CS 309s Problem We have sprinklers in our yard which need to be connected with pipe Assuming pipes may only intersect at sprinkler heads what is the minimum amount of pipe we need to connect the sprinklers ID: 646626
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Slide1
Graphs 2
Kevin Kauffman
CS 309sSlide2
Problem
We have sprinklers in our yard which need to be connected with pipe. Assuming pipes may only intersect at sprinkler heads, what is the minimum amount of pipe we need to connect the sprinklers?Slide3
Minimum Spanning Tree
Set of edges which
Connect all the nodes
Has the minimum total weight of all edges
Contains no cycles
If there are multiple MSTs for a given graph, they all contain the same individual edge weights
If on an
xy
plane, it will not self intersectSlide4
Calculating the MST
Create a set of connected nodes
Add a random node to “seed” the set
While the set doesn’t contain all nodes
Find the shortest edge which goes from a node in the set to a node out of the set
Add the edge to the MST and the other node to the setSlide5
Implementation
Naïve: iterate through all edges in each cycle, and take the minimum which satisfies the in/out requirement
O(V*E)
Better: maintain sorted list of edges which touch the nodes in the set, add to it as we go
O(E*log(E))Slide6
Maintaining Sorted List
Use
int
[2] or
java.awt.point
to represent edges
Place them on a
TreeSet
when we see them, which will sort them in the order specified by our:
Custom ComparatorSlide7
Custom Comparator
Takes input A and B; returns an
int
Returns positive if A comes before B
Returns 0 if A is the same as B
Returns negative if A comes after B
Rules:
Commutative: if comp(A,B)<0, then comp(B,A)>0
MUST break ties (or comparator will think A and B are the same and delete one of them)
This must still adhere to rule 1Slide8
Shortest Distance
BFS allows us to calculate distance in
unweighted
graphs, how do we do it in weighted graphs?
For an edge of length n, insert n-1 “fake” nodes, so all edges are length 1
Graph is now
unweighted
Run BFS!Slide9
A Better Way
If we add nodes proportional to edge length, runtime now depends on edge length (bad for long edges)
Instead of iterating over all the fake nodes can we calculate the “real” node which we are going to reach next?Slide10
Dijkstra’s Algorithm
Lemma: if we are at a node, we can calculate the next node BFS would visit by comparing edge lengths
Start at start node
Sort edges from start node to find next node visited (second node)
How do we find the
the
third node, since it could be connected to the second node, but not the start node?Slide11
Dijkstra’s Algorithm
Because of Lemma, we can calculate the distance from the second node to all the nodes its connected to
Total distance to those nodes is distance(2
nd
node) + weight(edge)
EXCEPT if we had a shorter distance directly from the start node
Sort all those distances and find the actual third node visited
REPEATSlide12
Dijkstra’s TL-DR
Tree set of nodes by distance
Pop shortest remaining distance (u), make it permanent
for all edges from u to an unvisited node (v), if dist(u)+dist(
uv
)<dist(v), update dist(v)
O((E+V)log(E)), doesn’t depend on edge lengthSlide13
Implementation
Array storing distances (must be final since we reference it in the
compartor
)
Treeset
sorting unvisited nodes by distance
Every time we pop a node and check all its edges, if we find that ends up being less than current distance:
Remove that node from the
treeset
Update its value
Add it to the
treeset
TREESET DOESN’T LIKE VALUES CHANGED UNDER IT