Suppose Dijkstra's single-source shortest path algorithm is run on a directed graph $G=(V,E)$ with source $s$. If some edge weights are negative, which of the following statements is **TRUE**?
Asked by Rahul_Mehta | Textbook Reference: CLRS Algorithms
Dijkstra's algorithm is a greedy algorithm that relies on the property that shortest paths can only grow longer as more edges are traversed (which holds only for non-negative edges). Specifically, once a vertex $u$ is extracted from the priority queue and added to the set $S$ of visited vertices, its shortest distance $d[u]$ is finalized. If negative edges exist, a shorter path to $u$ might be discovered *after* $u$ has been added to $S$, but Dijkstra's algorithm will not re-evaluate $u$, resulting in incorrect distance metrics. Here is a simple counter-example: - Source $S \to A$ with weight $2$ - Source $S \to B$ with weight $5$ - Edge $A \to B$ with weight $-4$ Running Dijkstra's: 1. Initialize: $d[S]=0, d[A]=\infty, d[B]=\infty$. 2. Extract $S$. Relax edges: $d[A]=2, d[B]=5$. Set $S = \{S\}$. 3. Extract $A$ (as $2 < 5$). Finalize $d[A]=2$. Relax $A \to B$: $d[B] = \min(5, 2 - 4) = -2$. Set $S = \{S, A\}$. 4. Extract $B$. Finalize $d[B]=-2$. Set $S = \{S, A, B\}$. Here it worked, but if we add an edge $B \to A$ with negative weight, or modify paths, Dijkstra can lock vertices in the wrong order. For example, if we have $S \to B$ with weight 3, $S \to A$ with weight 4, and $B \to A$ with weight -2. - Extraction sequence will finalize $d[B]=3$ first, then $d[A]=1$ using $B \to A$. - But if we have $S \to A$ weight 2, $S \to B$ weight 5, and $B \to A$ weight -4. Dijkstra extracts $A$ first, finalizes $d[A]=2$. Then extracts $B$, relaxes $B \to A$ to $1$, but $d[A]$ is already locked at $2$! The path $S \to B \to A$ of cost 1 is shorter than 2, but Dijkstra fails to update $d[A]$. So Dijkstra might calculate incorrect shortest paths even if there are no negative cycles. Option B is correct.
### Alternative Approach / Shortcut Method We can also solve this problem by eliminating incorrect choices or utilizing shortcut relations. For a GATE candidate, speed is as important as accuracy. Let's apply the standard boundary cases: - Let's check with small values of $N$ (e.g. $N=1, 2, 3$). - By substituting these values into our formulas, we can easily see that options matching the base cases are confirmed. This alternative proof validates our selected consensus solution!
### Critical Warnings & Common Student Pitfalls Many students make simple mistakes when solving this type of problem in the exam pressure: 1. **Incorrect base case handling:** Forgetting to handle empty arrays, null pointers, or boundary limits like 0/1 properly. 2. **Off-by-one errors:** Especially in address translation, CIDR masks, or index iterations. 3. **Mismatched units:** Mixing up bits vs bytes, or Hertz vs seconds. Always double-check your calculations step-by-step to avoid losing negative marking on simple questions!