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 ## Student Responses
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+---
 
-[Your solutions here]
+## Problem 1: Vertex Cover ≤ₚ Efficient Recruiting
 
+### Reduction
+
+Given an instance of Vertex Cover: a graph G = (V, E) and an integer k, we construct an instance of Efficient Recruiting as follows:
+
+- **Counselors (applicants):** Each vertex v ∈ V becomes a potential counselor. So m = |V|.
+- **Sports:** Each edge e = {u, v} ∈ E becomes a sport. So n = |E|.
+- **Qualifications:** For each sport corresponding to edge e = {u, v}, the counselors qualified in that sport are exactly u and v. That is, the subset of applicants qualified for sport e is A_e = {u, v}.
+- **Budget:** The hiring budget is k (same as the vertex cover budget).
+
+This reduction runs in O(|V| + |E|) time, which is polynomial.
+
+### Proof of Correctness
+
+**(Forward direction):** Suppose G has a vertex cover S ⊆ V with |S| ≤ k. We claim that hiring the set of counselors corresponding to S covers all sports. Consider any sport, which corresponds to some edge e = {u, v} ∈ E. Since S is a vertex cover, at least one of u or v belongs to S. Therefore, at least one counselor qualified in this sport has been hired. Since this holds for every sport (edge), hiring the counselors in S — at most k of them — covers all n sports. So the Efficient Recruiting instance is a YES instance.
+
+**(Backward direction):** Suppose the Efficient Recruiting instance is a YES instance, meaning we can hire a set H of at most k counselors that covers every sport. We claim that H (viewed as a set of vertices) is a vertex cover. Consider any edge e = {u, v} ∈ E. The sport corresponding to e has qualified applicants {u, v}, and since H covers this sport, at least one of u or v is in H. Since this holds for every edge, H is a vertex cover of size at most k.
+
+**(Conclusion):** G has a vertex cover of size ≤ k if and only if the constructed Efficient Recruiting instance can be solved with ≤ k hires. Since the reduction is computable in polynomial time, we have Vertex Cover ≤ₚ Efficient Recruiting. ∎
+
+---
+
+## Problem 2: Hitting Set ≤ₚ Plot Fulfillment
+
+### Reduction
+
+Given an instance of Hitting Set: a universe U = {u₁, u₂, ..., u_m}, a collection of subsets S₁, S₂, ..., S_n ⊆ U, and an integer k, we construct an instance of Plot Fulfillment as follows:
+
+**Construct the directed graph (hypertext) as a layered graph with k layers:**
+
+- **Pages:**
+  - A start page s and an end page t.
+  - For each layer l ∈ {1, 2, ..., k} and each element u_j ∈ U, create a page p_{l,j}. This gives km pages plus s and t.
+
+- **Links (directed edges):**
+  - From s to p_{1,j} for every j ∈ {1, ..., m} (start page links to all pages in layer 1).
+  - From p_{l,j} to p_{l+1,j'} for every l ∈ {1, ..., k-1} and every j, j' ∈ {1, ..., m} (every page in layer l links to every page in layer l+1).
+  - From p_{k,j} to t for every j ∈ {1, ..., m} (all pages in layer k link to the end page).
+
+- **Thematic elements:** For each subset S_i (i = 1, ..., n), define the thematic element T_i = {p_{l,j} : 1 ≤ l ≤ k, u_j ∈ S_i}. That is, T_i contains all pages across all layers that correspond to elements in S_i.
+
+This reduction runs in polynomial time: the graph has 2 + km pages, O(km²) links, and n thematic elements.
+
+### Proof of Correctness
+
+**Key observation:** Any trail from s to t in this layered graph has the form:
+
+s → p_{1, j₁} → p_{2, j₂} → ... → p_{k, j_k} → t
+
+The trail selects exactly one page per layer, corresponding to elements u_{j₁}, u_{j₂}, ..., u_{j_k} from U (with possible repetitions). The set of elements "chosen" is H = {u_{j₁}, ..., u_{j_k}}, which has |H| ≤ k.
+
+**(Forward direction):** Suppose (U, S₁, ..., S_n, k) is a YES instance of Hitting Set. Then there exists H = {u_{j₁}, ..., u_{j_{k'}}} with k' ≤ k such that H ∩ S_i ≠ ∅ for all i. Construct a trail by:
+- In layers 1 through k', visit pages p_{l, j_l} corresponding to elements u_{j₁}, ..., u_{j_{k'}}.
+- In layers k'+1 through k (if k' < k), visit p_{l, j₁} (repeating the first element).
+
+This is a valid trail from s to t. For each thematic element T_i: since H ∩ S_i ≠ ∅, some u_{j_l} ∈ S_i, so the page p_{l, j_l} ∈ T_i is visited. Hence the trail contains at least one page from each T_i.
+
+**(Backward direction):** Suppose there is a trail from s to t that contains at least one page from each T_i. The trail visits pages p_{1,j₁}, ..., p_{k,j_k}. Define H = {u_{j₁}, ..., u_{j_k}} ⊆ U. Then |H| ≤ k. For each S_i: since the trail hits T_i, some p_{l, j_l} ∈ T_i, meaning u_{j_l} ∈ S_i, so H ∩ S_i ≠ ∅. Therefore H is a hitting set of size at most k.
+
+**(Conclusion):** The Hitting Set instance is a YES instance if and only if the constructed Plot Fulfillment instance is a YES instance. Since the construction is polynomial-time, we have Hitting Set ≤ₚ Plot Fulfillment. ∎
+
+---
+
+## Problem 3: Independent Set ≤ₚ Path Selection
+
+### Reduction
+
+Given an instance of Independent Set: a graph G = (V, E) and an integer k, we construct an instance of Path Selection as follows:
+
+**Construct a directed graph G' and a set of path requests:**
+
+- **Nodes of G':**
+  - For each vertex v ∈ V, create a unique start node s_v and end node t_v.
+  - For each edge e ∈ E, create a single shared node n_e.
+  - Total: 2|V| + |E| nodes.
+
+- **Path requests:** For each vertex v ∈ V, create a path request P_v defined as follows. Let e₁, e₂, ..., e_{d_v} be the edges incident to v in some fixed ordering (where d_v is the degree of v). Then:
+
+  P_v: s_v → n_{e₁} → n_{e₂} → ... → n_{e_{d_v}} → t_v
+
+  If v is an isolated vertex (d_v = 0), then P_v: s_v → t_v.
+
+- **Edges of G':** All directed edges along the paths P_v (for all v ∈ V).
+
+- **Selection budget:** k (same as the independent set target).
+
+This reduction runs in polynomial time.
+
+### Proof of Correctness
+
+**Key property:** Two paths P_u and P_v share a node if and only if u and v are adjacent in G.
+
+*Proof of key property:*
+- The start and end nodes s_v, t_v are unique to path P_v, so they never cause sharing.
+- The only potentially shared nodes are the edge nodes n_e. Node n_e appears in path P_v if and only if edge e is incident to v. So n_e appears in both P_u and P_v if and only if e is incident to both u and v, which means e = {u, v} ∈ E, i.e., u and v are adjacent.
+
+**(Forward direction):** Suppose G has an independent set S ⊆ V with |S| ≥ k. Consider the set of paths {P_v : v ∈ S}. Since no two vertices in S are adjacent, no edge of G connects two vertices in S. By the key property, no two paths in {P_v : v ∈ S} share any node. We have selected |S| ≥ k paths with no shared nodes.
+
+**(Backward direction):** Suppose we can select at least k paths from {P_v : v ∈ V} such that no two selected paths share any node. Let S be the set of vertices corresponding to the selected paths. Then |S| ≥ k. We claim S is an independent set: if two vertices u, v ∈ S were adjacent (i.e., {u, v} ∈ E), then by the key property, P_u and P_v would share the node n_{{u,v}}, contradicting the selection constraint. So no two vertices in S are adjacent.
+
+**(Conclusion):** G has an independent set of size ≥ k if and only if we can select ≥ k node-disjoint paths in the constructed Path Selection instance. Since the reduction is polynomial-time, we have Independent Set ≤ₚ Path Selection. ∎
+
+---
+
+## Problem 4: 3-SAT ≤ₚ Half 3-SAT Satisfiability
+
+### Reduction
+
+Given a 3-SAT instance φ with n variables x₁, ..., x_n and m clauses C₁, ..., C_m, we construct a Half 3-SAT instance φ' as follows:
+
+- Introduce one fresh variable y (not appearing in φ).
+- φ' contains the following 2m clauses:
+  1. The m original clauses C₁, C₂, ..., C_m (unchanged).
+  2. m dummy clauses D₁, D₂, ..., D_m, where each D_i = (y ∨ y ∨ y).
+
+The formula φ' has n + 1 variables and 2m clauses. Since 2m is always even, the Half 3-SAT requirement that the number of clauses be even is satisfied.
+
+This reduction runs in O(n + m) time, which is polynomial.
+
+### Proof of Correctness
+
+**Preliminary lemma:** *If φ is unsatisfiable, then there is no assignment to x₁, ..., x_n that makes all m clauses of φ simultaneously false.*
+
+*Proof of lemma:* Suppose for contradiction that assignment α makes all m clauses false. Each clause C_i = (l_{i1} ∨ l_{i2} ∨ l_{i3}) being false means l_{i1} = l_{i2} = l_{i3} = false under α. Define α' by flipping every variable: α'(x_j) = ¬α(x_j) for all j. Under α', every literal that was false under α becomes true: if l = x_j was false (α(x_j) = F), then α'(x_j) = T so l is true; if l = ¬x_j was false (α(x_j) = T), then α'(x_j) = F so ¬x_j is true. Since all three literals in every clause were false under α, all three are true under α', making every clause true. So α' satisfies φ, contradicting the assumption that φ is unsatisfiable. ∎
+
+**Behavior of dummy clauses:** Under any assignment, all m dummy clauses D_i = (y ∨ y ∨ y) have the same truth value:
+- If y = true: all m dummy clauses are true.
+- If y = false: all m dummy clauses are false.
+
+**(Forward direction):** Suppose φ is satisfiable, with satisfying assignment α. Extend α to α' by setting y = false. Under α':
+- All m original clauses are true (since α satisfies φ).
+- All m dummy clauses are false (since y = false).
+- Total true clauses: m out of 2m, which is exactly half. ✓
+
+**(Backward direction):** Suppose there exists an assignment β to φ' that makes exactly m out of 2m clauses true. There are two cases:
+
+- *Case β(y) = false:* All dummy clauses are false, contributing 0 true clauses. So all m true clauses must come from the original clauses. This means all m original clauses are true under β, so β restricted to x₁, ..., x_n satisfies φ.
+
+- *Case β(y) = true:* All m dummy clauses are true. Since exactly m of 2m total clauses are true, exactly 0 original clauses are true. By the preliminary lemma (applied contrapositively), if all original clauses are simultaneously false under some assignment, then flipping all variables gives a satisfying assignment. So φ is satisfiable.
+
+In both cases, φ is satisfiable.
+
+**(No false positives for unsatisfiable φ):** If φ is unsatisfiable, we show no assignment to φ' achieves exactly m true clauses:
+- If y = false: all dummy clauses are false. Original true count j ≤ m - 1 (since φ is unsatisfiable). Total = j ≤ m - 1 < m.
+- If y = true: all dummy clauses are true (contributing m). Original true count j ≥ 1 (by the lemma, j = 0 is impossible for unsatisfiable φ). Total = j + m ≥ m + 1 > m.
+
+In neither case does the total equal m. So no assignment makes exactly half the clauses true.
+
+**(Conclusion):** φ is satisfiable if and only if φ' has an assignment making exactly half its clauses true. Since the reduction is polynomial-time, we have 3-SAT ≤ₚ Half 3-SAT. ∎