Finding Shortest Path after Meeting in the Middle

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Из курса от партнера University of California San Diego

Algorithms on Graphs

978 оценки

University of California San Diego

Algorithms on Graphs

978 оценки

Курс 3 из 6 — Specialization Data Structures and Algorithms

If you have ever used a navigation service to find optimal route and estimate time to destination, you've used algorithms on graphs. Graphs arise in various real-world situations as there are road networks, computer networks and, most recently, social networks! If you're looking for the fastest time to get to work, cheapest way to connect set of computers into a network or efficient algorithm to automatically find communities and opinion leaders in Facebook, you're going to work with graphs and algorithms on graphs. In this course, you will first learn what a graph is and what are some of the most important properties. Then you'll learn several ways to traverse graphs and how you can do useful things while traversing the graph in some order. We will then talk about shortest paths algorithms — from the basic ones to those which open door for 1000000 times faster algorithms used in Google Maps and other navigational services. You will use these algorithms if you choose to work on our Fast Shortest Routes industrial capstone project. We will finish with minimum spanning trees which are used to plan road, telephone and computer networks and also find applications in clustering and approximate algorithms.

Из урока

Advanced Shortest Paths Project (Optional)

In this module, you will learn Advanced Shortest Paths algorithms that work in practice 1000s (up to 25000) of times faster than the classical Dijkstra's algorithm on real-world road networks and social networks graphs. You will work on a Programming Project based on these algorithms. You will find the shortest paths on the real maps of parts of US and the shortest paths connecting people in the social networks. We encourage you not only to use the ideas from this module's lectures in your implementations, but also to come up with your own ideas for speeding up the algorithm! We encourage you to compete on the forums to see whose implementation is the fastest one :)

Programming Project: Introduction1:33

Bidirectional Search10:15

Six Handshakes6:49

Bidirectional Dijkstra5:54

Finding Shortest Path after Meeting in the Middle9:01

Computing the Distance2:31

Познакомьтесь с преподавателями

Alexander S. Kulikov
Visiting Professor
Department of Computer Science and Engineering
Michael Levin
Lecturer
Computer Science
Daniel M Kane
Assistant Professor
Department of Computer Science and Engineering / Department of Mathematics
Neil Rhodes
Adjunct Faculty
Computer Science and Engineering

It is not always the case that the shortest

path go through the meeting point.

So what exactly then do we need to do to compute the distance

when we found the common meeting point, and is it actually possible?

Well it is possible, but it is little bit more complicated.

So let's denote by dist of u the distance estimate in

the forward Dijkstras from source s in G and by dist R of u.

The same but the distance estimated in the backward Dijkstra from T in the GR.

So dist R is an estimate of the distance from u to the target vertex T.

So after some node v is processed in both G and GR.

I say that there exists some shortest path from s to t which passes though some node

U which is processed either is the forward search or backward search or in both.

And the distance from the s to t

is equal to the distance estimate of u in the forward search,

plus the distance estimate of the backwards search from u.

What it means is that there exist some vertex u that chooses path from s to t,

at least some of this path from s to t goes through this vertex u.

And this u is either processed in G, or is processed in GR.

And also, the distance estimates for

that vertex are already correct for both forward search and backward search.

So of course, if it is processed in the forward search, for

example, then we know that the distance estimated to this node is already correct.

That the distance from source vertex to this node u

is equal to it's distance estimate dist of u.

But if it's not yet processed in the backwards search, in the reverse search,

then this is not guaranteed in the general case for the dist r.

For the distance estimate in the backwards search.

But we can show that for

some node u is actually also a correct estimate

which is equal to the distance to the distance from U to the target vertex t.

So there exists some nodes u and some shortest path that this node u lies

on the shortest path and both distance estimates from source to u and u to target

are already correct by the moment our forward search meets our backward search.

Now, let's prove this dilemma.

So, we have these two gray circles, which show us

this can't vertices by the forward search from s, and by the backward search from t.

And they touch in the vertex v, which is the meeting point.

Now first, let's prove that if there is some node u which

is not processed by forward search and is not processed by the backward search.

And there is the shortest path from s to t going through this node u,

that then there is also shortest path which goes from s to t through v.

To see that, first let's compare the first half of the shortest path from

s to t going through u to the shortest path from s to v.

We know that v is already processed in the forward search and

it means that the distance from s to v is at most the distance from s to u.

So here the green part is less than or equal to the red part.

Also if we consider the backward the extra search, we know that v Is processed,

but u is still not processed in this backward search.

So, the right half,

the green right half is also less than or equal to the red half.

So, in both halves, the green parts are less than or equal to the red parts.

And this means that the path from s to t going through v is shorter or

of the same length as the shortest path from s to t going through u.

So it means that actually the path from s to t going through v is the shortest path.

And so in this case, our theorem is proved.

If there is at least some one vertex u which is not processed by forward and

not processed by backwards search.

And there's a shortest path going through the vertex then there is a shortest path,

another one which goes through the vertex v.

Which is already processed by both forward and backward search.

And of course both distance estimates of v are already correct.

Because it's already processed in both backwards and forward search.

Now let's consider another case when there are no

varices which are unprocessed by both forward and backward searches.

But we already meet a point v.

So let's consider then any shortest spot from s to t, and let's consider the last

vertex u on such shortest spot that is processed by the forward surge.

So u is the last vertex which is on the shortest path,

which is processed by the forward search.

And then the distance estimates of u in the forward search is already correct.

This is equal to the distance from s to u.

Let's consider the next edge of the shortest path.

It goes to some vertex w.

And this vertex w has to be processed by the backward search.

Why? Because w is not processed by the forward

search because u is the last vertex of the shortest path

which is processed by the forward surge.

And w cannot be unprocessed by both forward and

backward search, because it lies on the shortest path.

And so w is indeed inside the gray circle around t,

it is indeed already processed by the backward search.

And so the distance estimate of the backwards search for w is already correct.

It is already equal to the distance from w to t.

Now, what we know is that the distance between s and

t is equal to the distance from s to u.

Which is equal to the distance estimate of u,

dist of u plus length of the edge from u to w.

Plus distance from w to t which is already

equal to the distance estimate of the backward search, dist R of w.

And now what we want to prove is that this is also equal to the sum of distance

estimate of u and the reverse distance estimate of u.

So the first sum is the same.

And now I claim that the distance estimate in the reverse direction

is equal to the sum of the two other summons in the first equation.

Why is that?

Well from one point of view we know that the distance estimate of u

is always bigger or equal to the real distance from u to t.

Because the distance estimate in any moment

of Dijkstra's algorithm is greater or equal to the real distance.

From the other hand,

we know that the vertex w is already processed in the reversed search.

And when it was processed, in particular,

the edge from u to w was relaxed in the backwards search.

And when it was relaxed it made sure that the distance

estimate of u in reverse direction is at most this sum

of length of this edge plus the distance estimate for w.

So we know that distance estimate of u is at most.

A length of edge from u to w plus distance estimate of w.

But this sum is the length of the shortest path from u to t going through w.

So in fact, we know that distance estimate of u is both greater or

equal than the true shortest path from u to t.

And also is less than or equal to the shortest path from F from u to t.

And this means that it is exactly equal to the shortest path from u to t.

And that the distant from s to t is actually equal

to the distance estimate a u plus the reverse distance estimated of u.

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