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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

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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.