Dijkstra’s Algorithm

10th Friday Fun Session – 17th Mar 2017

Dijkstra’s algorithm is a Single-Source Shortest Path (SSSP) algorithm developed by Edsger Wybe Dijkstra. It uses a greedy process and yet finds the optimal solution. It looks similar to Breadth-first search.

Compare to Bellman-Ford

It is asymptotically the fastest SSSP algorithm, at a cost O(|E|+|V|log|V|), when min-priority queue implemented by Fibonacci heap is used.

That is quite cheap, given Bellman-Ford’s complexity of O(|V||E|) to find the same, something that can become prohibitively expensive for a dense graph having |V|2 edges.

However, while Bellman-Ford can work with negative edge and can detect negative cycle, Dijkstra’s algorithm cannot work with negative edge. Since it cannot work with negative edge, there is no question of detecting negative cycle at all.

Standard Algorithm

dist[] //shortest path vector

p[] //predecessor vector, used to reconstruct the path

Q //vertex set

for each vertex v in Graph
  dist[v] = ∞
  p[v] = undefined
  add v to Q

dist[s] = 0

while Q is not empty
  u = vertex with min dist[] value
  remove u from Q

  for each neighbor v of u
    alt = dist[u] + weight(u, v)
    if alt < dist[v]
      dist[v] = alt
      p[v] = u

return dist[], p[]

Given source vertex s, it finds the shortest distance from s to all other vertices. At first, it initializes dist[] vector to infinite; to mean that it cannot reach any other vertex. And sets dist[s] = 0; to mean it can reach itself at a cost of 0, the cheapest. All vertices including itself are added to the vertex set Q.

Then, it chooses the vertex with min dist[] value. At first, s (set to u) would be chosen. Then using each of the outgoing edges of u to v, it tries to minimize dist[v] by checking whether v can be reached via u using edge (u, v). If yes, dist[v] is updated. Then again it retrieves vertex u with the cheapest dist[u] value and repeats the same. This continues till Q is not empty. Whenever, a vertex u is removed from Q, it means that the shortest distance from s to u is found.

Since we are retrieving |V| vertices from Q, and for each vertex, trying with all its edges (=|V|, at max), to minimize distance to other vertices, the cost can be |V|2.

So, here we see a greedy process where it is retrieving the vertex with min dist[] value. Since retrieving a vertex u from Q means that we found the minimum distance from s to u, if we are solving shortest path from a single source s to a single destination d, then when u matches the destination d, we are done and can exit. It can also be noted that from source s, we find the shortest distances to all other vertices, in the ascending order of their distances.

Finally, we see that dist[] vector is continuously changing. And each time when we retrieve a vertex u, we choose the one with min dist[] value. That indicates using min-priority queue might be the right choice of data structure for this algorithm.

Using Fibonacci Heap

dist[] //shortest path vector
p[] //predecessor vector, used to reconstruct the path
Q //priority queue, implemented by Fibonacci Heap

dist[s] = 0

for each vertex v
  if(s != v)
    dist[v] = ∞
    p[v] = undefined
  
  Q.insert_with_priority(v, dist[v]) // insert

while Q.is_empty() = false
  u = Q.pull_with_min_priority() // find min and delete min
  
  for each neighbor v of u
    alt = dist[u] + weight(u, v)
    if alt < dist[v]
      dist[v] = alt
      p[v] = u
      Q.decrease_priority(v, alt) //decrease key

return dist[], p[]

In the above algorithm, we have used a function called decrease_priority(), something that is absent in standard priority queue but present in Fibonacci heap. So the above algorithm is implemented using Fibonacci heap.

Fibonacci heap is a special implementation of a priority queue that supports decrease key (decrease_priority()) operation. Meaning, we can decrease the value of a key while it is still inside the priority queue. And this can be achieved by using constant amortized time for insert, find min and decrease key operation and log (n) time for delete min operation.

As for cost, since we have called delete operation for each of the v vertices, and we have treated each of the |E| edges once, the cost here is O(|E|+|V|log|V|), as mentioned at the beginning of this post, as the cost of Dijkstra’s algorithm.

Using Standard Priority Queue

Standard priority queue implementation takes log (n) time for both insert and delete operation and constant time for find min operation. But there is no way to change the key value (decrease key) while the item is still in the priority queue, something Dijkstra’s algorithm might need to do quite frequently as we have already seen.

If standard priority queue is used, one has to delete the item from the priority queue and then insert into it again, costing log (n) each time, or an alternative to that effect. However, as long as standard priority queue is used, it is going to be slower than Fibonacci heap. With standard priority queue, the algorithm would look like below:

dist[] //shortest path vector
p[] //predecessor vector, used to reconstruct the path
Q //standard priority queue

for each vertex v
  dist[v] = ∞
  p[v] = undefined

dist[u] = 0
Q.insert_with_priority(u, dist[u]) // insert

while Q.is_empty() = false
  u = Q.pull_with_min_priority() // find min and delete min
  
  for each neighbor v of u
    alt = dist[u] + weight(u, v)
    if alt < dist[v]
      dist[v] = alt
      p[v] = u
      insert_with_priority(v, alt) //insert v 
                                     even if already exists 
return dist[], p[]

There are two differences from the earlier algorithm:

First, we have not inserted all vertices into the standard priority queue at first, rather inserted the source only.

Second, instead of decreasing priority that we cannot do using standard priority queue, we have kept on inserting vertex v when dist[v] decreases. That might mean, inserting a vertex v again while it is already there inside the queue with a higher priority/dist[v].

That is another way of pushing aside the old entry (same v but with higher priority) out of consideration for the algorithm. When shortest distances from source s to all other vertices v are found, those pushed aside vertices will be pulled one by one from the priority queue and removed. They will not affect dist[] vector anymore. And thus the queue will be emptied and the algorithm will exit.

Negative Edge

Please check Dijkstra’s Problem with Negative Edge for further details.

Index

Floyd-Warshall Algorithm

35th Friday Fun Session – 29th Sep 2017

Floyd-Warshall, also known as Roy-Warshall is an All-Pairs Shortest Path (APSP) algorithm developed by Robert Floyd, Bernard Roy, and Stephen Warshall. It is an example of dynamic programming that uses 3 nested loops. At a cost O(|V|3), it is quite impressive, given that Bellman-Ford might encounter the same cost (O(|V||E|)) to find only Single Source Shortest Path (SSSP) for dense graph having |V|2 edges. Floyd-Warshall can work with negative edges just like Bellman-Ford. After all, both are based on dynamic programming. As for detecting negative cycle, once again, both can detect it. However, in presence of negative cycle, results from both are invalid.

Three Nested Loops

dist[][] //shortest path matrix
p[][] //predecessor matrix, used to reconstruct the path

dist[][] = ∞

for each vertex i
  dist[i][i] = 0

for each edge (i, j)
  dist[i][j] = weight(i, j)
  p[i][j] = j

for k = 1 to |V|
  for i = 1 to |V|
    for j = 1 to |V|
      if dist[i][j] > dist[i][k] + dist[k][j]
        dist[i][j] = dist[i][k] + dist[k][j]
        p[i][j] = p[i][k]

To compute the shortest path between any pair (s, t), we have considered each of the |V| vertices as intermediate points k, and chosen the cheaper between i) existing (s, t) and ii) the sum of s to k and then from k to t, meaning s to t via k.

Short-circuiting an SSSP?

Does it mean that we can derive a SSSP solution for any pair (s, t), at a cost of O(|V|2)? To be precise, can we do the following?

for k = 1 to |V|
  if dist[i][j] > dist[i][k] + dist[k][j]
    dist[i][j] = dist[i][k] + dist[k][j]

After all, we have relaxed via all the intermediate nodes. Well, that will not work! Why?

Dynamic Programming

If we want to get the shortest path between (i, j) using k (1 to k) intermediate nodes then we have to choose the cheaper between the below paths:

  1. Without using k: dist[i][j] using intermediate nodes 1 to k-1.
  2. Using k: dist[i][k] + dist[k][j], where both dist[i][k] and dist[j][k] should make use of intermediate nodes 1 to k-1.

At k = 0, dist[][] is initialized using edge weights where exists, 0 for diagonals (dist[v][v]) and infinite for the rests.

An Example

Suppose, we want to compute dist[2][3] when k = 5.

Then, dist[2][3] = min { dist[2][3], dist[2][5] + dist[5][3] }

Here, all three distances – dist[2][3], dist[2][5] and dist[5][3] must already use intermediate nodes 1 to 4. Meaning, dist[2][5] is not the static cost set at k=0; possibly the edge cost, 0 or infinite. Rather, dist[2][5] is already computed using k from 1 to 4. Similarly, dist[5][3] (and dist[2][3] as well) is also computed using k from 1 to 4.

In other words, we cannot compute a certain dist[s][t] alone, using the intermediate nodes 1 to k. Rather for each intermediate node k, we need to compute dist[i][j] progressively, using the 3 nested loops, as shown earlier.

Obviously we can use recursion without the loops. That will not save any work for us. In fact, while using recursion, if we are not reusing existing solutions for the sub-problems, we will repeat the computation – something very expensive.

Path Reconstruction

The predecessor matrix p, keeps track of the shortest path. If we have to find the best path from s to t, we know for sure that we start with s. We print s. To know where we went from there, we have to look at p[s][t]. If that is t, we are done as that is the destination. However, if that is not the case, that means we find another node r. Then we know from s we went to an intermediate node r. So this becomes the new start s for the rest of the path. However, destination remains the same t. Again we look at p[s][t] and continue the same till we reach t, all along printing r (=p[s][t]).

Incremental Node Addition

Suppose as of now, we have 4 nodes and APSP is computed. At this point 5th node arrives, along with some edges connecting the existing nodes. Instead of computing APSP from the scratch, at a cost of O(|V|3) = O(125), we can use the already computed APSP and extend that to complete it for 5 nodes, at a cost of O(|V|2) = O(25).

Adjusting Edge Weight Changes

What if weight for an edge changes (increases or decreases)? Do we need to re-compute APSP from scratch? Or we can adjust the existing results using some partial computations?

Index

Currency Arbitrage

11th JLTi Code Jam – Jan 2018

Here we revisit Manipulating Money Exchange problem where we tried to find currency arbitrage using Bellman-Ford at a time complexity of O(|V||E|).

In general, this kind of graph can be dense. Suppose, there are 4 currencies: USD, SGD, GBP and INR. Usually, a rate is given from each currency to all other currencies, resulting in |V|2 edges. Hence, Bellman-Ford ends up with O(|V||E|) = O(|V||V|2) = O(|V|3), that is quite expensive. Specially, when you consider the fact that apart from the few hundred fiat currencies, there are 1000+ cryptocurrencies out there.

Also we should not forget that currency exchange rate is not a factor of solely the currency itself, rather it is tied with an exchange. For example, suppose, Moneycorp exchange has a USD to SGD rate 1.4 while for HiFX it is 1.396 for the same. So we see, USD appearing twice in the graph – once as part of Moneycorp and again as part of HiFX.

However, computing shortest paths, a prerequisite for finding arbitrage, is something quite expensive. In this problem, we need to incrementally compute shortest paths when a new vertex, nth one arrives, assuming we have pre-computed results for (n-1) vertices that we can re-use.

To be more specific, suppose, at this moment, we have 12344 vertices and we already know whether there is an arbitrage, after computing the necessary shortest paths. And then, a new currency, JioCoin arrives with some new rates (from JioCoin to some existing currencies, say, JioCoin to INR and from some existing currencies to JioCoin, say, SGD to JioCoin). Now we have 12345 vertices. Computing shortest paths for a dense graph with 12345 vertices would take a very long time (try running 3 nested for loops, each looping 12345 times), doing billions of computations.

At this moment, would it be not wise to use the existing results for 12344 vertices? And then incrementally adjust the new shortest paths and compute some new ones? That is precisely, this problem is all about. We need to incrementally, adjust/add shortest paths once a new vertex arrives. And this is to be done at a time complexity of O(|V|2), something that is comfortably manageable. After this, we have to now say, whether an arbitrage exists.

Input:

1 USD = 1.380 SGD

1 SGD = 3.080 MYR

1 MYR = 15.120 INR

1 INR = 0.012 GBP

1 GBP = 1.30 USD

I CAD = 0.57 GBP

Explanation: Whenever a rate arrives, starting from the first, for each new vertex, we need to incrementally adjust/add shortest paths, find whether an arbitrage exists or not and output the same. We have 6 inputs here. Each time an input comes, we need to output and hence, we have 6 lines of output. The first 4 did not result in any arbitrage, we output “No luck here”. From 5th we have an arbitrage and we output the same.

Once an arbitrage is found, it is going to last. Note that, there might exist more than one arbitrage. Printing any one will do.

An important thing: rate between a certain currency pair will not appear twice in the input. Meaning once, GBP to USD rate arrives at line 5, a new rate between the two won’t arrive again.

Output:

No luck here

No luck here

No luck here

No luck here

USD -> SGD -> MYR -> INR -> GBP -> USD

USD -> SGD -> MYR -> INR -> GBP -> USD

Input:

1 USD = 1.38295 SGD

1 SGD = 3.08614 MYR

1 MYR = 15.0996 INR

1 INR = 0.0119755 GBP

1 GBP = 1.295 USD

Output:

No luck here

No luck here

No luck here

No luck here

No luck here

Task: For each line of input, for each new vertex, incrementally adjust/add shortest paths at a cost (time) of O(|V|2), detect the presence of an arbitrage and output as specified.

Index