Difference between revisions of "CSC212 Lab 13 2014"

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(Problem #1)
(Problem #2)
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=Problem #2=
 
=Problem #2=
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* A graph is ''connected'' if there is a path from any vertex to any other vertex in the graph.
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* Create a new method called ''isConnected( )'' that is based on DFS, and that returns '''true''' if the graph is connected, and '''false''' otherwise.
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* The graph created by the '''init1()''' method is connected.  You will need to add a new method called, say, '''init2()''' that initializes the graph with several disconnected components.  (Hints: You can probably remove some edges from the graph generated by init1() to get a graph with several components!)
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=Problem #3=
 
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Revision as of 20:08, 17 November 2014

--D. Thiebaut (talk) 11:19, 17 November 2014 (EST)



Problem #1


  • Create a new program called Graph1.java and copy the code from this page.
  • Implement the DFS function, with the main function calling a recursive helper function.


Question 1
Make your new DFS start with a vertex of your choice, and display all the vertices it visits. Below is an example of the output for G.DFS( 0 ).


DFS starting on 0: 0, 1, 2, 4, 3, 7, 8, 5, 6,


Question 2
Same question, but this time make DFS print the edges of the graph it visits. Below is an example of the output for G.DFS(0).


DFS starting on 0:
visiting Edge (0)---(1)
visiting Edge (1)---(2)
visiting Edge (1)---(4)
visiting Edge (4)---(3)
visiting Edge (3)---(7)
visiting Edge (7)---(8)
visiting Edge (1)---(5)
visiting Edge (5)---(6)



Problem #2


  • A graph is connected if there is a path from any vertex to any other vertex in the graph.
  • Create a new method called isConnected( ) that is based on DFS, and that returns true if the graph is connected, and false otherwise.
  • The graph created by the init1() method is connected. You will need to add a new method called, say, init2() that initializes the graph with several disconnected components. (Hints: You can probably remove some edges from the graph generated by init1() to get a graph with several components!)


Problem #3


7-degrees of separation

Problem #3


All-Pairs Shortest Paths