---
title: "Dijkstra's Shortest Path Algorithm"
format:
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    toc: true
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---

[In-class notes](hand_written_notes/Dijkstra3.pdf)

## Learning Goals

- Describe Dijkstra's Algorithm
- Prove correctness of Dijkstra's Algorithm
- Analyze the runtime of Dijkstra's Algorithm

## Shortest Path

### Problem Definition

Recall from the [Bellman-Ford Shortest Path Notes](BellmanFord.qmd) the Shortest Path Problem, except now instead of $w:E\rightarrow \mathbb{R}$, we have $w:E\rightarrow \mathbb{R}^+$. (In other words, the weights on edge must all be non-negative.)

**Input:** A directed $G=(V,E)$, $w:E\rightarrow \mathbb{R}^+$, $s,t\in V$, (we denote $n:=|V|$ and $m:=|E|$)

**Output:** A path $P$ of directed edges from $s$ to $t$ in $G$ (e.g. $P=((s,u), (u,v), \dots, (r,t))$), such that $L(P)$ is minimized where
$$
L(P)=\sum_{e\in P}w(e).
$${#eq-pathLength}

**Example:**
Now the example has only positive weights:


```{dot}
//| label: fig-simple
//| fig-cap: "Example input to Shortest Path."
digraph G {
  node [shape=circle, width=0.4];
  nodesep=1.5;
  s -> u [label = "1"];
  s -> v [label = "4"];
  s -> t [label = "6"];
  u -> t [label = "2"];
  v -> t [label = "1"];
}
```

For the graph in @fig-simple, the shorted path is $P=((s,u),(u,t)).$

Note that now there can be no negative cycles, so we no longer have to worry about that subtlety.

### Approaches

As previously discusses, there 3 different paradigms to solve this problem. (And there are even more variants than those I list).

1. Greedy $\rightarrow$ Dijkstra's Algorithm.
    * Use when all edges have positive weights
2. Brute Force $\rightarrow$ Breadth First Search. 
    * Use when all edges have weight one
3. Dynamic Programming $\rightarrow$ Bellman-Ford.
    * Edges can have negative weight
    * Can take global or distributed description of $G$ (more on this later)
    * Fails if negative

Since we will be considering situations where edges have positive weights, we will be learning a greedy approach called Dijkstra's algorithm


## Algorithm  

### Pseudocode

> **HUFFMAN**   
>
> **Input**: $G=(V,E)$, $s\in V$, $w:E\rightarrow \mathbb{R}^+$, such that $|V|=n$.   
> **Output**: n dimensional arrays $L$ and $P$ such that $L[v]=$length of shortest path from $s$ to $v$ in $G$, and $P[v]=$shortest path from $s$ to $v$ in $G$.  
>  
> // Initialization:   
> $X\leftarrow\{s\}$    $\quad$ // $X$ is the set of visited/processed vertices   
> $L[s]\leftarrow0$   
> $P[s]\leftarrow\emptyset$   
> 
> // Greedily Processing Vertices   
> While there is an edge from $X$ to $\bar{X}=V-X$:   
> $\quad$ * $C\leftarrow\{(u,v):u\in X, v\in \bar{X}\}$   
> $\quad$ * $(u^*,v^*\leftarrow \textrm{argmin}_{(u,v)\in C}\{L[u]+w(u,v)\}$    
> $\quad$ * $L[v^*]\leftarrow L[u^*]+w(u^*,v^*)$    
> $\quad$ * $P[v^*]\leftarrow P[u^*]+(u^*,v^*)$ $\quad$  // $+$ means append    
> $\quad$ * $X\leftarrow X\cup \{v^*\}$   


We call $L[u]+w(u,v)$ the Dijkstra criterion of edge $(u,v)$, and we might say things like, "$(u,v)$ has minimal Dijkstra criterion."

### Working through an example:

Consider the following example graph:

```{dot}
//| label: fig-simple
//| fig-cap: "Example input to Shortest Path for stepping through Dijkstras."
digraph G {
  node [shape=circle, width=0.4];
  nodesep=1.5;
  s -> u [label = "1"];
  s -> v [label = "4"];
  u -> v [label = "2"];
}
```

The algorithm initializes:   
* $X\leftarrow \{s\}$   
* $L[s]\leftarrow 0$   
* $P[s]\leftarrow \emptyset$. 

Now in the first round of the while loop, $X=\{s\}$, so $\bar{X}=\{u,v\}$. This means $C=\{(s,u),(s,v)\}$. We next calculate the Dijkstra criterion for each edge in $C$:   
* $A[s]+w(s,u)=0+1=1$   
* $A[s]+w(s,v)=0+4=4$   
We see that $(s,u)$ has the smaller Dijkstra criterion, so we set   
* $X\leftarrow \{s,u\}$   
* $L[u]\leftarrow 0+w(s,u)=1$   
* $P[u]\leftarrow \emptyset+(s,u)=(s,u)$.

Now in the next round of the while loop, $X=\{s,u\}$, so $\bar{X}=\{v\}$. This means $C=\{(s,v),(u,v)\}$. We next calculate the Dijkstra criterion for each edge in $C$:    
* $A[s]+w(s,v)=0+4=4$      
* $A[v]+w(u,v)=1+2=3$
We see that $(u,v)$ has the smaller Dijkstra criterion, so we set   
* $X\leftarrow \{s,u,v\}$   
* $L[v]\leftarrow L[u]+w(u,v)=1+2=3$   
* $P[v]\leftarrow P[u]+(u,v)=((s,u),(u,v)$   


### Group Work

Show that Dijkstra's Algorithm can fail when there are negative-weight edges, as in the following graph

```{dot}
//| label: fig-simple
//| fig-cap: "Example input to Shortest Path with negative weights."
digraph G {
  node [shape=circle, width=0.4];
  nodesep=1.5;
  s -> u [label = "1"];
  s -> v [label = "-4"];
  u -> v [label = "-2"];
}
```
Under what conditions an Dijkstra's algorithm have negative weights but still be successful?

## Proving Correctness of Dijkstra's Algorithm

:::{#thm-Dijkstra}
Dijkstra's algorithm correctly returns the shortest path.
:::

:::{.proof}
We will prove using induction on $n=|X|$ that Dijkstra's algorithm correctly assigns $L[v]$ and $P[v]$ for all $v\in X$.

Base case: when $n=1$, $X=\{s\}$ and $L[s]=0$ and $P[s]=\emptyset$, which are correct because you can get from $s$ to itself with no edges and a $0$-length path.

Inductive step: Let $k\geq 1$. Assume for induction that Dijkstra's algorithm correctly assignes $L[v]$ and $P[v]$ $\forall v\in X$ when $|X|=k$. We want to show that it will correctly add the $(k+1)^{th}$ element. Let $(u^*,v^*\leftarrow \textrm{argmin}_{(u,v)\in C}\{L[u]+w(u,v)\}$ be the edge that minimizes the Dijkstra criterion at this point, so the algorithm chooses $v^*$ to be the $(k+1)^{th}$ element to add to $X$ and sets $P[v^*]=P[u^*]+(u^*,v^*)$ and sets $L[v^*]=L[u^*]+w(u^*,v^*)$. Our task is to prove that these assignments are the correct shortest path and correct length of the shortest path to $v^*.$

Suppose for contradiction that $P=P[u^*]+(u^*,v^*)$ is not the shortest path from $s$ to $v^*$. Let $P^*\neq P$ be the optimal path. Then $P^*$ looks like the path in @fig-Pstar.

![The assumed optimal path with the first edge in C highlighted.](quarto_note_pdfs/Pstar.png){width=50% #fig-Pstar}

In @fig-Pstar, the first edge in $C$ to appear in $P^*$ is highlighted - the edge $(x,y)$. We can divide up this path into two segments, the part of the path from $s$ to $y$, and the part of the path from $y$ to $v^*$. Then the total length of this path $L(P^*)$ is the sum of these two lengths:
$$
L(P^*)=L(s\rightarrow y)+L(y\rightarrow v^*).
$$
We have that
$$
L(s\rightarrow y)\geq ....???\\
L(y\rightarrow v^*)\geq ...???
$$

Thus $L(P^*)\geq L(P)$, contradicting the fact that $P$ is not optimal. Thus in fact, $P$ must have been optimal, and the assignments of the algorithm for up to $k+1$ elements in $X$ must be optimal.
:::

## Runtime of Dijkstra's Algorithm

### Group work

* What data structures should you use to improve the algorithm and why?   
 

### MinHeap Set-up

We will use a MinHeap and store Vertex objects in the heap.

We will create a Vertex object with the following attributes:   
* *name*    
* *key*    
* *prior*   

We will create such an object for each $v\in \bar{X}$. For a vertex $v$, the attributes should be set to:   
* *name:* $v$    
* *key:* $\textrm{min}_{u\in X}L[u]+w(u,v)$    
* *prior:* $\textrm{argmin}_{u\in X}L[u]+w(u,v)$     
 
Consider the following snapshot of variables in the algorithm:

![](quarto_note_pdfs/ObjectTest.png){width=50%}

In this situation, what should $v.prior$ be set to?   
A) $u$   
B) $w$   
C) $15$   
D) $8$  

We will put Vertex objects into the MinHeap according to their *key* values.

Recall the properties of a Min Heap:   
- You can initialize $n$ items in the heap in $O(n\log n)$ time    
- You can remove ~the item with the minimum key value~ **any item** in $O(\log n)$ time, **as long as have pointer to that element of the heap**.   
- You can insert a new item into the heap in $O(\log n)$ time.

### Modifying the Pseudocode to Incorporate the MinHeap


> **HUFFMAN**   
>
> **Input**: $G=(V,E)$, $s\in V$, $w:E\rightarrow \mathbb{R}^+$, such that $|V|=n$.   
> **Output**: n dimensional arrays $L$ and $P$ such that $L[v]=$length of shortest path from $s$ to $v$ in $G$, and $P[v]=$shortest path from $s$ to $v$ in $G$.  
>  
> // Initialization:   
> $X\leftarrow\{s\}$    $\quad$ // $X$ is the set of visited/processed vertices   
> $L[s]\leftarrow0$   
> $P[s]\leftarrow\emptyset$   
>  
> // Initializing Heap   
> Create empty Heap $H$   
> For $u\in V-\{s\}$:   
> $\quad$ * If $(s,u)\in E$:   
> $\qquad$ - $u.key\leftarrow w(s,u)$   
> $\qquad$ - $u.prior\leftarrow s$   
> $\quad$ * Else:   
> $\qquad$ - $u.key\leftarrow \infty$   
> $\qquad$ - $u.prior\leftarrow \emptyset$   
> $\quad$ * Push $u$ into $H$   
>  
> // Greedily Processing Vertices   
> While $H\neq \emptyset$:   
> $\quad$ // Process current vertex   
> $\quad$ * $u\leftarrow H.pop$   
> $\quad$ * $X\leftarrow X\cup \{u\}$    
> $\quad$ * $L[u]\leftarrow u.key$   
> $\quad$ * $P[u]\leftarrow P[u.prior]+(u,prior,u)$ // Remove! See pset   
>
> $\quad$ // Update Heap      
> $\quad$ * For $v:(u,v)\in E$:   
> $\qquad$ - Remove $v$ from $H$   
> $\qquad$ - If $v.key>L[u]+w(u,v)$:   
> $\qquad$$\qquad$ ~ $v.key\leftarrow L[u]+w(u,v)$   
> $\qquad$$\qquad$ ~ $v.prior\leftarrow u$   
> $\qquad$ - Reinsert $v$ into $H$  

Why do we need the section "Update Heap?" Each vertex $v$ in $\bar{X}$ is stored in the Heap, and its $key$ should hold the vertex in $X$ that results in the minimal Dijkstra criterion to the current vertex. However, after we have added a new vertex $u$ to $X$, it could be that now the vertex $(u,v)$ has the minimal Dijkstra criterion. We previously hadn't considered this edge because until this round, $u$ was not an element of $X$. But because $u$ is the only new vertex being added to $X$, only edges originating at $u$ might result in edges with better Dijkstra criterion, so we just need to go through and check the neighbors of $u$ to see if any of them do better than what we currently have for Dijkstra criterion.

Why is the runtime of this pseudocode $O((n+m)\log n)$ when the graph $G$ is given as an adjacency list?

Note that this is an improvement over the Bellman-Ford runtime, which is $O((n+m)n)$ with an adjacency list.