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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]
+## Part 1
 
+### Problem 1: Two-Sum with Divide and Conquer [10 points]
+
+**Algorithm (Pseudocode):**
+
+```
+TwoSum(S, x):
+    Sort S in ascending order                    // O(n log n)
+    left = 0
+    right = n - 1
+
+    while left < right:
+        sum = S[left] + S[right]
+        if sum == x:
+            return true
+        else if sum < x:
+            left = left + 1
+        else:
+            right = right - 1
+
+    return false
+```
+
+**Complexity Justification:**
+
+The algorithm has O(n log n) complexity:
+- **Sorting:** O(n log n) using merge sort or quicksort
+- **Two-pointer search:** O(n) - each element is visited at most once as the left pointer moves right and the right pointer moves left
+- **Total:** O(n log n) + O(n) = O(n log n)
+
+The two-pointer technique works because the sorted array has the property that if we have a sum that is too small, we need a larger number (move left pointer right), and if the sum is too large, we need a smaller number (move right pointer left). This ensures we examine each element at most once while maintaining correctness.
+
+---
+
+### Problem 2: Induction on Sum of Products [10 points]
+
+**Claim:** For every positive integer n,
+$$\sum_{i=1}^{n} i(i + 1)(i + 2) = \frac{n(n+1)(n+2)(n+3)}{4}$$
+
+**Proof by Mathematical Induction:**
+
+**Base Case (n = 1):**
+- LHS: 1(1+1)(1+2) = 1(2)(3) = 6
+- RHS: $\frac{1(2)(3)(4)}{4} = \frac{24}{4} = 6$
+- LHS = RHS ✓
+
+**Inductive Step:**
+Assume the claim is true for n = k, i.e.,
+$$\sum_{i=1}^{k} i(i + 1)(i + 2) = \frac{k(k+1)(k+2)(k+3)}{4}$$
+
+We must prove it for n = k + 1:
+$$\sum_{i=1}^{k+1} i(i + 1)(i + 2) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}$$
+
+Starting with the LHS:
+$$\sum_{i=1}^{k+1} i(i + 1)(i + 2) = \sum_{i=1}^{k} i(i + 1)(i + 2) + (k+1)(k+2)(k+3)$$
+
+By the inductive hypothesis:
+$$= \frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3)$$
+
+Factor out $(k+1)(k+2)(k+3)$:
+$$= (k+1)(k+2)(k+3) \left[\frac{k}{4} + 1\right]$$
+
+$$= (k+1)(k+2)(k+3) \left[\frac{k+4}{4}\right]$$
+
+$$= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$$
+
+This is exactly the RHS for n = k + 1. ✓
+
+By mathematical induction, the claim holds for all positive integers n.
+
+---
+
+### Problem 3: Sum of Cubes [10 points]
+
+**Claim:** For all n ≥ 0,
+$$\sum_{i=0}^{n} i^3 = \left(\sum_{i=0}^{n} i\right)^2$$
+
+**Proof by Mathematical Induction:**
+
+**Base Case (n = 0):**
+- LHS: $0^3 = 0$
+- RHS: $(0)^2 = 0$
+- LHS = RHS ✓
+
+**Inductive Step:**
+Assume the claim is true for n = k, i.e.,
+$$\sum_{i=0}^{k} i^3 = \left(\sum_{i=0}^{k} i\right)^2$$
+
+We know that $\sum_{i=0}^{k} i = \frac{k(k+1)}{2}$.
+
+We must prove for n = k + 1:
+$$\sum_{i=0}^{k+1} i^3 = \left(\sum_{i=0}^{k+1} i\right)^2$$
+
+Starting with the LHS:
+$$\sum_{i=0}^{k+1} i^3 = \sum_{i=0}^{k} i^3 + (k+1)^3$$
+
+By the inductive hypothesis:
+$$= \left(\sum_{i=0}^{k} i\right)^2 + (k+1)^3$$
+
+$$= \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$$
+
+$$= \frac{k^2(k+1)^2}{4} + (k+1)^3$$
+
+Now for the RHS:
+$$\left(\sum_{i=0}^{k+1} i\right)^2 = \left(\sum_{i=0}^{k} i + (k+1)\right)^2$$
+
+$$= \left(\frac{k(k+1)}{2} + (k+1)\right)^2$$
+
+$$= \left((k+1)\left[\frac{k}{2} + 1\right]\right)^2$$
+
+$$= \left((k+1)\cdot\frac{k+2}{2}\right)^2$$
+
+$$= \frac{(k+1)^2(k+2)^2}{4}$$
+
+Expanding:
+$$= \frac{(k+1)^2[(k+1)^2 + 4(k+1) + 4]}{4}$$
+
+Wait, let me recalculate more carefully:
+$$= \frac{(k+1)^2(k+2)^2}{4}$$
+
+We need to verify that $\frac{k^2(k+1)^2}{4} + (k+1)^3 = \frac{(k+1)^2(k+2)^2}{4}$.
+
+Factor out $(k+1)^2$:
+$$\frac{(k+1)^2}{4}[k^2 + 4(k+1)] = \frac{(k+1)^2}{4}[k^2 + 4k + 4] = \frac{(k+1)^2(k+2)^2}{4}$$ ✓
+
+By mathematical induction, the claim holds for all n ≥ 0.
+
+---
+
+## Part 2
+
+### Problem 4: Plane Separation by Lines [10 points]
+
+**Claim:** n lines separate the plane into $\frac{n^2 + n + 2}{2}$ regions, provided no two lines are parallel and no three lines pass through a common point.
+
+**Proof by Mathematical Induction:**
+
+**Base Case (n = 0):**
+- 0 lines create 1 region (the entire plane)
+- Formula: $\frac{0^2 + 0 + 2}{2} = \frac{2}{2} = 1$ ✓
+
+**Base Case (n = 1):**
+- 1 line divides the plane into 2 regions
+- Formula: $\frac{1^2 + 1 + 2}{2} = \frac{4}{2} = 2$ ✓
+
+**Inductive Step:**
+Assume n-1 lines separate the plane into $\frac{(n-1)^2 + (n-1) + 2}{2}$ regions.
+
+When we add the n-th line:
+- The n-th line is not parallel to any existing line (by assumption)
+- The n-th line intersects each of the previous n-1 lines at distinct points (no three lines through a common point)
+- These n-1 intersection points divide the n-th line into n segments (including two rays)
+- Each segment splits an existing region into two, creating exactly n new regions
+
+Therefore, the total number of regions with n lines is:
+$$\frac{(n-1)^2 + (n-1) + 2}{2} + n$$
+
+$$= \frac{(n-1)^2 + (n-1) + 2 + 2n}{2}$$
+
+$$= \frac{n^2 - 2n + 1 + n - 1 + 2 + 2n}{2}$$
+
+$$= \frac{n^2 + n + 2}{2}$$
+
+By mathematical induction, the formula holds for all n ≥ 0.
+
+---
+
+### Problem 5: Postage Stamp Problem [10 points]
+
+**Claim:** For all n ≥ 18, a postage of n cents can be formed using 4-cent and 7-cent stamps.
+
+**Proof by Mathematical Induction:**
+
+**Base Cases:**
+- n = 18: 18 = 4 + 7 + 7 ✓
+- n = 19: 19 = 4 + 4 + 4 + 7 ✓
+- n = 20: 20 = 4 + 4 + 4 + 4 + 4 ✓
+- n = 21: 21 = 7 + 7 + 7 ✓
+
+**Inductive Step:**
+Assume the claim holds for all k where 18 ≤ k ≤ n for some n ≥ 21.
+
+Consider n + 1. Since n + 1 ≥ 22, we have n + 1 - 4 ≥ 18, so (n+1) - 4 = n - 3 falls within our inductive hypothesis range [18, n].
+
+By the inductive hypothesis, n - 3 cents can be formed using 4-cent and 7-cent stamps.
+
+Therefore, n + 1 = (n - 3) + 4 cents can also be formed by adding one 4-cent stamp to the combination for n - 3 cents. ✓
+
+By mathematical induction, for all n ≥ 18, a postage of n cents can be formed using 4-cent and 7-cent stamps.
+
+---
+
+### Problem 6: Integer Powers Property [10 points]
+
+**Claim:** Let x be a real number such that $x + \frac{1}{x}$ is an integer. Then $x^n + \frac{1}{x^n}$ is an integer for all integers n ≥ 0.
+
+**Proof by Mathematical Induction:**
+
+Let $k = x + \frac{1}{x}$, where k is an integer.
+
+**Base Cases:**
+- n = 0: $x^0 + \frac{1}{x^0} = 1 + 1 = 2$, which is an integer ✓
+- n = 1: $x + \frac{1}{x} = k$, which is an integer by assumption ✓
+
+**Inductive Step:**
+Assume that $x^m + \frac{1}{x^m}$ is an integer for all 0 ≤ m ≤ n, where n ≥ 1.
+
+Consider:
+$$x^{n+1} + \frac{1}{x^{n+1}} = \left(x + \frac{1}{x}\right)\left(x^n + \frac{1}{x^n}\right) - \left(x^{n-1} + \frac{1}{x^{n-1}}\right)$$
+
+**Verification of the identity:**
+$$\left(x + \frac{1}{x}\right)\left(x^n + \frac{1}{x^n}\right) = x^{n+1} + x^{n-1} + \frac{1}{x^{n-1}} + \frac{1}{x^{n+1}}$$
+
+$$= \left(x^{n+1} + \frac{1}{x^{n+1}}\right) + \left(x^{n-1} + \frac{1}{x^{n-1}}\right)$$
+
+Therefore:
+$$x^{n+1} + \frac{1}{x^{n+1}} = \left(x + \frac{1}{x}\right)\left(x^n + \frac{1}{x^n}\right) - \left(x^{n-1} + \frac{1}{x^{n-1}}\right)$$
+
+By the inductive hypothesis:
+- $x + \frac{1}{x}$ is an integer (given)
+- $x^n + \frac{1}{x^n}$ is an integer (inductive hypothesis)
+- $x^{n-1} + \frac{1}{x^{n-1}}$ is an integer (inductive hypothesis, since n-1 ≥ 0)
+
+The product and difference of integers are integers, so $x^{n+1} + \frac{1}{x^{n+1}}$ is an integer. ✓
+
+By mathematical induction, $x^n + \frac{1}{x^n}$ is an integer for all integers n ≥ 0.
+
+---
+
+## Part 3
+
+### Problem 7: Sequence Bound [10 points]
+
+**Claim:** Let the sequence $\{a_n\}$ be defined by $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, and $a_n = a_{n-1} + a_{n-2} + a_{n-3}$ for all n ≥ 4. Then $a_n < 2^n$ for all positive integers n.
+
+**Proof by Mathematical Induction:**
+
+**Base Cases:**
+- n = 1: $a_1 = 1 < 2^1 = 2$ ✓
+- n = 2: $a_2 = 2 < 2^2 = 4$ ✓
+- n = 3: $a_3 = 3 < 2^3 = 8$ ✓
+
+**Inductive Step:**
+Assume $a_k < 2^k$ for all positive integers k ≤ n, where n ≥ 3.
+
+We must show $a_{n+1} < 2^{n+1}$.
+
+By the recurrence relation:
+$$a_{n+1} = a_n + a_{n-1} + a_{n-2}$$
+
+By the inductive hypothesis:
+$$a_{n+1} < 2^n + 2^{n-1} + 2^{n-2}$$
+
+Now we simplify the right-hand side:
+$$2^n + 2^{n-1} + 2^{n-2} = 2^{n-2}(2^2 + 2 + 1) = 2^{n-2} \cdot 7$$
+
+Since $7 < 8 = 2^3$, we have:
+$$2^{n-2} \cdot 7 < 2^{n-2} \cdot 2^3 = 2^{n+1}$$
+
+Therefore:
+$$a_{n+1} < 2^n + 2^{n-1} + 2^{n-2} < 2^{n+1}$$ ✓
+
+By mathematical induction, $a_n < 2^n$ for all positive integers n.
+
+---
+
+### Problem 8: Binary Strings with More 0's than 1's [10 points]
+
+**Recursive Definition:**
+
+Let S be the set of all binary strings with more 0's than 1's. S is the smallest set such that:
+
+1. **Base case:** "0" ∈ S
+2. **Recursive cases:**
+   - If w ∈ S, then "0w" ∈ S
+   - If w ∈ S, then "w0" ∈ S
+   - If w ∈ S and the number of 0's in w exceeds the number of 1's in w by at least 2, then "1w" ∈ S
+   - If w ∈ S and the number of 0's in w exceeds the number of 1's in w by at least 2, then "w1" ∈ S
+
+**Alternatively, more elegantly:**
+
+Let S be the set of all binary strings with more 0's than 1's. S is defined recursively as the smallest set such that:
+
+1. **Base case:** "0" ∈ S
+2. **Recursive cases:**
+   - If w ∈ S, then "0w" ∈ S and "w0" ∈ S
+   - If w ∈ S and the difference (count of 0's - count of 1's) in w is ≥ 2, then "1w" ∈ S and "w1" ∈ S
+
+This definition ensures that:
+- Every string in S has more 0's than 1's
+- Every binary string with more 0's than 1's can be constructed this way
+
+---
+
+### Problem 9: Balanced Parentheses [20 points]
+
+#### Part (a): Recursive Definition
+
+**Definition:** Let B be the set of all balanced strings of parentheses. B is the smallest set such that:
+
+1. **Base case:** The empty string ε ∈ B
+2. **Recursive cases:**
+   - If x ∈ B, then "(" + x + ")" ∈ B (wrap x with parentheses)
+   - If x ∈ B and y ∈ B, then xy ∈ B (concatenate two balanced strings)
+
+**Examples:**
+- ε ∈ B (base case)
+- "()" ∈ B (from ε using rule 2a)
+- "()()" ∈ B (from "()" and "()" using rule 2b)
+- "(())" ∈ B (from "()" using rule 2a)
+- "((()))" ∈ B (from "(())" using rule 2a)
+
+#### Part (b): Proof that Left Parentheses Equal Right Parentheses
+
+**Claim:** If x is a balanced string of parentheses, then the number of left parentheses equals the number of right parentheses in x.
+
+**Proof by Structural Induction:**
+
+We prove this by induction on the structure of balanced strings as defined above.
+
+**Base Case:**
+The empty string ε has 0 left parentheses and 0 right parentheses, so they are equal. ✓
+
+**Inductive Step - Case 1:**
+Assume x ∈ B and x has equal numbers of left and right parentheses, say k left and k right.
+
+Consider the string "(" + x + ")".
+- Number of left parentheses: 1 (from the prefix) + k (from x) = k + 1
+- Number of right parentheses: k (from x) + 1 (from the suffix) = k + 1
+
+The numbers are equal. ✓
+
+**Inductive Step - Case 2:**
+Assume x ∈ B and y ∈ B, where:
+- x has $m_1$ left and $m_1$ right parentheses (by inductive hypothesis)
+- y has $m_2$ left and $m_2$ right parentheses (by inductive hypothesis)
+
+Consider the concatenation xy.
+- Number of left parentheses: $m_1$ (from x) + $m_2$ (from y) = $m_1 + m_2$
+- Number of right parentheses: $m_1$ (from x) + $m_2$ (from y) = $m_1 + m_2$
+
+The numbers are equal. ✓
+
+By structural induction, every balanced string of parentheses has equal numbers of left and right parentheses.
+
+---