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IEOR266 © 2008 1 Classification of MCNF problems
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IEOR266 © 2008 2 The Bridges of Koenigsberg: Euler 1736 “Graph Theory” began in 1736 Leonard Euler –Visited Koenigsberg –People wondered whether it is possible to take a walk, end up where you started from, and cross each bridge in Koenigsberg exactly once –Generally it was believed to be impossible
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IEOR266 © 2008 3 The Bridges of Koenigsberg: Euler 1736 A D C B 12 4 3 7 6 5 Is it possible to start in A, cross over each bridge exactly once, and end up back in A?
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IEOR266 © 2008 4 The Bridges of Koenigsberg: Euler 1736 A D C B 12 4 3 7 6 5 Conceptualization: Land masses are nodes
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IEOR266 © 2008 5 The Bridges of Koenigsberg: Euler 1736 12 4 3 7 6 5 Conceptualization: Bridges are edges A C D B
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IEOR266 © 2008 6 The Bridges of Koenigsberg: Euler 1736 12 4 3 7 6 5 Translation to graphs or networks: Is there a walk starting at A and ending at A and passing through each edge exactly once? Why isn’t there such a walk? A C D B
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IEOR266 © 2008 7 Adding two bridges creates such a walk A, 1, B, 5, D, 6, B, 4, C, 8, A, 3, C, 7, D, 9, B, 2, A 12 4 3 7 6 5 A C D B 8 9 Here is the walk. Note: the number of edges incident to B is twice the number of times that B appears on the walk.
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IEOR266 © 2008 8 Eulerian cycle: a closed walk that passes through each edge exactly once Degree of a node = number of edges incident to the node Necessary condition: each node has an even degree. Why necessary? The degree of a node j is twice the number of times j appears on the walk (except for the initial and final node of the walk.) This condition is also nearly sufficient. If every node has even degree, and if the graph is connected, then there is an eulerian cycle.
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IEOR266 © 2008 9 Eulerian cycles Eulerian cycles and extensions are used in practice Mail Carrier routes: –visit each city block at least once –minimize travel time –other constraints in practice? Trash pickup routes –visit each city block at least once –minimize travel time –other constraints in practice?
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IEOR266 © 2008 10 Eulerian cycle: a closed walk that passes through each arc exactly once Degree of a node = number of edges incident to the node Necessary condition: each node has an even degree. Why necessary? The degree of a node j is twice the number of times j appears on the walk (except for the initial and final node of the walk.) Theorem.A graph has an eulerian cycle if and only if the graph is connected and every node has even degree.
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IEOR266 © 2008 11 Eulerian path: a walk that is not closed and passes through each edge exactly once Theorem. A graph has an eulerian path if and only if exactly two nodes have odd degree and the graph is connected.
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IEOR266 © 2008 12 Exercise: Does the graph below have an Eulerian path or cycle? If so, find it.
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IEOR266 © 2008 13 Eulerian cycles Eulerian cycles and extensions are used in practice Mail Carrier routes: –visit each city block at least once –minimize travel time –other constraints in practice? Trash pickup routes –visit each city block at least once –minimize travel time –other constraints in practice?
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IEOR266 © 2008 14 The Traveling Salesman Problem : find a tour that visit all cities and minimizes the total distance traveled.
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IEOR266 © 2008 15 More on cycles A hamiltonian cycle is a cycle that passes through each node of the graph exactly once.
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IEOR266 © 2008 16 Hamilton’s Around the World Game In 1857, the Irish mathematician, Sir William Rowan Hamilton invented a puzzle that he hoped would be very popular. The objective was to make what we just called a hamiltonian cycle. The game was not a commercial success, especially the 3D version. But the mathematics of hamiltonian cycles is very popular today.
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IEOR266 © 2008 17 Hamilton’s Around the World Game We will see this problem again when we generalize it to be the traveling salesman problem.
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IEOR266 © 2008 18 The knight’s tour problem Can a knight visit all squares of a chessboard exactly once, starting at some square, and by making 63 legitimate moves? The knight’s tour problem is a special case of the hamiltonian tour problem. The answer is yes!
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