## Friday, 24 August 2012

### Maximum Flow / Maximum Bipartite Matching

Recently I was trying to understand a maximum flow algorithm well enough to apply it. Most resources I found online were not that easy to follow, or were 1+ hour long videos. So, I decided to write about it for future reference, and to hopefully make it easier for others.

## Maximum Flow

### The problem:

Imagine that you work in central London. Every morning you take the tube to get to work. Once the train arrives, a lot of people try to get from the platform to the exit. The problem is, everyone walks very slowly cause it's so busy. There are multiple possible paths from the platform to the exit, each consisting of a number of corridors and there are points where these paths intersect forming crossroads. Each corridor can only be walked in one direction and has a limited capacity so that only up to a number of people can walk in it at the same time.

Because you're smart and like algorithms, TFL decided to give you the task of calculating the optimal way in which people can be sent along the various paths, so that the maximum number of people can get from the platform to the exit at the same time.

 A simple network flow graph

A network flow graph G = (V, E) is a directed graph with two special vertices: the source vertex s, and the sink (destination) vertex t. Each other vertex represents a crossroad. An edge (u,v) in the graph means that there is a corridor from u to v. Each edge has an associated capacity which is finite, representing the amount of people that can use the corridor.

### A first attempt:

So, how can you approach this problem? A straightforward solution is to do the following:
1. keep finding paths from the platform to the exit where you can send people along
2. send as many people as possible along each path
3. update the flow graph afterwards to account for the used space.

Let's randomly choose the path s -> u -> v -> t. The capacities along this path are 3, 3, 4 respectively. But we can only send as many people as the smallest corridor can take, which is 3. This is the bottleneck capacity.
Now we send 3 people along this path and try to update the graph to reflect that we've used up 3 out of the total capacity. A way to do that is to decrease the capacity of each corridor we used by 3.

 Every edge is annotated with how much flow we sent out of the total capacity

Now the only other path left is s->v->t. The edge (s,v) has capacity 2, and the edge (v,t) now has capacity 1, because we have already sent 3 people along this corridor.
With the same update procedure we obtain this graph:
 Algorithm ends, but this is not optimal

Now our algorithm ends because there aren't any other paths from s to t with free space.

Is our solution the optimal?

It turns out we could have done better. If we had only sent 2 people on (u,v) and diverted the third person to (u,t), then we would have opened up a new space in both edges (u,v) and (v,t). We could then send one more person on the path s->v->t, increasing our total flow to 5.
The optimal solution is the following:
 The optimal solution

What went wrong with our algorithm? Let's take some time to observe what is exactly the difference between the suboptimal and the optimal solution.

One way to interpret the difference is this:

We try to send one more person along s->v. He gets to v easily as there is available capacity. But then at the v crossroad, we have 2 people coming from s, and 3 people coming from u. As (v,t) has only a capacity of 4,  they can't all continue forward.

So, we push back ask politely one of the three people who came from u, to go back to u and find a new path. The only other available edge is (u,t).

### A better attempt:

It turns out that this fix yields a correct solution to the maximum flow problem.

I'll summarize the algorithm:
1. Initially, the flow on every edge is 0. We associate a capacity function along each edge c(u,v), that tells us how many units of flow can go from u to v. Initially, c(u,v) is set to the maximum capacity for each edge (u,v) and the flow f(u,v) on every edge is 0.
2. Keep finding paths from s to t using two types of edges: (these are called augmenting paths)
1. Forward edge (u,v) if f(u,v) < c(u,v).
That's the normal case where we can send a person along an edge as long as it's not filled to capacity.
2. Backward edge (v,u) if f(u,v) > 0.
That's the case where a person that originally went from u to v (thus f(u,v)>0), has to be sent back to u.
3. Send flow along the path equal to the bottleneck capacity. The bottleneck capacity is the minimum capacity of all edges in the path. That is
1. for forward edges: c(u,v) - f(u,v)
2. for backward edges: f(v,u)
4. Update the graph:
1. increase flow on forward edges
2. decrease flow on backward edges

#### Example implementation:

We use a BFS to find an augmenting path in each step from s to t. In prev[] we store the parent of each node, and -1 means it's unvisited. Once we find a path to t, we trace it back to s using prev[] in order to calculate the bottleneck.

This algorithm is called the Ford-Fulkerson Algorithm after its inventors.

### Time complexity

The running time of Ford-Fulkerson is O((n+m) * F), where m is the number of edges, n the number of nodes, and F is the maximum flow in the graph. This is because we can do at most F augmenting paths. Each one being a BFS needs O(n+m) time. (In the snippet above though the BFS takes O(n^2) because it uses an adjacency matrix).

However Edmonds and Karp proved that if we always find the shortest augmenting path first, then we can achieve a better bound. Since BFS already finds the shortest path in an unweighted graph, the implementation above is already the Edmonds-Karp algorithm. This runs in O(m^2 * n) time with an adjaceny list and O(m*n^3) with an adjacency matrix. There are better and more complicated algorithms that solve the maximum flow problem in O(m*n^2) and even O(n^3) that can be found in CLRS.

## Maximum Bipartite Matching

### The problem:

Let's consider another problem. You finally get out of the tube and go to your office ready for some coding. But today everyone must pair program! There are two development teams, A and B, and in order to increase collaboration between the teams, today people from team A must pair with people from team B. The problem though is that every developer from team A, only likes to pair with a subset of developers from team B, and vice versa.
Your boss knowing you like graph algorithms, gives you a list of people that each developer likes to pair with, and asks you to find what is the maximum number of developer pairs between team A and B, so that in each pair both developers like to pair program with each other.

#### Some terminology:

A bipartite graph is a graph G=(V,E) in which it is possible to split the set of vertices into two non-empty sets, L and R, such that all of the edges in E have one endpoint in L and the other in R.
A matching in any graph is any subset, S, of the edges chosen in such a way that no two edges in S share an endpoint.
A maximum matching is a matching of maximum size.

### The (same) solution:

In general, the problem of finding a maximum matching in a graph is NP-hard, but if the graph is bipartite there is a polynomial solution.

The problem above can be modelled with a bipartite graph. On the left side we have developers from team A, on the right side we have developers from team B, and there is an edge between two of them if they both like to pair program with each other. The answer we are looking for is the maximum matching.

This problem reduces to Maximum Flow. Draw the bipartite graph  with team A developers on the left, and team B developers on the right. Give each edge a capacity of 1. Add a vertex s on the far left and connect it to every developer of team A by an edge of capacity 1. Add a vertex t on the far right and connect it to every developer of team B by an edge of capacity 1. Finally, find the maximum flow from s to t.

#### Example implementation

We could have just used the same code as in the maximum flow problem, but this is a simpler problem because every capacity is 1, so we can simplify the code a lot. Most of the work is done by the bpm(u) function that tries to match the left vertex u to something on the right side. It tries all the right vertices v and assigns u to v if v is unassigned, or if v's match on the left side can be reassigned to some other vertex on the other side. This is just another way of implementing DFS.

This concludes this post on Maximum Flow, I hope you enjoyed it.

## Sunday, 22 July 2012

### Factorial base numbers and permutations

This is something I came across while doing a project euler problem. The problem was asking you to find what is the millionth lexicographic permutation of the digits 0 to 9. After solving the problem by writing a method that generated the next permutation of a number and calling it 999,999 times, I was browsing the thread with other people's solutions, and came across someone who used factoradic numbers to solve it. I was quite intrigued as I hadn't heard of this number system before and had no idea how it can help us generate the millionth permutation. I'm sure you are intrigued too, dear reader.

The idea is simple. In this number system, the ith number from the right has base i.

This sounds simple, but it makes one rethink how numbers work. We are used to constant base numbers, where the ith digit from the right has a place value of base^(i-1). However in factoradic, the ith digit from the right has a place value of (i-1)!. Why, is not immediately obvious, but if we go back and think why do decimal numbers' digits have the place values they do, we can see the parallelism.

#### How base ten numbers work

In decimal, if you had just one digit, you could only represent the numbers from 0 to 9. If you wanted to keep counting further than 9, one way would be to use another, different kind of digit, one that you would use to count the number of Tens. Why Tens though and not Nines or Elevens? Because 10 is 1 more than the maximum value you could represent with the previously existing digits. If we decided to count the number of Nines instead with that new digit, then we would end up with two different representations for the number nine: 09 and 10 would be the same. On the other hand, if we decided to count the number of Elevens with that new digit, then we would have no way to represent the number ten. So now you  have two digits, each one with 10 possible values and you can represent up to 10*10 different numbers, from 0 to 99. If you wanted to keep going further, you would have to add another yet different kind of digit to count the 100s, and so on.

#### How factorial base numbers work

Let's adapt that to factoradic. The first digit has base 1, with only one possible value: 0. That's not so useful, as you can't count anything. It can only be zero, nothing else. So we add another kind of digit to count the number of Ones. But this new digit will have a different base: base 2. So it has two possible values: 0 and 1. As we have a digit with 2 possible values, and another with 1 possible value, we can create a total of 2*1 = 2! different numbers. Their values range from 0 to 2!-1.

If we want to count further than that, we add another digit, which will have base 3 this time, and possible values: 0, 1, 2. With this digit, we will count the number of 2!s, which is the previously maximum representable value plus 1. Now we can represent 3*2*1 = 3! different numbers ranging from 0 to 3!-1.
And so on.
• The ith digit from the right has place value (i-1)! as it counts the number of (i-1)!s.
• The maximum value we can represent with n digits is :  (n-1)*(n-1)! +  (n-2)*(n-2)! +.. + 3*3! + 2*2! + 1*1!
• To convert a 7 digit number like 4533210 from factoradic to decimal we do  4 * 6!  + 5 * 5! + 3 * 4! + 3 * 3! + 2 * 2! +1 * 1! + 0 * 0! = 3575
Another fact which makes this number system work, and which can be proven by induction is that
which means that the biggest number you can represent with n digits in factoradic, is equal to the smallest number you can represent with n+1 digits minus 1. This establishes that there is a unique representation for every integer in factoradic.

Now the reason any of this is relevant to the permutations project Euler problem, is that there is a natural mapping between the numbers with n digits in factorial representation, and the permutations of n elements in lexicographical order.

A permutation of a set of objects is a rearrangement of these objects into a particular order.

For example, the set {0, 1, 2} has 3! permutations. In lexicographical (or alphabetical) order, they are the following:

```Lexicographic permutation   Index (decimal)  Index (factoradic)
```
`012                         0                000`
`021                         1                010`
`102                         2                100`
`120                         3                110`
`201                         4                200`
`210                         5                210`

There is a way to calculate a permutation by its index without enumerating all the previous ones! Let's say  you want to know what is the 4th lexicographical permutation of the digits in the set {0, 1, 2}. That is, the permutation with index 3, as they are zero indexed.

The first thing to do is convert decimal 3 to its factoradic representation and get 110. Then in order to calculate the permutation we follow these steps:

 Available set elements in lexicographic order {0, 1, 2} Factoradic index 110 Permutation ?

• The leftmost digit of the factoradic number, is the same as the leftmost digit of the permutation. In our case that's "1". We use this digit as the first digit in the permutation, and delete it from the set of available elements.

 Available set elements in lexicographic order {0, 1, 2} Factoradic index 110 Permutation 1..

• At this point our available set elements are {0, 2} and the unused digits in the factoradic number are 110. Now, the cool bit is that if we view the set {0, 2} as a zero-indexed list, and if we view the remaining digits in the factoradic number as indices to this list, then each successive digit defines the next number of the permutation!
So, looking at the factoradic number  110 the next digit to use is 1, which means that the next number in the permutation is the number with index 1 from the set {0, 1, 2}. That is 2. We add this number to the permutation, delete it from the set elements, and mark it as used in the factoradic number.Now we have:

 Available set elements in lexicographic order {0, 1, 2} Factoradic index 110 Permutation 12..

• Looking at the factoradic number 11the next digit in the permutation is the one with index 0 in the list  {0, 1, 2}.  That is 0. We delete this element from the set and put it in the permutation.

 Available set elements in lexicographic order {0, 1, 2} Factoradic index 110 Permutation 120

And that's it! We calculated that the permutation with index 3 is 120 which can be verified from our table.

### Converting decimal to factorial base

When we did the conversion above, the first step was to convert the index of the wanted permutation from decimal to factorial base. But how do we do that?

Fortunately, the old method of successive integer divisions by the place value for each position still works.

Let's see an example. Let's convert 81 decimal to factoradic.

81 is less than 5! (=120) but more than 4!(=24) which means it should be possible to represent it using 5 digits in factoradic, because as we saw, the ith digit from the right has place value (i-1)!

Since our number has 5 digits, it will be of the form d4 * 4! + d3 * 3! + d2 * 2! + d1 * 1! + d0 * 0!

Performing successive divisions by the place value for each position we get
81 = 3 * 4! + 9
9 = 1 * 3! + 3
3 = 1 * 2! + 1
1 = 1 * 1! + 0
0 = 0 * 0! + 0

So the result is 31110 in factoradic.

Let's convert it back to decimal to check we got the right number:

3*4! + 1*3!  + 1*2! + 1*1! + 0*0!  equals 81 indeed.

### So how can you find the millionth permutation fast?

Going back to the original problem of finding the millionth lexicographic permutation of the digits 0-9, now with our newly acquired factoradic skills, we can find the answer by simply doing