Add in-progress notes for 2026.07.21

.gitignore

@@ -0,0 +1,3 @@
+.idea
+
+

notes-and-examples/2026.05.27/homework.py

@@ -60,8 +60,12 @@ def rotate_left(node: Node) -> Node:
     new_right_child = node.right.left
 
     node.right = new_right_child
-    node.parent = new_parent
+    if new_right_child is not None:
+        new_right_child.parent = node
+
     new_parent.left = node
+    node.parent = new_parent
+    new_parent.parent = original_parent
 
     if original_parent is not None:
         if original_parent.left == node:
@@ -81,8 +85,12 @@ def rotate_right(node: Node) -> Node:
     new_left_child = node.left.right
 
     node.left = new_left_child
-    node.parent = new_parent
+    if new_left_child is not None:
+        new_left_child.parent = node
+
     new_parent.right = node
+    node.parent = new_parent
+    new_parent.parent = original_parent
 
     if original_parent is not None:
         if original_parent.left == node:
@@ -115,48 +123,86 @@ def trees_equal(a: Node | None, b: Node | None) -> bool:
 
 
 if __name__ == "__main__":
+    # note that matching is done by value for simplicity here
 
     def test_rotate_right_example():
         # The worksheet example: right rotate on 10.
         root = Node(
             10,
-            left=Node(5, left=Node(3, left=Node(2)), right=Node(7)),
+            left=Node(
-            right=Node(15),
+                5,
+                left=Node(3, left=Node(2, parent=Node(3)), parent=Node(5)),
+                right=Node(7, parent=Node(5)),
+                parent=Node(10),
+            ),
+            right=Node(15, parent=Node(10)),
         )
         expected = Node(
             5,
-            left=Node(3, left=Node(2)),
+            left=Node(3, left=Node(2, parent=Node(3)), parent=Node(5)),
-            right=Node(10, left=Node(7), right=Node(15)),
+            right=Node(
+                10,
+                left=Node(7, parent=Node(10)),
+                right=Node(15, parent=Node(10)),
+                parent=Node(5),
+            ),
         )
         return rotate_right(root), expected
 
+
     def test_rotate_left_example():
         # The worksheet example: left rotate on 16.
-        root = Node(16, right=Node(33, left=Node(22), right=Node(55)))
+        root = Node(
-        expected = Node(33, left=Node(16, right=Node(22)), right=Node(55))
+            16,
+            right=Node(
+                33,
+                left=Node(22, parent=Node(33)),
+                right=Node(55, parent=Node(33)),
+                parent=Node(16),
+            ),
+        )
+        expected = Node(
+            33,
+            left=Node(16, right=Node(22, parent=Node(16)), parent=Node(33)),
+            right=Node(55, parent=Node(33)),
+        )
         return rotate_left(root), expected
 
+
     def test_rotate_right_two_nodes():
         # Smallest case: a root with only a left child.
-        return rotate_right(Node(2, left=Node(1))), Node(1, right=Node(2))
+        return rotate_right(Node(2, left=Node(1, parent=Node(2)))), Node(1, right=Node(2, parent=Node(1)))
+
 
     def test_rotate_left_two_nodes():
         # Smallest case: a root with only a right child.
-        return rotate_left(Node(1, right=Node(2))), Node(2, left=Node(1))
+        return rotate_left(Node(1, right=Node(2, parent=Node(1)))), Node(2, left=Node(1, parent=Node(2)))
+
 
     def test_round_trip():
         # A left rotate undone by a right rotate restores the original tree.
-        original = Node(1, left=Node(0), right=Node(3, left=Node(2), right=Node(4)))
+        original = Node(1, left=Node(0, parent=Node(1)),
-        expected = Node(1, left=Node(0), right=Node(3, left=Node(2), right=Node(4)))
+                        right=Node(3, left=Node(2, parent=Node(3)), right=Node(4, parent=Node(3)), parent=Node(1)))
+        expected = Node(1, left=Node(0, parent=Node(1)),
+                        right=Node(3, left=Node(2, parent=Node(3)), right=Node(4, parent=Node(3)), parent=Node(1)))
         return rotate_right(rotate_left(original)), expected
 
+
     def test_rotate_with_parent():
         # run it on 33 instead of the root
         root = Node(16, right=Node(33, left=Node(22), right=Node(55)))
+        root.right.parent = root
+        root.right.left.parent = root.right
+        root.right.right.parent = root.right
         original = root.right
-        expected = Node(55, left=Node(33, left=Node(22)))
+        expected = Node(
+            55,
+            left=Node(33, left=Node(22, parent=Node(33)), parent=Node(55)),
+            parent=Node(16),
+        )
         return rotate_left(original), expected  # pyright: ignore[reportArgumentType]
 
+
     tests = [
         ("rotate_right on worksheet example (10)", test_rotate_right_example),
         ("rotate_left on worksheet example (16)", test_rotate_left_example),

notes-and-examples/2026.05.27/overview.md

@@ -0,0 +1,16 @@
+## Outline
+
+- Review homework and do some tree exercises
+- Introduce tree balancing
+- Introduce red-black trees
+- Basic properties
+    - Root property: the root is black
+    - External property: every nil child pointer is considered a black node
+    - Internal property: children and parents of a red node are black
+    - Depth property: all nodes have the same black depth
+- Insertion and balancing algorithms
+- Assignment: implement red-black tree balancing
+
+## Resources
+
+- [Red-black tree - Wikipedia](https://en.wikipedia.org/wiki/Red–black_tree)

notes-and-examples/2026.06.12/overview.md

@@ -0,0 +1,41 @@
+## Outline
+
+- Write a red-black tree with rebalancing
+- Assignment: start on the red-black tree (moved to 2026.07.01)
+    - Requirement: you must use the left- and right-rotation code in the previous assignment
+    - You may change the code to fit the class
+- Next week: introduce graphs and graph algorithms
+
+## Git reference
+
+There are two repositories to keep track of:
+
+- [My code](https://gitea.bchen.dev/brendan/dsa-tutoring)
+  - Get notes and sample code
+- [Your code](https://gitea.bchen.dev/ethan/dsa-homework)
+  - Submit your homework
+
+To submit homework, add all new files and changes to your code:
+
+```
+git add .
+```
+
+Commit with a message:
+
+```
+git commit -m "Hello World"
+```
+
+Push changes:
+
+```
+git push
+```
+
+
+To pull new changes to the notes and sample code, run this in the `dsa-tutoring` directory:
+
+```
+git pull
+```

notes-and-examples/2026.07.01/overview.md

@@ -0,0 +1,41 @@
+## Outline
+
+- Review tree rotations and red-black trees
+- Properties of a red-black tree
+    - Top (root) node is black
+    - The children and parent of a red node are black
+    - Null nodes are colored black
+    - The path from any particular node to a null node must contain the
+      same number of black nodes
+- Dive into the steps for rebalancing a red-black tree
+
+Also see the accompanying whiteboard pictures.
+
+![Rotations](rotations.png)
+
+![Red-black trees](red-black-trees.png)
+
+## Assignment
+
+**Clarifications on the red-black tree example:**
+
+- In your assignment, you should first color a node _red_ when inserting it.
+    - Then, check for red-red violations (the second rule). Then, rebalance if necessary.
+- If the uncle of the inserted node (parent → parent → right child) is red,
+  you'll want to do two things:
+    - Color the parent, uncle, and grandparent in a way which preserves all the rules
+      (which ones are colored which, are an exercise left to you)
+    - _Recursively_ run the rebalancing algorithm on the grandparent
+- Otherwise, your task is to figure out the correct conditions in which
+  each of the rebalancing algorithms apply.
+
+Copy 2026.07.01/homework.py into your own [Git repository](https://gitea.bchen.dev/ethan/dsa-homework),
+with a new folder for the date. Implement the `rebalance_from_just_inserted`
+function.
+
+Don't change the test cases, but use them to inform how you implement your
+code. Try to pass all the test cases.
+
+Follow the instructions from last session (2026.06.12) to commit and push
+your changes to Gitea.
+

notes-and-examples/2026.07.01/red_black_tree.py

@@ -0,0 +1,494 @@
+import traceback
+
+class Node:
+    def __init__(self, value, left=None, right=None, parent=None, is_red=False):
+        self.value = value
+        self.left: Node | None = left
+        self.right: Node | None = right
+        self.is_red = is_red
+
+        # required to traverse the tree and perform various checks
+        self.parent: Node | None = parent
+
+class RedBlackTree:
+    def __init__(self):
+        self.root: Node | None = None
+
+    def insert(self, value):
+        # assumption: inserting the same value twice will
+        # insert nothing the second time
+
+        node = Node(value)
+        if self.root is None:
+            # the root is always black (Node defaults to black)
+            self.root = node
+            return
+
+        # insert like a normal binary search tree
+        prev: Node | None = None
+        current = self.root
+        while current:
+            if current.value == value:
+                return
+            prev = current
+            if current.value > value:
+                current = current.left
+            elif current.value < value:
+                current = current.right
+
+        # a newly inserted (non-root) node is always red, and it needs a
+        # parent pointer so rebalancing can walk back up toward the root
+        node.is_red = True
+        node.parent = prev
+        if prev and prev.value > value:
+            prev.left = node
+        elif prev and prev.value < value:
+            prev.right = node
+
+        self.rebalance_from_just_inserted(node)
+
+    def delete(self, value, root=None, parent=None):
+        # does nothing if the value doesn't exist
+        if parent is None:
+            root = self.root
+            if root is None:
+                return
+
+        if root.value == value:
+            new_root: Node | None
+            if root.left is None:
+                new_root = root.right
+            elif root.right is None:
+                new_root = root.left
+            else:
+                # both left and right subtrees exist
+                # make left subtree root the tree root, and re-attach
+                # the right subtree
+                right_subtree_root = root.right
+                new_root = root.left
+
+                prev = None
+                current = new_root
+                while current:
+                    prev = current
+                    current = current.right
+                prev.right = right_subtree_root
+
+            if parent is None:
+                self.root = new_root
+            else:
+                if parent.value > root.value:
+                    parent.left = new_root
+                else:
+                    parent.right = new_root
+        elif root.value > value:
+            self.delete(value, root.left, root)
+        elif root.value < value:
+            self.delete(value, root.right, root)
+
+    def rebalance_from_just_inserted(self, node: Node | None):
+        if node is None:
+            return
+
+        parent = node.parent
+        if parent is not None and not parent.is_red:
+            return
+        elif parent is None:
+            node.is_red = False
+            return
+
+        grandparent = parent.parent
+        if grandparent is None:
+            return
+
+        uncle = grandparent.right if grandparent.left == parent else grandparent.left
+
+        if uncle is not None and uncle.is_red:
+            uncle.is_red = False
+            parent.is_red = False
+            grandparent.is_red = True
+
+            self.rebalance_from_just_inserted(grandparent)
+            return
+
+        if grandparent.left == parent and parent.right == node:
+            node, parent = parent, self.rotate_left(parent)
+        elif grandparent.right == parent and parent.left == node:
+            node, parent = parent, self.rotate_right(parent)
+
+        new_grandparent: Node | None = None
+        if grandparent.left == parent and parent.left == node:
+            new_grandparent = self.rotate_right(grandparent)
+        elif grandparent.right == parent and parent.right == node:
+            new_grandparent = self.rotate_left(grandparent)
+
+        if grandparent == self.root and new_grandparent is not None:
+            self.root = new_grandparent
+            self.root.is_red = False
+            grandparent.is_red = True
+
+
+    def rotate_left(self, node: Node) -> Node:
+        if node.right is None:
+            return node
+
+        original_parent = node.parent
+        new_parent = node.right
+        new_right_child = node.right.left
+
+        node.right = new_right_child
+        if new_right_child is not None:
+            new_right_child.parent = node
+
+        new_parent.left = node
+        node.parent = new_parent
+        new_parent.parent = original_parent
+
+        if original_parent is not None:
+            if original_parent.left == node:
+                original_parent.left = new_parent
+            elif original_parent.right == node:
+                original_parent.right = new_parent
+
+        return new_parent
+
+    def rotate_right(self, node: Node) -> Node:
+        if node.left is None:
+            return node
+
+        original_parent = node.parent
+        new_parent = node.left
+        new_left_child = node.left.right
+
+        node.left = new_left_child
+        if new_left_child is not None:
+            new_left_child.parent = node
+
+        new_parent.right = node
+        node.parent = new_parent
+        new_parent.parent = original_parent
+
+        if original_parent is not None:
+            if original_parent.left == node:
+                original_parent.left = new_parent
+            elif original_parent.right == node:
+                original_parent.right = new_parent
+
+        return new_parent
+
+    def visualize(self) -> str:
+        # renders the tree top-down with the root on top and branches
+        # ( / and \ ) drawn down to each child. spacing is computed so that
+        # subtrees never overlap, no matter the shape of the tree.
+        if self.root is None:
+            return "<empty tree>"
+
+        # render() returns, for the subtree rooted at `node`:
+        #   lines  - the block of text drawing the subtree
+        #   width  - how many characters wide that block is
+        #   height - how many lines tall that block is
+        #   middle - the column where this node's value is centred (so the
+        #            caller knows where to attach its branch)
+        def render(node: Node) -> tuple[list[str], int, int, int]:
+            # label shows the value plus its color: R (red) or B (black)
+            label = f"{node.value}{'R' if node.is_red else 'B'}"
+            label_width = len(label)
+
+            # leaf: just the value on a single line
+            if node.left is None and node.right is None:
+                return [label], label_width, 1, label_width // 2
+
+            # only a left child
+            if node.right is None:
+                lines, width, height, mid = render(node.left)
+                first = (mid + 1) * " " + (width - mid - 1) * "_" + label
+                second = mid * " " + "/" + (width - mid - 1 + label_width) * " "
+                shifted = [line + label_width * " " for line in lines]
+                return [first, second] + shifted, width + label_width, height + 2, width + label_width // 2
+
+            # only a right child
+            if node.left is None:
+                lines, width, height, mid = render(node.right)
+                first = label + mid * "_" + (width - mid) * " "
+                second = (label_width + mid) * " " + "\\" + (width - mid - 1) * " "
+                shifted = [label_width * " " + line for line in lines]
+                return [first, second] + shifted, width + label_width, height + 2, label_width // 2
+
+            # two children: render each side, then place this node between them
+            left_lines, left_w, left_h, left_mid = render(node.left)
+            right_lines, right_w, right_h, right_mid = render(node.right)
+            first = (
+                (left_mid + 1) * " "
+                + (left_w - left_mid - 1) * "_"
+                + label
+                + right_mid * "_"
+                + (right_w - right_mid) * " "
+            )
+            second = (
+                left_mid * " "
+                + "/"
+                + (left_w - left_mid - 1 + label_width + right_mid) * " "
+                + "\\"
+                + (right_w - right_mid - 1) * " "
+            )
+            # pad the shorter side so the two blocks line up row-for-row
+            if left_h < right_h:
+                left_lines += [left_w * " "] * (right_h - left_h)
+            elif right_h < left_h:
+                right_lines += [right_w * " "] * (left_h - right_h)
+            merged = [l + label_width * " " + r for l, r in zip(left_lines, right_lines)]
+            return (
+                [first, second] + merged,
+                left_w + right_w + label_width,
+                max(left_h, right_h) + 2,
+                left_w + label_width // 2,
+            )
+
+        lines, _, _, _ = render(self.root)
+        return "\n".join(lines)
+
+
+# =============== TEST CASES ===================
+#
+# These run automatically:   python red_black_tree.py
+#
+# Each case inserts ONE value into a hand-built red-black tree and checks the
+# result, printing the tree BEFORE, the ACTUAL tree after your rebalance, and
+# the EXPECTED tree. A case passes when the actual tree matches the expected
+# tree AND the result is still a valid red-black tree.
+#
+# HOW TO ADD YOUR OWN CASE:
+#   Append a Case(...) to the CASES list below with four fields:
+#       name     - short description
+#       before   - a function returning the tree BEFORE the insert (None = empty)
+#       insert   - the value to insert
+#       expected - a function returning the tree you expect AFTERWARD
+#   Build trees with node(value, "R" or "B", left=..., right=...). Any child
+#   you leave out is treated as a (black) nil leaf.
+
+from dataclasses import dataclass
+from typing import Callable
+
+
+def node(value, color, left=None, right=None):
+    """Build a Node with an explicit color ("R"/"B"), wiring parent pointers."""
+    n = Node(value, is_red=(color == "R"))
+    n.left = left
+    n.right = right
+    if left is not None:
+        left.parent = n
+    if right is not None:
+        right.parent = n
+    return n
+
+
+def tree(root):
+    """Wrap a root Node (or None) in a RedBlackTree."""
+    t = RedBlackTree()
+    t.root = root
+    return t
+
+
+@dataclass
+class Case:
+    name: str
+    before: Callable      # () -> Node | None
+    insert: int
+    expected: Callable    # () -> Node | None
+
+
+CASES = [
+    Case(
+        "Empty tree -> black root",
+        before=lambda: None,
+        insert=25,
+        expected=lambda: node(25, "B"),
+    ),
+    Case(
+        "Black parent -> new red child, no rebalancing",
+        before=lambda: node(25, "B"),
+        insert=15,
+        expected=lambda: node(25, "B", left=node(15, "R")),
+    ),
+    Case(
+        "Duplicate insert -> tree unchanged",
+        before=lambda: node(25, "B", left=node(15, "R")),
+        insert=15,
+        expected=lambda: node(25, "B", left=node(15, "R")),
+    ),
+    Case(
+        "Red uncle -> recolor (grandparent is the root)",
+        before=lambda: node(25, "B", left=node(15, "R"), right=node(35, "R")),
+        insert=10,
+        expected=lambda: node(
+            25, "B",
+            left=node(15, "B", left=node(10, "R")),
+            right=node(35, "B"),
+        ),
+    ),
+    Case(
+        "LL -> right-rotate the grandparent",
+        before=lambda: node(30, "B", left=node(20, "R")),
+        insert=10,
+        expected=lambda: node(20, "B", left=node(10, "R"), right=node(30, "R")),
+    ),
+    Case(
+        "RR -> left-rotate the grandparent",
+        before=lambda: node(30, "B", right=node(40, "R")),
+        insert=50,
+        expected=lambda: node(40, "B", left=node(30, "R"), right=node(50, "R")),
+    ),
+    Case(
+        "LR -> left-rotate parent, then right-rotate grandparent",
+        before=lambda: node(30, "B", left=node(20, "R")),
+        insert=25,
+        expected=lambda: node(25, "B", left=node(20, "R"), right=node(30, "R")),
+    ),
+    Case(
+        "RL -> right-rotate parent, then left-rotate grandparent",
+        before=lambda: node(30, "B", right=node(40, "R")),
+        insert=35,
+        expected=lambda: node(35, "B", left=node(30, "R"), right=node(40, "R")),
+    ),
+    Case(
+        "Cascade: recolor propagates up, then rotate near the root",
+        before=lambda: node(
+            11, "B",
+            left=node(
+                2, "R",
+                left=node(1, "B"),
+                right=node(7, "B", left=node(5, "R"), right=node(8, "R")),
+            ),
+            right=node(14, "B", right=node(15, "R")),
+        ),
+        insert=4,
+        expected=lambda: node(
+            7, "B",
+            left=node(
+                2, "R",
+                left=node(1, "B"),
+                right=node(5, "B", left=node(4, "R")),
+            ),
+            right=node(
+                11, "R",
+                left=node(8, "B"),
+                right=node(14, "B", right=node(15, "R")),
+            ),
+        ),
+    ),
+]
+
+
+def validate_rb(t):
+    """Return a list of red-black property violations ([] means valid)."""
+    problems = []
+    root = t.root
+    if root is None:
+        return problems
+    if root.is_red:
+        problems.append("root is red")
+    if root.parent is not None:
+        problems.append("root has a non-None parent pointer")
+
+    black_heights = set()
+    seen = set()
+
+    def check(n, low, high, black_count):
+        if n is None:
+            black_heights.add(black_count + 1)  # nil leaves count as black
+            return
+        if id(n) in seen:
+            # a proper tree never reaches the same node twice
+            problems.append(f"not a tree: node {n.value} reached twice (cycle or shared subtree)")
+            return
+        seen.add(id(n))
+        if low is not None and n.value <= low:
+            problems.append(f"BST order broken at {n.value}")
+        if high is not None and n.value >= high:
+            problems.append(f"BST order broken at {n.value}")
+        if n.is_red and ((n.left and n.left.is_red) or (n.right and n.right.is_red)):
+            problems.append(f"red node {n.value} has a red child")
+        if n.left is not None and n.left.parent is not n:
+            problems.append(f"broken parent pointer: left child {n.left.value} does not point back to {n.value}")
+        if n.right is not None and n.right.parent is not n:
+            problems.append(f"broken parent pointer: right child {n.right.value} does not point back to {n.value}")
+        nb = black_count + (0 if n.is_red else 1)
+        check(n.left, low, n.value, nb)
+        check(n.right, n.value, high, nb)
+
+    check(root, None, None, 0)
+    if len(black_heights) > 1:
+        problems.append(f"unequal black-heights on paths to nil: {sorted(black_heights)}")
+    return problems
+
+
+def _indent(text: str) -> str:
+    return "\n".join("    " + line for line in text.split("\n"))
+
+
+def run_tests():
+    passed = 0
+    for i, case in enumerate(CASES, 1):
+        print("=" * 64)
+        print(f"CASE {i}: {case.name}")
+        print("=" * 64)
+
+        t = tree(case.before())
+        print("BEFORE:")
+        print(_indent(t.visualize()))
+        print(f"\n  insert({case.insert})\n")
+
+        # Everything the student's code can affect is inside this guard, so a
+        # buggy rebalance (even one that builds a cyclic/broken tree) fails just
+        # this case instead of aborting the whole suite.
+        try:
+            t.insert(case.insert)
+            actual = t.visualize()
+            expected_tree = tree(case.expected())
+            expected = expected_tree.visualize()
+            violations = validate_rb(t)
+            expected_problems = validate_rb(expected_tree)
+        except RecursionError:
+            print("  RESULT: ERROR - hit maximum recursion depth.")
+            print("    Your rebalance likely created a cycle or otherwise broke the")
+            print("    tree structure (a child pointing back up at an ancestor).\n")
+            continue
+        except Exception:
+            print(f"  RESULT: ERROR - see stack trace below\n")
+            print(traceback.format_exc())
+            print("")
+            continue
+
+        # Guard the author (you) against a typo when adding a new case: the
+        # 'expected' tree should itself be a valid red-black tree.
+        if expected_problems:
+            print("  RESULT: BAD TEST - the 'expected' tree is not a valid red-black tree:")
+            for p in expected_problems:
+                print(f"    - {p}")
+            print()
+            continue
+
+        print("ACTUAL (what your code produced):")
+        print(_indent(actual))
+        print("\nEXPECTED:")
+        print(_indent(expected))
+        print()
+
+        if actual == expected and not violations:
+            print("  RESULT: PASS")
+            passed += 1
+        else:
+            print("  RESULT: FAIL")
+            if actual != expected:
+                print("    - actual tree does not match the expected tree")
+            for v in violations:
+                print(f"    - red-black property broken: {v}")
+        print()
+
+    print("=" * 64)
+    print(f"{passed}/{len(CASES)} cases passed")
+    print("=" * 64)
+
+
+if __name__ == "__main__":
+    run_tests()

notes-and-examples/2026.07.08/overview.md

@@ -0,0 +1,10 @@
+## Outline
+
+Review the algorithm for a red-black tree rebalancing.
+
+Next week, we will begin an introduction of graphs.
+
+## Assignment
+
+See the `red_black_tree` assignment from last week 2026.07.01.
+

notes-and-examples/2026.07.15/graphs-intro.md

@@ -0,0 +1,70 @@
+## Fundamentals
+
+- A *node* or *vertex* contains a data point
+- An *edge* is what connects one node to another
+
+A graph is just a collection of vertices and edges.
+
+> [!QUESTION] What is a real-life example of a simple graph with only vertices and edges?
+
+Some additional properties we can put on a graph:
+
+- A *weight* is a numerical value that can be assigned to an edge
+- We can also assign a *direction* to an edge, such that it only points from one node to another, not the other way around
+
+We can, of course, combine both of these properties too.
+
+> [!QUESTION] What's something we can model with weights in a graph? What about with directional edges?
+
+Going forward, we'll use $V$ to represent the number of vertices in a graph, and $E$ to represent the number of edges.
+
+## Representing a graph
+
+The *adjacency list* stores a list of connected vertices for each node, and it can fit in a dictionary or hash map structure.
+
+```
+1 -> 2, 4
+2 -> 1, 3
+3 -> 2
+4 -> 1
+```
+
+> [!QUESTION] How would you draw out this graph?
+
+> [!QUESTION] What would this mapping look like if we wanted to add weights? What about directional edges?
+
+The *adjacency matrix* is a 2D array where each position `arr[x][y]` represents an edge, and `x` and `y` each represent a node.
+
+```
+[
+    [0, 1, 0, 1],
+    [1, 0, 1, 0],
+    [0, 1, 0, 0],
+    [1, 0, 0, 0]
+]
+```
+
+> [!QUESTION] How would you draw out this graph? What would it look like as an adjacency list?
+
+> [!QUESTION] How do we add weights and/or directional edges to this graph?
+
+## Categorizing graphs
+
+We say a graph is *directed and acyclic*, or a *directed acyclic graph (DAG)*, if there are no cycles formed using the directional edges.
+
+> [!QUESTION] What's something we can model with a DAG?
+
+> [!QUESTION] What's another data structure we went over that also classifies as a DAG?
+
+A graph is *dense* if $E$ is closer to $V^2$, and *sparse* if $E$ is closer to $V$.
+
+> [!QUESTION] What's the implication for the graph if $E$ is closer to $V^2$, i.e. how is it different compared to a sparse graph?
+
+> [!QUESTION] What's the maximum number of edges we can have in a graph with no self-loops, relative to the number of vertices $V$?
+
+A graph can contain *self-loops* (a loop from a vertex to itself). A graph can also have *multiple edges* going from one vertex to another.
+
+> [!QUESTION] How do we represent a self-loop using an adjacency matrix?
+
+Finally, a graph is *connected* if every vertex is reachable from every other vertex.
+

notes-and-examples/2026.07.15/overview.md

@@ -0,0 +1,18 @@
+## Outline
+
+- Go over the homework
+- Go over the basics of graphs, see [Introduction to graphs](./graphs-intro.md)
+
+![Graphs introduction whiteboard](./graphs.png)
+
+To open the `.excalidraw` file:
+
+- Launch https://excalidraw.com
+- Click the top left menu, then select "Open"
+- Select the `.excalidraw` file from the file picker
+
+## Assignment
+
+- Try to complete the red-black implementation from last week
+- Complete the exercise for the 2D array (see the picture)
+

notes-and-examples/2026.07.21/breadth-first-search.md

@@ -0,0 +1,13 @@
+## Breadth-first search (BFS)
+
+In the notes on [depth-first search](./depth-first-search.md), we mention that the difference between DFS and BFS is the order in which nodes are traversed.
+
+Using the same example, what would a breadth-first traversal look like if we start at vertex 0 in this graph?
+
+```
+    0 --- 1
+    |     |
+    3 --- 2
+```
+
+Can we also formalize an algorithm for breadth-first search?

notes-and-examples/2026.07.21/depth-first-search.md

@@ -0,0 +1,46 @@
+## Traversal introduction
+
+Many algorithms that involve graphs must involve some way to traverse the elements of a graph. The two simplest ways of traversal are depth-first search (DFS) and [breadth-first search (BFS)](https://en.wikipedia.org/wiki/Breadth-first_search). The major difference here is the *order* in which nodes are traversed.
+
+If we start from vertex 0 in the tree, in what order would you expect depth-first search to traverse the nodes? (There are multiple correct answers!)
+
+```
+        0
+       / \
+      1   2
+     / \
+    3   4
+```
+
+Note that a single traversal step checks for already-visited nodes. So, if the path is `0 -> 1 -> 3`, the path cannot become `0 -> 1 -> 3 -> 1`.
+
+What about starting from vertex 0 in this graph?
+
+```
+    0 --- 1
+    |     |
+    3 --- 2
+```
+
+What about this one?
+
+```
+        0
+       / \
+      1   5
+     / \   \
+    2   3   6
+     \ /   /
+      4   7
+       \ /
+        8
+```
+
+## Formalizing the algorithm
+
+Based on these examples, can we create a formal algorithm that takes a starting node, producing a valid traversal path for all three graphs?
+
+Some starter questions:
+
+- Could recursion help us here?
+- What are some ways we can track already-visited nodes? What's the most *time-efficient* way to do so?

notes-and-examples/2026.07.21/main.py

@@ -0,0 +1,25 @@
+# Adjacency list
+first_graph = {
+    1: [3, 20],
+    2: [20],
+    3: [1, 20],
+    20: [2, 3]
+}
+
+# First exercise: go from 1, to 20, to 2
+first_graph[first_graph[1][1]][0]
+
+# pass visited by reference
+def depth_first_search(graph: dict, start, visited=set()):
+    print(start)
+    if len(visited)==len(graph):
+        return visited
+
+    for x in graph[start]:
+        if x not in visited:
+            visited.add(x)
+            return depth_first_search(graph, x, visited)
+    pass
+    
+
+depth_first_search(first_graph, 1)

notes-and-examples/2026.07.21/overview.md

@@ -0,0 +1,11 @@
+## Outline
+
+To make our knowledge of graphs useful, we'll go over our first traversal method today: [depth-first search](./depth-first-search.md). We will also go over [breadth-first search](./breadth-first-search.md), and how both algorithms can be useful.
+
+Assignment: write functions for depth-first search and breadth-first search. Because there is no starter file, the constraints are listed below:
+
+- You may use either an adjacency list or adjacency matrix to represent your graph
+- You should demonstrate that your algorithm works by running it through some test cases
+    - For each test case, specify the graph, starting point, and expected output(s)
+
+Bonus: can you output *all* the valid paths for both DFS and BFS in a particular case?