Graphs 2

1. Strongly Connected Components

In a directed graph, an SCC is a maximal set of vertices where every vertex can reach every other. Upon compressing all SCC's into their respective nodes, the resulting graph is a DAG.

• Turning directed graphs into DAGs enable other techniques we have seen before such as DP/topo sort.

• Algorithms to perform the compression:

  • Kosaraju: two DFS passes (on graph and reversed graph)
  • Tarjan: one DFS pass with low-link values

• For SCC's and 2-SAT, you can check out the implementation in KACTL

Complexity

• O(n + m)

Practice:

• Planets and Kingdoms (CSES)
• Coin Collector (CSES)
• Grass Cownoisseur (USACO)

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2. 2-SAT

A 2-SAT instance in CNF can be stated as a bunch of OR clauses with 2 variables AND'd together. For example a possible instance of 2-SAT is:

Determine whether there is an assignment of truth values to $x$, $y$, and $z$ such that the following statement is true:

$$(x \lor y) \land (x \lor \neg z) \land (\neg y \lor \neg z)$$

We can transform 2-SAT by creating an implication graph and determining the SCC's of this graph. For each variable in the 2-SAT instance we create $2$ nodes - one for itself and one for its complement. In the example above, the graph would have $6$ nodes: $x, \neg x, y, \neg y, z, \neg z$.

Then we write each OR clause as an implication and its contrapositive. For example $x \lor y$ is equivalent to $(\neg x \rightarrow y)$ which is equivalent to $(\neg y \rightarrow x)$. We draw these two edges as directed edges in the original graph for each of the clauses.

Now, one thing that would clearly prove a 2-SAT instance is unsatisfiable is if for some variable $x$, both $x \rightarrow \neg x$ and $\neg x \rightarrow x$ via some chain of implications. Confirming that this doesn't happen turns out to be sufficient to show that the 2-SAT instance is satisfiable. A rough proof idea of this is that if a variable $x$ implies $\neg x$ via a chain, we set $x$ to be false and $\neg x$ to be true. This assignment will never cause a contradiction to occur due to the nature of how the implication graph is constructed (details to be covered in class).

Problems:

• Giant Pizza (CSES)
• Corporate Retreat (Rocky Mountain Regional 2025)
• Coprime Solitaire (AC)
• Babysitting (CF)

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3. Union-Find (Disjoint Set Union, DSU)

DSU maintains a partition of nodes into connected components under merges.

KACTL Implementation

• Union by rank or size gives a complexity of $O(\log n)$.
• Use path compression as well for near-constant time. (amortized ~ O(α(n))) per operation

Practice:

• Road Construction (CSES)
• Graph Destruction (AC)
• Ada and Branches (SPOJ)

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4. Minimum Spanning Trees (MST)

Given a connected weighted undirected graph, an MST is a tree that connects all nodes with minimum sum of all edge weights.

Cut property: For any cut, one of the lightest edges crossing it must be in the MST. Moreover taking any of the lightest edges will result in a MST. This is the main property used to reason about MST algorithms and problems.

Other properties:

Cycle property: For any cycle in the graph, if an edge is the unique maximum, it will not be included in any MST.

Edge Removal: The two trees created by removing an edge from an MST are MST's of their respective subgraphs.

Two standard algorithms:

Kruskal: Process edges in increasing order by weight and include it in the MST if it connects two different components. Use a DSU to maintain the connected components.

Prim:: Start with a single node and grow the MST one edge/vertex at a time. Repeatedly pick the smallest edge connected a vertex in the current tree to one outside it. Use a heap over the cut edges to efficiently query the minimum at each step.

Can you use the cut property to prove the correctness of both algorithms?

Complexity:

• Kruskal: O(m log m) (sorting dominates)
• Prim: O(m log n) with heap

Practice:

• Road Reparation (CSES)
• Moo Network (USACO)
• MST Edge Cost