Let's break down the problem: **Given BFS Order:** Vertices are visited in the order: **a, b, d, c, g, f, e** We need to determine which graph (A, B, C, or D) could produce this order with BFS, starting from vertex **a**. --- ## **Step-by-step BFS Simulation** Let's simulate BFS on each graph and match the order. --- ## **Graph A** Edges: a → b, a → d b → c, b → g d → g c → g g → e, g → f ### BFS from a: 1. Start at a: **[a]** 2. Visit neighbors: b, d → **[a, b, d]** (queue: b, d) 3. Visit b: neighbors are c, g (queue: d, c, g) 4. Visit d: neighbor g (already in queue), skip (queue: c, g) 5. Visit c: neighbor g (already in queue), skip (queue: g) 6. Visit g: neighbors e, f (queue: e, f) 7. Visit e: (queue: f) 8. Visit f: done Order: **a, b, d, c, g, e, f** **Does not match the given order.** --- ## **Graph B** Edges: a → b, a → d b → c, b → g d → e, d → f ### BFS from a: 1. Start at a: **[a]** 2. Visit neighbors: b, d → **[a, b, d]** (queue: b, d) 3. Visit b: neighbors c, g → (queue: d, c, g) 4. Visit d: neighbors e, f → (queue: c, g, e, f) 5. Visit c: (queue: g, e, f) 6. Visit g: (queue: e, f) 7. Visit e: (queue: f) 8. Visit f: done Order: **a, b, d, c, g, e, f** **Does not match the given order.** --- ## **Graph C** Edges: a → b b → g, b → d g → c, g → f d → e, d → f ### BFS from a: 1. Start at a: **[a]** 2. Visit b: neighbors g, d (queue: g, d) 3. Visit g: neighbors c, f (queue: d, c, f) 4. Visit d: neighbors e, f (queue: c, f, e, f) 5. Visit c: (queue: f, e, f) 6. Visit f: (queue: e, f) 7. Visit e: (queue: f) 8. Visit f: done Order: **a, b, g, d, c, f, e** **Does not match the given order.** --- ## **Graph D** Edges: a → b, a → d, a → g b → d d → c c → g g → f f → e ### BFS from a: 1. Start at a: **[a]** 2. Visit neighbors: b, d, g (queue: b, d, g) 3. Visit b: neighbor d (already in queue), skip (queue: d, g) 4. Visit d: neighbor c (queue: g, c) 5. Visit g: neighbor f (queue: c, f) 6. Visit c: neighbor g (already visited), skip (queue: f) 7. Visit f: neighbor e (queue: e) 8. Visit e: done Order: **a, b, d, g, c, f, e** **Does not match the given order.** --- ## **Analysis** Let's double-check **Graph A**, as it was close. Here's the queue order after each step: - Start at a: [a] - a neighbors: b, d (queue: b, d) - b neighbors: c, g (queue: d, c, g) - d neighbor: g (already in queue) (queue: c, g) - c neighbor: g (already in queue) (queue: g) - g neighbors: e, f (queue: e, f) - e: (queue: f) - f: done Order: **a, b, d, c, g, e, f** Given order: **a, b, d, c, g, f, e** **Graph A** is the only graph that matches the order closely, but the positions of f and e are swapped. This can happen if there is no order preference between adding e and f to the queue from g, since BFS could visit them in any order. But the question says: **"Assume that there is no particular preference for which edge to follow when given a choice of edges."** So, **Graph A** is correct, as BFS could visit either e or f after g, matching the required order. --- ## **Final Answer** **Circle Graph A.** --- **Explanation:** BFS on Graph A from node a can produce the order **a, b, d, c, g, f, e** if, after visiting g, we enqueue f before e. Since there's no specified preference, this is allowed. **Final Answer:** **A** (Circle the letter under Graph A)

Let's carefully analyze the problem step-by-step. --- ### **Given BFS traversal order:** **a, b, d, c, g, f, e** We are to identify which of the four graphs (A, B, C, D) could produce this BFS order starting from vertex **a**. --- ### **Understanding BFS traversal:** - BFS explores neighbors level by level. - When multiple neighbors are available, the order of visiting them depends on the order in which edges are considered. - Since the problem states there is no preference for which edge to follow, any order consistent with BFS's level-by-level approach is acceptable. --- ### **Step 1: Simulate BFS on each graph** --- ### **Graph A** Edges: - a → b, a → d - b → c, b → g - d → g - c → g - g → e, g → f **BFS starting at a:** 1. **Start with a:** Queue: [a] Visited order: a 2. **Visit neighbors of a:** b, d Queue: [b, d] Order: a 3. **Dequeue b:** Neighbors: c, g - Enqueue c, g (assuming order: c, g) Queue: [d, c, g] Order: a, b 4. **Dequeue d:** Neighbors: g (already in queue) Queue: [c, g] Order: a, b, d 5. **Dequeue c:** Neighbors: g (already visited or in queue) Queue: [g] Order: a, b, d, c 6. **Dequeue g:** Neighbors: e, f Enqueue e, f (order: e, f) Queue: [e, f] Order: a, b, d, c, g 7. **Dequeue e:** No new neighbors. Queue: [f] Order: a, b, d, c, g, e 8. **Dequeue f:** No new neighbors. Queue: [] Order: a, b, d, c, g, e, f **Comparison with the given order:** Our order: **a, b, d, c, g, e, f** Given order: **a, b, d, c, g, f, e** **Difference:** The positions of e and f are swapped. Since BFS can choose the order of enqueuing neighbors when multiple options are available, and there is no preference specified, **this order is possible** if e is enqueued before f after visiting g. --- ### **Conclusion for Graph A:** **Yes,** BFS can produce the order **a, b, d, c, g, f, e** in this graph if the neighbors of g are enqueued as **f** before **e**. --- ### **Check other graphs for possibility** --- ### **Graph B** Edges: - a → b, a → d - b → c, b → g - d → e, d → f **Simulation:** Start at a: 1. [a] 2. Neighbors of a: b, d → [b, d] 3. Dequeue b: neighbors c, g → [d, c, g] 4. Dequeue d: neighbors e, f → [c, g, e, f] 5. Dequeue c: no new neighbors 6. Dequeue g: no new neighbors 7. Dequeue e 8. Dequeue f Order: a, b, d, c, g, e, f **Given order:** a, b, d, c, g, f, e The order of e and f is swapped again. **Since no preference,** it is possible to enqueue e before f after g, so this order could also be produced. --- ### **Graph C** Edges: - a → b - b → g, b → d - g → c, g → f - d → e, d → f Simulation starting at a: Order of visiting: a → b → g, d (neighbors of b) → c, f (neighbors of g) → e, f (neighbors of d) Order: a, b, g, d, c, f, e **Given order:** a, b, d, c, g, f, e **Not matching** the BFS order from this graph directly. --- ### **Graph D** Edges: - a → b, a → d, a → g - b → d - d → c - c → g - g → f - f → e Simulation: a → b, d, g Order after first step: a, b, d, g Next: - From b: neighbor d (already visited) - From d: neighbor c - From g: neighbor f - From c: neighbor g (already visited) - From f: neighbor e Order: a, b, d, g, c, f, e **Given order:** a, b, d, c, g, f, e **Close, but note that g appears after c in BFS order in our simulation, which differs from the given.** --- ### **Final Decision:** Considering the flexibility in BFS order (neighbors can be enqueued in any order), **Graph A** is the best candidate because: - The order **a, b, d, c, g, f, e** can be obtained if **f** is enqueued before **e** after visiting **g**. - The other graphs produce orders that are less consistent with the given BFS sequence or require more assumptions. --- ## **Final Answer:** **Circle Graph A.** --- **Summary:** - BFS order **a, b, d, c, g, f, e** can be produced in **Graph A** when neighbors are enqueued in the order that visits **f** before **e** after node **g**. - Since the question states there's no preference, **Graph A** is the correct choice. --- **You should circle the letter: A**.