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Step 1: Choose a vertex in the graph and make it the source vertex & mark it visited.
Step 2: Determine a vertex which is adjacent to the source vertex and begun a new search if it is not already visited.
Step 3: Repeat step 2 via a new source vertex. While all adjacent vertices are visited, return to earlier source vertex and continue search from there.
If n refer to the number of vertices in the graph & the graph is represented through an adjacency matrix, then the total time taken to carry out DFS is O(n2). If G is revel by an adjacency list and the number of edges of G are e, then the time taken to carry out DFS is O(e).
Illustrate the Back Face Detection Method A single polyhedron is a convex solid, which has no external angle between faces less than 180° and there is a simple object space me
When writing a code for a program that basically answers Relative Velocity questions how do you go at it? How many conditions should you go through?
Merge sort: Merge sort is a sorting algorithm that uses the idea of split and conquers. This algorithm splits the array into two halves, sorts them separately and then merges t
How branching takes place in Instruction pipeline. Explain with suitable examples
A Sort which relatively passes by a list to exchange the first element with any element less than it and then repeats with a new first element is called as Quick sort.
how I can easily implement the bubble,selection,linear,binary searth algorithms?
Properties of colour Colour descriptions and specifications generally include three properties: hue; saturation and brightness. Hue associates a colour with some position in th
Example: Assume the following of code: x = 4y + 3 z = z + 1 p = 1 As we have been seen, x, y, z and p are all scalar variables & the running time is constant irrespective
Taking a suitable example explains how a general tree can be shown as a Binary Tree. Conversion of general trees to binary trees: A general tree can be changed into an equiv
How do collisions happen during hashing? Usually the key space is much larger than the address space, thus, many keys are mapped to the same address. Assume that two keys K1 an
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