right, or add an AND i != k to that if statement bool condition in isSafe()

the node is adjacent to it self , you should consider that .

when u backtrack at k'th node, k-1'th level will have x[k] = c but at k-1 we are limiting ourselves to only k-1 nodes so it doesnt matter what x[k] is coz we will replace it with another color later

yup. The magic happens at graphColor(k+1). if it returns to parent function, the parent function will try another color.

@@prateekkej2506 i think that return statement at last wont allow backtracking/looping for another color. If the return statement is written outside of the safecheck, then only it will backtrack and try for another color

@@jinxblaze but in isSafe they check all the nodes from 0 to n.

You're right. And m^n is what is shown in the video. Look again

This one uses m. If it fails, you can try m+1

when we call graphColor(k+1) if it fails then it will return to the next iteration in its parent graphColor()'s for loop where it will continue to try other colors and if its does not returns that means its color was right from the first call.

+Ilmer Oliveira Good question. It becomes a much more computationally challenging problem this way, and you have to try the problem with m = 1, m = 2, m = 3... until you find the min value for m where all of the nodes can take colour.

To add, this is a known NP-Complete problem meaning there does not exist a solution that is polynomial time for large enough n.

+CSBreakdown If I make a function that travels the entire array seeking for blank nodes and make my looop runs until it's false(therefore all the nodes are colored)... Might work?

+Ilmer Oliveira With the backtracking algorithm, isSafe will return false if there aren't enough m (colours) to satisfy. The algorithm will reach the very end without colouring all of the nodes. In the 'if' statement for isSafe on the left block of code, add an else block and check to see if c = m. So if 'isSafe' = false, and c = m, you know that you don't have enough colours for m. Run the algorithm again, but this time make m 1 bigger.

+CSBreakdown Can you say if I understood the idea? pastebin.com/7KbaYKGF

+Sarbottam Chatterjee Hi, so x[i] is an array that holds colors if the nodes that have already been colored. So c == x[i] is checking if the current color that you are trying to place (c) is equal to the color value at x[i] (the already colored nodes). If 2 nodes are adjacent (meaning if G[k][i] == 1 AND the adjacent node at x[i] has the same color, then it returns false because we can't place the same colour next to it.

Thant clears my doubt. Thanks for making time to reply to my comment. Do you have a Traveling Salesman Problem solution using Dynamic Programming?

+Sarbottam Chatterjee I don't unfortunately. The travelling salesman problem is NP-Complete. If I had a true solution to the problem, I would be the worlds richest man! Travelling salesman is best done with approximation algorithms using current technologies. With a large enough value of N, travelling salesman cannot be accurately solved. Maybe there are solutions out there for small enough values of N, but I'm not aware of one and haven't attempted the problem.

actually there is .when we call graphColor(k+1) if it fails then it will return to the next iteration in its parent graphColor()'s for loop where it will continue to try other colors and if its does not returns that means its color was right from the first call.