Homework6_Solutions

ECE360C: Algorithms Homework Assignment #6 University of Texas at Austin Due: October 13, 2023 (11:59pm) SOLUTIONS STUDENT CHOICE: Homework Assignment #6 – Solutions Solutions are for individual use only by students registered for ECE 360C in Spring 2023 semester. They should not be otherwise shared or posted. Problem 1: Remember [20 points] Answer the following true/false questions. True False □ ⊠ In the interval scheduling problem, the greedy choice that considers the intervals in decreasing order by end time (rather than increasing order by end time) does result in an optimal algorithm. ⊠ □ In Dijkstra’s algorithm, once you remove a node from the queue (because it has the current minimum distance estimate), you are guaranteed to know the shortest path from the source to that node. □ ⊠ Dijkstra’s algorithm computes the shortest cost path from every vertex in a graph to every other vertex in a graph. □ ⊠ Dijkstra’s algorithm will work, without modification, on a graph that has negative weight edges, as long as there are no negative weight cycles. □ ⊠ For a dense graph (when m >> n ), the running time of Prim’s algo- rithm (which is O ( m log n )) is asymptotically faster than Kruskal’s algorithm (which is O ( m log m )). ⊠ □ In a graph where each edge has a unique weight, there is only a single minimum spanning tree. □ ⊠ In the Interval Scheduling Problem we discussed in class (where our goal is to schedule the maximum number of compatible jobs), the job with the earliest finishing time will be present in every optimal solution. ⊠ □ If all edges in a graph have equal weight, then the shortest path between two nodes is the path with the least number of edges between those two nodes. □ ⊠ Let T be the MST of a graph G , then the path from u to v in T is the shortest path from u to v in G . □ ⊠ The first step of every greedy algorithm is to sort the input in some way using some metric.

Homework Assignment #6: October 13, 2023 (11:59pm) 2 Problem 2: Understand Consider the problem of matching a set of available skis to a set of skiers. The input consists of n skiers with heights p 1 , . . . p n and n sets of skis with lengths s 1 , . . . s n . The problem is to assign each skier a ski to minimize the average difference between the height of a skier and the length of his or her assigned set of skis. That is, if the i th skier is given the α ( i ) th pair of skis, then you want to minimize: 1 n n X i =1 | p i − s α ( i ) | Consider the following greedy algorithm: Find the difference between the height of each skier and the length of each set of skis. Choose the smallest difference first, and assign those skis to that skier. Continue to make the next greediest choice until all skiers have been assigned skis Prove (or disprove) that this greedy choice is optimal: Solution This choice is not optimal. Consider the following example: skiers: { 1 , 6 } , skis: { 5 , 10 } . This choice would pair (6 , 5) and (1 , 10) for an average difference of 6.5 vs. the optimal average difference of 4.

Homework Assignment #6: October 13, 2023 (11:59pm) 4 (b) Give the runtime of your algorithm and briefly justify it. O ( nk ) or O ( m ), because iterating over every station for every home, in other words iterating over all edges Alternate: The runtime of Prim’s algorithm is O ( m log( n + k )). Then we have to go through each home (there are n of them), doing O ( m ) work, in aggregate . So the total running time is O ( m log( n + k )) + O ( m + n ), which is upperbounded by the runtime of Prim’s. Working solution: (c) Suppose there is ultimately only the ability to install k − 1 stations in the village. These stations must be installed at the k predetermined locations considered above, but, in the end, one location will be without a station. Assuming the same home-to-station connectivity constraints as above, give an efficient algorithm that determines whether k − 1 stations are sufficient to connect all of the homes to electricity and, if it is possible, outputs the optimal connectivity graph (i.e., the graph that connects all homes to stations with the least amount of wire). Again, your algorithm does not have to be asymptotically optimal, but it must be efficient and it must result in the optimal connectivity graph. Run algorithm from part a) k times. Each time, remove a different station. Return the lowest cost station removal, or return false if it cannot be done. (d) Give the runtime of your algorithm and briefly justify it. O ( nk 2 ) or O ( mk ). We execute algorithm from part a), which is O ( m ), and we do it k times. Alternate: O ( km log( n + k )) .

Homework Assignment #6: October 13, 2023 (11:59pm) 5 Problem 4: Analyze Consider a situation where you have to find available classrooms for n different lectures. Of course, you must avoid scheduling two or more overlapping lecture in the same room. Each lecture i begins at s i and ends at t i . (a) Find an algorithm that assigns the smallest number of rooms possible. Solution The greedy choice we will make to solve this problem is to schedule the earliest compatible classes first. That is to say, we will take the earliest class that has not been scheduled, assign it to a new classroom. Then we will take the earliest class that doesn’t overlap with this class, and assign it to the same classroom. We will repeat this until no more classes can be assigned to this room, then we will create a new room, and repeat the process. We will sort the list of classes, c by starting time before running the following algorithm on it. GreedyRooms ( c ) 1 d ← 0 (number of allocated classrooms) 2 for j = 1 to n 3 do if class j is compatible with a classroom, k 4 then schedule j in classroom k 5 update the finish time of classroom k 6 else allocate a new classroom d + 1 7 schedule class j in classroom d + 1 8 d ← d + 1 Each classroom k maintains the finish time of the last job added. Classrooms are kept in a priority queue. The running time of this algorithm is O ( n log( n )) (b) Show that your algorithm is optimal. Solution In our algorithm, we start a new classroom d because we need to schedule a job j that is incompatible with all other classrooms 1 ...d − 1. Because we chose to schedule greedily by start time, the incompatibilities in classrooms 1 ...d − 1 are caused by lectures with start times ≤ s j . Thus, there are d lectures that overlap, and we would need ≥ d classrooms to schedule all of them.

Homework Assignment #6: October 13, 2023 (11:59pm) 7 Problem 6: Create [20 points] A thief is given the choice of n objects to steal, but only has one knapsack with a capacity of taking M weight. Each object i has weight w i , and profit p i . (a) First, suppose that the objects are divisible (e.g., the thief is in a cheese shop and the items are rolls of cheese that can be cut). For each object i , if a fraction x i , 0 ≤ x i ≤ 1 (taking x i = 1 would be taking the entire object) is placed in the knapsack, then the profit earned is p i x i . Come up with an efficient greedy algorithm for maximizing the profit of the thief. Prove the correctness and running time. Solution Efficient Greedy Algorithm. If all objects fit in the knapsack, take them all. Otherwise: ∀ i , set x i = 0. Index the objects as 1 , ..., n s.t. for any two objects with indices i and j respectively, i < j ⇐⇒ p i /w i ≥ p j /w j , i.e. the items are sorted in decreasing order of the ratio p i /w i . Let k be the maximum non-negative integer s.t. ∑ k i =1 w i ≤ M . Then, ∀ i ∈ { 1 , 2 , .., k } , set x i = 1. Finally, if there is still space in the knapsack, i.e. if M − ∑ k i =1 w i = α > 0, set x k +1 = a/w k +1 (i.e. fill the remaining space with the largest possible quantity of object k + 1). Correctness. The key argument for the algorithm’s correctness is that an optimal solution would have to take as much quantity as possible of object 1 (i.e. the object i with the highest p i /w i ratio). We can prove this by contradiction: Assume that ∃ O = { x 1 = o 1 , ..., x n = o n } s.t. O is an optimal solution with value P ( O ) and O doesn’t take the greatest amount of object 1 possible, i.e. o 1 < min { 1 , M/w i } . There are two possible cases for solution O . Either the knapsack is full or it isn’t. If it isn’t, we can increase the amount of item i that we take, and thus have a solution of value greater than that of O . Since we assumed that O is optimal, this is a contradiction. If the knapsack is full, ∃ j ̸ = 1 s.t. o j > 0. By removing ϵ weight of item j , s.t. ϵ ∈ (0 , min { w 1 − o 1 , w j ∗ o j } ], and adding ϵ weight of item 1 in its place, we get a solution that has value P ( O ) − ϵ ∗ p j /w j + ϵ ∗ p 2 /w 2 = P ( O )+ ϵ ∗ ( p 2 /w 2 − p j /w j ) > P ( O ). Since we assumed that O is optimal, this again is a contradiction. Applying this argument recursively, we get that our greedy algorithm outputs an optimal solution. Running Time. Sorting takes O ( n log n ) time. Iterating through our list of objects to figure out how many of them can be added to the solution takes O ( n ) time. Total O ( n log n ).

Homework Assignment #6: October 13, 2023 (11:59pm) 8 (b) Now, suppose the items cannot be taken fractionally. In other words, the thief can either take an entire item or leave it behind. Suppose we also know the following: the order of these items when sorted by increasing weight is the same as their order when sorted by decreasing value. Give a greedy algorithm to find an optimal solution to this variant of the knapsack problem. Prove the correctness and running time. Solution Very similar to part (a). Efficient Greedy Algorithm. If all objects fit in the knapsack, take them all. Otherwise: ∀ i , set x i = 0. Index the objects as 1 , ..., n s.t. for any two objects with indices i and j respectively, i < j ⇐⇒ p i ≥ p j . Let k be the maximum non-negative integer s.t. ∑ k i =1 w i ≤ M . Then, ∀ i ∈ { 1 , 2 , .., k } , set x i = 1. Correctness. Proof by contradiction. Assume there is an optimal solution O that does not include the first k objects. Let i ∈ 1 , ..., k be an object that is not included in O . There either is an item j > i in O , or then isn’t. If there isn’t, we can add i to the knapsack and get a higher value than that of O . This is a contradiction. If there is such an item j , we can remove j and add i in its place, since w j ≥ w i , and get a solution with a higher value than that of O , since v j ≤ v i . Again, this is a contradiction to our initial assumption. Hence, the optimal solution must include the first k objects, so our algorithm is correct. Running Time. Sorting takes O ( n log n ) time. Iterating through our list of objects to figure out how many of them can be added to the solution takes O ( n ) time. Total O ( n log n ).