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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: Understanding Graph-theoretic Definitions and Terminology
 
+### 1.1 Degree Sequences
+
+**K_4 (complete graph on 4 nodes):**
+Every node is connected to all other 3 nodes, so each node has degree 3.
+Degree sequence: **[3, 3, 3, 3]**
+
+**C_5 (cycle on 5 nodes):**
+Every node is connected to exactly its two neighbors in the cycle, so each node has degree 2.
+Degree sequence: **[2, 2, 2, 2, 2]**
+
+**K_{m,n} (complete bipartite graph):**
+K_{m,n} has two disjoint sets: m nodes (call them the "left" set) each connected to all n nodes on the right, and n nodes (the "right" set) each connected to all m nodes on the left.
+- Each left-set node has degree n.
+- Each right-set node has degree m.
+
+Assuming n ≥ m, the degree sequence (non-increasing order) is:
+**[n, n, ..., n (m times), m, m, ..., m (n times)]**
+
+If m > n, swap the roles of m and n above to get [m, ..., m (n times), n, ..., n (m times)].
+
+---
+
+### 1.2 Planar Graphs
+
+**K_2:** Two nodes with one edge. Trivially planar (no crossings possible).
+Drawing: `A --- B`
+
+**K_3:** Three nodes forming a triangle. Planar.
+Drawing:
+```
+    A
+   / \
+  B---C
+```
+
+**K_4:** Planar. Place three nodes A, B, C as a triangle, and place D inside the triangle, connecting D to each of A, B, C. All 6 edges are drawn without crossings.
+Drawing:
+```
+    A
+   /|\
+  / | \
+ B--D--C
+  \   /
+   \ /
+    (B-C edge on bottom)
+```
+Edges: {A,B}, {A,C}, {B,C} (triangle), {A,D}, {B,D}, {C,D} (D inside) — no crossings.
+
+**K_5:** **Not planar.** Proof by Euler's formula: For a connected planar simple graph, V - E + F = 2. K_5 has V = 5, E = 10, so F = 2 - 5 + 10 = 7 faces. However, every face must be bounded by at least 3 edges, and each edge borders exactly 2 faces, so 3F ≤ 2E, giving 3(7) = 21 ≤ 2(10) = 20, a contradiction. Therefore K_5 is not planar.
+
+**Summary:** K_2, K_3, and K_4 are planar; K_5 is not.
+
+---
+
+### 1.3 Course Prerequisites
+
+**(a) Precedence Graph**
+
+Nodes represent courses. A directed edge A → B means "A is a direct prerequisite for B."
+
+Edges (direct prerequisites only):
+- MATH 111 → MATH 112
+- COMP 140 → COMP 182
+- MATH 211 → COMP 182
+- COMP 182 → COMP 215
+- COMP 215 → COMP 310
+- MATH 112 → STAT 310
+- MATH 211 → STAT 310
+- COMP 182 → COMP 311
+- COMP 182 → COMP 382
+- STAT 310 → COMP 382
+- ELEC 220 → COMP 321
+- COMP 215 → COMP 321
+- COMP 321 → COMP 410
+- COMP 310 → COMP 410
+- COMP 311 → COMP 410
+
+Diagram (text representation):
+```
+MATH 111 ---> MATH 112 ---> STAT 310 ---> COMP 382
+                               ^               ^
+MATH 211 ---> COMP 182 --------+               |
+                |   |--------> COMP 311 --------+
+                |                               |
+                +-------> COMP 215 --> COMP 310 --> COMP 410
+                                  \                     ^
+ELEC 220 --------------------------> COMP 321 ----------+
+                                                        |
+COMP 140 -> COMP 182 (already shown above)       COMP 311 ^
+```
+
+**(b) Is the graph acyclic?**
+
+Yes, the graph is acyclic (it is a directed acyclic graph, or DAG).
+
+It makes sense for it to be acyclic. If there were a directed cycle, it would mean some course A has (directly or transitively) itself as a prerequisite, making it impossible to ever take course A. In any real course-scheduling scenario, prerequisites must form an acyclic structure so that students can actually complete all courses.
+
+**(c) Minimum number of semesters**
+
+Assign each course a "level" equal to the length of the longest prerequisite chain ending at that course (counting courses in the chain):
+
+- **Level 1** (no prerequisites): COMP 140, MATH 111, MATH 211, ELEC 220
+- **Level 2**: MATH 112 (needs MATH 111), COMP 182 (needs COMP 140, MATH 211)
+- **Level 3**: COMP 215 (needs COMP 182), STAT 310 (needs MATH 112, MATH 211), COMP 311 (needs COMP 182)
+- **Level 4**: COMP 310 (needs COMP 215), COMP 321 (needs ELEC 220, COMP 215), COMP 382 (needs COMP 182, STAT 310)
+- **Level 5**: COMP 410 (needs COMP 321, COMP 310, COMP 311)
+
+The minimum number of semesters is **5**.
+
+**General algorithm:** For an arbitrary precedence DAG, compute the minimum number of semesters by finding the length of the longest path in the DAG. This can be done by:
+1. Perform a topological sort of the DAG.
+2. For each node v in topological order, set level(v) = 1 + max(level(u)) over all direct prerequisites u of v; set level(v) = 1 if v has no prerequisites.
+3. The minimum number of semesters = max level(v) over all nodes v.
+
+This works because the level of a course equals the minimum semester in which it can first be taken, and the maximum level is the semester when the last course is completed.
+
+---
+
+### 1.4 Matchings
+
+**(a) Graph with ≥ 5 nodes and a matching of size 2**
+
+Let G = (V, E) be the path graph P_5 with V = {1, 2, 3, 4, 5} and edges {1,2}, {2,3}, {3,4}, {4,5}.
+
+The set M = {{1,2}, {3,4}} is a matching of size 2: edges {1,2} and {3,4} share no node.
+
+**(b) Graph with ≥ 5 nodes and no matching of size 2**
+
+Let G be the star graph K_{1,4} with nodes {c, 1, 2, 3, 4} and edges {c,1}, {c,2}, {c,3}, {c,4} (node c is the center).
+
+Every edge in G is incident to c. Therefore, any two edges in G share the node c. No two edges are disjoint, so the maximum matching has size 1—there is no matching of size 2.
+
+**(c) Size of a perfect matching for a graph with n nodes (n even)**
+
+If a perfect matching exists, its size is **n/2**.
+
+**Reasoning:** In a perfect matching, every node must be incident to at least one matching edge. Since each edge covers exactly 2 nodes and no two matching edges share a node, n/2 edges are needed to cover all n nodes. Moreover, n/2 edges can cover exactly n nodes, so the size of any perfect matching is exactly n/2.
+
+---
+
+## Problem 2: Graph-theoretic Problem Formulation
+
+### 2.1 Jury Selection
+
+**Graph:** Construct an undirected graph G = (V, E) where:
+- V = the set of n individuals in the city
+- {u, v} ∈ E if and only if individuals u and v know each other
+
+**Input:** Graph G and integer m.
+
+**Output:** A subset S ⊆ V with |S| = m such that no two nodes in S are adjacent in G (i.e., S is an **independent set** of size m).
+
+This is the **Independent Set problem**: find m individuals such that no two of them know each other, which corresponds to finding an independent set of size m in the "acquaintance graph."
+
+---
+
+### 2.2 Student Groups
+
+**Graph:** Construct an undirected graph G = (V, E) where:
+- V = the set of students in COMP 182
+- {u, v} ∈ E if and only if students u and v like each other
+
+**Input:** Graph G.
+
+**Output:** A partition of V into the minimum number of subsets (groups) such that each subset is a **clique** in G (every pair of students in the group like each other).
+
+This is the **minimum clique cover problem**. Equivalently, it is the **graph coloring problem** on the complement graph G' = (V, E'): form G' by placing an edge {u,v} in E' whenever {u,v} ∉ E (students who do NOT like each other). A proper coloring of G' assigns the same color only to students who are mutually non-adjacent in G', i.e., who all like each other—forming a clique in G. The minimum number of groups equals the chromatic number χ(G').
+
+---
+
+### 2.3 Course Scheduling
+
+**Graph:** Construct an undirected graph G = (V, E) where:
+- V = the set of courses offered in Fall 2023
+- {u, v} ∈ E if and only if courses u and v have at least one student in common (i.e., they conflict)
+
+**Input:** Graph G and integer k (number of available time slots).
+
+**Output:** A proper coloring of V using the minimum number of colors, where each color represents a time slot. A coloring is proper if no two adjacent nodes (conflicting courses) receive the same color.
+
+This is the **graph coloring problem**: the minimum number of time slots needed equals the chromatic number χ(G) of the conflict graph. Each color class (set of courses with the same color/slot) contains only mutually non-conflicting courses that can be scheduled simultaneously.
+
+---
+
+## Problem 3: Proving Graph-theoretic Results
+
+### 3.1 Bipartite Bit Vector Graph
+
+**Claim:** B_n is bipartite for all n ≥ 2.
+
+**Proof:**
+
+We explicitly construct a bipartition of V.
+
+For any bit vector v ∈ V, let wt(v) denote the number of 1-bits in v (its Hamming weight). Define:
+- L = {v ∈ V : wt(v) is even}
+- R = {v ∈ V : wt(v) is odd}
+
+Clearly L and R are disjoint and L ∪ R = V (every bit vector has either even or odd Hamming weight), so (L, R) is a partition of V.
+
+**Claim:** Every edge in E_n connects a node in L to a node in R.
+
+Let {u, v} ∈ E_n. By definition of B_n, u and v differ in exactly one bit position. Flipping one bit either:
+- Changes a 0 to a 1: wt increases by 1, changing its parity.
+- Changes a 1 to a 0: wt decreases by 1, changing its parity.
+
+In either case, wt(u) and wt(v) have opposite parity. Therefore, one of u, v is in L (even parity) and the other is in R (odd parity).
+
+Since every edge in E_n connects a node in L to a node in R, B_n is bipartite with bipartition (L, R). □
+
+---
+
+### 3.2 Bridges and Even Degrees
+
+**Claim:** If the degrees of all nodes in g are even, then g does not contain a bridge.
+
+**Proof:**
+
+It suffices to prove the claim for connected graphs, since each connected component of g inherits the even-degree property, and any bridge must lie within some connected component.
+
+**Assume g is connected with all even degrees.**
+
+By Euler's theorem (Theorem 2 in Section 10.5): a connected graph has an Euler circuit if and only if every vertex has even degree. Since all degrees in g are even, g has an Euler circuit—a closed walk that traverses every edge of g exactly once.
+
+**Suppose for contradiction** that e = {u, v} is a bridge in g. Then g' = g - e has two connected components: C_u (containing u) and C_v (containing v), with the only edge between them in g being e.
+
+Consider the Euler circuit EC of g, starting at some vertex s. EC traverses every edge exactly once. At some point, EC traverses e. Say EC goes from u to v via e (the direction is symmetric).
+
+- After traversing e, the circuit is at v ∈ C_v and must eventually return to s.
+- **Case 1 (s ∈ C_u):** After arriving at v, the circuit must reach s ∈ C_u. But the only edge between C_u and C_v is e, which has already been used. EC cannot reuse edges—contradiction.
+- **Case 2 (s ∈ C_v):** Before traversing e (going u → v), the circuit must have reached u ∈ C_u from s ∈ C_v at some earlier step. But the only edge between C_v and C_u is e, and we haven't used it yet (we're about to use it). So to reach C_u from C_v, the circuit would have to use e going v → u, then later use e again going u → v—but that means e is used twice, contradicting the Euler circuit property.
+
+In both cases we reach a contradiction. Therefore, no bridge exists in g. □
+
+---
+
+### 3.3 Edge Coloring and Hamiltonian Circuits
+
+**Claim:** If every node in g has degree 3 and g has a Hamiltonian circuit, then there exists a 3-coloring of g's edges.
+
+**Proof:**
+
+**Step 1: n must be even.**
+
+Since every node has degree 3, the sum of all degrees is 3n. By the handshaking lemma, 3n = 2|E|. Since the right side is even, 3n must be even. Since gcd(3, 2) = 1, n must be even.
+
+**Step 2: Partition edges into H and M.**
+
+Let H = (v_1, v_2, \ldots, v_n, v_1) be a Hamiltonian circuit in g. Let E_H = {e_1, e_2, \ldots, e_n} denote the edges of H, where e_i = {v_i, v_{i+1}} for i < n and e_n = {v_n, v_1}.
+
+Since every node has degree 3 in g and the Hamiltonian circuit accounts for exactly 2 edges at each node (entering and leaving), each node has exactly **one** edge in E \ E_H. Thus M = E \ E_H is a **perfect matching** (no two edges of M share a node).
+
+**Step 3: Define the 3-coloring f : E → {1, 2, 3}.**
+
+- For each e_i ∈ E_H: assign f(e_i) = 1 if i is odd, f(e_i) = 2 if i is even.
+- For each e ∈ M: assign f(e) = 3.
+
+**Step 4: Verify f is a valid 3-coloring.**
+
+We must show that any two edges sharing a node receive different colors.
+
+**(i) Two edges both in E_H sharing a node:**
+Adjacent edges in H are e_i and e_{i+1} (sharing v_{i+1}), or e_n and e_1 (sharing v_1).
+- e_i and e_{i+1}: one has odd index and one has even index, so they receive colors 1 and 2 (different). ✓
+- e_n and e_1: since n is even, f(e_n) = 2 and f(e_1) = 1—different colors. ✓
+
+**(ii) Two edges both in M sharing a node:**
+M is a perfect matching, so no two edges in M share a node. There are no such pairs. ✓
+
+**(iii) One edge from E_H and one from M sharing a node:**
+Every edge in E_H receives color 1 or 2; every edge in M receives color 3. Since {1, 2} ∩ {3} = ∅, these edges always receive different colors. ✓
+
+Therefore f is a valid 3-coloring of the edges of g. □
+
+---
+
+### 3.4 Cartesian Product Graphs
+
+Recall: C = (V_C, E_C) where V_C = V_1 × V_2 and {(u,x),(v,y)} ∈ E_C iff {u,v} ∈ E_1 OR {x,y} ∈ E_2.
+
+#### (a) Hamiltonian Circuits — TRUE
+
+**Claim:** If g_1 and g_2 both contain Hamiltonian circuits, then C contains a Hamiltonian circuit.
+
+**Proof:**
+
+Let H_1 = (u_1, u_2, \ldots, u_n, u_1) be a Hamiltonian circuit of g_1, and H_2 = (x_1, x_2, \ldots, x_m, x_1) be a Hamiltonian circuit of g_2.
+
+We construct a Hamiltonian circuit of C using a **boustrophedon (snake) traversal** of the n × m node grid:
+
+Traverse the rows in alternating directions:
+- **Row i (odd):** (u_i, x_1) → (u_i, x_2) → ··· → (u_i, x_m)  [left to right]
+- **Row i (even):** (u_i, x_m) → (u_i, x_{m-1}) → ··· → (u_i, x_1)  [right to left]
+- **Transition from row i to row i+1:** jump from the last node of row i to the first node of row i+1.
+- **Closing edge:** after row n, return to (u_1, x_1).
+
+**Validity of each edge type:**
+
+*Intra-row edge* (within row i, moving from (u_i, x_j) to (u_i, x_{j±1})):
+The condition requires {u_i, u_i} ∈ E_1 (impossible, self-loop) **OR** {x_j, x_{j±1}} ∈ E_2. The latter holds because consecutive nodes in H_2 are adjacent in g_2 by definition of a Hamiltonian circuit. ✓
+
+*Inter-row transition* (from last node of row i to first node of row i+1):
+- If i is odd: jump from (u_i, x_m) to (u_{i+1}, x_m). Requires {u_i, u_{i+1}} ∈ E_1 **OR** {x_m, x_m} ∈ E_2. The first holds since consecutive nodes in H_1 are adjacent. ✓
+- If i is even: jump from (u_i, x_1) to (u_{i+1}, x_1). Requires {u_i, u_{i+1}} ∈ E_1. ✓
+
+*Closing edge* (from the end of row n back to (u_1, x_1)):
+- If n is odd: closing edge is from (u_n, x_m) to (u_1, x_1). Requires {u_n, u_1} ∈ E_1 (true, closing edge of H_1) **OR** {x_m, x_1} ∈ E_2 (true, closing edge of H_2). ✓
+- If n is even: closing edge is from (u_n, x_1) to (u_1, x_1). Requires {u_n, u_1} ∈ E_1. ✓
+
+**Coverage:** Each of the n rows contributes exactly m nodes, and each node (u_i, x_j) appears exactly once in the traversal. The total count is n × m = |V_C|. The traversal starts and ends at (u_1, x_1), forming a closed circuit.
+
+Therefore C has a Hamiltonian circuit. □
+
+---
+
+#### (b) Euler Circuits — TRUE
+
+**Claim:** If g_1 and g_2 both contain Euler circuits, then C contains an Euler circuit.
+
+We use the criterion (Theorem 2, Section 10.5): a connected graph has an Euler circuit if and only if every vertex has even degree.
+
+**Step 1: Compute the degree of each node in C.**
+
+For node (u, x) ∈ V_C, its neighbors are all (v, y) such that {u,v} ∈ E_1 **OR** {x,y} ∈ E_2 (with (v,y) ≠ (u,x)). By inclusion-exclusion:
+
+$$\deg_C(u,x) = |\{(v,y) : \{u,v\} \in E_1\}| + |\{(v,y) : \{x,y\} \in E_2\}| - |\{(v,y) : \{u,v\} \in E_1 \text{ and } \{x,y\} \in E_2\}|$$
+
+$$= \deg_1(u) \cdot |V_2| + |V_1| \cdot \deg_2(x) - \deg_1(u) \cdot \deg_2(x)$$
+
+(Note: (u,x) itself is excluded from the neighbor count since self-loops do not exist in g_1 or g_2.)
+
+**Step 2: Show all degrees in C are even.**
+
+Since g_1 has an Euler circuit, all degrees in g_1 are even: deg_1(u) is even for all u. Similarly, deg_2(x) is even for all x. Therefore:
+- deg_1(u) · |V_2| is even (product with an even factor).
+- |V_1| · deg_2(x) is even (product with an even factor).
+- deg_1(u) · deg_2(x) is even (product of two even numbers).
+
+Thus deg_C(u, x) = even + even − even = **even** for every (u, x) ∈ V_C.
+
+**Step 3: Show C is connected.**
+
+Let (u, x) and (v, y) be any two nodes in V_C. Since g_1 is connected, there exists a path u = u_0, u_1, \ldots, u_k = v in g_1. Since {u_i, u_{i+1}} ∈ E_1, the edge {(u_i, x), (u_{i+1}, x)} ∈ E_C (OR condition: first component uses E_1). So
+
+$$(u, x) = (u_0, x) \to (u_1, x) \to \cdots \to (u_k, x) = (v, x)$$
+
+is a path in C. Similarly, since g_2 is connected, there is a path x = x_0, x_1, \ldots, x_ℓ = y in g_2, and {(v, x_j), (v, x_{j+1})} ∈ E_C (OR condition: second component uses E_2). So
+
+$$(v, x) = (v, x_0) \to (v, x_1) \to \cdots \to (v, x_ℓ) = (v, y)$$
+
+is a path in C. Concatenating, (u, x) is connected to (v, y) in C. Since (u, x) and (v, y) were arbitrary, C is connected.
+
+**Conclusion:**
+
+C is connected and every node has even degree. By Euler's theorem, C contains an Euler circuit. □