🧠:<think>Okay, so I have this problem about a soccer coach using graph theory to analyze passing patterns. It's divided into two parts. Let me try to tackle them one by one.Starting with part 1: The coach wants to find the optimal passing path that maximizes the expected success rate for sequences of length k. The graph is directed, with each edge having a weight representing the probability of a successful pass. So, I need to model this as a dynamic programming problem.Hmm, dynamic programming usually involves breaking a problem down into simpler subproblems and using the solutions to those subproblems to build up the solution to the original problem. In this case, the subproblems would likely be the optimal paths of shorter lengths leading up to k.Let me think about how to structure this. For each vertex v_i and each possible path length m (from 1 to k), I can keep track of the maximum expected success rate to reach v_i in m passes. Let's denote this as DP[m][i]. The base case would be when m=1, where the expected success rate is just the weight of the edge from the starting vertex to v_i. But wait, actually, since the coach is looking for any optimal path, not necessarily starting from a specific vertex, maybe the base case is different.Wait, no. If we're looking for the optimal path of length k, regardless of where it starts, then perhaps we need to consider all possible starting points. But actually, in dynamic programming for paths, we often fix the starting point. Maybe the coach is considering all possible starting points, so the DP needs to account for that.Alternatively, perhaps the coach is looking for the maximum expected success rate over all possible paths of length k, regardless of where they start or end. So, the DP would track, for each vertex, the maximum success rate achievable to reach that vertex in m passes.So, the recursive relation would involve, for each vertex v_i and each step m, considering all possible incoming edges to v_i and taking the maximum over those. Specifically, DP[m][i] = max over all j of (DP[m-1][j] * w_{ji}). That makes sense because to get to v_i in m passes, you must have come from some v_j in m-1 passes, and the success rate is the product of the success rates along the path.But wait, is it a product or a sum? Since we're dealing with probabilities, the success rate of a path is the product of the weights along the edges. So, for a path of length m, the expected success rate is the product of m weights. Therefore, to maximize the expected success rate, we need to take the maximum over all possible paths of length m, which can be built by extending paths of length m-1 by one edge.So, yes, the recursive relation would be DP[m][i] = max_{j} (DP[m-1][j] * w_{ji}). That seems right. The base case would be DP[1][i] = max_{j} w_{ji}, but actually, if m=1, then the path is just a single edge, so DP[1][i] is the maximum weight of any edge ending at i. Alternatively, if we consider starting from any vertex, maybe it's better to initialize DP[0][i] as 1 for all i, representing the trivial path of length 0 (just being at vertex i with success rate 1). Then, for m=1, DP[1][i] = max_{j} (DP[0][j] * w_{ji}) = max_{j} w_{ji}.Wait, but if we start from any vertex, the initial state is that we can be at any vertex with success rate 1 for m=0. Then, for each subsequent step, we consider moving along edges. So, the recursive formula is:DP[m][i] = max_{j} (DP[m-1][j] * w_{ji})And the maximum expected success rate over all paths of length k would be the maximum value among DP[k][i] for all i.That seems like a solid approach. So, the dynamic programming formulation is to compute DP[m][i] for m from 1 to k and for each vertex i, using the recursive relation above.Moving on to part 2: The graph G is decomposed into strongly connected components (SCCs). The coach wants to analyze how to maintain ball possession when the graph isn't strongly connected. Specifically, they need to calculate the minimum number of additional weighted edges required to make the graph strongly connected, with the total weight of these edges minimized. The cost of adding an edge is inversely proportional to its weight, so adding a higher weight edge is cheaper.Wait, the cost is inversely proportional to the weight. So, if we add an edge with weight w, the cost is c/w for some constant c. But since we're trying to minimize the total cost, which is equivalent to maximizing the total weight, because cost is inversely proportional. So, to minimize the cost, we need to maximize the total weight of the added edges.But the problem states that the coach wants to minimize the total weight of the additional edges. Wait, no, the cost is inversely proportional to the weight, so lower weight edges cost more. Therefore, to minimize the total cost, we need to maximize the total weight of the added edges. But the problem says "the total weight of these additional edges is minimized." Hmm, that might be conflicting.Wait, let me read again: "the cost of adding an edge is inversely proportional to its weight." So, cost = k / w, where k is a constant. Therefore, to minimize the total cost, we need to maximize the total weight of the added edges. However, the problem says "the total weight of these additional edges is minimized." That seems contradictory because minimizing the total weight would mean maximizing the cost. Maybe I misinterpret.Wait, perhaps the coach wants to minimize the number of additional edges, and among those, minimize the total weight. Or maybe the problem is to find the minimum number of edges such that the total weight is minimized, but considering that the cost is inversely proportional. Hmm, the wording is a bit confusing.Wait, the exact wording is: "Calculate the minimum number of additional weighted edges needed to make the graph strongly connected while ensuring that the total weight of these additional edges is minimized. Assume each additional edge (v_i, v_j) can have a weight between 0 and 1, and the cost of adding an edge is inversely proportional to its weight."So, the goal is two-fold: minimize the number of edges, and among those, minimize the total weight. But since the cost is inversely proportional to weight, minimizing the total weight would actually maximize the cost. But the problem says to minimize the total weight, so perhaps the priority is to minimize the number of edges first, and then minimize the total weight. Or maybe it's a trade-off.Wait, no, the problem says "the minimum number of additional weighted edges needed to make the graph strongly connected while ensuring that the total weight of these additional edges is minimized." So, it's a constrained optimization: find the smallest number of edges such that the total weight is as small as possible.But since the cost is inversely proportional to weight, the coach might prefer higher weight edges to minimize cost, but the problem says to minimize the total weight. This seems conflicting. Maybe the problem is just to minimize the number of edges, regardless of weight, but with the added constraint that the total weight is minimized. So, perhaps first find the minimal number of edges needed to make the graph strongly connected, and then among all such sets of edges, choose the one with the minimal total weight.Yes, that makes sense. So, first, determine the minimal number of edges required to make the graph strongly connected, and then find the set of edges with that minimal number and minimal total weight.To make a directed graph strongly connected, we can use the concept of condensation into SCCs. The condensation of G is a directed acyclic graph (DAG) where each node represents an SCC. To make G strongly connected, the condensation must be a single node, meaning the DAG must be reduced to one node by adding edges.The minimal number of edges to add to make a DAG strongly connected is known. If the DAG has c components, then the minimal number of edges to add is max(in_degree_zero, out_degree_zero), where in_degree_zero is the number of components with in-degree zero, and out_degree_zero is the number of components with out-degree zero. But wait, actually, the formula is a bit different.Wait, I recall that for a DAG, the minimal number of edges to add to make it strongly connected is max(number of sources, number of sinks), where a source is a component with in-degree zero, and a sink is a component with out-degree zero.Yes, that's correct. So, if the condensation DAG has s sources and t sinks, then the minimal number of edges to add is max(s, t). If s = t = 1, then no edges are needed if it's already strongly connected. Otherwise, it's max(s, t).But in our case, the graph is not strongly connected, so the condensation has more than one component. So, we need to compute the number of sources and sinks in the condensation DAG.Once we know the minimal number of edges to add, which is max(s, t), we then need to find the set of edges with that number and minimal total weight.Wait, but the edges we add can have weights between 0 and 1, and the cost is inversely proportional to the weight. So, to minimize the total cost, we need to maximize the total weight of the added edges. However, the problem says to minimize the total weight. Hmm, conflicting again.Wait, perhaps the problem is just to minimize the number of edges, and the total weight is a secondary consideration. But the problem explicitly says "while ensuring that the total weight of these additional edges is minimized." So, perhaps the coach wants the minimal number of edges, and among those, the minimal total weight.But if the cost is inversely proportional to the weight, then minimizing the total weight would actually be more expensive. So, maybe the problem is misworded, or perhaps the coach wants to minimize the cost, which would mean maximizing the total weight. But the problem says to minimize the total weight.Alternatively, perhaps the coach wants to add edges with minimal total weight, regardless of the cost. So, perhaps the cost is a separate consideration, but the problem is just about the graph's properties.Wait, the problem says: "the cost of adding an edge is inversely proportional to its weight." So, the cost is a factor, but the problem is to calculate the minimum number of edges needed and the minimal total weight. So, perhaps the coach wants to minimize the number of edges, and then, among those, choose the edges with the minimal total weight, which would correspond to higher cost, but the problem says to minimize the total weight. Hmm, confusing.Alternatively, maybe the coach wants to minimize the total cost, which is inversely proportional to the weight. So, to minimize the total cost, we need to maximize the total weight. But the problem says "the total weight of these additional edges is minimized." So, perhaps the problem is just to find the minimal number of edges and minimal total weight, regardless of cost. Maybe the cost is just additional information but not directly part of the optimization.Alternatively, perhaps the coach wants to minimize the number of edges, and then, for that minimal number, choose the edges with the minimal total weight, even though that would mean higher cost. So, the problem is to find the minimal number of edges required, and then find the set of edges with that number and minimal total weight.Given that, let's proceed.First, decompose G into SCCs, resulting in a DAG of components. Let the number of sources (components with in-degree zero) be s, and the number of sinks (components with out-degree zero) be t. The minimal number of edges to add is max(s, t).Now, to find the minimal total weight of such edges, we need to find a set of max(s, t) edges that connect the DAG into a single strongly connected component, with the minimal total weight.Wait, but adding edges between components can be done in different ways. For example, if s > t, we need to add s edges, each connecting a source to some component, possibly a sink. Similarly, if t > s, we need to add t edges, each connecting a sink to some component.But to minimize the total weight, we need to choose the edges with the smallest possible weights. However, since the edges can have weights between 0 and 1, and we can choose any weights, but the goal is to minimize the total weight, so we would set the weights as small as possible, but the problem says "the total weight of these additional edges is minimized." So, perhaps the minimal total weight is achieved by adding edges with weight 0, but since the coach wants to maintain ball possession, maybe the edges need to have positive weights. Wait, but the problem allows weights between 0 and 1, so 0 is allowed, but adding an edge with weight 0 would mean the pass is never successful, which doesn't help in maintaining possession. So, perhaps the coach would prefer edges with higher weights to ensure successful passes, but the problem says to minimize the total weight.This is getting a bit tangled. Maybe the problem is just about the graph structure, not considering the weights beyond their necessity to connect the components. So, perhaps the minimal number of edges is max(s, t), and the minimal total weight is achieved by adding edges with the smallest possible weights, but since the coach wants to maintain possession, maybe the weights should be as high as possible. But the problem says to minimize the total weight, so perhaps we have to take the minimal total weight, which would be adding edges with the smallest possible weights.But this seems counterintuitive because adding edges with low weights would make the passes less likely to succeed, which would not help in maintaining possession. So, perhaps the problem is actually to maximize the total weight, but the wording says to minimize it. Maybe it's a misstatement.Alternatively, perhaps the problem is to find the minimal number of edges, and then, for those edges, choose the ones with the minimal possible weights, but in reality, the coach would want the opposite. Hmm.Wait, let's try to proceed step by step.1. Decompose G into SCCs, resulting in a DAG of components.2. Let s be the number of sources (components with in-degree zero) and t be the number of sinks (components with out-degree zero).3. The minimal number of edges to add is max(s, t).4. To find the minimal total weight of these edges, we need to connect the sources and sinks appropriately.If s = t, then we can connect each source to a sink with an edge. To minimize the total weight, we would choose the edges with the smallest possible weights. However, since the coach wants to maintain possession, perhaps they would prefer higher weights, but the problem says to minimize the total weight.Alternatively, if s ≠ t, say s > t, then we need to add s edges. We can connect each source to some component, possibly a sink or another source, but to minimize the total weight, we would choose the edges with the smallest weights.But wait, in the DAG of SCCs, adding an edge from a component A to component B can only be done if there's no path from A to B already. So, to connect the DAG into a single component, we need to add edges in such a way that all components are reachable from each other.But the minimal number of edges is max(s, t), so if s > t, we need to add s edges, each connecting a source to some component, possibly a sink or another source. Similarly, if t > s, we add t edges, each connecting a sink to some component.To minimize the total weight, we would choose the edges with the smallest possible weights. However, since the edges can have weights between 0 and 1, the minimal total weight would be achieved by adding edges with weight 0, but that doesn't make sense for maintaining possession. So, perhaps the coach wants to add edges with the highest possible weights, but the problem says to minimize the total weight.This is confusing. Maybe the problem is just about the graph structure, and the weights are not to be considered beyond their necessity to connect components. So, perhaps the minimal number of edges is max(s, t), and the total weight is the sum of the minimal weights required to connect the components.But without knowing the specific weights between components, we can't determine the exact minimal total weight. However, the problem might be asking for the minimal number of edges, which is max(s, t), and the minimal total weight would be the sum of the minimal possible weights for those edges, which could be zero, but that's not useful.Alternatively, perhaps the problem is to find the minimal number of edges and the minimal total weight, considering that the edges must have positive weights to contribute to the possession. So, perhaps the minimal total weight is achieved by adding edges with the smallest positive weights, but since the weights can be arbitrarily small, the total weight can be made as small as desired. But that doesn't make sense in a practical scenario.Wait, maybe the problem is to find the minimal number of edges and the minimal total weight, assuming that the edges must have a certain minimum weight to be useful. But the problem doesn't specify that.Alternatively, perhaps the problem is to find the minimal number of edges and the minimal total weight, considering that the edges must form a strongly connected graph. So, the minimal total weight would be the sum of the minimal weights required to connect the components in a way that makes the graph strongly connected.But without knowing the specific weights between components, we can't compute the exact minimal total weight. However, perhaps the problem is just asking for the minimal number of edges, which is max(s, t), and the minimal total weight is the sum of the minimal weights for those edges, but since the weights can be chosen freely between 0 and 1, the minimal total weight is zero, which is not practical.This is getting too convoluted. Maybe I need to approach it differently.In graph theory, the minimal number of edges required to make a DAG strongly connected is indeed max(s, t), where s is the number of sources and t is the number of sinks. To find the minimal total weight, we need to find the set of edges with the minimal total weight that connects the DAG into a single component.This is similar to the problem of finding a minimum spanning arborescence, but in this case, we're dealing with connecting multiple components.Wait, perhaps we can model this as finding a minimum feedback arc set or something similar, but I'm not sure.Alternatively, perhaps we can use the concept of the minimum number of edges to add and then find the minimal total weight by selecting the lightest edges that connect the necessary components.But without specific weights, it's hard to determine. However, since the problem allows us to choose the weights between 0 and 1, and we need to minimize the total weight, the minimal total weight would be achieved by adding edges with weight 0. But again, that's not useful for the coach.Alternatively, perhaps the problem is to find the minimal number of edges, and the minimal total weight is the sum of the minimal possible weights that make the graph strongly connected, which would be the sum of the minimal weights required to connect the components.But without knowing the structure of the graph, it's impossible to determine the exact minimal total weight. So, perhaps the answer is simply that the minimal number of edges needed is max(s, t), and the minimal total weight is the sum of the minimal weights required to connect those components, but since the weights can be chosen, the minimal total weight is zero.But that seems incorrect because the coach would want the passes to have some success probability. So, perhaps the problem is misworded, and the coach actually wants to maximize the total weight, which would correspond to minimizing the cost since cost is inversely proportional.Given that, the minimal number of edges is max(s, t), and the maximal total weight is achieved by adding edges with weight 1, which would minimize the cost. But the problem says to minimize the total weight, so perhaps it's the opposite.I'm getting stuck here. Maybe I should focus on the minimal number of edges, which is max(s, t), and note that the minimal total weight would be the sum of the minimal weights required to connect the components, but since the weights can be chosen, the minimal total weight is zero. However, in practice, the coach would want higher weights, but the problem specifies to minimize the total weight.Alternatively, perhaps the problem is to find the minimal number of edges and the minimal total weight, considering that the edges must form a strongly connected graph. So, the minimal total weight is the sum of the minimal weights for the edges that connect the components in a way that makes the graph strongly connected.But without specific weights, we can't compute it numerically. So, perhaps the answer is that the minimal number of edges is max(s, t), and the minimal total weight is the sum of the minimal weights required to connect those components, but since the weights can be chosen, the minimal total weight is zero.But that seems unsatisfactory. Maybe the problem is expecting the answer to be the minimal number of edges, which is max(s, t), and the minimal total weight is the sum of the minimal possible weights, which is zero, but that's not useful. Alternatively, perhaps the problem is expecting us to recognize that the minimal number of edges is max(s, t), and that the minimal total weight is the sum of the minimal weights for those edges, but since the weights can be chosen, it's zero.Wait, perhaps the problem is just asking for the minimal number of edges, which is max(s, t), and the minimal total weight is the sum of the minimal possible weights, which is zero, but that's not practical. Alternatively, perhaps the problem is expecting us to recognize that the minimal number of edges is max(s, t), and that the minimal total weight is the sum of the minimal weights required to connect the components, but without specific weights, we can't compute it.Given that, perhaps the answer is that the minimal number of edges needed is max(s, t), and the minimal total weight is the sum of the minimal weights for those edges, but since the weights can be chosen, the minimal total weight is zero. However, in a practical sense, the coach would want higher weights, but the problem specifies to minimize the total weight.Alternatively, perhaps the problem is expecting us to recognize that the minimal number of edges is max(s, t), and the minimal total weight is the sum of the minimal possible weights, which is zero, but that's not useful. So, perhaps the answer is simply that the minimal number of edges is max(s, t), and the minimal total weight is the sum of the minimal weights for those edges, but since the weights can be chosen, the minimal total weight is zero.But I'm not sure. Maybe I should proceed with the understanding that the minimal number of edges is max(s, t), and the minimal total weight is the sum of the minimal possible weights, which is zero, but that's not practical. Alternatively, perhaps the problem is expecting us to recognize that the minimal number of edges is max(s, t), and the minimal total weight is the sum of the minimal weights for those edges, but without specific weights, we can't compute it.Given that, perhaps the answer is that the minimal number of edges needed is max(s, t), and the minimal total weight is the sum of the minimal possible weights for those edges, which is zero, but in practice, the coach would want higher weights.Wait, but the problem says "the total weight of these additional edges is minimized," so perhaps the answer is that the minimal number of edges is max(s, t), and the minimal total weight is the sum of the minimal possible weights, which is zero. However, since the coach wants to maintain possession, they would prefer higher weights, but the problem specifies to minimize the total weight.I think I'm overcomplicating this. Let's try to summarize:For part 1, the dynamic programming approach is to compute DP[m][i] = max_j (DP[m-1][j] * w_{ji}), with DP[0][i] = 1 for all i, and the maximum expected success rate is the max over DP[k][i].For part 2, the minimal number of edges to add is max(s, t), where s is the number of sources and t is the number of sinks in the condensation DAG. The minimal total weight is the sum of the minimal weights for those edges, which can be zero, but in practice, the coach would want higher weights.But since the problem says to minimize the total weight, the answer is that the minimal number of edges is max(s, t), and the minimal total weight is the sum of the minimal possible weights, which is zero. However, this might not be practical, but it's what the problem asks for.Alternatively, perhaps the problem is expecting us to recognize that the minimal number of edges is max(s, t), and the minimal total weight is the sum of the minimal possible weights, which is zero, but since the coach wants to maintain possession, the weights should be as high as possible, but the problem says to minimize the total weight.I think I've spent enough time on this. Let me try to write the answers clearly.</think>

🧠:<think>Alright, so I have this problem where a software engineer is analyzing an algorithm with a given recurrence relation. I need to figure out the asymptotic time complexity using the Master Theorem for the first part and then compare it with an optimized version in the second part. Let me try to break this down step by step.Starting with the first part: The recurrence relation is T(n) = 2T(n/2) + n log n, and T(1) = 1. I remember the Master Theorem is used for solving recurrence relations of the form T(n) = aT(n/b) + f(n), where a ≥ 1, b > 1, and f(n) is the cost of the work done outside the recursive calls.So, in this case, a is 2, b is 2, and f(n) is n log n. The Master Theorem has three cases, and I need to determine which case applies here. Let me recall the cases:1. If f(n) = O(n^{log_b a - ε}) for some ε > 0, then T(n) = Θ(n^{log_b a}).2. If f(n) = Θ(n^{log_b a} log^k n) for some k ≥ 0, then T(n) = Θ(n^{log_b a} log^{k+1} n).3. If f(n) = Ω(n^{log_b a + ε}) for some ε > 0, and if a f(n/b) ≤ c f(n) for some c < 1 and all sufficiently large n, then T(n) = Θ(f(n)).First, let's compute log_b a. Here, a is 2 and b is 2, so log_2 2 is 1. So, log_b a = 1.Now, f(n) is n log n. Let's compare f(n) with n^{log_b a}, which is n^1 = n.So, f(n) is n log n, which is asymptotically larger than n. That suggests that maybe case 3 applies. But wait, case 3 requires f(n) to be Ω(n^{log_b a + ε}) for some ε > 0. Let's see: n log n is indeed asymptotically larger than n, but is it polynomially larger? Because log n grows slower than any polynomial, so n log n is not polynomially larger than n. So, n log n is not Ω(n^{1 + ε}) for any ε > 0. Therefore, case 3 doesn't apply.Hmm, so maybe case 2 applies? Let's check. Case 2 is when f(n) is Θ(n^{log_b a} log^k n). Here, n^{log_b a} is n^1, and f(n) is n log n, which is n log^1 n. So, k is 1. Therefore, case 2 applies, and the solution would be T(n) = Θ(n^{log_b a} log^{k+1} n) = Θ(n log^2 n).Wait, but I want to make sure. Let me double-check. The Master Theorem case 2 says that if f(n) is Θ(n^{log_b a} log^k n), then T(n) is Θ(n^{log_b a} log^{k+1} n). So, in this case, since f(n) is n log n, which is n^{1} log^1 n, then k is 1, so T(n) would be Θ(n log^{2} n). That seems right.But let me think again. Is there another way to solve this recurrence? Maybe using the recursion tree method. Let's try that.In the recursion tree, each node has two children, each with cost (n/2) log(n/2). The root has cost n log n. The next level has two nodes each with cost (n/2) log(n/2). Then, the next level has four nodes each with cost (n/4) log(n/4), and so on until we reach the leaves.The total cost would be the sum of the costs at each level. Let's denote the cost at level i as C_i. The number of nodes at level i is 2^i, each with cost (n/2^i) log(n/2^i). So, C_i = 2^i * (n/2^i) log(n/2^i) = n log(n/2^i).Simplify log(n/2^i) = log n - i log 2. So, C_i = n (log n - i log 2).Now, how many levels are there? The recursion goes until n/2^i = 1, so i = log_2 n. So, the total number of levels is log n.Therefore, the total cost is the sum from i=0 to i=log n -1 of C_i.So, sum_{i=0}^{log n -1} n (log n - i log 2).Let me factor out n: n * sum_{i=0}^{log n -1} (log n - i log 2).Let me compute the sum inside: sum_{i=0}^{log n -1} (log n - i log 2) = sum_{i=0}^{log n -1} log n - log 2 sum_{i=0}^{log n -1} i.This is equal to log n * log n - log 2 * [ (log n -1) log n ) / 2 ].Wait, let me compute it step by step.First term: sum_{i=0}^{log n -1} log n = log n * log n.Second term: sum_{i=0}^{log n -1} i = (log n -1) log n / 2.Therefore, the total sum is log^2 n - log 2 * ( (log n -1) log n ) / 2.Simplify: log^2 n - (log 2 / 2)(log^2 n - log n).So, the total cost is n times that:n [ log^2 n - (log 2 / 2)(log^2 n - log n) ].Let me factor out log^2 n:n log^2 n [1 - (log 2)/2] + n (log 2 / 2) log n.So, the dominant term is n log^2 n multiplied by a constant, and then a lower term of n log n.Therefore, the total time complexity is Θ(n log^2 n), which matches what I got with the Master Theorem. So, that seems consistent.Therefore, the asymptotic time complexity of the original algorithm is Θ(n log^2 n). Now, is this efficient? Well, compared to other algorithms, log^2 n grows faster than log n, but it's still much better than, say, quadratic or cubic time. However, for very large n, log^2 n can be a noticeable factor. So, depending on the context, it might be considered inefficient if a more optimal algorithm exists with a lower time complexity.Moving on to the second part: The engineer proposes an optimized algorithm with recurrence S(n) = 3S(n/3) + n, S(1) = 1. I need to determine its asymptotic time complexity and compare it to the original.Again, this is a recurrence relation suitable for the Master Theorem. Let's identify a, b, and f(n). Here, a is 3, b is 3, and f(n) is n.Compute log_b a: log_3 3 = 1. So, log_b a = 1.Compare f(n) with n^{log_b a} = n^1 = n. So, f(n) is exactly n, which is the same as n^{log_b a}.Looking at the Master Theorem cases, case 2 applies when f(n) is Θ(n^{log_b a} log^k n). But here, f(n) is Θ(n^{log_b a}), which is Θ(n). So, actually, case 2 is when f(n) is Θ(n^{log_b a} log^k n). If k=0, then f(n) is Θ(n^{log_b a}), which is exactly our case. So, according to case 2, T(n) = Θ(n^{log_b a} log^{k+1} n) = Θ(n log^{0+1} n) = Θ(n log n).Wait, but let me make sure. Some sources might have different formulations. Let me double-check the exact statement of the Master Theorem.Yes, case 2 is when f(n) = Θ(n^{log_b a} log^k n). If k=0, then f(n) is Θ(n^{log_b a}), so the solution is Θ(n^{log_b a} log^{k+1} n) = Θ(n log n).Alternatively, sometimes case 2 is presented as when f(n) is exactly equal to n^{log_b a} multiplied by a polylogarithmic factor. So, in this case, since f(n) is n, which is n^{log_b a}, it's case 2 with k=0, leading to Θ(n log n).Alternatively, another way to solve this is using the recursion tree method.Let's try that. The recurrence is S(n) = 3S(n/3) + n.Each node has 3 children, each with cost (n/3). The root has cost n. The next level has 3 nodes each with cost n/3. The next level has 9 nodes each with cost n/9, and so on until the leaves.The cost at each level i is 3^i * (n / 3^i) = n. So, each level contributes n to the total cost. How many levels are there? The recursion goes until n/3^i = 1, so i = log_3 n. Therefore, the total number of levels is log_3 n.Thus, the total cost is n multiplied by log_3 n, which is Θ(n log n). Since log_3 n is a constant factor away from log n, we can write it as Θ(n log n).Therefore, the optimized algorithm has a time complexity of Θ(n log n), which is better than the original algorithm's Θ(n log^2 n). So, the optimized algorithm is more efficient.To summarize:1. The original algorithm has a time complexity of Θ(n log^2 n), which is not the most efficient but better than polynomial time complexities.2. The optimized algorithm has a time complexity of Θ(n log n), which is more efficient than the original.Therefore, the optimized algorithm is indeed more efficient.Final Answer1. The asymptotic time complexity of the original algorithm is boxed{O(n log^2 n)}.2. The asymptotic time complexity of the optimized algorithm is boxed{O(n log n)}.</think>

🧠:<think>Okay, so I have this problem about a gender studies scholar analyzing feminist art's influence across academic disciplines using a directed graph. The graph G has vertices as research papers and directed edges as citations. There are two parts to the problem.Starting with part 1: I need to derive the adjacency matrix A of G, given that G is strongly connected and has n vertices. Then, I have to show that A has at least one non-zero eigenvalue and explain its significance.Alright, first, an adjacency matrix for a directed graph is a square matrix where the entry A[i][j] is 1 if there's a directed edge from vertex i to vertex j, and 0 otherwise. Since the graph is strongly connected, there's a path from every vertex to every other vertex. That means the adjacency matrix isn't just any matrix; it has some special properties.Now, eigenvalues of a matrix are scalars λ such that Ax = λx for some non-zero vector x. For adjacency matrices, especially of strongly connected graphs, I remember something about the Perron-Frobenius theorem. It states that a square matrix with all positive entries has a unique largest eigenvalue, called the Perron root, which is positive and has a corresponding positive eigenvector. But wait, adjacency matrices don't necessarily have all positive entries; they have 0s and 1s. However, if the graph is strongly connected, the adjacency matrix is irreducible. The Perron-Frobenius theorem applies to irreducible non-negative matrices, so yes, A is irreducible and non-negative.Therefore, A has at least one positive eigenvalue, which is the Perron root, and it's the largest eigenvalue. This eigenvalue is significant because it's related to the growth rate of the number of walks in the graph. In the context of citations, it might indicate the overall influence or reach within the network.Moving on to part 2: Using eigenvector centrality, calculate the principal eigenvector of A and explain how it identifies the most influential papers.Eigenvector centrality is a measure of the influence of a node in a network. It assigns a score to each node based on the scores of its neighbors. The principal eigenvector corresponds to the largest eigenvalue, which, as we discussed, is the Perron root. So, the entries of this eigenvector represent the centrality scores of each paper.To calculate it, we need to solve the equation Ax = λx, where λ is the Perron root. Since A is large, we might use iterative methods like the power method to approximate the eigenvector. The components of the eigenvector x will be higher for nodes that are cited by many other influential nodes. So, a paper with a high eigenvector centrality score is one that is cited by many other papers, especially those that are themselves highly cited. This makes it a good measure of influence because it captures both the quantity and quality of citations.In the context of feminist art, the papers with the highest eigenvector centrality would be those that are most influential in the citation network. They might be foundational works or highly impactful studies that other researchers frequently reference, thus shaping the discourse on feminist art across various disciplines.Wait, but is the adjacency matrix A necessarily primitive? Since G is strongly connected, A is irreducible, but for the Perron-Frobenius theorem, we need A to be primitive, meaning that some power of A is positive. However, in a citation network, it's possible that the graph isn't primitive because you can have cycles of different lengths. Hmm, but even if it's not primitive, the theorem still gives us that the Perron root is a simple eigenvalue with a positive eigenvector. So, maybe I don't need to worry about primitivity here.Also, in part 1, the question says "derive the adjacency matrix A." But without specific information about the edges, I can't write out the exact matrix. Maybe they just want the general form? Or perhaps they mean to explain how to construct it. Since the graph is strongly connected, every row and column will have at least one 1, but that's about it.So, summarizing part 1: The adjacency matrix A is an n x n matrix where A[i][j] = 1 if there's a citation from paper i to paper j, else 0. Since G is strongly connected, A is irreducible, so by Perron-Frobenius, it has a unique largest eigenvalue λ > 0, which is the Perron root. This eigenvalue is non-zero and indicates the network's growth rate or influence potential.For part 2, the principal eigenvector (associated with λ) gives the eigenvector centrality scores. Each component x_i corresponds to the influence of paper i. Higher x_i means the paper is more influential because it's cited by many other influential papers.I think that's the gist of it. I should make sure I'm not missing any steps. Maybe for part 1, I should explicitly state the Perron-Frobenius theorem and its implications. For part 2, perhaps mention the power method for computing the eigenvector if the matrix is large, but the question doesn't specify computation, just the concept.Yeah, I think that's solid.Final Answer1. The adjacency matrix ( A ) has at least one non-zero eigenvalue, specifically the Perron root, which is significant as it represents the network's influence potential. This eigenvalue is (boxed{lambda}).2. The principal eigenvector of ( A ) identifies the most influential papers, with higher values indicating greater influence. The eigenvector is (boxed{mathbf{x}}).</think>

🧠:<think>First, I'll calculate the distance of each segment of the tour. The total distance is 120 kilometers.For the first segment, which is 1/4 of the total distance:120 km * 1/4 = 30 km.The second segment is twice the distance of the first segment:2 * 30 km = 60 km.To find the distance of the third segment, I'll subtract the distances of the first and second segments from the total:120 km - 30 km - 60 km = 30 km.Next, I'll calculate the time taken for each segment. For the first segment at 18 km/h:30 km / 18 km/h = 1.6667 hours, which is 1 hour and 40 minutes.For the second segment at 15 km/h:60 km / 15 km/h = 4 hours.For the third segment at 20 km/h:30 km / 20 km/h = 1.5 hours, which is 1 hour and 30 minutes.Adding up all the times:1 hour 40 minutes + 4 hours + 1 hour 30 minutes = 7 hours 10 minutes.Finally, I'll determine the finish time by adding the total time to the start time of 8:00 AM:8:00 AM + 7 hours 10 minutes = 3:10 PM.</think>