Floyd’s Algorithm (shortest-path problem) Section 8.2

Shortest-path Problem ► Given: ► A set of objects (called vertices) and ► A set of distances between objects (called edges) ► Find: ► The shortest path from a designated starting point ► The shortest path between any pair of vertices

Shortest-path Problem ► Given: ► A set of objects (called vertices) and ► A set of distances between objects (called edges) ► Find: ► The shortest path from a designated starting point ► The shortest path between any pair of vertices ► Floyd’s algorithm solves this problem

Shortest-path Problem ► Given: ► A set of objects (called vertices) and ► A set of distances between objects (called edges) ► Find: ► The shortest path from a designated starting point ► BTW, Dijkstra’s algorithm solves this one ► The shortest path between any pair of vertices

Floyd’s Algorithm ► Dynamic Programming Algorithm ► It uses an N X N matrix ► N is the # of vertices ► BTW, Pink Floyd’s The Wall is great movie to watch if you want to…

Floyd’s Algorithm ► Recall a graph can be represented using an N X N matrix AB C DE ABCDE A0 B0 C0 D0 E    

Floyd’s Algorithm ► Dynamic Programming  Solves one step of the problem  Stores it, so it doesn’t have to be recomputed  Uses stored step to solve the next step  Avoids re-computing progressive steps ► Many Algorithms (Brute Force, even Divide-and Conquer) re-compute the same information over and over again.

Floyd’s Algorithm ► Dynamic Programming  Uses extra memory to store steps (or sub- problems)  Avoids re-computation  However, the extra memory can be expensive.

Floyd’s Algorithm ► Famous Example of Dynamic Programming  Optimal sequence alignment algorithm ► Most famous version is called the Smith-Waterman Algorithm  Align two sequences of length N ► Previous algorithm O(2 N ) comparisons O(N) memory ► Smith-Waterman Algorithm O(N 2 ) comparisons O(N 2 ) memory.  In this case the trade-off was worth it.

Floyd’s Algorithm ► Back to the shortest-path problem. AB C DE ABCDE A0 B0 C0 D0 E    

Floyd’s Algorithm ► First step, where can we go without using an intermediate vertex ► M k, where k the set of intermediate vertices ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA0831 B8042 C34011 D1108 E2180 M {}

Floyd’s Algorithm ► Second step, where can we go if we use A as an intermediate vertices ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA0831 B80492 C34011 D19108 E2180 M {A}

Floyd’s Algorithm ► Third step, where can we go if we use A and B as intermediate vertices ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA B80492 C34011 D19108 E M {A,B}

Floyd’s Algorithm ► Fourth step, where can we go if we use A, B, and C as intermediate vertices ► You tell me! ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA B80492 C34011 D19108 E M {A,B,C}

Floyd’s Algorithm ► Fourth step, where can we go if we use A, B, and C as intermediate vertices ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA07314 B70452 C34011 D15102 E42120 M {A,B,C}

Floyd’s Algorithm ► Fifth step, where can we go if we use A, B, C, and D as intermediate vertices ► You tell me! ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA07214 B70452 C24011 D15102 E42120 M {A,B,C,D}

Floyd’s Algorithm ► Fifth step, where can we go if we use A, B, C, and D as intermediate vertices ► OK, here is the answer. ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA06213 B60452 C24011 D15102 E32120 M {A,B,C,D}

Floyd’s Algorithm ► Final step, where can we go if we use all the vertices as intermediates. ► You tell me! ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA06213 B60452 C24011 D15102 E32120 M {A,B,C,D,E}

Floyd’s Algorithm ► Final step, where can we go if we use all the vertices as intermediates. ► You tell me! ABCDE A0831 B8042 C34011 D1108 E2180 AB C DE ABCDEA05213 B50352 C23011 D15102 E32120 M {A,B,C,D,E}

Floyd’s Algorithm ► So how do we actually compute (update) the matrix? ► First of all… ► Given N vertices how many steps are there? ► i.e., how many times must the matrix be updated?

Floyd’s Algorithm ► for (int k = 0; k < N; k++)  Update the matrix, i.e. compute M {k} ► Next… ► How would you actually update each matrix cell? ► i.e., how would you iterate through the matrix?

Floyd’s Algorithm AB C DE ABCDE A0831 B8042 C34011 D1108 E2180 Allowing no hops…M[D][E] = 8 What if we allow a hop through C? M[D][C] = 1 and M[C][E] = 1 If we allow a hop through C, what can we compare to see if M[D][E] needs to be updated?

Floyd’s Algorithm ► for (int k = 0; k < N; k++)  for (int i = 0; i < N; i++) ► for (int j = 0; j < N; j++)  update M {k} [i][j] ► What do we have to compare to see if M {k} [i][j] needs to be updated?

Floyd’s Algorithm ► for (int k = 0; k < N; k++)  for (int i = 0; i < N; i++) ► for (int j = 0; j < N; j++)  if (M[i][k]+M[k][j] < M[i][j]) then ► How do we do the update?

Floyd’s Algorithm ► for (int k = 0; k < N; k++)  for (int i = 0; i < N; i++) ► for (int j = 0; j < N; j++)  if (M[i][k]+M[k][j] < M[i][j]) then ► M[i][j] = M[i][k]+M[k][j]