Semester 10
1 / 14

Semester 10 - PowerPoint PPT Presentation

  • Uploaded on

Semester 10. Time sure flies. PotW Solution. One possible solution is to randomly search the grid: At each point in your search, look at the (up to four) empty neighboring spaces For each neighbor, conduct a flood-fill, and count the number of empty spaces in that region

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

PowerPoint Slideshow about 'Semester 10' - spence

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Semester 10

Semester 10

Time sure flies.

Potw solution
PotW Solution

  • One possible solution is to randomly search the grid:

    • At each point in your search, look at the (up to four) empty neighboring spaces

    • For each neighbor, conduct a flood-fill, and count the number of empty spaces in that region

      • This tells you the best possible result if you go in that direction

      • You might call this the "heuristic" function

    • Out of the neighboring spaces with the maximal heuristic, pick a random one.

    • If you have no empty spaces, you restart the search, and keep looking until time runs out

Potw solution example
PotW Solution - Example

  • Orange is travelled path

  • Red is suboptimal – green is better

General shortest path info
General Shortest Path Info

  • A shortest path between two nodes on a graph is exactly what it sounds like

    • "Distance" in this case is measured as the sum of edge weights

  • Note that this measure is still well-defined if edges are:

    • unidirectional: you can only travel across certain edges in one direction

    • negative: travelling across the edge sends you backwards in time (!)

Dijkstra s shortest path algorithm
Dijkstra’s ShortestPath Algorithm!

 not the shortest title

  • Given a source node s, Dijkstra's can quickly tell you the distance from a single source node to every other node

  • Dijkstra's Algorithm works for all graphs, except those that have negative edges

  • At each step of Dijkstra's, the algorithm maintains a list of distances, dist, from the source to every other node

    • dist is not quite finalized until the end of the algorithm

    • It is instead improved over time

    • dist(source,source) is initialized as 0

    • dist(source, x) is initalized as infinity for all other x

Dijkstra s algorithm cont
Dijkstra’s Algorithm (cont.)

  • It is, in essence, a greedy algorithm

    • At each step of the algorithm, it finds the closest node, cur, and marks it as visited

      • This node's distance is now finalized

    • It then updates all neighbors of cur, neigh

      • Update dist(source,neigh) using cur as the intermediary node

    • This greediness only works if all edge weights are nonnegative

  • "Updating" the distance from a to b given an intermediary m:

    • dist(a,b)=min(dist(a,b),dist(a,m)+dist(m,b))

    • Note that order matters: dist(x,y) may not be the same as dist(y,x)


  • Number of nodes n

    • Number of edges: m

  • Note that you visit each node exactly once when you perform updates

  • O(n2 + m) with looping to find node with minimum distance

    • Look over each node O(n) times

  • O((m + n) logn) with binary heaps - found in Java's PriorityQueue

  • O(m + nlogn) with complex data structures

    • No advantage in practice over previous option

    • This is essentially optimal in terms of theoretical complexity

Using priorityqueue
Using PriorityQueue

  • O(n2+m) can be turned into O((m+n) logn):

    • The PriorityQueue data structure helps you maintain the smallest item in a dynamic (changing) set

    • In this case, it helps you find the closest node quickly

    • Every time you update a node's distance, add it to the queue: this takes O(mlogn) time

      • Note that this allows duplicate items

      • This is necessary because only the top item in the queue is accessible

      • Just make sure you skip over duplicates when you query (peek) later on

    • Every time you query the closest node, pop it from the queue: this takes O(nlogn) time

Examples extensions

  • To reconstruct the shortest path:

    • For each node, remember its "parent" node

    • Parent = the node that was last used to update its distance

  • Usually, in contest problems, constructing the graph is the hardest part of the problem

    • What if you're allowed to magically skip (teleport across) up to k edges?

    • What if you multiply edge weights instead of adding?

  • Extension: In 1994, David Eppstein published an elegant solution for the kth shortest path

    • Amazingly, its complexity is basically the same as Dijkstra's: O(m+nlogn+k)


  • Today is the last day to take the January USACO!

    • 4 hours instead of 3! (supposedly b/c problems are tougher than usual)

    • As usual, participation is worth 5 points of PotW credit

  • December and January USACO problems will be covered next meeting

Potw round trip
PotW: Round Trip

Today, Bessie wants to take a trip to a city, but doesn't know where yet. She has x dollars to spend, and the airline she is using charges $1 per mile. Tell her how many cities she might be able to visit, if she starts at node #1. Please note that all possible flights are one-way only, and after visiting her destination, she must also return. She can also take as many flights as she wants. Please also include node #1 in your count.

Round trip details
Round Trip: Details

  • Assume that every city is visitable from every other city given an infinite amount of money

  • Also assume that all plane flight distances are less than 1000

  • Let N=# of cities, and M=# of flights

  • For 25 points, solve for N<100 and M<100

  • For 40 points, solve for N<10000 and M<100000

Sample input output
Sample Input/Output

  • Input:11 (x)4 6 (# of cities, # of flights)1 2 1 (flight #1: city #1, city #2, then distance in miles)2 3 3 (flight #2)1 3 6 (etc.)3 1 73 4 64 1 2

  • Output:3 (she can visit nodes #1,2, or 3, but not 4)


  • Note that this is very similar to a standard Dijkstra's problem

    • cities = nodes, and flights = unidirectional edges

    • However, how do you take into account Bessie having to come back from her destination?

  • Run Dijkstra's twice:

    • Once with the given edges, and once with the given edges, but reversed

    • For every city, add these two distances up, and if they're less than or equal to x, you can indeed visit the city.