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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: Efficient Recruiting [25 points]
 
+**Claim:** Vertex Cover ≤ₚ Efficient Recruiting.
+
+### Reduction
+
+Given a Vertex Cover instance (G = (V, E), k), we construct an Efficient Recruiting instance as follows:
+
+- **Sports:** One sport for each edge e ∈ E (so there are |E| sports total).
+- **Applicants:** One applicant for each vertex v ∈ V (so there are |V| applicants total).
+- **Qualifications:** Applicant v is qualified in sport e if and only if v is an endpoint of e.
+- **Budget:** k remains k (hire at most k counselors).
+
+This construction runs in polynomial time (O(|V| + |E|)).
+
+### Proof of Correctness
+
+**(→) If G has a vertex cover S with |S| ≤ k, then the Efficient Recruiting instance has a solution.**
+
+Consider the set of applicants S (one applicant per vertex in S). We have |S| ≤ k. For every sport (edge) e = (u, v) ∈ E: since S is a vertex cover, at least one of u or v is in S. The corresponding applicant (u or v) is qualified in sport e (since that vertex is an endpoint of e). Therefore S covers all sports with at most k applicants.
+
+**(←) If the Efficient Recruiting instance has a solution A with |A| ≤ k, then G has a vertex cover of size ≤ k.**
+
+Let A be a set of at most k applicants covering all sports. Consider the corresponding set of vertices S = A ⊆ V. We have |S| ≤ k. For every edge e = (u, v) ∈ E: sport e must be covered by some applicant in A, and an applicant is qualified in sport e only if they correspond to an endpoint of e. Therefore either u ∈ S or v ∈ S. Since this holds for every edge, S is a vertex cover.
+
+Since both directions hold, (G, k) ∈ Vertex Cover ⟺ the constructed instance ∈ Efficient Recruiting. ∎
+
+---
+
+## Problem 2: Plot Fulfillment [25 points]
+
+**Claim:** Hitting Set ≤ₚ Plot Fulfillment.
+
+### Reduction
+
+Recall the Hitting Set problem: given sets S₁, ..., Sₙ over universe U = {u₁, ..., uₘ} and integer k, does there exist H ⊆ U with |H| ≤ k such that H ∩ Sᵢ ≠ ∅ for all i?
+
+Given a Hitting Set instance (S₁, ..., Sₙ, U, k), we construct a Plot Fulfillment instance as follows:
+
+**Pages:** Create the following pages:
+- A start page s and an end page t.
+- For each layer l ∈ {1, ..., k} and each element j ∈ {1, ..., m}, a page v_{l,j}.
+
+(Total: km + 2 pages.)
+
+**Links (directed edges):**
+- s → v_{1,j} for all j ∈ {1, ..., m}.
+- v_{l,j} → v_{l+1,j'} for all l ∈ {1, ..., k−1} and all j, j' ∈ {1, ..., m}.
+- v_{k,j} → t for all j ∈ {1, ..., m}.
+
+This forms a layered DAG where any trail from s to t visits exactly one page per layer.
+
+**Thematic elements:** For each set Sᵢ (i = 1, ..., n), define
+$$T_i = \{ v_{l,j} : 1 \le l \le k,\; u_j \in S_i \}.$$
+
+**Start/End:** s and t as defined above.
+
+This construction runs in polynomial time (O(k·m·n) thematic elements and O(k·m²) links).
+
+### Proof of Correctness
+
+**Observation:** The constructed hypertext graph is a layered DAG. Any trail from s to t must visit exactly one page from each layer 1 through k, say v_{1,j₁}, v_{2,j₂}, ..., v_{k,jₖ} for some indices j₁, ..., jₖ ∈ {1, ..., m}. Furthermore, any such sequence of indices yields a valid trail from s to t (all required links exist).
+
+**(→) If there is a hitting set H with |H| ≤ k, then there is a fulfilling trail.**
+
+Since |H| ≤ k, write H = {u_{j₁}, ..., u_{jₖ}} (padding with repetitions if |H| < k). Construct the trail:
+$$s \to v_{1,j_1} \to v_{2,j_2} \to \cdots \to v_{k,j_k} \to t.$$
+For each thematic element Tᵢ: since H is a hitting set, some u_{jₗ} ∈ H ∩ Sᵢ. Then v_{l,jₗ} ∈ Tᵢ (by definition), and the trail visits v_{l,jₗ}. Thus the trail contains at least one page from every Tᵢ.
+
+**(←) If there is a trail from s to t visiting all Tᵢ, then there is a hitting set of size ≤ k.**
+
+Any such trail visits exactly one page per layer, say v_{1,j₁}, ..., v_{k,jₖ}. Define H = {u_{j₁}, ..., u_{jₖ}}. Then |H| ≤ k. For each set Sᵢ: the trail visits some page in Tᵢ, say v_{l,jₗ} ∈ Tᵢ. By definition of Tᵢ, u_{jₗ} ∈ Sᵢ, so H ∩ Sᵢ ∋ u_{jₗ} ≠ ∅. Thus H is a hitting set.
+
+Since both directions hold, the Hitting Set instance is a YES-instance ⟺ the Plot Fulfillment instance is a YES-instance. ∎
+
+---
+
+## Problem 3: Path Selection [25 points]
+
+**Claim:** Independent Set ≤ₚ Path Selection.
+
+### Reduction
+
+Given an Independent Set instance (G = (V, E), k), we construct a Path Selection instance as follows:
+
+**Directed graph H = (V_H, E_H):**
+- For each vertex v ∈ V: create unique nodes sᵥ (source) and tᵥ (sink).
+- For each edge e ∈ E: create a unique node nₑ (edge node).
+
+**Paths:** For each vertex v ∈ V, let e₁, e₂, ..., eₐ be the edges of G incident to v, listed in some fixed order (e.g., sorted by edge index). Define the path:
+$$P_v = s_v \to n_{e_1} \to n_{e_2} \to \cdots \to n_{e_a} \to t_v.$$
+(If v is isolated, Pᵥ = sᵥ → tᵥ.) Add the directed edges of each Pᵥ to H.
+
+**Budget:** k' = k (select at least k paths).
+
+This construction runs in polynomial time: O(|V| + |E|) nodes and O(|V| · max_degree) edges.
+
+### Proof of Correctness
+
+**Key observation:** Two paths Pᵤ and Pᵥ share a node if and only if (u, v) ∈ E.
+
+*Proof of observation:*
+- The nodes in Pᵤ are {sᵤ, tᵤ} ∪ {nₑ : e is incident to u}, and similarly for Pᵥ.
+- Source/sink nodes are unique: sᵤ ≠ sᵥ and tᵤ ≠ tᵥ for u ≠ v.
+- An edge node nₑ appears in Pᵤ iff e is incident to u, and in Pᵥ iff e is incident to v. For a simple graph, the only edge incident to both u and v is the edge e = (u,v) itself (if it exists).
+- Therefore Pᵤ and Pᵥ share the node n_{(u,v)} iff (u, v) ∈ E, and share no nodes if (u, v) ∉ E. ∎
+
+**(→) If G has an independent set S with |S| ≥ k, then k node-disjoint paths can be selected.**
+
+Consider {Pᵥ : v ∈ S}. This is a set of |S| ≥ k paths. For any u, v ∈ S with u ≠ v: since S is independent, (u, v) ∉ E. By the key observation, Pᵤ and Pᵥ share no nodes. Thus these paths are pairwise node-disjoint.
+
+**(←) If k pairwise node-disjoint paths can be selected from {Pᵥ : v ∈ V}, then G has an independent set of size ≥ k.**
+
+Let {P_{v₁}, ..., P_{vₖ}} be pairwise node-disjoint paths, and let S = {v₁, ..., vₖ}. We have |S| = k. For any two distinct vᵢ, vⱼ ∈ S: since P_{vᵢ} and P_{vⱼ} are node-disjoint, by the key observation (contrapositive), (vᵢ, vⱼ) ∉ E. Therefore no two vertices of S are adjacent, so S is an independent set of size k.
+
+Since both directions hold, (G, k) ∈ Independent Set ⟺ the constructed Path Selection instance is a YES-instance. ∎
+
+---
+
+## Problem 4: Half 3-SAT Satisfiability [25 points]
+
+**Claim:** 3-SAT ≤ₚ Half 3-SAT Satisfiability.
+
+### Key Lemma
+
+**Lemma:** For any 3-CNF formula φ, if φ is unsatisfiable, then for every assignment A to the variables of φ, at least one clause of φ is satisfied by A.
+
+**Proof:** Suppose for contradiction that some assignment A satisfies no clause of φ. Then for every clause (l₁ ∨ l₂ ∨ l₃) of φ, we have A(l₁) = A(l₂) = A(l₃) = 0. Consider the complementary assignment Ā defined by Ā(x) = 1 − A(x) for every variable x. For any literal l:
+- If l = xⱼ: A(l) = A(xⱼ) = 0, so Ā(xⱼ) = 1, hence Ā(l) = 1.
+- If l = ¬xⱼ: A(l) = 1 − A(xⱼ) = 0, so A(xⱼ) = 1 and Ā(xⱼ) = 0, hence Ā(¬xⱼ) = 1.
+
+Thus Ā(l) = 1 for every literal l appearing in any clause of φ, which means Ā satisfies every clause of φ. This contradicts the assumption that φ is unsatisfiable. ∎
+
+### Reduction
+
+Given a 3-SAT formula φ with m clauses C₁, ..., Cₘ (over variables x₁, ..., xₙ), construct a Half 3-SAT formula φ' with 2m clauses as follows:
+
+- **Variables:** x₁, ..., xₙ (original) plus fresh variables w₁, w₂, w₃.
+- **Clauses:** The original m clauses C₁, ..., Cₘ, plus m additional clauses D₁, ..., Dₘ, each equal to (¬w₁ ∨ ¬w₂ ∨ ¬w₃).
+
+So φ' = C₁ ∧ ··· ∧ Cₘ ∧ (¬w₁ ∨ ¬w₂ ∨ ¬w₃) ∧ ··· ∧ (¬w₁ ∨ ¬w₂ ∨ ¬w₃) (2m clauses total, which is even).
+
+This construction runs in polynomial time (O(m) additional clauses, O(1) additional variables).
+
+### Proof of Correctness
+
+**Observation:** All m copies of (¬w₁ ∨ ¬w₂ ∨ ¬w₃) are satisfied or unsatisfied simultaneously. The clause (¬w₁ ∨ ¬w₂ ∨ ¬w₃) is **false** iff w₁ = w₂ = w₃ = true, and **true** otherwise.
+
+Thus, the total number of satisfied clauses in φ' under any assignment is:
+$$\text{total} = (\text{# of } C_i \text{ satisfied}) + m \cdot \mathbf{1}[\neg w_1 \vee \neg w_2 \vee \neg w_3].$$
+
+We need total = m (exactly half of 2m). There are two cases:
+
+**Case 1: w₁ = w₂ = w₃ = true.** The indicator equals 0. Then total = (# of Cᵢ satisfied). For total = m, we need all m original clauses satisfied, i.e., φ is satisfiable.
+
+**Case 2: Not all of w₁, w₂, w₃ are true.** The indicator equals 1. Then total = (# of Cᵢ satisfied) + m. For total = m, we need (# of Cᵢ satisfied) = 0, i.e., no clause of φ is satisfied.
+
+**(→) If φ is satisfiable:** Take a satisfying assignment A* to x₁, ..., xₙ (so all m clauses Cᵢ are satisfied). Extend with w₁ = w₂ = w₃ = true. Then:
+- All m clauses Cᵢ are satisfied.
+- (¬w₁ ∨ ¬w₂ ∨ ¬w₃) = false, so all m clauses Dᵢ are unsatisfied.
+- Total = m + 0 = m = 2m/2. ✓
+
+So φ' ∈ Half 3-SAT Satisfiability.
+
+**(←) If φ is not satisfiable:** We show no assignment achieves total = m in φ'.
+- **Case 1** (w₁ = w₂ = w₃ = true): total = (# of Cᵢ satisfied) ≤ m − 1 < m, since φ is unsatisfiable.
+- **Case 2** (not all true): total = (# of Cᵢ satisfied) + m. By the Key Lemma (since φ is unsatisfiable), (# of Cᵢ satisfied) ≥ 1. Therefore total ≥ m + 1 > m.
+
+In both cases, total ≠ m. So φ' ∉ Half 3-SAT Satisfiability.
+
+Since both directions hold, φ ∈ 3-SAT ⟺ φ' ∈ Half 3-SAT Satisfiability. ∎