I would like to know, what would be the most efficient way (w.r.t., Space and Time) to solve the following problem:
Given an undirected Graph G = (V, E), a positive number N and a vertex S in V. Assume that every vertex in V has a cost value. Find the N highest cost vertices that is connected to S.
For example:
G = (V, E)
V = {v1, v2, v3, v4},
E = {(v1, v2),
(v1, v3),
(v2, v4),
(v3, v4)}
v1 cost = 1
v2 cost = 2
v3 cost = 3
v4 cost = 4
N = 2, S = v1
result: {v3, v4}
This problem can be solved easily by the graph traversal algorithm (e.g., BFS or DFS). To find the vertices connected to S, we can run either BFS or DFS starting from S. As the space and time complexity of BFS and DFS is same (i.e., time complexity: O(V+E), space complexity: O(E)), here I am going to show the pseudocode using DFS:
Parameter Definition:
* G -> Graph
* S -> Starting node
* N -> Number of connected (highest cost) vertices to find
* Cost -> Array of size V, contains the vertex cost value
procedure DFS-traversal(G,S,N,Cost):
let St be a stack
let Q be a min-priority-queue contains <cost, vertex-id>
let discovered is an array (of size V) to mark already visited vertices
St.push(S)
// Comment: if you do not want to consider the case "S is connected to S"
// then, you can consider commenting the following line
Q.push(make-pair(S, Cost[S]))
label S as discovered
while St is not empty
v = St.pop()
for all edges from v to w in G.adjacentEdges(v) do
if w is not labeled as discovered:
label w as discovered
St.push(w)
Q.push(make-pair(w, Cost[w]))
if Q.size() == N + 1:
Q.pop()
let ret is a N sized array
while Q is not empty:
ret.append(Q.top().second)
Q.pop()
Let's first describe the process first. Here, I run the iterative version of DFS to traverse the graph starting from S. During the traversal, I use a priority-queue to keep the N highest cost vertices that is reachable from S. Instead of the priority-queue, we can use a simple array (or even we can reuse the discovered array) to keep the record of the reachable vertices with cost.
Analysis of space-complexity:
To store the graph: O(E)
Priority-queue: O(N)
Stack: O(V)
For labeling discovered: O(V)
So, as O(E) is the dominating term here, we can consider O(E) as the overall space complexity.
Analysis of time-complexity:
DFS-traversal: O(V+E)
To track N highest cost vertices:
By maintaining priority-queue: O(V*logN)
Or alternatively using array: O(V*logV)
The overall time-complexity would be: O(V*logN + E) or O(V*logV + E)
Related
I'm pretty sure this problem is P and not NP, but I'm having difficulty coming up with a polynomially bound algorithm to solve it.
You can :
check that number of edges in the graph is n(n-1)/2.
check that each vertice is connected to exaclty n-1 distinct vertices.
This will run in O(V²), which is polynomial.
Hope it helped.
Here's an O(|E|) algorithm that also has a small constant.
It's trivial to enumerate every edge in a complete graph. So all you need to do is scan your edge list and verify that every such edge exists.
For each edge (i, j), let f(i, j) = i*|V| + j. Assuming vertices are numbered 0 to |V|-1.
Let bitvec be a bit vector of length |V|2, initialized to 0.
For each edge (i, j), set bitvec[f(i, j)] = 1.
G is a complete graph if and only if every element of bitvec == 1.
This algorithm not only touches E once, but it's also completely vectorizable if you have a scatter instruction. That also means it's trivial to parallelize.
Here is an O(E) algorithm:
Use O(E) as it is input time, to scan the graph
Meanwhile, record each vertex p's degree, increase degree only if the neighbor is not p itself (self-connecting edge) and is not a vertex q where p and q has another edge counted already (multiple edge), these checking can be done in O(1)
Check if all vertex's degree is |V|-1, this step is O(V), if Yes then it is a complete graph
Total is O(E)
For a given graph G = (V,E), check for each pair u, v in the V, and see if edge (u,v) is in E.
The total number of u, v pairs are |V|*(|V|-1)/2. As a result, with a time complexity of O(|V|^2), you can check and see if a graph is complete or not.
The time complexity to go over each adjacent edge of a vertex is, say, O(N), where N is number of adjacent edges. So, for V numbers of vertices the time complexity becomes O(V*N) = O(E), where E is the total number of edges in the graph. Since removing and adding a vertex from/to a queue is O(1), why is it added to the overall time complexity of BFS as O(V+E)?
I hope this is helpful to anybody having trouble understanding computational time complexity for Breadth First Search a.k.a BFS.
Queue graphTraversal.add(firstVertex);
// This while loop will run V times, where V is total number of vertices in graph.
while(graphTraversal.isEmpty == false)
currentVertex = graphTraversal.getVertex();
// This while loop will run Eaj times, where Eaj is number of adjacent edges to current vertex.
while(currentVertex.hasAdjacentVertices)
graphTraversal.add(adjacentVertex);
graphTraversal.remove(currentVertex);
Time complexity is as follows:
V * (O(1) + O(Eaj) + O(1))
V + V * Eaj + V
2V + E(total number of edges in graph)
V + E
I have tried to simplify the code and complexity computation but still if you have any questions let me know.
Considering the following Graph we see how the time complexity is O(|V|+|E|) but not O(V*E).
Adjacency List
V E
v0:{v1,v2}
v1:{v3}
v2:{v3}
v3:{}
Operating How BFS Works Step by Step
Step1:
Adjacency lists:
V E
v0: {v1,v2} mark, enqueue v0
v1: {v3}
v2: {v3}
v3: {}
Step2:
Adjacency lists:
V E
v0: {v1,v2} dequeue v0;mark, enqueue v1,v2
v1: {v3}
v2: {v3}
v3: {}
Step3:
Adjacency lists:
V E
v0: {v1,v2}
v1: {v3} dequeue v1; mark,enqueue v3
v2: {v3}
v3: {}
Step4:
Adjacency lists:
V E
v0: {v1,v2}
v1: {v3}
v2: {v3} dequeue v2, check its adjacency list (v3 already marked)
v3: {}
Step5:
Adjacency lists:
V E
v0: {v1,v2}
v1: {v3}
v2: {v3}
v3: {} dequeue v3; check its adjacency list
Step6:
Adjacency lists:
V E
v0: {v1,v2} |E0|=2
v1: {v3} |E1|=1
v2: {v3} |E2|=1
v3: {} |E3|=0
Total number of steps:
|V| + |E0| + |E1| + |E2| +|E3| == |V|+|E|
4 + 2 + 1 + 1 + 0 == 4 + 4
8 == 8
Assume an adjacency list representation, V is the number of vertices, E the number of edges.
Each vertex is enqueued and dequeued at most once.
Scanning for all adjacent vertices takes O(|E|) time, since sum of lengths of adjacency lists is |E|.
Hence The Time Complexity of BFS Gives a O(|V|+|E|) time complexity.
The other answers here do a great job showing how BFS runs and how to analyze it. I wanted to revisit your original mathematical analysis to show where, specifically, your reasoning gives you a lower estimate than the true value.
Your analysis goes like this:
Let N be the average number of edges incident to each node (N = E / V).
Each node, therefore, spends O(N) time doing operations on the queue.
Since there are V nodes, the total runtime is the O(V) · O(N) = O(V) · O(E / V) = O(E).
You are very close to having the right estimate here. The question is where the missing V term comes from. The issue here is that, weirdly enough, you can't say that O(V) · O(E / V) = O(E).
You are totally correct that the average work per node is O(E / V). That means that the total work done asympotically is bounded from above by some multiple of E / V. If we think about what BFS is actually doing, the work done per node probably looks more like c1 + c2E / V, since there's some baseline amount of work done per node (setting up loops, checking basic conditions, etc.), which is what's accounted for by the c1 term, plus some amount of work proportional to the number of edges visited (E / V, times the work done per edge). If we multiply this by V, we get that
V · (c1 + c2E / V)
= c1V + c2E
= Θ(V + E)
What's happening here is that those lovely lower-order terms that big-O so conveniently lets us ignore are actually important here, so we can't easily discard them. So that's mathematically at least what's going on.
What's actually happening here is that no matter how many edges there are in the graph, there's some baseline amount of work you have to do for each node independently of those edges. That's the setup to do things like run the core if statements, set up local variables, etc.
Performing an O(1) operation L times, results to O(L) complexity.
Thus, removing and adding a vertex from/to the Queue is O(1), but when you do that for V vertices, you get O(V) complexity.
Therefore, O(V) + O(E) = O(V+E)
Will the time complexity of BFS be not O(V) considering we only have to traverse the vertices in the adjacency list? Am I missing something here?
For the below graph represented using adjacency list for ex:
0 ->1->2->null
1->3->4->null
3->null
4->null
While creating the graph we have the adjacency list which is an array of linked lists. So my understanding is during traversal this array is available to us and it's enough if we only traverse all the elements of this array and mark each element as visited to not visit it twice. What am I missing here?
I would just like to add to above answers that if we are using an adjacency matrix instead of a adjacency list, the time complexity will be O(V^2), as we will have to go through a complete row for each vertex to check which nodes are adjacent.
You are saying that total complexity should be O(V*N)=O(E). Suppose there is no edge between any pair of vertices i.e. Adj[v] is empty for all vertex v. Will BFS take a constant time in this case? Answer is no. It will take O(V) time(more accurately θ(V)). Even if Adj[v] is empty, running the line where you check Adj[v] will itself take some constant time for each vertex. So running time of BFS is O(V+E) which means O(max(V,E)).
One of the ways that I grasped the intuition of the time complexity
O ( V + E)
is that when we traverse the graph (let's take BFS pseudocode in Java):
for(v:V){ // segment 1
if(!v.isVisited) {
q = new Queue<>();
q.add(v);
v.isVisited = true
while(!q.isEmpty) {
curr = q.poll()
for(u: curr.adjacencyList ){ //Segment 2
//do some processing
u.isVisited = true
}
}
}
}
As, we can see there are two important segments 1 and 2 which determines the time complexity.
Case 1: Consider a graph with only vertices and a few edges, sparsely connected graph (100 vertices and 2 edges).
In that case, the segment 1 would dominate the course of traversal.
Hence making, O(V) as the time complexity as segment 1 checks all vertices in graph space once.
Therefore, T.C. = O(V) (since E is negligible).
Case 2: Consider a graph with few vertices but a complete graph (6 vertices and 15 edges) (n C 2).
Here the segment 2 will dominate as the number of edges are more and the segment 2 gets evaluated 2|E| times for an undirected graph.
T.C. of first vertex processing would be,
O(1) * O(2|E|) = O(E)
The rest of the vertex will not be evaluated for the segment 1 and would just add V-1 times of processing (since they are already visited in segment 2 which is O(V).
Thus, in this case its better to say T.C. = O(E) + O(V)
So, in the worst/best case of number of edges, we have
TC(taversing) O(E) + O(V) or
= O(E+V)
As per my understanding, I have calculated time complexity of Dijkstra Algorithm as big-O notation using adjacency list given below. It didn't come out as it was supposed to and that led me to understand it step by step.
Each vertex can be connected to (V-1) vertices, hence the number of adjacent edges to each vertex is V - 1. Let us say E represents V-1 edges connected to each vertex.
Finding & Updating each adjacent vertex's weight in min heap is O(log(V)) + O(1) or O(log(V)).
Hence from step1 and step2 above, the time complexity for updating all adjacent vertices of a vertex is E*(logV). or E*logV.
Hence time complexity for all V vertices is V * (E*logV) i.e O(VElogV).
But the time complexity for Dijkstra Algorithm is O(ElogV). Why?
Dijkstra's shortest path algorithm is O(ElogV) where:
V is the number of vertices
E is the total number of edges
Your analysis is correct, but your symbols have different meanings! You say the algorithm is O(VElogV) where:
V is the number of vertices
E is the maximum number of edges attached to a single node.
Let's rename your E to N. So one analysis says O(ElogV) and another says O(VNlogV). Both are correct and in fact E = O(VN). The difference is that ElogV is a tighter estimation.
Adding a more detailed explanation as I understood it just in case:
O(for each vertex using min heap: for each edge linearly: push vertices to min heap that edge points to)
V = number of vertices
O(V * (pop vertex from min heap + find unvisited vertices in edges * push them to min heap))
E = number of edges on each vertex
O(V * (pop vertex from min heap + E * push unvisited vertices to min heap)). Note, that we can push the same node multiple times here before we get to "visit" it.
O(V * (log(heap size) + E * log(heap size)))
O(V * ((E + 1) * log(heap size)))
O(V * (E * log(heap size)))
E = V because each vertex can reference all other vertices
O(V * (V * log(heap size)))
O(V^2 * log(heap size))
heap size is V^2 because we push to it every time we want to update a distance and can have up to V comparisons for each vertex. E.g. for the last vertex, 1st vertex has distance 10, 2nd has 9, 3rd has 8, etc, so, we push each time to update
O(V^2 * log(V^2))
O(V^2 * 2 * log(V))
O(V^2 * log(V))
V^2 is also a total number of edges, so if we let E = V^2 (as in the official naming), we will get the O(E * log(V))
let n be the number of vertices and m be the number of edges.
Since with Dijkstra's algorithm you have O(n) delete-mins and O(m) decrease_keys, each costing O(logn), the total run time using binary heaps will be O(log(n)(m + n)). It is totally possible to amortize the cost of decrease_key down to O(1) using Fibonacci heaps resulting in a total run time of O(nlogn+m) but in practice this is often not done since the constant factor penalties of FHs are pretty big and on random graphs the amount of decrease_keys is way lower than its respective upper bound (more in the range of O(n*log(m/n), which is way better on sparse graphs where m = O(n)). So always be aware of the fact that the total run time is both dependent on your data structures and the input class.
In dense(or complete) graph, E logV > V^2
Using linked data & binary heap is not always best.
That case, I prefer to use just matrix data and save minimum length by row.
Just V^2 time needed.
In case, E < V / logV.
Or, max edges per vertex is less than some constant K.
Then use binary heap.
I find it easier to think at this complexity in the following way:
The nodes are first inserted in a priority queue and the extracted one by one leading to O(V log V).
Once a node is extracted, we iterate through its edges and update the priority queue accordingly. Note that every edge is explored only once, moreover, updating the priority queue is O(log V), leading to an overall O(E log V).
TLDR. You have V extractions from the priority queue and E updates to the priority queue, leading to an overall O((V + E) log V).
Let's try to analyze the algorithm as given in CLRS book.
for each loop in line 7: for any vertex say 'u' the number of times the loop runs is equal to the number of adjacent vertices of 'u'.
The number of adjacent vertices for a node is always less than or equal to the total number of edges in the graph.
If we take V (because of while loop in line 4) and E (because of for each in line 7) and compute the complexity as VElog(V) it would be equivalent to assuming each vertex has E edges incident on it, but in actual there will be atmost or less than E edges incident on a single vertex. (the atmost E adjacent vertices for a single vertex case happens in case of a star graph for the internal vertex)
V:Number of Vertices,
E:Number of total_edges
Suppose the Graph is dense
The complexity would be O(V*logV) + O( (1+2+...+V)*logV)
1+2+....+(V-1) = (v)*(v+1)/2 ~ V^2 ~ E because the graph is dense
So the complexity would be O(ElogV).
The sum 1+2+...+ V refers to: For each vertex v in G.adj[u] but not in S
If you think about Q before a vertex is extracted has V vertices then it has V-1 then V-2
... then 1.
E is edges and V is vertices. Number of edges
(V *(V-1)) / 2
approximately
V ^ 2
So we can add maximum V^2 edges to the min heap. So sorting the elements in min heap will take
O(Log(V ^ 2))
Every time we insert a new element into min heap, we are going to sort. We will have E edges so we will be sorting E times. so total time complexity
O(E * Log(V ^ 2)= O( 2 * E * Log(V))
Omitting the constant 2:
O( E * Log(V))
I have to implement Prim's algorithm using a min-heap based priority queue. If my graph contained the vertices A, B, C, and D with the below undirected adjacency list... [it is sorted as (vertex name, weight to adjacent vertex)]
A -> B,4 -> D,3
B -> A,4 -> C,1 -> D,7
C -> B,1
D -> B,7 -> A,3
Rough Graph:
A-4-B-1-C
| /
3 7
| /
D
What would the priority queue look like? I have no idea what I should put into it. Should I put everything? Should I put just A B C and D. I have no clue and I would really like an answer.
Prim's: grow the tree by adding the edge of min weight with exactly one end in the tree.
The PQ contains the edges with one end in the tree.
Start with vertex 0 added to tree and add all vertices connected to 0 into the PQ.
DeleteMin() will give you the min weight edge (v, w), you add it to the MST and add all vertices connected to w into the PQ.
is this enough to get you started?
---
so, in your example, the in the first iteration, the MST will contain vertex A, and the PQ will contain the 2 edges going out from A:
A-4-B
A-3-D
Here's prim's algorithm:
Choose a node.
Mark it as visited.
Place all edges from this node into a priority queue (sorted to give smallest weights first).
While queue not empty:
pop edge from queue
if both ends are visited, continue
add this edge to your minimum spanning tree
add all edges coming out of the node that hasn't been visited to the queue
mark that node as visited
So to answer your question, you put the edges in from one node.
If you put all of the edges into the priority queue, you've got Kruskal's algorithm, which is also used for minimum spanning trees.
It depends on how you represent your graph as to what the running time is. Adjacency lists make the complexity O(E log E) for Kruskal's and Prim's is O(E log V) unless you use a fibonacci heap, in which case you can achieve O(E + V log V).
You can assign weights to your vertices. Then use priority queue based on these weights. This is a reference from the wiki: http://en.wikipedia.org/wiki/Prim's_algorithm
MST-PRIM (G, w, r) {
for each u ∈ G.V
u.key = ∞
u.parent = NIL
r.key = 0
Q = G.V
while (Q ≠ ø)
u = Extract-Min(Q)
for each v ∈ G.Adj[u]
if (v ∈ Q) and w(u,v) < v.key
v.parent = u
v.key = w(u,v)
}
Q will be your priority queue. You can use struct to hold the information of the vertices.
This is a problem from a practice exam that I'm struggling with:
Let G = (V, E) be a weighted undirected connected graph, with positive
weights (you may assume that the weights are distinct). Given a real
number r, define the subgraph Gr = (V, {e in E | w(e) <= r}). For
example, G0 has no edges (obviously disconnected), and Ginfinity = G
(which by assumption is connected). The problem is to find the
smallest r such that Gr is connected.
Describe an O(mlogn)-time algorithm that solves the problem by
repeated applications of BFS or DFS.
The real problem is doing it in O(mlogn). Here's what I've got:
r = min( w(e) ) => O(m)
while true do => O(m)
Gr = G with edges e | w(e) > r removed => O(m)
if | BFS( Gr ).V | < |V| => O(m + n)
r++ (or r = next smallest w(e))
else
return r
That's a whopping O(m^2 + mn). Any ideas for getting it down to O(mlogn)? Thanks!
You are iterating over all possible edge costs which results in the outer loop of O(m). Notice that if the graph is disconnected when you discard all edges >w(e), it is also disconnected for >w(e') where w(e') < w(e). You can use this property to do a binary search over the edge costs and thus do this in O(log(n)).
lo=min(w(e) for e in edges), hi=max(w(e) for e in edges)
while lo<hi:
mid=(lo+hi)/2
if connected(graph after discarding all e where w(e)>w(mid)):
lo=mid
else:
hi=mid-1
return lo
The binary search has a complexity of O(log (max_e-min_e)) (you can actually bring it down to O(log(edges)) and discarding edges and determining connectivity can be done in O(edges+vertices), so this can be done in O((edge+vertices)*log(edges)).
Warning: I have not tested this in code yet, so there may be bugs. But the idea should work.
How about the following algorithm?
First take a list of all edges (or all distinct edge lengths, using ) from the graph and sort them. That takes O(m*log m) = O(m*log n) time: m is usually less than n^2, so O(log m)=O(log n^2)=O(2*log n)=O(log n).
It is obvious that r should be equal to the weight of some edge. So you can do a binary search on the index of the edge in the sorted array.
For each index you try, you take the length of the correspondong edge as r, and check the graph for connectivity, only using the edges of length <= r with BFS or DFS.
Each iteration of the binary search takes O(m), and you have to make O(log m)=O(log n) iterations.