单源最短路径(SSSP)

1. Introduction

In the Single-Source Shortest Paths (SSSP) problem, we aim to find the shortest paths weights (and the actual paths) from a particular single-source vertex to all other vertices in a directed weighted graph (if such paths exist).


The SSSP problem is a(nother) very well-known Computer Science (CS) problem that every CS students worldwide need to be aware of and hopefully master.


The SSSP problem has several different efficient (polynomial) algorithms (e.g., Bellman-Ford, BFS, DFS, Dijkstra — 2 versions, and/or Dynamic Programming) that can be used depending on the nature of the input directed weighted graph, i.e. weighted/unweighted, with/without (negative weight) cycle, or structurally special (a tree/a DAG).

1-1. 动机

SSSP is one of the most frequent graph problem encountered in real-life. Every time we want to move from one place (usually our current location) to another (our destination), we will try to pick a short — if not the shortest — path.


SSSP algorithm(s) is embedded inside various map software like Google Maps and in various Global Positioning System (GPS) tool.

1-2. SSSP 问题 - 两个输入值

输入 1: 一个有向加权图 G(V, E), 不一定是连接着的, 其中 V/顶点可用于描述交叉点,交汇点 ,房屋,地标等,还有 E/边可用于描述街道,道路 ,大道,且具有适当的方向和权重/成本。


输入 2: 顾名思义, SSSP 问题有另一个输入:一个源顶点 sV.

1-3. SSSP 问题 - 输出

SSSP 问题的目的是找到从 s 到每个顶点 uV 的最短路径权,表示为 δ(s, u) ( δ 发音为 'delta' ) 以及从 su 的实际最短路径。

路径 p 仅仅是沿改路径的边的权重的总和。

ss 的最短路径的权重是微不足道的:0。从 s 到任何不可到达顶点的最短路径也是微不足道的:+∞。

PS: 从 sv 的最短路径的权重,其中 (s, v) ∈ E ,不一定是 w(s, v) 的权重。请参阅下面的几张幻灯片来理解这一点。

1-4. SSSP问题 - 输出变量

The outputs of all six (6) SSSP algorithms for the SSSP problem discussed in this visualization are these two arrays/Vectors:

  1. An array/Vector D of size V (D stands for 'distance')
    Initially, D[u] = 0 if u = s; otherwise D[u] = +∞ (a large number, e.g. 109)
    D[u] decreases as we find better (shorter) paths
    D[u]δ(s, u) throughout the execution of SSSP algorithm
    D[u] = δ(s, u) at the end of SSSP algorithm
  2. An array/Vector p of size V (p stands for 'parent'/'predecessor'/'previous')
    p[u] = the predecessor on best path from source s to u
    p[u] = NULL (not defined, we can use a value like -1 for this)
    This array/Vector p describes the resulting SSSP spanning tree

1-5. s = 0 的例子

Initially, D[u] = +∞ (practically, a large value like 109) ∀uV\\{s}, but D[s] = D[0] = 0.
Initially, p[u] = -1 (to say 'no predecessor') ∀uV.


Now click Dijkstra(0) — don't worry about the details as they will be explained later — and wait until it is over (approximately 10s on this small graph).


At the end of that SSSP algorithm, D[s] = D[0] = 0 (unchanged) and D[u] = δ(s, u)uV
e.g. D[2] = 6, D[4] = 7 (these values are stored as red text under each vertex).
At the end of that SSSP algorithm, p[s] = p[0] = -1 (the source has no predecessor), but p[v] = the origin of the orange edges for the rest, e.g. p[2] = 0, p[4] = 2.


Thus, if we are at s = 0 and want to go to vertex 4, we will use shortest path 0 → 2 → 4 with path weight 7.

1-6. 定义不明确的案例

Some graphs contain negative weight edge(s) (not necessarily cyclic) and/or negative weight cycle(s). For example (fictional): Suppose you can travel forward in time (normal, edges with positive weight) or back in time by passing through time tunnel (special wormhole edges with negative weight), as the example shown above.


On that graph, the shortest paths from the source vertex s = 0 to vertices {1, 2, 3} are all ill-defined. For example 1 → 2 → 1 is a negative weight cycle as it has negative total path (cycle) weight of 15-42 = -27. Thus we can cycle around that negative weight cycle 0 → 1 → 2 → 1 → 2 → ... forever to get overall ill-defined shortest path weight of -∞.


However, notice that the shortest path from the source vertex s = 0 to vertex 4 is ok with δ(0, 4) = -99. So the presence of negative weight edge(s) is not the main issue. The main issue is the presence of negative weight cycle(s) reachable from source vertex s.

1-7. 主要操作: 松弛 (u, v, w(u, v))

The main operation for all SSSP algorithms discussed in this visualization is the relax(u, v, w(u, v)) operation with the following pseudo-code:

relax(u, v, w_u_v)
if D[v] > D[u]+w_u_v // if the path can be shortened
D[v] = D[u]+w_u_v // we 'relax' this edge
p[v] = u // remember/update the predecessor
// update some other data structure(s) as necessary

For example, see relax(1,2,4) operation on the figure below: relax operation example

2. 输入图表

对于指定一个输入图,有两种不同的方法:

  1. 绘制图: 您可以绘制任何有向加权图作为输入图。
  2. 示例图: 您可以从我们选择的示例图列表中进行挑选,以帮助您入门。这些示例图具有不同的特征。

3. SSSP算法

In this visualization, we will discuss 6 (SIX) SSSP algorithms.


We will start with the O(V×E) Bellman-Ford algorithm first as it is the most versatile (but also the slowest) SSSP algorithm. We will then discuss 5 (FIVE) other algorithms (including two variants of Dijkstra's algorithm) that solve special-cases of SSSP problem in a much faster manner.

4. O(V×E) Bellman-Ford Algorithm

The general purpose Bellman-Ford algorithm can solve all kinds of valid SSSP problem variants (expect one — the one that is ill-defined anyway, to be discussed soon), albeit with a rather slow O(V×E) running time. It also has an extremely simple pseudo-code:

for i = 1 to |V|-1 // O(V) here, so O(V×E×1) = O(V×E)
for each edge(u, v) ∈ E // O(E) here, e.g. by using an Edge List
relax(u, v, w(u, v)) // O(1) here

Without further ado, let's see a preview of how it works on the example graph above by clicking BellmanFord(0) (≈30s, and for now, please ignore the additional loop at the bottom of the pseudo-code).

4-1. 优化形式: O(k×E)

Bellman-Ford algorithm can be made to run slightly faster on normal input graph, from the worst case of O(V×E) to just O(k×E) where k is the number of iterations of the outer loop of Bellman-Ford.


Discussion: How to do this? Is the speed-up significant?

4-2. 答案

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4-3. T.1: 最短路径是简单路径*

To convince the worldwide audience that Bellman-Ford algorithm works, let's temporarily move from visualization mode to proof mode for a few slides.


Theorem 1: If G = (V, E) contains no negative weight cycle, then the shortest path p from source vertex s to a vertex v must be a simple path.


Recall: A simple path is a path p = {v0, v1, v2, ..., vk}, (vi, vi+1) ∈ E, ∀ 0 ≤ i ≤ (k-1) and there is no repeated vertex along this path.

4-4. 反证法 - 第1部分

  1. Suppose the shortest path p is not a simple path
  2. Then p must contains one (or more) cycle(s) (by definition of non-simple path)
  3. Suppose there is a cycle c in p with positive weight (e.g., greenbluegreen on the left image) cycle
  4. If we remove c from p, then we will have a shorter 'shortest path' than our shortest path p
  5. A glaring contradiction, so p must be a simple path

4-5. 反证法 - 第2部分

  1. Even if c is actually a cycle with zero (0) total weight — it is possible according to our Theorem 1 assumption: no negative weight cycle (see the same greenbluegreen but on the right image), we can still remove c from p without increasing the shortest path weight of p cycle
  2. In conclusion, p is a simple path (from point 5) or can always be made into a simple path (from point 6)

In another word, shortest path p has at most |V|-1 edges from the source vertex s to the 'furthest possible' vertex v in G (in terms of number of edges in the shortest path — see the Bellman-Ford Killer example above).

4-6. T.2: Bellman-Ford Algorithm is Correct

Theorem 2: If G = (V, E) contains no negative weight cycle, then after Bellman-Ford algorithm terminates, we will have D[u] = δ(s, u), ∀ uV.


For this, we will use Proof by Induction and here are the starting points:


Consider the shortest path p from source vertex s to vertex vi where vi is defined as a vertex which the actual shortest path to reach it requires i hops (edges) from source vertex s. Recall from Theorem 1 that p will be simple path as we have the same assumption of no negative weight cycle.

4-7. 归纳证明法

  1. Initially, D[v0] = δ(s, v0) = 0, as v0 is just the source vertex s
  2. After 1 pass through E, we have D[v1] = δ(s, v1)
  3. After 2 pass through E, we have D[v2] = δ(s, v2)
  4. ...
  5. After k pass through E, we have D[vk] = δ(s, vk)
  6. When there is no negative weight cycle, the shortest path p is a simple path (see Theorem 1), thus the last iteration should be iteration |V|-1
  7. After |V|-1 pass through E, we have D[v|V|-1] = δ(s, v|V|-1), regardless the ordering of edges in E — see the Bellman-Ford Killer example above

4-8. 最坏情况的表现

Try running BellmanFord(0) on the 'Bellman-Ford Killer' example above. There are V = 7 vertices and E = 6 edges but the edge list E is configured to be at its worst possible order. Notice that after (V-1)×E = (7-1)*6 = 36 operations (~40s, be patient), Bellman-Ford will terminate with the correct answer and there is no way we can terminate Bellman-Ford algorithm earlier.

4-9. 在有负权重循环的图上

The only input graph that Bellman-Ford algorithm has issue is the input graph with negative weight cycle reachable from the source vertex s.


However, Bellman-Ford can be used to detect if the input graph contains at least one negative weight cycle reachable from the source vertex s by using the corollary of Theorem 2: If at least one value D[u] fails to converge after |V|-1 passes, then there exists a negative-weight cycle reachable from the source vertex s.


Now run BellmanFord(0) on the example graph that contains negative edges and a negative weight cycle. Please concentrate on the loop at the bottom of the pseudo-code.

5. 五种简化假设

Sometimes, the actual problem that we face is not the general form of the original problem. Therefore in this e-Lecture, we want to highlight five (5) special cases involving the SSSP problem. When we encounter any one of them, we can solve it with different and (much) faster algorithm than the generic O(V×E) Bellman-Ford algorithm. They are:

  1. On Unweighted Graphs: O(V+E) BFS,
  2. On Graphs without negative weight: O((V+E) log V) Dijkstra's algorithm,
  3. On Graphs without negative weight cycle: O((V+E) log V) Modified Dijkstra's,
  4. On Tree: O(V+E) DFS/BFS,
  5. On Directed Acyclic Graphs (DAG): O(V+E) Dynamic Programming (DP)

6. 广度优先搜索

O(V+E) 广度优先搜索 (BFS) 算法可以解决有特殊情况的SSSP问题,当输入图是未加权时(所有边都具有单位权重1,在上面的例子 'CP3 4.3' 上尝试 BFS(5)),或者正常数加权(所有有边都具有相同的恒定权重,例如,你可以在上面的例图中用你所选择的任何正恒定权重去更改所有的边的权重)。

6-1. 解释说明

When the graph is unweighted — this appears quite frequently in real life — the SSSP problem can be viewed as a problem of finding the least number of edges traversed from the source vertex s to other vertices.


The BFS spanning tree from source vertex s produced by the fast O(V+E) BFS algorithm — notice the + sign — precisely fits the requirement.


Compared with the O(V×E) of Bellman-Ford — notice the × sign — it is a no-brainer to use BFS for this special case of SSSP problem.

6-2. 次要实施调整

Compared to the standard BFS in Graph Traversal module, we need to perform simple modifications to make BFS able to solve the unweighted version of the SSSP problem:

  1. First, we change the Boolean array visited into an Integer array D.
  2. At the start of BFS, instead of setting visited[u] = false, we set D[u] = 1e9 (a large number to symbolise +∞ or even -1 to symbolise 'unvisited' state, but we cannot use 0 as D[0] = 0) ∀uV\{s}; Then we set D[s] = 0
  3. We change the BFS main loop from
    if (visited[v] = 0) { visited[v] = 1 ... } // v is unvisited
    to
    if (D[v] = 1e9) { D[v] = D[u]+1 ... } // v is 1 step away from u

6-3. 在一般图表上是错误答案

However, BFS will very likely produce wrong answer when run on weighted graphs as BFS is not actually designed for to solve the weighted version of SSSP problem. There may be a case that taking a path with more number of edges used produces lower total overall path weight than taking a path with minimum number of edges used — which is the output of BFS algorithm.


In this visualization, we will allow you to run BFS even on 'wrong' input graph for pedagogical purpose, but we will display a warning message at the end of the algorithm. For example, try BFS(0) on the general graph above and you will see that vertices {3,4} will have wrong D[3] and D[4] values (and also p[3] and p[4] values).


We will soon see Dijkstra's algorithm (2 implementation variants) for solving certain weighted SSSP problems in a faster way than the general Bellman-Ford algorithm.

7. 戴克斯特拉算法(原始的实现)

The O((V+E) log V) Dijkstra's algorithm is the most frequently used SSSP algorithm for typical input: Directed weighted graph that has no negative weight edge at all, formally: ∀ edge(u, v) ∈ E, w(u, v) ≥ 0. Such weighted graph is very common in real life as travelling from one place to another always use positive time unit(s). Try Dijkstra(0) on one of the Example Graphs: CP3 4.17 shown above.

7-1. 关键想法

Dijkstra's algorithm maintains a set S (Solved) of vertices whose final shortest path weights have been determined. Initially S = {s}, the source vertex s only.


Then, it repeatedly selects vertex u in {V\S} with the minimum shortest path estimate, adds u to S, and relaxes all outgoing edges of u. Detailed proof of correctness of this Dijkstra's algorithm is usually written in typical Computer Science algorithm textbooks. For a simpler intuitive visual explanation on why this greedy strategy works, see this.


This entails the use of a Priority Queue as the shortest path estimates keep changing as more edges are processed. The choice of relaxing edges emanating from vertex with the minimum shortest path estimate first is greedy, i.e. use the "best so far", but we will see later that it can be proven that it will eventually ends up with an optimal result — if the graph has no negative weight edge.

7-2. O((V+E) log V) 时间复杂度 - 第 1部分

In Dijkstra's algorithm, each vertex will only be extracted from the Priority Queue (PQ) once. As there are V vertices, we will do this maximum O(V) times.


ExtractMin() operation runs in O(log V) whether the PQ is implemented using a Binary Min Heap or using a balanced BST like AVL Tree.


Therefore this part is O(V log V).

7-3. O((V+E) log V) 时间复杂度 - 第 2部分

Every time a vertex is processed, we relax its neighbors. In total, E edges are processed.


If by relaxing edge(u, v), we have to decrease D[v], we call the O(log V) DecreaseKey() operation in Binary Min Heap (harder to implement as C++ STL priority_queue/Python heapq/Java PriorityQueue does not support this operation efficiently yet) or simply delete the old entry and then re-insert a new entry in balanced BST like AVL Tree (which also runs in O(log V), but this is much easier to implement, just use C++ STL set/Java TreeSet — unfortunately not natively supported in Python).


Therefore, this part is O(E log V).


Thus in overall, Dijkstra's algorithm runs in O(V log V + E log V) = O((V+E) log V) time, which is much faster than the O(V×E) Bellman-Ford algorithm.

7-4. 正确性的证明

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7-5. 在有负权重的表上是错误答案

When the input graph contains at least one negative weight edge — not necessarily negative weight cycle — Dijkstra's algorithm can produce wrong answer.


Try Dijkstra(0) on one of the Example Graphs: CP3 4.18.


At the end of the execution of Dijkstra's algorithm, vertex 4 has wrong D[4] value as the algorithm started 'wrongly' thinking that subpath 0 → 1 → 3 is the better subpath of weight 1+2 = 3, thus making D[4] = 6 after calling relax(3,4,3). However, the presence of negative weight -10 at edge 2 → 3 makes the other subpath 0 → 2 → 3 eventually the better subpath of weight 10-10 = 0 although it started worse with path weight 10 after the first edge 0 → 2. This better D[3] = 0 is never propagated further due to the greedy nature of Dijkstra's algorithm, hence D[4] is wrong.

8. Modified Dijkstra's Algorithm

Dijkstra's algorithm can also be implemented differently. The O((V+E) log V) Modified Dijkstra's algorithm can be used for directed weighted graphs that may have negative weight edges but no negative weight cycle.


Such input graph appears in some practical cases, e.g., travelling using an electric car that has battery and our objective is to find a path from source vertex s to another vertex that minimizes overall battery usage. As usual, during acceleration (or driving on flat/uphill road), the electric car uses (positive) energy from the battery. However, during braking (or driving on downhill road), the electric car recharges (or use negative) energy to the battery. There is no negative weight cycle due to kinetic energy loss.


For example, try ModifiedDijkstra(0) on one of the Example Graphs: CP3 4.18 that has troubled the original version of Dijkstra's algorithm (see previous slide).

8-1. 关键想法

The key idea is the 'usage modification' done to C++ STL priority_queue/Python heapq/Java PriorityQueue to allow it to perform the required 'DecreaseKey' operation efficiently, i.e., in O(log V) time.


The technique is called 'Lazy Update' where we leave the 'outdated/weaker/bigger-valued information' in the Min Priority Queue instead of deleting it straight-away. As the items are ordered from smaller values to bigger values in a Min PQ, we are guaranteeing ourself that we will encounter the smallest/most-up-to-date item first before encountering the weaker/outdated item(s) later - which by then can be easily ignored.

8-2. O((V+E) log V) 时间复杂度?

On non-negative weighted graphs, the behavior of Modified Dijkstra's implementation is exactly the same as the Original Dijkstra's so we can use the same time complexity analysis of O((V+E) log V).


PS: We note that when we use the Modified Dijkstra's algorithm, there can be more items (up to E) in the Priority Queue than if we use the Original Dijkstra's algorithm (up to V). However, since O(log E) = O(log V^2) = O(2 log V) = O(log V), we still treat the Priority Queue operations as O(log V).


However, if the graph has at least one negative weight edge, the analysis is harder.

8-3. 在没有负权重循环的图上更正

当输入图包含至少一个负权重边但没有负权重周期时 - 修改后的Dijkstra算法产生正确的答案。

尝试使用其中一个示例图表ModifiedDijkstra(0):CP3 4.18导致Dijkstra(0)出现问题。

在ModifiedDijkstra算法执行结束时,顶点4具有正确的D [4]值,因为虽然修改后的Dijkstra算法也开始“错误地”认为子路径0→1→3是权重1 + 2 = 3的更好的子路径,因此在调用relax(3,4,3)后使D [4] = 6。这里,修改后的Dijkstra算法在发现其他子路径0→2→3最终是权重10-10 = 0的更好子路径之后继续传播D [3] = 0.因此D [4]最终再次正确。然而,这是以比O((V + E)log V)可能运行(更多)操作为代价的。

8-4. 具有负权重循环的图的问题

Unfortunately, running ModifiedDijkstra(0) on the graph with negative weight cycle as shown on one of the Example Graphs: CP3 4.17 above will cause an endless loop (the animation is very long but we limit the number of loop to be 100 edges processed so your web browser will not hang).

8-5. 最差的案例输入

Try ModifiedDijkstra(0) on the extreme corner case above that is very hard to derive without proper understanding of this algorithm and was part of Asia Pacific Informatics Olympiad (APIO) 2013 task set by Steven.


The Modified Dijkstra's algorithm will terminate with correct answer, but only after running exponential number of operations (each carefully constructed triangle raises the number of required operations by another power of two). Thus we cannot prematurely terminate Modified Dijkstra's in this worst case input situation.


However, such extreme corner case is rare and thus in practice, Modified Dijkstra's algorithm can be used on directed graphs that have some negative weighted edges as long as the graph has no negative weight cycle reachable from the source vertex s.

9. 深度优先搜索

The O(V) Depth-First Search (DFS) algorithm can solve special case of SSSP problem, i.e. when the input graph is a (weighted) Tree.


In a Tree, there is only one unique and acylic path that connects two distinct vertices. Thus the unique path that connects the source vertex s to any another vertex uV is actually also the shortest path. For example, try DFS(0) on the Tree above.


Notice that for a (weighted) Tree, we can also use BFS. For example, try BFS(0) on the same Tree above.


Discussion: Why DFS (and also BFS) runs in O(V) instead of O(V+E) if the input is a (weighted) Tree?

9-1. 答案

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9-2. Wrong Answer on Non Tree Graphs


在不是Tree的任何其他图形上运行时,DFS很可能会产生错误的答案。我们将为此类案例显示警告消息,但我们不会阻止您出于教学目的而尝试此功能。

例如,在上面的一般图表上尝试DFS(0),您将看到顶点{4}将具有错误的D [4]值(以及错误的p [4]值),因为DFS(0)深入0→1→首先是3→4,一直回溯到顶点0并且最终访问0→2但是由于DFS之前已经访问过顶点4,所以不能处理边缘2→4。

10. 动态规划(DP)

The O(V+E) Dynamic Programming algorithm can solve special case of SSSP problem, i.e. when the input graph is a Directed Acyclic Graph (DAG) thus we can find at least one topological order of the DAG and process the edge relaxation according to this topological order.


For example, try DP(0) on the example DAG above. First, it computes one (there are other) possible topological order using either the O(V+E) DFS or the BFS/Kahn's algorithm outlined in Graph Traversal module. For example, assume one topological order is {0,2,1,3,4,5}. Then, it relaxes the outgoing edges of vertices listed in that topological order. After just one O(V+E) pass, we will have correct D[u] values ∀uV.

10-1. 长处与缺点

On the Modified Dijkstra's killer example shown above, DP(0) works fast as the graph is actually a DAG, albeit having negative weight edge. As the graph is a DAG, there will not be any negative weight cycle to worry about.


However, DP will not work for any non DAG as non DAG contains at least one cycle and thus no topological order can be found within that cycle.