DIjakstra algorithim and bellaman ford Practice questions
Duration: 4 min
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The video addresses a conceptual question regarding shortest paths in weighted graphs. The central problem asks whether doubling the weight of every edge in a graph preserves the identity of the shortest path from a source to any destination, altering only the total weight. The instructor analyzes this by first reading the problem statement and underlining critical conditions. He then constructs a specific directed graph with four vertices (A, B, C, D) and assigns positive integer weights to the edges. Through step-by-step calculation, he compares the total weights of different paths before and after the modification to verify if the minimum weight path remains unchanged.
Chapters
0:00 – 2:00 00:00-02:00
The instructor introduces the problem statement: "Given a graph, suppose we have calculated shortest path from a source to all other vertices. If we modify the graph such that weights of all edges is becomes double of the original weight, then the shortest path remains same only the total weight of path changes." He underlines phrases like "shortest path from a source to all other vertices" and "weights of all edges is becomes double" to focus attention. To test the validity of this statement, he begins drawing a graph on the whiteboard. He sketches four nodes labeled A, B, C, and D. He draws directed edges from A to B with weight 2, A to C with weight 3, B to D with weight 5, and C to D with weight 7. This visual setup serves as a concrete test case for the question.
2:00 – 4:14 02:00-04:14
The instructor proceeds to calculate the shortest paths for the drawn graph. Initially, he computes the path from A to D. He identifies two possible paths: A->B->D with total weight 2+5=7, and A->C->D with total weight 3+7=10. Since 7 is less than 10, the shortest path is A->B->D. Next, he applies the modification: doubling all edge weights. The new weights become 4, 6, 10, and 14. He recalculates the path sums: A->B->D becomes 4+10=14, and A->C->D becomes 6+14=20. He observes that 14 is still less than 20, meaning the shortest path A->B->D remains the same. He generalizes this observation by writing inequalities showing that if w1 <= w2, then 2w1 <= 2w2. Consequently, he concludes the statement is valid and selects option (A) True.
The lecture demonstrates a fundamental property of shortest path algorithms: scaling all edge weights by a positive constant factor does not change the shortest path structure. By using a concrete example where path weights are compared before and after doubling, the instructor proves that the relative order of path costs is preserved. This confirms that the shortest path remains the same, validating the statement. The mathematical generalization provided at the end solidifies the understanding that scalar multiplication maintains inequality relationships between path sums.