L16. M-Coloring Problem | Backtracking
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- Опубліковано 22 сер 2024
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Please comment if you understand, a comments means a lot to me :)
Eid Mubarak
bhaiyaa I was not able to understand the brute force approach solution of this at gfg please explain that also in some other problems
Striever why we don't use any of graph tarversal?
Why TC will be M^N ?? it should be (M^N) * N, Because we traverse adjacency list as well in isSafe method everytime and maximum cost of adjList will be of N So. TC will be (M^N) * N @takeUforward
Tried solving without looking at any explanation for the first time in this playlist, took me almost three hours but did it on my own❤
Do u just solved the algo or full. Coding
@@arnobsaha7071 Full code
Heads up: If you are practicing on gfg the graphs are stored in a different manner for different programing languages.
For c++ : in matrix
For Java: in an adjacency list
For Python : in matrix
For Javascript : in matrix
Just posting because when I started implementing with python I was access the graph as if it is an adjacency matrix and it was resulting in wrong submission so, I thought if anyone else is trying to do the same can save some time by not repeating my mistake.
posting python version of same code :
def is_safe(node, c, graph, color, n):
for i in range(n):
if i != node and graph[i][node] == 1 and color[i] == c:
return False
return True
def solve(node, graph, color, m, n):
if node == n:
return True
for c in range(1, m + 1):
if is_safe(node, c, graph, color, n):
color[node] = c
if solve(node + 1, graph, color, m, n):
return True
color[node] = 0
return False
color = [0] * V
return solve(0, graph, color, k, V)
I was looking for this problem explanation for the past few days but could'nt find a proper one. Now I can surely say, this one was the best amongst all. Thanks for this.🙌
Thankyouu :)
Good explanation like always, thanks🙂❤️
Wanted to get clarified of two things here:
🤔0. Isn't the complexity M^N, because there are N places to fill with M choices for each, so wouldn't M be multiplied N times making it M^N?
🤔1. We haven't considered the time complexity for checking the possibility of filling the color(isSafe), which can be of order N at worse, but shouldn't we?
total time (M^N)*N(issafe)
@@bharathkumar5870 yes, that's what I thought it should be.
@@bharathkumar5870 Yes, That's what I believe is correct
@@bharathkumar5870 why not (M*N) ^ N ?
@@shivamnegi7552 yes checking will be taken for each color pick so (M*N)^N
Thank you brother! I'm preparing for my placements following your sde sheet and I'm getting so much confidence, please continue doing the great work 🙏👍
how is the progress? Placed yet?
Sir placed?
Please reply ✨
I solved the problem myself just with your explanation upto 13 minutes. Thanks Striver. The way you spoonfeed us nobody does, it just stays in my head.
time complexity = O(m ^ n) not o(n^m) thanks bhaiya .
as we have m choice for each node .
yes right
U are doing a great job for students sir. Thank you so much for your efforts.
If the graph is connected simple dfs based recursion also works but one can only appreciate this if he wrote code for connected and realises if there are more than one components what should be done
True, I skipped this part. Attaching the DFS solution which handles all cases
bool solve(int sv, bool graph[101][101], int V, vector& visited, int colors) {
unordered_map visitedColors;
for(int i = 0; i < V; i++) {
if(graph[sv][i] == true && visited[i] != -1) visitedColors[visited[i]] = 1;
}
if(visitedColors.size() == colors) return false;
for(int i = 0; i < colors; i++) {
if(visitedColors[i] > 0) continue;
visited[sv] = i;
bool isAll = true;
int j = 0;
for(j = 0; j < V; j++) {
if(graph[sv][j] == true && visited[j] == -1) {
if(solve(j, graph, V, visited, colors) == false) break;
}
}
if(j == V) return true;
}
visited[sv] = -1;
return false;
}
bool graphColoring(bool graph[101][101], int m, int V) {
vector visited(V, -1);
for(int i = 0; i < V; i++) {
if(visited[i] == -1) {
if(solve(i, graph, V, visited, m) == false) return false;
}
}
return true;
}
@@vankshubansal6495 can you pls explain how your code is working?
Very detailed and easy to understand explanation.. 10 times better than the so called paid courses.. Thank you so much bhaiya ❤
Correction in Time-Complexity discussed in striver's vedio
T.C = O(M^N)*N(isSafe-fxn)
For one node ,we have m different colors.
For n node we have m*m*m.....n times =M^N
Also isSafe() takes O(N) in worst case.
Let me know if I was wrong.
It N^m* N
Beautifully explained a tough problem very simply and in an easy to understand way. Thank you so much!
Amazing explanation as well as the code. I haven't seen so much clear explanation on any other channel. Thankyou for this❤️
Correction:
Time complexity is O(M^N) not O(N^M) (which solves the P vs NP problem and he would have been a millionaire by now)
What do you mean by him being a millionaire?
@@priyanshudwivedi7535 P vs NP problems are problems can truly revolutionize the world of computing and maths. and have reward of a million dollars for the person who solve them.
The time complexity has to be M^N ?
Because the height of tree will go till N and for each node we have M choices.
correct, it confused me
even i think so
The best explanation I have found on youtube. Thank you so much.
Shuru majburi mei kiye the ab mazza aa rha hai...
i was not able to draw the recursion tree, now as I understood the approach, I just coded it by myself, thanks bhaiya
Understood............Thanks a ton for this wonderful video..........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Small Correction In Time Complexity
T.C = ~M^N not N^M
Because we have M-Color choice for every nodes. Tell me am i correct or not..
I Think it's Complexity is M^n because m is no.of choice like it happen in printing all sub sets 2^n.......please correct me if i'm wrong
I think the time complexity should be O(m^N)
Yupp it will be
Badhiya bhai, woh tumney strategy same rakha h parallel recursion wala and saarey backtracking solve kar liye.
How do we get the intuition that here we had to traverse serially on the nodes and not initiate dfs for every connected component?
Had been doing that and got stuck at what the base case would be for dis-connected components as backtracking was erasing everything :|
Yes I was also stucked at the same point
Hello bhaiya i hope you are doing extremely well
Great video! I have one doubt though. Shouldn't the time complexity be O(m^N)?
Can u explain why n^m seems correct as for every node there are m possibility
@@abhinavpandey3356 Here's how i devised the TC for this:
At each node, there will be at max there will be m operations and for each operation on a node, their childs will have their own respective m operations.
i.e if a graph was:
1
/ \
2 3
then at node 1 there will be m operations but for each operation on node 1, there will be m operations on its successive nodes(here on the node 2 and then on node 3). i.e m*m*m = m^3 == m^n
thus on graphs having n nodes, starting from source node, there will be: m*m*m*...........*m(n terms) == m^n.
Thus, TC will be O(m^n).
I hope it helps!
@@abhinavpandey3356 easy o(n) approach is there
Yes yr right cause the height of the tree will be N and at each level of rec we will have m nodes.
Yes, imo that should be the correct time complexity
Thank you striver, after watching your previous videos of this playlist, I could do this on my own;
I have doubt , time complexity should be N*(N^m) because we have that IsSafe function also there and at max a node can be connected to all the others node So O(N) for chacking
btw Great explanation
Yes it should be there
@@tushar7305 that will just be O(N^(m+1)) which will be O(N^m) only, i think?
C++ Code for M coloring Problem-->
TC-- M^V v is no of vertices and M is max no if colors that can be used. Exponential TC due to Recursive nature of algorithm
SC-- O(V) as color vector stores color assigned to each vertex. Size is proportional to no of vertices in the graph.But it is at most O(V) due to depth of recursion.
----- Code Follows ------
#include
#include
using namespace std;
class Graph
{
private:
int V; // No of Vertices
vector adj; // adjacent list;
public:
// constructor
Graph(int Vertices) : V(Vertices), adj(Vertices) {}
// Funciton to add an edge to the graph
void addEdge(int u, int v)
{
adj[u].push_back(v);
adj[v].push_back(u);
}
// func to check if it is possible to color the graph with at most M colors
bool graphColoring(int M);
private:
// utility function for graph coloring;
bool graphColoringUtil(int vertex, vector &color, int M);
};
bool Graph :: graphColoringUtil(int vertex, vector &color, int M)
{
// check if we have assigned colors to all vertices
if (vertex == V)
return true;
// try assigning different colors to current vertex
for (int c = 1; c
One more optimization,if m>=4 then it is always possible to colour a graph by 4 colour theorem. So if m>=4 return true.
Your videos make problem look very simple.Thanks
you make all the problems easy⭐
Great explanation man❤️
python code for anyone struggling with it:
def graphColoring(graph, k, V):
color=[0]*V
def mcolor(vertex,graph,k,V):
if vertex==V:
return True
for i in range(1,k+1):
flag=1
for j in range(V):
if graph[vertex][j]==1 and color[j]==i:
flag=0
break
if flag==1:
color[vertex]=i
if mcolor(vertex+1,graph,k,V):
return True
color[vertex]=0
return False
return mcolor(0,graph,k,V)
hey can u tell me why are we doing => graph[vertex][j]==1 ?
@@riyakumari8377to check if the node j is connected to vertex node
What if I don't put colour [node]=0 Does it affect anything ,as I'm looking colr of negbour node to match with the colr I'm am trying to color for a particular node for deciding it's color
for skip ads go to 4:01
Time Complexity: O(M^N).
The best what you get
i have done the recursion and backtracking from another paid courser but it is not worth it
but the striver recursion playlist is bast of class
i have done all recursion from 1 to last and now the recursion feel like game not very easy but so far so good
explanation is good but time complexity is (n-1)*m^n by the way.
Can't we just count the maximum number of adjacent nodes present in the graph? if the maximum nodes is greater than the number of colors, return false, else return true?
I had the same thought and tried applying it on the bipartite graph problem but one test case failed which is this one:
[[4,1],[0,2],[1,3],[2,4],[3,0]]
Here every node has 2 adjacent but we still cannot color the graph using 2 colors.
So i think it wont work for this problem as well
Best explanation....learning for my practical exams
Wonderful explanation sir!! Thank you !!!
A Detailed And A Great Explanation .
This question is exactly similar to n queen problem for those who have problem can first try that one and then come to this question
Bro why we have backtracked in this as for loop itself change its colour suppose if statement gives u false then ultimately it will go to another colour why does it makes difference whether we have done 0 previously or not
In this problem, we are trying every color, if and only if the color is possible to take, i.e. we are filtering and taking, we are not waiting to mismatch!
For adjacency list, shouldn't we check:
public boolean isColorPossible(int node, int length, int colorTochoose,boolean graph[][], int[] color )
{
for(int i=0;i
Thanks
Thanks brother. Understood
Hi All, I was really confused between M coloring problem and bi partite graph problem. The Bi partite graph problem uses DFS/BFS for checking if no 2 adjacent nodes are filled with the same color(2 colors). So I was wondering if we could use DFS/BFS here as well. Turns out no, as in the GFG practice problem, in one TC. we have disconnected nodes as well in the graph and this cant be covered with DFS/BFS. Please let me know if my understanding is correct.
I wonder why the BFS solution is not working! Could you tell me what's wrong with BFS?
Amazing thought process sir !!
clear and straight-forward explanation. Thanks bro :)
thanks!!
again solved on my own even without the explanation... i think im getting better : ")
After watching this series , I understood why striver is god of coding.
C++ code starts at 20:53
Thanks for the playlist sir......it's really helpful
hi in this problem we can not simply check for each node. instead we must create adjacent node for each node and then check for possibility of coloring
Eid Mubarak striver Bhai
Eid mubarak bhai ❤️
shouldn't the graph be List where every index is considered as node and the list that is there in that index are the adjacent nodes??
Do i need to learn graphs for this problem?
Amazing explanation!
Thanks you sir for this video. Sir can you please clarify whether for SDE profile it is mandatory to know C++ or Java ? Python is not sufficient for cracking test , I am practicing with Python only? Sir please guide..... 🤷♂️
Time complexity will be O(M^N) isn't it?
(worst case)M colors for every node(N).
Thank you so much sirr
thanks for the best explanation in yt
can someone tell me why we have taken G[node] in for loop for(int it:G[node]){
//Here y we have taken G[node] i m stuck here please clarify my doubt someone.
Thank you sir
"Definitely" some great stuff Striver. Thanks a lot.
explanation ek no.!!!!!!!!!!!!!!!!!
why do we have to check all possible safe colors for every node?
why cant we just assign any 1 of the colors which are not adjacent to current node.
why we are calling the function for next serial node, shouldn't we call for the nodes attached to this node ?
Thank you!
Hello Everyone, Can any one tell me that why k!=node is required in issafe function in c++ code ?
Code will work fine without that (k!=node) , as in the case when (k == node) , graph[k][node] will always be 0 i.e the node itself which obviously can't be neighbour of itself
thanks!
Hello striver , I see there are no articles linked to bit manipulation , linked list problems . Plus there are no javascript code on striver website .
Can we not use a greedy approach for this problem? Would it fail, if yes why?
Thanks bro for amazing content .
I havent done Graphs yet, do i need to know graphs for this question or Recursion is sufficient alone?
You should have very basic knowledge of graphs for this
Understood
@takeUforward Why does this code give me a wrong ans?
bool graphColoring(bool graph[101][101], int m, int n)
{
vector color(n, 0);
for(int i=0;i
I haven't learned about graph yet
Why in this problem, we didn't follow standard dfs and used as node numbers as a part of traversal?
same doubt
understood.
isnt time complexity m^n?
as m choices of colour available in n recursive calls?
Understood, valuable content
Bhaiya Time complexity M^N ayegi na ?
Hnn
@@takeUforward Ok bhaiya
@@takeUforward Bhaiya it should be N*(N^m) because we have that IsSafe function also there and at max a node can be connected to all the others node So O(N) for chacking
Great job anna.. luv from TN 😍
Brother cant we just check for bipartiteness of the graph? Like if it is bipartite, we can always color using m>=2 and if its not then anything m>=3... Wont this work?
Exactly what i was thinking.
Consider when you are asked to tell if a graph can be colored using 5 colours and suppose it is not biparatite. Then what will you return. You are not sute about 3,4,5 colors. So only checking biparatiye won't work every time.
Hope it is helpful in some way
If a graph is not paratite m>=3 might work
i think time complexity will be m^n since each fxn is calling m another fxn so... m*m*m*m*.....n times
will this algorithm work if there is more than one connected components?
i think we cant do it by graph traversal beacause it give tle in case of cycle and also hard to manage backtracking right?
Understood ❤
God level explanation🔥....really appreciate your efforts❤️....what was that song in the end??
understood
I have 1questions:
1) For Java you are just checking 1 condition in the loop
if (color[it] == col) return false;
2) Whereas in C++ you are checking multiple conditions:
if (k != node && graph[k][node] == 1 && color[k] == col)
Is this because you have done some preprocessing in Java to have graph as an adjacency list but are passing the graph itself in C++?
In cpp he used adjacency Matrix nd in Java adjacency list
In Matrix you will check if there is an edge with every other vertex (g[i][j]==1) but in adjacency list only edges are present so no need to check if there's an edge
@@SanthoshSunny21 Thanks
*Tippy tippy tip top which color do you choose?*
c++ 20:58
can you make some videos on BFS approach of this ques ,its bit confusing for me