Breadth First Search grid shortest path | Graph Theory
ฝัง
- เผยแพร่เมื่อ 12 พ.ค. 2024
- Finding the shortest path on a grid using the Breadth First Search (BFS) algorithm on an unweighted graph.
Algorithms repository:
github.com/williamfiset/algor...
Video slides:
github.com/williamfiset/Algor...
Dungeon master problem link:
open.kattis.com/problems/dungeon
Breadth-first search Intro:
• Breadth First Search A...
Personal website:
www.williamfiset.com
0:00 Previous Breadth first search video
0:44 Motivation for BFS on a grid
1:36 Converting your grid to an adjacency list/matrix
3:55 Grid vectors
6:09 Dungeon problem
7:51 Breadth first search on grid example
9:12 Using multiple queues for state representation
11:07 Pseudocode
15:56 Recap
=====================================
Practicing for interviews? I have used, and recommend `Cracking the Coding Interview` which got me a job at Google. Link on Amazon: amzn.to/3cvMof5
A lot of the content on this channel is inspired by the book `Competitive Programming` by Steven Halim which I frequently use as a resource and reference. Link on Amazon: amzn.to/3wC2nix
Support me by purchasing the full graph theory course on Udemy which includes additional problems, exercises and quizzes not available on TH-cam:
www.udemy.com/course/graph-th...
If you are beginner in graph, watch his videos twice or thrice because then you will feel like no one can explain better than him and you will understand each and every point mentioned in the video. Thank you Sir for this playlist.
What a great tutorial! A great explanation complemented by incredibly clear slides and animations. Thanks.
Finally, i'm waiting for this kind of tutorial
You're great dude
Amazing explanation, thanks for the video! I've been always avoiding BFS and going with DFS since I wasn't comfortable with it, but not anymore! Subscribed to your channel... Will be waiting for more tutorials from you!
William your explanation and animation creates a clear picture in my head about the underlying Concept, Thanks alot man!
really like the summary in the end, as well. it's a nice touch.
Great video. thank you very much. This is exactly what I needed to understand how to solve graph problem using adjacency matrix. Thank you once again!
this is exactly what i needed to know to solve graph problems, thanks for the video :) watching your tutorial for the first time, should have watched it earlier.
I was hammering about how to solve these ki9nd of problems. You are a savior! Please make these kind of tutorials more. I cannot thank you enough. Subscribed!
YOU ARE A LIFE SAVER I COULD NEVER FIND A VIDEO LIKE THIS THANK YOU SO MUCH FOR THE GIFT YOUVE BESTOWED UPON THIS LAND
Best explanation for coding problems and approaches!! Thanks for making such high quality content on TH-cam
i love how slow you speak and how thoroughly you explain everything. thanks!
I am preparing for IOI and this video was very Helpful, continue to make great contents like this. Thanks!
Thanks a lot Will! This is from cracking the coding interview and your walkthrough is well explained. I'm subbing to your channel now
This channel is a gem. very high quality content in these tutorial videos!
Bravo! you are one of a kind, keep rocking!!!
Thank you William for the great video. Fellow Canadian here. thank you for the high-quality content!
You are just brilliant man! I hope your channel grows big!
This is nice.I have seen this trick used by competitive programmers for traversing in the grid.
Thank you so much! You definitely have talent in explaining!
Thank you so much for simplifying it
excellent explanation. I particularly liked your concept for using direction vectors
An alternative to using 'nodes_left_in_layer' and 'nodes_in_next_layer' variables is to store the distance/level along with the coordinates into your queue.
It does have the downside of using an Object or Array to hold the coordinates and the level, but is easier to understand.
Thanks for this approach! Very intuitive indeed!
Your videos are amazing, but there is one possible improvement:
run through the pseudo-code with the animation for the algorithm, so visualizing the pseudo-code becomes easier
Video for visualization!!
Watch till end for better visualization.
Used exactly same concept as explained!!
th-cam.com/video/EdJa84ymIXg/w-d-xo.html
Best video on 2d Grid Thank You for the explanation
This is probably one of the most helpful 'Algorithms & Data Structures' channels I've yet encountered. The idea of maintaining multiple queues for each dimension is really handy.
Can you please explain why its better to have separate queues for each dimension? I cant see how its better than having one queue with the dimensions encapsulated into one object..
@@alikhansmt It's just a personal preference of mine. I've found this approach quicker to implement, and reuse over larger code structures, where I don't like to maintain a vast amount of structs, dot callbacks etc... Speaking in terms of competitive programming of course.
@@brunokawka I see thanks!
I rewrote your pseudocode in JavaScript adjusted it a bit to suit my needs. Worked perfectly right out of the box! Awesome work, William!
Can you share it?
@@saiffyros
// th-cam.com/video/KiCBXu4P-2Y/w-d-xo.html
// Globals
// number of rows and columns
const R = 10;
const C = 14;
// start cell values
const sr = 4;
const sc = 0;
// row and column queue
const rq = [];
const cq = [];
// save the directions found
const dir = [];
// variables to track the steps taken
let move_count = 0;
let nodes_left_in_layer = 1;
let nodes_in_next_layer = 0;
// variable to check if we've reached the end
let reached_end = false;
const map = {
cols: 14,
rows: 10,
sSize: 64,
tsize: 40
}
const gameGrid = [
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0],
[ 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0],
[ 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0],
[ 8, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0],
[ 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0],
[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]
// R x C matrix of false values aka visited positions
const visited = [
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]
// north, south, east, west direction vectors
const dr = [ -1, +1, 0, 0];
const dc = [ 0, 0, +1, -1];
function findPath(){
rq.push(sr);
cq.push(sc);
visited[sr][sc] = true;
// Keep going as long as there are items in the queue
while(rq.length > 0 || cq.length > 0){
let r = rq.shift();
let c = cq.shift();
if(gameGrid[r][c] === 'E'){
reached_end = true;
break;
}
check_neighbors( r, c)
nodes_left_in_layer --;
if(nodes_left_in_layer === 0){
nodes_left_in_layer = nodes_in_next_layer;
nodes_in_next_layer = 0;
move_count ++;
}
}
if(reached_end){
return move_count;
}
return -1;
}
function check_neighbors( r, c){
for(let i = 0; i < 4; i++){
let rr = r + dr[i];
let cc = c + dc[i];
//skip out of bounds cells
if(rr < 0 || cc < 0){
continue;
}
if(rr >= R || cc >= C){
continue;
}
// skip blocked or visited cells
if(visited[rr][cc]){
continue;
}
if(gameGrid[rr][cc] === 0){
continue;
}
let move = [dc[i], dr[i]];
dir.push(move);
rq.push(rr);
cq.push(cc);
visited[rr][cc] = true;
nodes_in_next_layer ++;
}
}
findPath();
function getTile(c, r){
return gameGrid[r][c];
}
export { dir, map, getTile };
This is amazing ♥️
Please keep solving that kind of problems ♥️♥️😍
A lot of you are asking about how to reconstruct the path, so let me just explain it here.
To reconstruct the path from the start to the end you need to maintain additional information, namely a 2D matrix which tracks the cell which was used to reach the current cell. Let me call this matrix "prev", short for previous. Every time you advance to the next cell, keep track of which cell you came from in the prev matrix. Once you reach the end of the BFS, start reconstructing the path by beginning at the end node in the prev matrix and work your way to the start node. The obtained path will be in reverse order, so you will need to reverse it.
This is explained in more detail in the BFS video except that the prev matrix is 1D for the general case: th-cam.com/video/oDqjPvD54Ss/w-d-xo.html
Hi William - @14:47, shouldn't we enqueue prior to "continue" on the # equality check? Otherwise certain positions would never be processed.
@@austind649 Hi Austin, whenever we reach a '#' cell, that cell can not be visited as it's an obstacle so can not be in our path. so we don't need to process it and we need to find another path through a reachable cell. The whole point here is only go through the path that is reachable.
Can you please share the code for saving the actual path along with the moves?
@@ss4036 Here is my code in C++, implemented by the williamfiset approach, which stores the the path and also shows BFS path taken in the matrix.
pastebin.com/TeMDhspF
Hi, what about if you can move diagonally? Mine calculates the distance correctly and I had the path reconstruction working for the cardinal directions but once I add diagonal vectors then the path reconstruction works strangely. I know its probably hard to say without looking at my code but how do you modify the above algorithm to account for diagonal? Or should it "just work" if you add the new vectors?
thank you so much for these videos!
Quality content. Really appreciate it👍
Such a great explanation!!
Great buddy! Got it in the first go!
Great tutorial! Congrats
My mind is blown at how many things can be modeled as a graph. I go out into the world and just see graphs now.
Thank you.. this is helpful for making games as well
Amazing tutorial! Keep filming please :)
keep with the good job!
Спасибо вам, за хорошее объяснение
Thank you very much. It helped me a lot.
Awesome video, thank you so much!
Good explanation, thanks a lot
Brilliant ! Thank you :)
This is some of the best fucking content I've ever seen. Absolutely fucking incredible. 10/10 would peek again!
This helped me so much you don't even know TYSM
A fellow student from hyperskill, psted your video link. On the lee algorithm course conent.
It is helpful.
Amazing video! Thanks a ton!
Awesome video!!! Thanks bro!
Very clear approach! Just would like to share something regarding using 2 queues for 2D Matrixes, it just affects the performance very badly single Queue implementation is much more efficient. Other than this, great explanation thanks.
Efficient in terms of memory complexity, right?
(Because handling so many queues in n-dimensional space would require a lot of memory I presume).
THANK YOU!
Good stuff, for my dumb brain it wasn't clear in the first glance why we needed the nodes_in_next_layer and nodes_left_in_layer and how it's used. Had to replay and follow through the algorithm to understand its use.
Try to think this way: BFS is visiting graph layer by layer. Initially the nodes_left_in_layer = 1; supposed that there're 2 adjacent nodes (A, B)connecting the starting node(S), so the nodes_in_next_layer will be incremented to 2 after the explorer_neighbours(), then the nodes_left_in_layer is decremented to 0. Here, we reset the nodes_left_in_layer to 2 and nodes_in_next_layer to 0 and increment the move_count to 1. It means after visiting layer 1, we are expecting to visit 2 nodes in layer 2. We can image node A has 3 adjacent nodes (C, D, E) and node B has 1 adjacent nodes (F). Now try to go-thru layer 2 with the imaginary graph - we visit A, nodes_in_next_layer is set to 3, nodes_left_in_layer is set to 1, then visit B, nodes_in_next_layer is set to 4 and nodes_left_in_layer to 0. Here, we are finishing visit layer 2 and expecting to visit layer 3 with 4 coming nodes.
snap shots of queue:
(S)
(A) (B) // nodes_in_next_layer = 2, after visiting layer 1;
(B) (C) (D) (E) // pop (A) and push all neighbours (C) (D) (E)
(C) (D) (E) (F) // pop (B) and push neighbour (E)
// nodes_in_next_layer = 4, after visiting layer 2;
Hope this helps.
i think level would be better name compared to layer as BFS processes nodes which are at same level/distance from current node. Once we process all the nodes in current level, we move onto the next level. BFS is also called commonly referred to as level order traversal
@@jsarvesh yea, level is a much better name. I also find easier to code and understand passing an additional information into the object/struct of the queue, that is the level so when i push neighbors to the queue i increment this level/distance like this:
q.push((position){.level=nextLevel, .row=nextRow, .col=nextCol, .distance=pos.distance + 1});
thanks ,,, that was helpfull
thanks very useful video can someone please tell me how i can solve a maze that does not have walls (boundaries) which means a maze that allows wrapping
Your videos are amazing, but how do I get the short path if reached end ?, because my move_count data has already been calculated
Could anyone explain what is the use of 'nodes_left_in_layer' and 'nodes_in_next_layer' variables?
Your videos are amazing, really!
Can I know where do you run your animations, please?
They're slides. Linked them in the description.
11:16 does this pseudo code applies for a 3d grid graph? im confused you mention to use 3 queues for 3ds grid graph and then went straight back to 2d pseudo code
Thanks a lot
You are the best
thanks a lot.
I created a ArrayList variable to store the steps taken, but it store all the block the BFS explored instead of the green block that shown in the video.
How can I only store the shortest block need to be explored only?
what is the time complexity for this?
Now that I know the shortest path in value. How do i highlight the path it took.
Big like for you. Thanks a lot :))
thanks it s really great bt how can I get the source code for this?
What about the Robot Vacuum Cleaner problem, a double array 3x3 a robot is somewhere in the grid and in some cells there is dirt and the robot must find the optimal path in order to clean the cells and doing that with the BFS algorithm. Any suggestions?
Damn this was good
thanks a lot
what happen if we are allowed to go in 8 direction diagonal also included then i think only BFS will not work anybody??
8:36 I understand the "Fill" method to get all the tiles, yet how are you getting this path from that ?
How is it possible to get returned the coords for the path? I dont think the queue in the end contains the path or my implementation is wrong.
This guy is insane
8:55 how do we find the number of steps without tracing the path itself?
Man, wish I could see this before. Encountered this same problem in my First Round of Google Interview For Software Engineer.
Dude, their on site is all day variations of BFS problems, there's no hope lol
@@kickhuggy really? thats actually pretty cool lol
What do you mean by a 'layer'? (I am sorry if I missed it but I think its missing from the video)
I don't understand the nodes left in layer and nodes in next layer part. Can someone link to a C++ solution to this problem using STL?
Hi William, shouldn't the move count be move_count+1(if reached_end is true) as we are breaking the loop and not incrementing move_count for the last layer(where end node was found)?
No. When reaching the layer containing the E, the step to get to that layer is already accounted for.
Thanks for the multiple queue trick. Storing them as pairs does not scale well to higher dimensions and p.first and p.second looks really ugly and ambiguous.
Hi Can someone please help me on this doubt. like if there are 2 routes from Cell Start to Cell End then how we are keeping track of steps for the 2 routes and taking minimum steps.
ok. Got it
for the dungeon problem, why do we have to use bfs? why not dfs?
can someone please help me understand, why is it we choose BFS over DFS to find "shortest path" ? i believe it has to be the way BFS traverses a graph as compared to DFS, but looking for a better explanation
I didnt understand how r u calculating nodes left in layer and nodes in next layer
Same
how do you show the path once its done?
@1.30 gr8 video
I have one question about these can you help me please
hey could we get the complete source code file?
Looking at the order of direction vectors which is (-1,0), (+1,0), (0,+1), (0, -1)...it looks like you are interested to move row wise first and then column wise. Then how does it become breadth first algorithm. As BFA says, we to move across the row first right.
Hi Shiva, it doesn't matter which direction vector is applied first because you're doing the BFS layer by layer and adding newly discovered nodes to the end of the queue.
That's exactly is my question.. how does this approach taking breadth first. With the use of vectors we go on finding the next node to traverse and looking at the these vectors, traversing need not be breadth wise. Because you are looking for next node above and below the current node first. I hope it makes sense.
Let me try and make this as clear as possible. For any given node, we add all the neighbors of that node to the list of nodes that need to be visited (unless the neighboring node has already been visited, is not traversable or something weird). The list of nodes to be visited is stored in a queue data structure. The queue has the property that the most recent node added to the queue is found at the end and the oldest node which has been in the queue the longest at the beginning. This means that we can add nodes to the queue and know that they will eventually get processed at a later stage. When the BFS starts you add the starting node to the queue to indicate that it should be visited. The algorithm begins by dequeuing (removing) the start node from the beginning of the queue, after which you add the left, right, down and up neighbors of the start node to the queue in that order (which shouldn't matter). Then the starting node's left node is first in the queue, so you add its left, right, down and up neighbors to the queue, then the starting node's right node if first in the queue, so you add its left, right, down and up neighbors to the queue, and etc... So the algorithm circles around if you will and the graph is explored layer by layer
Thanks for the detailed explanation ....
Hi, I have a (maybe so stupid question). BFS cannot use as a shortest path finding with weighted graph right? (it just work with unweight - or all the edge have the same priority). I am not sure about this because at the video Overview, you said it can be use as shortest path finding algorithms. (But to my knowlege, it is impossible). Maybe i got mistake, thanks.
what if we are allowed to take off one obstacle???
is there a related leetcode problem to practice?
I was wondering how I can solve the same problem using DFS (bearing in mind I want to get any path not the shortest one)
You can using backtracking, however I wouldn't recommend it since it'll be quite slow. If you need a faster shortest path algorithm have a look at my Dijkstra video
How do we know that this algorithm gives the shortest path ?
Sir, one doubt.
At first u said about using Adjacency List/Matrix.
Later u worked directly on the grid.
How to do it using Adjacency List/Matrix???
You can use a regular BFS with an adjacency list (or matrix), see th-cam.com/video/oDqjPvD54Ss/w-d-xo.html.
Ok, thank you sir
Hello!
Have you solved the Muddy Hike problem from Kattis?
Indeed I have! The creator of that problem happens to be a very close friend of mine :)
@@WilliamFiset-videos wow awesome! Could you please provide me with a hint or some advice as to how one can solve it? I've been struggling along time with it. My idea was to turn the grid into a graph and find the shortest path through it using the Dijkstra algorithm but this proved to be inefficient. How do you recommend one should solve this problem? Thank you!
Modified Dijkstras I believe should work, or binary search + BFS too :)
@@WilliamFiset-videos thank you for the answer. But just to be a little more specific: if we go with the second alternative for instance, how would the algorithm look like (if you wouldn't mind explaining it briefly)? I imagine that I'd sort the grid ascending at first using (as you recommend) binary search and then apply the BFS algorithm. But how is that guaranteed to give me the path with the minimum height?
No sorting is required for approach #2. You need to binary search on the height h to determine the minimum height required. More specifically, during each BFS subroutine of the binary search iteration only take squares where the height is < h (
may i have a code plz
انت وحش حقيقي
awsome
Notable
You da man
can you explain why we use BFS algorithm? why you did not solve it by DFS or any other?
I could have solved this using DFS, but the BFS has the advantage that it'll find the shortest path because all edges are unweighted
@@WilliamFiset-videos but what if the cost of moving left, right and diagonal cell cost us. then which algo you will prefer?
@@luqmanahmad3153 Dijkstras algorithm. I also have a video on it.
What about uniform cost search? It will find cheapest path