could use a function to manually check every tile around C, and for every floor tile around C, I check again every tile around, until I find a consistent path to get to E, but that doesn't seem very optimal.
Can I have a clue or some sort of direction in which to orient myself?
Upvotes: 5
Views: 38192
Answers (6)
Reputation: 11
You can utilize depth-first search or breadth first search as the simplest ways to solve this. Also look at Dijkstra algorithm or A* algorithm specifically for path/route finding, but they are too complex for a simple problem.
BFS for maze (in python, as it was in my AI lab work):
from collections import deque
class MazeGraph:
def __init__(self, maze):
self.maze = maze
self.rows = len(maze)
self.cols = len(maze[0])
self.graph = self.construct_graph()
def construct_graph(self):
graph = {}
for i in range(self.rows):
for j in range(self.cols):
if self.maze[i][j] == 0: # Only consider open spaces
neighbors = []
if i - 1 >= 0 and self.maze[i - 1][j] == 0:
neighbors.append((i - 1, j))
if i + 1 < self.rows and self.maze[i + 1][j] == 0:
neighbors.append((i + 1, j))
if j - 1 >= 0 and self.maze[i][j - 1] == 0:
neighbors.append((i, j - 1))
if j + 1 < self.cols and self.maze[i][j + 1] == 0:
neighbors.append((i, j + 1))
graph[(i, j)] = neighbors
return graph
def bfs(self, source, destination):
if source not in self.graph or destination not in self.graph:
print("Source or destination not in the valid open spaces.")
return
visited = set()
queue = deque([source])
path = {}
while queue:
current_node = queue.popleft()
if current_node == destination:
# If destination is reached, construct and print the path
path_result = self.construct_path(source, destination, path)
print(f"Path from {source} to {destination}: {path_result}")
return
if current_node not in visited:
print(current_node, end=' ')
visited.add(current_node)
for neighbor in self.graph[current_node]:
if neighbor not in visited:
queue.append(neighbor)
path[neighbor] = current_node
def construct_path(self, source, destination, path):
# Reconstruct the path from source to destination
current = destination
path_result = [current]
while current != source:
current = path[current]
path_result.append(current)
path_result.reverse()
return ' -> '.join(map(str, path_result))
maze_array = [
[1, 1, 1, 1, 1, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 0, 1],
[1, 0, 0, 0, 1, 0, 0, 1],
[1, 0, 0, 1, 1, 0, 1, 1],
[1, 0, 0, 0, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1, 1],
[1, 0, 0, 0, 1, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1, 1]
]
maze_graph = MazeGraph(maze_array)
source_node = (1, 1)
destination_node = (0, 6)
maze_graph.bfs(source_node, destination_node)
Upvotes: 0
Reputation: 421
"the most optimal way to figure out if there's a way for the character to reach the exit"
Well... blurring comes to mind, depending on the size of your maps.
Bit-blurring in small or dense maps/mazes
Imagine 'flattening' your map to a long string of bits, where a 1 is passable and a zero is impassable. The Width of your map can be used as a 'stride' ... that is, for any bit position -W bits is up and +W bits is down. +1 and -1 are left and right. Thus a 2D map is represented in 1D form.
Let's call this flat bitwise representation of the map 'M'
Now reserve an identical string of uint64's and set them to zero. This will be your maps working space.
Let's call this empty bitmap 'S', for Search.
Now, to blur you simply need to set a 1 bit where your player is, in 'S'
Then, OR 'S' with S>>W, S<<W, S>>1 and S<<1. This causes a + shape. Next, AND it with the M bitmap... this will remove any parts of the + which intersect obstructions.
Repeat this operation, until the bit in S which matches your exit is set (or S has not changed from its previous value, indicating that the search is complete)
This is fast in graphs with many paths. Using uint64's, you're evaluating 64 squares of your map at a time. If you turn on compiler optimisations you can bring SIMD to bear, evaluating 256 map squares per cycle.
Advanced bit-manipulation
There is a faster method, which involves some bit tricks to set entire horizontal runs in one operation, and to cascade the set bits downwards till they hit obstructions.
Unfortunately, the second method is probably a little complicated to explain here. Both methods are suitable for inline ASM as well as in your HLL of choice... and both can leverage SIMD.
Of course, these approaches only tell you if the exit is reachable.
Bytewise addition of fields
A similar scheme with bytes is possible, which allows you to blur 1's and add them back, generating a distance field that can be followed in reverse. So, you'd set 1's at the exit... and once done your player could follow the ascending values for up to 255 steps, towards the exit.
Better approaches: Precomputation
Precomputation is your friend. If you have a number of set levels you can bake these precomputed paths and distances into the level map itself... allowing it to be navigated efficiently during game time. Each square can have an (as small as 2-bit) value which encodes U,D,L and R ... indicating the direction to some feature. Multiple features can be encoded efficiently this way.
... after all, there's nothing faster than a precomputed lookup ; )
Pre-built, heavily simplified, graphs
Lastly, there are predefined graphs. Many mazes have long, possibly winding, passages that can effectively be thought of as a single square, or node. By saving such a highly simplified node structure with your level, you allow some very fast pathfinding on the fly.
Hope any of that helped someone think about the problem a different way. If not, there's A*, Dijkstra, BFS, etc...
Upvotes: 1
Reputation: 41
#include<bits/stdc++.h>
using namespace std;
#define ROW 9
#define COL 10
// Creating a shortcut for int, int pair type
typedef pair<int, int> Pair;
// Creating a shortcut for pair<int, pair<int, int>> type
typedef pair<double, pair<int, int> > pPair;
// A structure to hold the neccesary parameters
struct cell
{
// Row and Column index of its parent
// Note that 0 <= i <= ROW-1 & 0 <= j <= COL-1
int parent_i, parent_j;
// f = g + h
double f, g, h;
};
// A Utility Function to check whether given cell (row, col)
// is a valid cell or not.
bool isValid(int row, int col)
{
// Returns true if row number and column number
// is in range
return (row >= 0) && (row < ROW) &&
(col >= 0) && (col < COL);
}
// A Utility Function to check whether the given cell is
// blocked or not
bool isUnBlocked(int grid[][COL], int row, int col)
{
// Returns true if the cell is not blocked else false
if (grid[row][col] == 1)
return (true);
else
return (false);
}
// A Utility Function to check whether destination cell has
// been reached or not
bool isDestination(int row, int col, Pair dest)
{
if (row == dest.first && col == dest.second)
return (true);
else
return (false);
}
// A Utility Function to calculate the 'h' heuristics.
double calculateHValue(int row, int col, Pair dest)
{
// Return using the distance formula
return ((double)sqrt ((row-dest.first)*(row-dest.first)
+ (col-dest.second)*(col-dest.second)));
}
// A Utility Function to trace the path from the source
// to destination
void tracePath(cell cellDetails[][COL], Pair dest)
{
printf ("\nThe Path is ");
int row = dest.first;
int col = dest.second;
stack<Pair> Path;
while (!(cellDetails[row][col].parent_i == row
&& cellDetails[row][col].parent_j == col ))
{
Path.push (make_pair (row, col));
int temp_row = cellDetails[row][col].parent_i;
int temp_col = cellDetails[row][col].parent_j;
row = temp_row;
col = temp_col;
}
Path.push (make_pair (row, col));
while (!Path.empty())
{
pair<int,int> p = Path.top();
Path.pop();
printf("-> (%d,%d) ",p.first,p.second);
}
return;
}
// A Function to find the shortest path between
// a given source cell to a destination cell according
// to A* Search Algorithm
void aStarSearch(int grid[][COL], Pair src, Pair dest)
{
// If the source is out of range
if (isValid (src.first, src.second) == false)
{
printf ("Source is invalid\n");
return;
}
// If the destination is out of range
if (isValid (dest.first, dest.second) == false)
{
printf ("Destination is invalid\n");
return;
}
// Either the source or the destination is blocked
if (isUnBlocked(grid, src.first, src.second) == false ||
isUnBlocked(grid, dest.first, dest.second) == false)
{
printf ("Source or the destination is blocked\n");
return;
}
// If the destination cell is the same as source cell
if (isDestination(src.first, src.second, dest) == true)
{
printf ("We are already at the destination\n");
return;
}
// Create a closed list and initialise it to false which means
// that no cell has been included yet
// This closed list is implemented as a boolean 2D array
bool closedList[ROW][COL];
memset(closedList, false, sizeof (closedList));
// Declare a 2D array of structure to hold the details
//of that cell
cell cellDetails[ROW][COL];
int i, j;
for (i=0; i<ROW; i++)
{
for (j=0; j<COL; j++)
{
cellDetails[i][j].f = FLT_MAX;
cellDetails[i][j].g = FLT_MAX;
cellDetails[i][j].h = FLT_MAX;
cellDetails[i][j].parent_i = -1;
cellDetails[i][j].parent_j = -1;
}
}
// Initialising the parameters of the starting node
i = src.first, j = src.second;
cellDetails[i][j].f = 0.0;
cellDetails[i][j].g = 0.0;
cellDetails[i][j].h = 0.0;
cellDetails[i][j].parent_i = i;
cellDetails[i][j].parent_j = j;
/*
Create an open list having information as-
<f, <i, j>>
where f = g + h,
and i, j are the row and column index of that cell
Note that 0 <= i <= ROW-1 & 0 <= j <= COL-1
This open list is implenented as a set of pair of pair.*/
set<pPair> openList;
// Put the starting cell on the open list and set its
// 'f' as 0
openList.insert(make_pair (0.0, make_pair (i, j)));
// We set this boolean value as false as initially
// the destination is not reached.
bool foundDest = false;
while (!openList.empty())
{
pPair p = *openList.begin();
// Remove this vertex from the open list
openList.erase(openList.begin());
// Add this vertex to the open list
i = p.second.first;
j = p.second.second;
closedList[i][j] = true;
/*
Generating all the 8 successor of this cell
N.W N N.E
\ | /
\ | /
W----Cell----E
/ | \
/ | \
S.W S S.E
Cell-->Popped Cell (i, j)
N --> North (i-1, j)
S --> South (i+1, j)
E --> East (i, j+1)
W --> West (i, j-1)
N.E--> North-East (i-1, j+1)
N.W--> North-West (i-1, j-1)
S.E--> South-East (i+1, j+1)
S.W--> South-West (i+1, j-1)*/
// To store the 'g', 'h' and 'f' of the 8 successors
double gNew, hNew, fNew;
//----------- 1st Successor (North) ------------
// Only process this cell if this is a valid one
if (isValid(i-1, j) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i-1, j, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i-1][j].parent_i = i;
cellDetails[i-1][j].parent_j = j;
printf ("The destination cell is found\n");
tracePath (cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i-1][j] == false &&
isUnBlocked(grid, i-1, j) == true)
{
gNew = cellDetails[i][j].g + 1.0;
hNew = calculateHValue (i-1, j, dest);
fNew = gNew + hNew;
// If it isn’t on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i-1][j].f == FLT_MAX ||
cellDetails[i-1][j].f > fNew)
{
openList.insert( make_pair(fNew,
make_pair(i-1, j)));
// Update the details of this cell
cellDetails[i-1][j].f = fNew;
cellDetails[i-1][j].g = gNew;
cellDetails[i-1][j].h = hNew;
cellDetails[i-1][j].parent_i = i;
cellDetails[i-1][j].parent_j = j;
}
}
}
//----------- 2nd Successor (South) ------------
// Only process this cell if this is a valid one
if (isValid(i+1, j) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i+1, j, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i+1][j].parent_i = i;
cellDetails[i+1][j].parent_j = j;
printf("The destination cell is found\n");
tracePath(cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i+1][j] == false &&
isUnBlocked(grid, i+1, j) == true)
{
gNew = cellDetails[i][j].g + 1.0;
hNew = calculateHValue(i+1, j, dest);
fNew = gNew + hNew;
// If it isn’t on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i+1][j].f == FLT_MAX ||
cellDetails[i+1][j].f > fNew)
{
openList.insert( make_pair (fNew, make_pair (i+1, j)));
// Update the details of this cell
cellDetails[i+1][j].f = fNew;
cellDetails[i+1][j].g = gNew;
cellDetails[i+1][j].h = hNew;
cellDetails[i+1][j].parent_i = i;
cellDetails[i+1][j].parent_j = j;
}
}
}
//----------- 3rd Successor (East) ------------
// Only process this cell if this is a valid one
if (isValid (i, j+1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i, j+1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i][j+1].parent_i = i;
cellDetails[i][j+1].parent_j = j;
printf("The destination cell is found\n");
tracePath(cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i][j+1] == false &&
isUnBlocked (grid, i, j+1) == true)
{
gNew = cellDetails[i][j].g + 1.0;
hNew = calculateHValue (i, j+1, dest);
fNew = gNew + hNew;
// If it isn’t on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i][j+1].f == FLT_MAX ||
cellDetails[i][j+1].f > fNew)
{
openList.insert( make_pair(fNew,
make_pair (i, j+1)));
// Update the details of this cell
cellDetails[i][j+1].f = fNew;
cellDetails[i][j+1].g = gNew;
cellDetails[i][j+1].h = hNew;
cellDetails[i][j+1].parent_i = i;
cellDetails[i][j+1].parent_j = j;
}
}
}
//----------- 4th Successor (West) ------------
// Only process this cell if this is a valid one
if (isValid(i, j-1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i, j-1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i][j-1].parent_i = i;
cellDetails[i][j-1].parent_j = j;
printf("The destination cell is found\n");
tracePath(cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i][j-1] == false &&
isUnBlocked(grid, i, j-1) == true)
{
gNew = cellDetails[i][j].g + 1.0;
hNew = calculateHValue(i, j-1, dest);
fNew = gNew + hNew;
// If it isn’t on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i][j-1].f == FLT_MAX ||
cellDetails[i][j-1].f > fNew)
{
openList.insert( make_pair (fNew,
make_pair (i, j-1)));
// Update the details of this cell
cellDetails[i][j-1].f = fNew;
cellDetails[i][j-1].g = gNew;
cellDetails[i][j-1].h = hNew;
cellDetails[i][j-1].parent_i = i;
cellDetails[i][j-1].parent_j = j;
}
}
}
//----------- 5th Successor (North-East) ------------
// Only process this cell if this is a valid one
if (isValid(i-1, j+1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i-1, j+1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i-1][j+1].parent_i = i;
cellDetails[i-1][j+1].parent_j = j;
printf ("The destination cell is found\n");
tracePath (cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i-1][j+1] == false &&
isUnBlocked(grid, i-1, j+1) == true)
{
gNew = cellDetails[i][j].g + 1.414;
hNew = calculateHValue(i-1, j+1, dest);
fNew = gNew + hNew;
// If it isn’t on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i-1][j+1].f == FLT_MAX ||
cellDetails[i-1][j+1].f > fNew)
{
openList.insert( make_pair (fNew,
make_pair(i-1, j+1)));
// Update the details of this cell
cellDetails[i-1][j+1].f = fNew;
cellDetails[i-1][j+1].g = gNew;
cellDetails[i-1][j+1].h = hNew;
cellDetails[i-1][j+1].parent_i = i;
cellDetails[i-1][j+1].parent_j = j;
}
}
}
//----------- 6th Successor (North-West) ------------
// Only process this cell if this is a valid one
if (isValid (i-1, j-1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination (i-1, j-1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i-1][j-1].parent_i = i;
cellDetails[i-1][j-1].parent_j = j;
printf ("The destination cell is found\n");
tracePath (cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i-1][j-1] == false &&
isUnBlocked(grid, i-1, j-1) == true)
{
gNew = cellDetails[i][j].g + 1.414;
hNew = calculateHValue(i-1, j-1, dest);
fNew = gNew + hNew;
// If it isn’t on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i-1][j-1].f == FLT_MAX ||
cellDetails[i-1][j-1].f > fNew)
{
openList.insert( make_pair (fNew, make_pair (i-1, j-1)));
// Update the details of this cell
cellDetails[i-1][j-1].f = fNew;
cellDetails[i-1][j-1].g = gNew;
cellDetails[i-1][j-1].h = hNew;
cellDetails[i-1][j-1].parent_i = i;
cellDetails[i-1][j-1].parent_j = j;
}
}
}
//----------- 7th Successor (South-East) ------------
// Only process this cell if this is a valid one
if (isValid(i+1, j+1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i+1, j+1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i+1][j+1].parent_i = i;
cellDetails[i+1][j+1].parent_j = j;
printf ("The destination cell is found\n");
tracePath (cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i+1][j+1] == false &&
isUnBlocked(grid, i+1, j+1) == true)
{
gNew = cellDetails[i][j].g + 1.414;
hNew = calculateHValue(i+1, j+1, dest);
fNew = gNew + hNew;
// If it isn’t on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i+1][j+1].f == FLT_MAX ||
cellDetails[i+1][j+1].f > fNew)
{
openList.insert(make_pair(fNew,
make_pair (i+1, j+1)));
// Update the details of this cell
cellDetails[i+1][j+1].f = fNew;
cellDetails[i+1][j+1].g = gNew;
cellDetails[i+1][j+1].h = hNew;
cellDetails[i+1][j+1].parent_i = i;
cellDetails[i+1][j+1].parent_j = j;
}
}
}
//----------- 8th Successor (South-West) ------------
// Only process this cell if this is a valid one
if (isValid (i+1, j-1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i+1, j-1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i+1][j-1].parent_i = i;
cellDetails[i+1][j-1].parent_j = j;
printf("The destination cell is found\n");
tracePath(cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i+1][j-1] == false &&
isUnBlocked(grid, i+1, j-1) == true)
{
gNew = cellDetails[i][j].g + 1.414;
hNew = calculateHValue(i+1, j-1, dest);
fNew = gNew + hNew;
// If it isn’t on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i+1][j-1].f == FLT_MAX ||
cellDetails[i+1][j-1].f > fNew)
{
openList.insert(make_pair(fNew,
make_pair(i+1, j-1)));
// Update the details of this cell
cellDetails[i+1][j-1].f = fNew;
cellDetails[i+1][j-1].g = gNew;
cellDetails[i+1][j-1].h = hNew;
cellDetails[i+1][j-1].parent_i = i;
cellDetails[i+1][j-1].parent_j = j;
}
}
}
}
// When the destination cell is not found and the open
// list is empty, then we conclude that we failed to
// reach the destiantion cell. This may happen when the
// there is no way to destination cell (due to blockages)
if (foundDest == false)
printf("Failed to find the Destination Cell\n");
return;
}
// Driver program to test above function
int main()
{
/* Description of the Grid-
1--> The cell is not blocked
0--> The cell is blocked */
int grid[ROW][COL] =
{
{ 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 },
{ 1, 1, 1, 0, 1, 1, 1, 0, 1, 1 },
{ 1, 1, 1, 0, 1, 1, 0, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 0, 0, 0, 0, 1 },
{ 1, 1, 1, 0, 1, 1, 1, 0, 1, 0 },
{ 1, 0, 1, 1, 1, 1, 0, 1, 0, 0 },
{ 1, 0, 0, 0, 0, 1, 0, 0, 0, 1 },
{ 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 },
{ 1, 1, 1, 0, 0, 0, 1, 0, 0, 1 }
};
// Source is the left-most bottom-most corner
Pair src = make_pair(8, 0);
// Destination is the left-most top-most corner
Pair dest = make_pair(0, 0);
aStarSearch(grid, src, dest);
return(0);
}
you can try A* algo made for 9 row and 10 cols update it as per your requirement
Upvotes: 4
Reputation: 11
Lets assume the ground as:
1 1 1 1 1 E 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 C 1 1 1 1 1
But always try to let the program know about properties of shapes e.g. square is equal on all sides so that's why a perfect diagonal way is shortest so if there are walls u could let your program choose the nearest part of the diagonal as
1 1 1 1 1 E 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 C 1 1 1 1 1
the part next to C (1) and then again continuing diagonally will help. Apologize for any grammatical or spelling mistakes.
Upvotes: 1
Reputation: 239653
There are so many algorithms which can find paths between two points. There are three algorithms which are easy to implement and understand.
- Depth first search (DFS)
- Breadth first search (BFS)
- Dijkstra's algorithm
Depth first search
This algorithm, takes the current node, finds all the neighbors, puts them in a stack, pops by one and traverses till the end or it finds the path.
Breadth first search
This algorithm, takes the current node, finds all the neighbors, puts them in a queue, dequeues one by one and traverses till the end or it finds the path.
The difference between DFS and BFS is that, DFS cannot guarantee optimal solution. Consider this case.
S 1 1
1 1 1
1 1 E
Assuming S is (0,0) and E is (2, 2). There are many optimal solutions to this maze. Since, DFS checks the path of its neighbours till the end, it might take S -> (1,0) -> (2,0) -> (2,1) -> (1,1) -> (1,2) -> E and it will return 6 as the cost of the path. Whereas, BFS finds all the neighbours, neighbours of all neighbours and goes on. If one of neighbors is E, it returns the cost. That would be guaranteed to be optimal. So, BFS might go like this. S -> (1,0) -> (2,0) -> (2,1) -> E (It finds neighbours of neighbours, it does not go till the end with each of the neighbours).
Dijkstra's algorithm
It is similar to BFS, but it can have weights. In the example, we assumed that the cost of moving from one node to another costs us 1 unit. In Dijkstra's algorithm, it allows us to use any positive integer as the cost, and it can be different for each and every link.
Conclusion
If you want optimal result, go for BFS or Dijkstra's algorithm. For your case, BFS should work.
Upvotes: 9
Reputation: 10837
Take a look at the most used path finding algorithms.
http://qiao.github.io/PathFinding.js/visual/
This is done in JS, but you should be able to find a C++ implementation of the one that fits your needs, or write your own.
Upvotes: 6
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