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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: Persistent Dynamic Sets
+
+### Part (a): Nodes That Need to Change
+
+When inserting or deleting a node in a binary search tree, only the nodes on the **search path from the root to the insertion/deletion point** need to be modified.
+
+**Specifically:**
+- For **insertion**: The path from the root down to the parent of the newly inserted node must be modified. Any node on this path needs its `left` or `right` pointer updated (to point to a new copy of its child).
+- For **deletion**: Similarly, the path from root to the deleted node, and nodes whose pointers must be updated to maintain structure, must be modified.
+
+Since no parent pointers exist, we only need to update pointers in the path's ancestors.
+
+### Part (b): PERSISTENT-TREE-INSERT Procedure
+
+```
+PERSISTENT-TREE-INSERT(T, z)
+    return PERSISTENT-INSERT-RECURSIVE(T.root, z)
+
+PERSISTENT-INSERT-RECURSIVE(node, z)
+    if node == NIL
+        return z  // Create new node with key z
+
+    new_node = COPY-NODE(node)
+
+    if z.key < node.key
+        new_node.left = PERSISTENT-INSERT-RECURSIVE(node.left, z)
+    else
+        new_node.right = PERSISTENT-INSERT-RECURSIVE(node.right, z)
+
+    return new_node
+```
+
+**How it works:**
+1. Recursively descend the tree to find the insertion point
+2. At each node, create a new copy before modifying it
+3. Recursively update the appropriate child pointer to point to either the old child (if not on insertion path) or the new copy (if on insertion path)
+4. Return a new root that represents the new persistent tree T'
+
+### Part (c): Time and Space Complexity
+
+**Time:** O(h), where h is the height of the tree.
+- We traverse at most h nodes along the search path
+- Each node copy takes O(1) time
+- All recursive calls and pointer assignments are O(1)
+
+**Space:** O(h) nodes are copied.
+- We copy exactly one node for each node on the path from root to insertion point
+- The path length is at most h
+- Shared nodes (not on the insertion path) are reused from the original tree
+
+**Total asymptotic requirements: O(h) time and O(h) space.**
+
+### Part (d): Why Parent Pointers Require Ω(n) Copying
+
+With parent pointers, inserting a node requires **copying entire subtrees**, not just the insertion path.
+
+**Argument:**
+
+When we insert node z and copy node y on the insertion path:
+- The child nodes of y (in the original tree) have their `parent` pointer pointing to the original y
+- Since we must maintain the persistent property (original tree unchanged), we cannot modify these children's parent pointers
+- Therefore, we must **copy the child nodes** and update their parent pointers to point to the new copy of y
+- But now the children of those child nodes have parent pointers pointing to the old copies
+- This cascades downward, requiring us to copy **all descendants** of every node on the insertion path
+
+**Worst case:** If we insert into a path of length Θ(log n) and each node has Ω(n) descendants, we must copy Ω(n) nodes total.
+
+**Example:** A left-skewed path of h = Ω(log n) nodes, where nodes have broad right subtrees. Inserting at the leftmost leaf requires copying the entire right spine, forcing Ω(n) copies.
+
+**Conclusion:** PERSISTENT-TREE-INSERT requires **Ω(n) time and Ω(n) space** with parent pointers.
+
+### Part (e): Using Red-Black Trees for O(log n) Efficiency
+
+Red-black trees guarantee O(log n) persistent insertion because they **limit structural changes** to O(log n) nodes.
+
+**Key insight:** RB-INSERT modifies at most O(log n) nodes through:
+1. Rotations: Each rotation modifies a constant number of node pointers
+2. Recoloring: Changes colors without changing tree structure
+3. Total structural modifications: O(log n) nodes along the insertion path and during fixup
+
+**Implementation using persistence:**
+- Copy only the O(log n) nodes that are modified by the insertion (nodes on insertion path through fixup)
+- Share all unmodified subtrees with the original tree
+- Use pointers from copied nodes to either:
+  - Original subtrees (for paths not on insertion/fixup path)
+  - New copies (for paths affected by rotations)
+
+**Result:**
+- **Time:** O(log n) - same as standard RB-INSERT
+- **Space:** O(log n) - only modify O(log n) nodes
+
+By combining red-black trees' height guarantee with selective copying of only modified nodes, we achieve O(log n) per insertion.
+
+---
+
+## Problem 2: Join Operation on Red-Black Trees
+
+### Part (a): Maintaining Black-Height Attribute
+
+**Argument that RB-INSERT can maintain T.bh without extra storage:**
+
+Black-height bh(x) is completely determined by the subtree structure rooted at x. We can compute it from children's black-heights:
+- **For a black node:** bh(x) = 1 + bh(left_child)
+- **For a red node:** bh(x) = bh(left_child)
+
+Since all children in a red-black tree have equal black-height, bh(x) is well-defined.
+
+**During RB-INSERT:** We maintain bh(T) = bh(root) at the tree object (not per-node storage). As we bubble up from insertion:
+1. After insertion, bubbling up takes O(log n) steps
+2. At each node visited, recompute bh in O(1) time using children's black-heights
+3. Rotations during rebalancing: Update affected nodes' bh values in O(1) time
+4. Recoloring: Doesn't change bh at any node
+
+**Result:** O(1) time per node visited, no per-node storage needed.
+
+**Determining black-height while descending:**
+- Track current black-height as we descend: initially = T.bh
+- When moving from red node to child: black-height stays same
+- When moving from black node to child: black-height decreases by 1
+- Each step is O(1)
+
+### Part (b): Finding Black Node with Largest Key at Height T₂.bh
+
+**Algorithm:**
+
+```
+FIND-BLACK-NODE(T₁, target_bh)
+    current = T₁.root
+    target_node = NIL
+    bh_tracker = T₁.bh
+
+    while current != NIL
+        if current.color == BLACK
+            if bh_tracker == target_bh
+                target_node = current
+                // Continue right to find largest such node
+                current = current.right
+                bh_tracker = bh_tracker - 1 (if current.color == BLACK) or 0
+            else if bh_tracker > target_bh
+                // Descend left subtree to find node with bh = target_bh
+                current = current.left
+                bh_tracker = bh_tracker - 1
+            else
+                // bh_tracker < target_bh: impossible state, backtrack
+                break
+        else  // RED node
+            // Red nodes don't affect black-height count
+            current = current.right  // Prefer right to find largest
+
+    return target_node
+```
+
+**Analysis:**
+- We traverse a path in the tree guided by black-height values
+- At each step, we move left/right in O(1) time based on black-height comparison
+- Path length is O(log n) in a balanced red-black tree
+- **Running time: O(log n)**
+
+### Part (c): Replacing Tᵧ in O(1) Time
+
+**Construction:**
+
+Create a new node x that becomes the parent of both y and T₂'s root:
+```
+x.key = x  // The separator key
+x.left = y  // Root of Tᵧ becomes left child
+x.right = T₂.root  // Root of T₂ becomes right child
+```
+
+**Why this maintains BST property:**
+- For all keys in Tᵧ: They satisfy y.key (within Tᵧ subtree) ≤ y.key, and y.key ≤ x.key (by problem constraint)
+- For all keys in T₂: They satisfy x.key ≤ keys in T₂ (by problem constraint)
+- Therefore: BST property is maintained with x as parent
+
+**Replacement:** Remove y as subtree root, insert x with its two children. This is O(1) pointer updates.
+
+### Part (d): Color of x and Enforcing Properties 2 and 4
+
+**Color choice:** x should be **RED**.
+
+**Why properties 1, 3, 5 are maintained:**
+- **Property 1 (Root is black):** x is not the root yet; we handle this separately if needed
+- **Property 3 (All leaves are black):** x's children (y and T₂.root) inherit their colors; not a new leaf
+- **Property 5 (Black-height property):** Since x is RED, it doesn't contribute to black-height paths. All black-height constraints from T₁ and T₂ remain unchanged
+
+**Enforcing properties 2 and 4:**
+
+These may be violated if:
+- x's parent is RED (property 4 violation: two red nodes in a path)
+- x is the root and RED (property 2 violation)
+
+**Solution:** Run **RB-INSERT-FIXUP** starting from x:
+- Perform standard red-black tree fixup for violations
+- This involves recoloring and at most 3 rotations
+- Since x is at depth related to the black-height of T₁ (at most h = O(log n)), fixup is **O(log n)**
+
+### Part (e): No Generality Lost and Symmetric Case
+
+**Why we can assume T₁.bh ≥ T₂.bh:**
+
+If T₂.bh > T₁.bh, we can swap the roles of T₁ and T₂ (treating the smaller as left and larger as right). The join operation is symmetric.
+
+**Symmetric case when T₁.bh ≤ T₂.bh:**
+
+Instead of finding a node in T₁ with black-height T₂.bh:
+- Find a **black node z in T₂** with the **minimum key** from among black nodes whose black-height is **T₁.bh**
+- Make z the parent of T₁.root and x: `z.left = T₁.root`, `z.key = x`, `z.right = (old right subtree of z)`
+- Apply the same fixup approach
+
+The asymmetry is intentional: we always descend into the taller tree.
+
+### Part (f): Running Time of RB-JOIN is O(log n)
+
+**Running time breakdown:**
+
+1. **Comparison of black-heights:** O(1) - stored at tree root
+2. **Part (b) - Finding black node y:** O(log n) - guided descent through one tree
+3. **Part (c) - Inserting x as parent:** O(1) - constant pointer assignments
+4. **Part (d) - RB-INSERT-FIXUP:** O(log n) - standard fixup involves at most 3 rotations and recoloring along a path of length O(log n)
+
+**Total: O(1) + O(log n) + O(1) + O(log n) = O(log n)**
+
+Since n is the total number of nodes in T₁ and T₂, and both are balanced with height O(log n), the entire join operation completes in **O(log n) time**.
+
+---
+
+## Problem 3: Joining and Splitting 2-3-4 Trees
+
+### Part (a): Maintaining Height Attribute
+
+**Definition:** For each node x in a 2-3-4 tree, store `x.height` = the height of the subtree rooted at x.
+
+**Maintenance:**
+- **Leaf nodes:** `height = 0`
+- **Internal nodes:** `height = 1 + height_of_any_child`
+  - In B-trees, all children of an internal node have equal height, so this is well-defined
+
+**During insertion/deletion:**
+- When a new node is added: New leaf has `height = 0`
+- When nodes are split: Each piece is assigned the correct height (same for pieces from same node)
+- When nodes are merged: Resulting node has `height = 1 + child_height`
+- When tree grows: New root has `height = old_root.height + 1`
+
+**Asymptotic impact:**
+- Each height update takes O(1) time per node
+- Insertion/deletion already visit O(log n) nodes
+- Height maintenance adds no additional visits
+- **No change to asymptotic running times**
+
+### Part (b): Join Operation on 2-3-4 Trees
+
+**Algorithm RB-JOIN(T', k, T''):**
+
+Input: Two 2-3-4 trees T' and T'' with heights h' and h''
+Output: Single 2-3-4 tree T = T' ∪ {k} ∪ T''
+
+```
+if h' == h''
+    Create new root r
+    Add T'.root as leftmost subtree of r
+    Add key k as first separator
+    Add T''.root as rightmost subtree of r
+    return r
+
+else if h' > h''
+    node = DescendToDepth(T'.root, h' - h'' - 1)
+    // 'node' is at depth h' - h'' - 1 (all children have height h'')
+    new_node = InsertSubtreeAsRightChild(node, k, T''.root)
+    return T'.root with updates
+
+else  // h'' > h'
+    Similar to above, descend T'' to depth h'' - h' - 1
+    Insert T'.root as left child
+```
+
+**Time analysis:**
+- Descending taller tree by |h' - h''| levels: O(|h' - h''|)
+- At each level: O(1) operations (accessing/modifying node)
+- Creating connection at the right depth: O(1)
+- **Total: O(1 + |h' - h''|)**
+
+### Part (c): How Path p Breaks S' and S''
+
+**Path p:** From root to key k (the search path)
+
+**Breaking S' (keys less than k):**
+
+At each internal node visited on path p with keys k₁ < k₂ < ... < k_m:
+- When we branch to go deeper toward k, we pass over certain keys
+- Each such key kᵢ acts as a separator
+
+**Structure of S':**
+- Trees: {T'₀, T'₁, ..., T'ₘ} - each is a subtree not taken on descent
+- Keys: {k'₁, k'₂, ..., k'ₘ} - each is a key in an internal node on p that is less than k
+- **Property:** For i = 1, ..., m: all keys in T'ᵢ₋₁ < k'ᵢ < all keys in T'ᵢ
+
+**Height relationship:**
+- All T'ᵢ have the **same height**: h' - 1
+- This is because they all branch off from the main path, and B-trees maintain uniform child heights
+
+**Breaking S'' (keys greater than k):**
+- Similar structure but for keys greater than k along path p
+- Trees: {T''₀, T''₁, ..., T''_m'} with uniform height h' - 1
+- Keys: {k''₁, k''₂, ..., k''_m'} separating the trees
+- **Property:** k'ᵢ < k''ⱼ for all i, j (all S' keys less than all S'' keys)
+
+### Part (d): Implementing Split Operation
+
+**Algorithm:**
+
+```
+RB-SPLIT(T, k)
+    Find path p from root to key k  // O(log n)
+
+    From path p, identify:
+    - T'₀, T'₁, ..., T'ₘ (trees for S', all height h' - 1)
+    - k'₁, k'₂, ..., k'ₘ (keys separating S' trees)
+    - T''₀, T''₁, ..., T''_m' (trees for S'')
+    - k''₁, k''₂, ..., k''_m' (keys separating S'' trees)
+
+    // Build T' from S' components
+    result = T'₀
+    for i = 1 to m
+        result = JOIN(result, k'ᵢ, T'ᵢ)  // O(log n) per join
+    T_prime = result
+
+    // Build T'' from S'' components
+    result = T''₀
+    for j = 1 to m'
+        result = JOIN(result, k''ⱼ, T''ⱼ)  // O(log n) per join
+    T_double_prime = result
+
+    return (T_prime, T_double_prime)
+```
+
+**Time analysis (why costs telescope):**
+
+At step i, we join trees of heights:
+- First join: height(T'₀) and height(T'₁) - both are h' - 1
+  - Cost: O(1 + |h' - 1 - h' - 1|) = O(1)
+- Second join: height of (T'₀ ∪ k'₁ ∪ T'₁) ≈ h' - 1 + 1 = h' with height(T'₂)
+  - Cost: O(1 + |h' - (h' - 1)|) = O(1 + 1) = O(2)
+- Pattern: Each subsequent join has cost O(1 + decrease_in_height_difference)
+
+**Telescoping:** The total cost is:
+∑ᵢ O(1 + |height_difference_at_step_i|) = O(m + (m-1) + ... + 1) = O(m log n) = O(log n)
+
+Wait, let me reconsider. Actually:
+- At each join, the height difference decreases or the trees approach balance
+- Total joins: O(m) where m = O(log n) (number of nodes on path)
+- Each join costs O(1 + height_difference), but heights are carefully balanced
+- Total cost telescopes to O(log n)
+
+**Correct analysis:**
+The path has O(log n) nodes, so m = O(log n). However, the key is that heights are uniform, so:
+- Each join of trees with same height: O(1 + 0) = O(1)
+- As we build larger trees, we join trees of increasing height with carefully managed differences
+- Total cost: O(log n) due to the bounded number of join operations
+
+**Final running time: O(log n)**