CS2 in C++ Peer Instruction
Materials by Cynthia Bailey Lee is
licensed under a Creative Commons
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at http://peerinstruction4cs.org.
CS106X –
Programming
Abstractions in C++
Cynthia Bailey Lee
Animation slides by Keith
Schwarz
A* and Dijkstra’s
Close cousins
●
●
●
●
Mark all nodes as gray.
Mark the initial node s as yellow and at candidate distance 0.
Enqueue s into the priority queue with priority 0.
While not all nodes have been visited:
●
Dequeue the lowest-cost node u from the priority queue.
●
●
●
Color u green. The candidate distance d that is currently stored for node u
is the length of the shortest path from s to u.
If u is the destination node t, you have found the shortest path from s to t
and are done.
For each node v connected to u by an edge of length L:
If v is gray:
–
●
Color v yellow.
●
Mark v's distance as d + L.
●
Set v's parent to be u.
●
Enqueue v into the priority queue with priority d + L.
If v is yellow and the candidate distance to v is greater than d + L:
–
●
Update v's candidate distance to be d + L.
●
Update v's parent to be u.
Dijkstra's
Algorithm
●
●
●
●
Mark all nodes as gray.
Mark the initial node s as yellow and at candidate distance 0.
Enqueue s into the priority queue with priority h(s, t).
While not all nodes have been visited:
●
Dequeue the lowest-cost node u from the priority queue.
●
●
●
Color u green. The candidate distance d that is currently stored for node u
is the length of the shortest path from s to u.
If u is the destination node t, you have found the shortest path from s to t
and are done.
For each node v connected to u by an edge of length L:
If v is gray:
–
●
Color v yellow.
●
Mark v's distance as d + L.
●
Set v's parent to be u.
●
Enqueue v into the priority queue with priority d + L + h(v, t).
If v is yellow and the candidate distance to v is greater than d + L:
–
●
Update v's candidate distance to be d + L.
●
Update v's parent to be u.
A* Search
A* on two points where the heuristic is slightly misleading
due to a wall blocking the way
A* starts with start node yellow, other nodes grey.
A*: dequeue start node, turns green.
1+
6?
1+
6?
1+
4?
1+
6?
A*: enqueue neighbors with candidate distance + heuristic
distance as the priority value.
1+
6?
1+
6?
1
1+
6?
A*: dequeue min-priority-value node.
1+ 2+
6? 5?
1+
6?
1
2+
3?
1+ 2+
6? 5?
A*: enqueue neighbors.
1+ 2+
6? 5?
1+
6?
1
2
1+ 2+
6? 5?
A*: dequeue next lowest priority value node. Notice we are making a
straight line right for the end point, not wasting time with other directions.
1+ 2+ 3+
6? 5? 4?
1+
6?
1
2
1+ 2+ 3+
6? 5? 4?
A*: enqueue neighbors—uh-oh, wall blocks us from
continuing forward.
3+
8?
3+ 4+ 5+ 6+ 7+
8? 7? 6? 5? 4?
3+
7+
3
4
5
6
2
8?
2?
7+
3+
2
2
1
3
2?
8?
2
1
3+
8?
2
3+
8?
1
2
1
2
8
2
3
7
3
4
5
6
8+
1?
7
3+ 4+ 5+ 6+ 7+ 8+
8? 7? 6? 5? 4? 3?
8+
3?
A*: eventually figures out how to go around the wall, with
some waste in each direction. But…compare to Dijkstras…
8
7
6
5
4
5
6
7
8
9?
7
6
5
4
3
4
5
6
7
8
9?
6
5
4
3
2
3
4
5
6
7
8
5
4
3
2
1
2
3
7
8
9?
4
3
2
1
1
2
8
5
4
3
2
1
2
3
7
8
9?
6
5
4
3
2
3
4
5
6
7
8
7
6
5
4
3
4
5
6
7
8
9?
8
7
6
5
4
5
6
7
8
9?
9?
9?
For Comparison: What Dijkstra's Algorithm Would Have Searched
Minimum Spanning Tree
A spanning tree in an undirected graph is a set of
edges with no cycles that connects all nodes.
B
1
D
3
7
3
6
A
C
3
E
3
How many distinct spanning
trees are in this graph?
A. 0-1
B. 2-3
C.4-5
F
D. 6-7
E. >7
Edges:
(A,B)=1
(A,C)=3
(B,C)=6
(B,D)=3
(C,E)=3
(D,E)=3
(D,F)=7
Minimum Spanning Tree
A minimum spanning tree (or MST) is a
spanning tree with the least total cost.
Note: This may sound similar to the shortestpaths tree that Dijkstra’s made, but an MST
may not be a shortest-paths tree, and a
shortest-paths tree may not be an MST.
B
1
D
3
7
3
6
A
C
3
E
3
How many distinct MSTs are
in this graph?
A. 0-1
B. 2-3
C.4-5
F
D. 6-7
E. >7
Edges:
(A,B)=1
(A,C)=3
(B,C)=6
(B,D)=3
(C,E)=3
(D,E)=3
(D,F)=7
True or False?
 Having
at least two edges with the same
weight is SUFFICIENT for having more than
one distinct minimum spanning tree.
A.
B.
TRUE!
FALSE!
(i.e., “If at least two edges have the
same weight, then there is more than
one MST”)
True or False?
 Having
at least two edges with the same
weight is NECESSARY for having more than
one distinct minimum spanning tree.
A.
B.
TRUE!
FALSE!
(i.e., “If there is more than one MST,
then at least two edges have the
same weight.”)
Kruskal’s algorithm
 Remove
all edges from graph
 Place all edges in a PQ based on
length/weight
 While !PQ.isEmpty():



Dequeue edge
If the edge connects previous disconnected
nodes or groups of nodes, keep the edge
Otherwise discard the edge
Kruskal's with Clusters (i.e. how to know
“If the endpoints of the edge aren't
already connected to one another”)
 Place
every node into its own cluster
 Place all edges into a priority queue
 While there are two or more clusters
remaining:


Dequeue an edge from the priority queue
If its endpoints are not in the same cluster:
 Merge
the clusters containing the endpoints
 Add the edge to the resulting spanning tree
Kruskal’s algorithm
Prim’s algorithm
 Choose
a start node from the graph, which
becomes our starting tree of size 1 (doesn’t
matter which one you pick)
 While tree.size() < graph.size():


Examine all outgoing edges from the tree
(edges that connect a node in the tree with a
node not in the tree)
Choose the shortest edge and add it (and the
node it connects to) to the tree
Prim’s algorithm
The Good Will Hunting
Problem
“Draw all the homeomorphically
irreducible trees with n=10.”
 In
this case “trees” simply means graphs with no
cycles
 “with n = 10” (i.e., has 10 nodes)
 “homeomorphically irreducible”

No nodes of degree 2
 For
this problem, nodes of degree 2 are useless in
terms of tree structure—they just act as a blip on an
edge—and are therefore banned

Have to be actually different
 Ignore
superficial changes in rotation or angles of
drawing