Lecture Notes: Digraphs and Relations

Preliminaries

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Tuesday Review

Question: Draw digraphs which are:

• Strongly connected and acyclic.
• Strong connected and not acyclic.
• Not strongly connected and acyclic.
• Not strongly connected and not acyclic.

Thursday Review

Question: Draw a digraph representing the relation $\text{mod}_3(m, n)$ if and only if $m \% 3 = n$.

Is this graph: Reflexive? Symmetric? Transitive?

Motivation

• Last week, we focused on finite digraphs.
• This week, you will see examples of infinite digraphs.
• You will see that relations and functions can be interpreted as graphs.
• We will also discuss more graph properties.

Infinite Digraphs

• Last week, we focused on finite digraphs: a finite number of vertices and arcs.
• This week, we will see some examples of infinite digraphs.
• Relations or functions can be interpreted as graphs.
• Consider the predecessor relation: $f(x) = x + 1 \text{ for } f: \mathbb{N} \rightarrow \mathbb{N}$
• The relation forms arcs: $x \rightarrow f(x)$
• If we draw the graph: $0 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow \ldots$
• Ask: Is this graph a DAG? What vertex is the source? Is there a sink?
• Remember that all functions are relations, so we can write the predecessor relation using relation notation.
• Write: $m$ is the predecessor of $n$, $\text{pred}(m, n)$ if and only if $n = m + 1$.
• Let’s consider a different binary relation, and interpret it as a graph.
• Write: $m$ is less than $n$, $\text{lt}(m, n)$ if and only if $m < n$.
• Draw:
  
• In the full digraph of the less-than relation, every vertex would have infinite out-degree (since for any number m, there are infinitely many n greater than m) and finite in-degree (because n is greater than exactly n natural numbers: $0, \ldots, n − 1$).
• Last week, we talked about defining a strict linear order on a set using a binary relation.
• Write: Definition: A strict linear order, denoted by $\prec$, is a binary relation on a finite set $S$ with the property that the elements of $S$ can be listed $S = \{ s_0, \ldots, s_n \}$ in such a way that $s_i \prec s_j$ if and only if $i < j$.
• Let’s think about whether each of these binary relations establishes a strict linear order on the set of natural numbers.
• For $\text{pred}(m, n)$ and $\text{lt}(m, n)$:
  • $m = 0, n = 1$? Yes. Yes.
  • $m = 1, n = 2$? Yes. Yes.
  • $m = 0, n = 2$? No. Yes.
  • $m = 1, n = 1$? No. No.
• The predecessor relation doesn’t establish a strict linear order. The less than relation does.
• Later in the week, we will explain how strict linear order can be defined in terms of graph properties.
• Today, we will see a connection between these two relations.

Transitive Relations

• Write: Definition: A binary relation $R \subseteq S \times S$ is transitive if for any elements $x, y, z \in S$, if $R(x, y)$ and $R(y, z)$ both hold, then $R(x, z)$ also holds.
• The less-than relation on natural numbers is transitive, since if $x < y$ and $y < z$ then $x < z$.
• Draw: In a graph of a transitive relation, you will see that:
  
• All linear orders are transitive, but many relations that are not linear orders are also transitive.
• “For example, consider the relations child and descendant between people, where child(x, y) is true if and only if x is a child of y, and descendant(x, y) is true if and only if x is a descendant of y.
• So descendant(x, y) holds if child(x, y) holds.
• The relation descendant(x, y) also holds if x is a grandchild of y; that is, there is a z such that child(x, z) and child(z, y). In general, descendant is transitive, because a descendant of a descendant of a person is a descendant of that person.”
• Draw: A family tree to illustrate.
• In any directed graph G, reachability is a transitive relation, since if there is a walk from u to v and a walk from v to w, then there is a walk from us to w: just connect the end fo the first walk to the beginning of the second.
• Write: Definition: The transitive closure of a binary relation $R$ is the smallest relation that contains $R$ and is transitive. It is denoted $R^+$.
• Credit: Wikipedia. I think it does a better job explaining transitive closure than the textbook.
• For example, consider the $\text{flight}(x, y)$ relation, showing it is possible to fly from airport x to airport y.
• Draw: ORH → JFK, ORH → LGA, JFK → BOS, BOS → DUB
• Ask: Has anyone taken any fights from Boston or NYC recently? Add them to the graph.
• The transitive closure of the flight relation, $\text{flight}^+(x, y)$, shows whether it is possible to fly from airport x to y.
• Draw: Additional arrows to show the transitive closure.
• Let’s figure out the transitive closures of the relations we talk about earlier.
  • $\text{child}^+ = \text{descendant}$
  • $\text{pred}^+ = \text{lt}$
• Draw: Vertices: a, b. Edges: a → b, b → a.
• Ask: Is this graph transitive?
• Answer: No. To be transitive, it must also include a → a and b → b. Refer to the definition. What if x and z refer to the same vertex? Then you need self-loops.

Reflexive Relations

• Write: Definition: A binary relation $R$ is reflexive if $R(x, x)$ holds for every $x \in S$.
• Ask: Are these relations reflexive?
  • =
  • ≠
  • ≤
  • <
  • pred
• Write: Definition: A binary relation $R$ is irreflexive if it is never the case that $R(x, x)$ for any $x \in S$.
• Ask: Are these relations reflexive?
  • =
  • ≠
  • ≤
  • <
  • pred
• Write: Definition: The reflexive closure of a binary relation $R$ adds to $R$ any ordered pairs $\langle x, x \rangle$ that are not already in $R$: $$R \cup \{ \langle x, x \rangle : x \in S \}$$
• This means every vertex will have a self-loop.
• Write: Definition: The reflexive transitive closure of $R$ is denoted $R^*$, and is formed by taking the reflexive closure of the transitive closure.
• For example: $\text{pred}^*$ is ≤
• Thinking about digraph representations: there is an arc $x \rightarrow y$ in $R^*$ if and only if there is a path (trivial or nontrivial) from $x$ to $y$ in $R$.
• The reflexive, transitive closure of a digraph is also referred to as its reachability relation.

Computer State as a Graph

• Consider a graph of the states of a computer.
• For a general-purpose computer, this graph would be enormous, since all possible states of all memory cells are represented as different vertices.
• Arcs represent changing from one state to another as a result of computation.
• Draw: A vertex for freshly powered-on. A vertex for crashed. Vertices in between.
• An important question is whether running a program will crash the computer, run forever, or give a valid output.
• You will explore this concept in CSCI 180: Automata Theory

Symmetric Relations

• Write: Definition: A binary relation $R \subseteq S \times S$ is symmetric if $R(y, x)$ holds whenever $R(x, y)$ holds.
• In terms of the digraph representation, a symmetric relation has an arc x←y whenever it has an arc x→y.
• For notational convenience, this can be depicted as x↔y.
• A graph containing two-way roads is symmetric. A graph containing at least one one-way road is not symmetric.
• Write: Definition: A relation is asymmetric if it is never the case that both x→y and y→x for any elements x and y.
• Ask: Can a reflexive relation be asymmetric? No, because self-loops would violate the asymmetry condition. Asymmetric relations are irreflexive.
• Write: Definition: A relation is antisymmetric if x→y and y→x both hold only if x=y.
• Ask: Can a reflexive relation be antisymmetric? Yes. For example, ≤ on integers.

Equivalence Relations

• Transitivity, reflexivity, and symmetry are independent of each other.
• Write: Definition: A binary relation that is symmetric, reflexive, and transitive is called an equivalence relation.
• For example, assume every person has only one place of birth (POB). The binary relation between people with the same POB is an equivalence relation:
  • It is symmetric: if your POB is the same as mine then mine is the same as yours
  • It is reflexive: since I was born in the same place as myself
  • It is transitive: if x and y share a POB and y and z share a POB then x and z also have the same POB.
• An equivalence relation can be used to split a set into disjoint subsets.
• Write: Definition: A collection of disjoint subsets of set S is called a partition of S. Each set in the partition is called a block. Symbolically, $\mathcal{T} \subseteq \mathcal{P}(S)$ is a partition of $S$ if and only if:
  • Each block is nonempty. That is, if $X \in \mathcal{T}$, then $X \neq \emptyset$.
  • The blocks are disjoint. That is, if $X_1, X_2 \in \mathcal{T}$, and $X_1 \neq X_2$, then $X_1 \cap X_2 = \emptyset$.
  • The blocks exhaust S. That is, if $\mathcal{T} = X_1, X_2, \ldots, X_n$ then $S = X_1 \cup X_2 \cup \ldots \cup X_n$.
• Recall that the binary relation between people with the same POB is an equivalence relation.
• Intuitively, you can use this binary relation to form a partition of all people into blocks within which people share the same POB.
• We will formalize this with a theorem.
• Write: Theorem: Let ↔ be any equivalence relation on a set S. Then the equivalence classes of ↔ are a partition of S. Conversely, any partition of S, say $\mathcal{T}$, yields an equivalence relation ↔ on S in which all elements in the same block of $\mathcal{T}$ are equivalent to each other under ↔.
• Proof, part 1: If ↔ is an equivalence relation on S, then $$\mathcal{T} = \{ \{ y : y \leftrightarrow x \} : x \in S \}$$ is a partition of S, since every element of S is in one and only one block in $\mathcal{T}$.
• Proof, part 2: If $\mathcal{T}$ is a partition of S, then the relation ↔ such that x↔y if and only if x and y belong to the same block of $\mathcal{T}$ is reflexive, symmetric, and transitive, and is therefore an equivalence relation:
  • It is symmetric: x and y are in the same block, y and x are in the same block
  • It is reflexive: since x is in the same block as x
  • It is transitive: if x and y share a block and y and z share a block then x and z also share the same block.
• Mention implementing __eq__ in Python

Strongly Connected Components

• Let’s review a definition from last week:
• Write: Definition: A digraph in which every vertex is reachable from every other vertex is strongly connected.
• Reachability is a transitive relation, but is not necessarily symmetric.
• Draw: Demonstrate using this graph:
  
• However, mutual reachability is transitive and symmetric.
• Demonstrate using the graph.
• Furthermore, each vertex is trivially reachable from itself, so mutual reachability is reflexive.
• Ask: Since mutual reachability is symmetric, reflexive, and transitive, what does that mean about the relation? It is an equivalence relation.
• So we can partition a digraph using the mutual reachability relation.
• Draw: Circle the blocks in the graph.
• Write: Definition: The subgraphs induced by the blocks of the mutual reachability relation are called the strongly connected components of the digraph, or simply strong components. Each strongly connected component is a strongly connected digraph, and no two vertices in different strong components are mutually reachable.
• For time, skip Theorem 14.3.

Partial Order Relations

• Recall from last week:
• Write: Definition: A linear order, denoted by $\preceq$, is a binary relation on a finite set $S$ with the property that the elements of $S$ can be listed $S = \{ s_0, \ldots, s_n \}$ in such a way that $s_i \preceq s_j$ if and only if $i \leq j$.
• We will see an alternate definition based on graph properties.
• Write: Definition: A partial order is a reflexive, antisymmetric, transitive binary relation.
• For example:
  • ≤ is a partial order on natural numbers
  • ⊆ is a partial order on $\mathcal{P}(\mathbb{N})$
• Draw: ≤ and ⊆ as graphs.
• Write: Definition: A linear order is a partial order such that any two distinct elements are related one way or the other.
• All linear orders are partial orders. Not every partial order is a linear order.
  • ≤ is a partial order and a linear order
  • ⊆ is a partial order, but not a linear order. For example, ${0} \not\subseteq {1}$ and ${1} \not\subseteq {0}$.
• Write: Definition: A strict partial order is an irreflexive, antisymmetric, transitive binary relation. No self-loops.
• For example:
  • < is a strict partial order on natural numbers
  • $\subsetneq$ is a strict partial order on $\mathcal{P}(\mathbb{N})$
• Draw: < and $\subsetneq$ as graphs.
• Write: Definition: A strict linear order is a strict partial order such that any two distinct elements are related one way or the other.
• The graph of any strict partial order is a DAG.
• Any partial order is the reflexive, transitive closure of a DAG.
• Let’s turn the CS prerequisite DAG into a partial order.
• Draw:
  
• Add self-loops
• Add arcs to form transitive closure.
• Ask: Is this graph a linear order? No. For example, there isn’t an arc between 121 and 122.
• Write: Definition: In a partial order, a minimal element is one that is not greater than any other element.
• Ask: What is the minimal element in this graph?
• Ask: Is it possible for a partial order to have multiple minimal elements? Erase 120. Then 121 and 122 are both minimal elements.
• Mention implementing __lt__ in Python