Algebraic Cost's Algorithm The maximal weighted matching issue in a weighted network can be solved in a different way using the algorithm attributed to Preis. The following stages make up the algorithm. 1. Input: A weighted graph G = (V, E, w) 2. Output: A maximal weighted matching M of G 3. M ← Ø 4. E ← E 5. V ← V 6. while E = Ø 7. select at random any v ∈ V 8. let e ∈ E be the heaviest edge incident to v 9. M ← M ∪ e 10. V ← V {v} 11. E ← E \ {e and all adjacent edges to e} We'll demonstrate two Python implementation methods for this technique.

3. Suppose that we want to know which agents are connected by a walk of length two in a given network. That is, want to know the set of ordered pairs (i, j) such that there exists a walk of length 2 between any i and j. Define n as the number of nodes and m as the number of links. (a) One way to do this is to input the adjacency matrix A, calculate A², then output all pairs (i, j) for which A(i) > 0. Defining scalar multiplication and addition as "basic operations," up to a constant c, how many "basic operations" are required to calculate A², in terms of n and m? (Hint: First think about how many operations are required to calculate each entry of A², then multiply by the number of entries.) (b) Suppose that we want to know the number of agents who are connected by a walk of length k in a given network, where k≥ 1. Up to a constant, how many basic operations are required to calculate Ak, in terms of n and m? (c) (Harder) We showed in class (Lecture 4) that a simple adaptation of…

10 by 10 grid, . The agent in the top right corner, fast route to left bottom corner, some cells = easy terrain = 1 time unit. Some cells = rough terrain = 5 time units. Some cells = impassible terrain and cannot be traversed. -Write a Breathe first search Python function: A function that generates a list of lists to represent the 2D grid, where each grid cell contains the number of time units taken to traverse through it. Each cell should have a probability of 0.4 of being easy terrain, 0.4 of being rough terrain and 0.2 of being untraversable. E.g, map = RandomMap()