🧠:<think>Okay, so I have this problem about optimizing a data network between government departments. It's divided into two sub-problems. Let me try to tackle them one by one.Starting with Sub-problem 1: The network is initially a complete graph ( K_n ), which means every department is connected to every other department. The goal is to reduce redundancy by removing edges while still keeping the network connected. They want a spanning tree because that's the minimal connected graph, right? So, a spanning tree on ( n ) nodes has exactly ( n - 1 ) edges. But wait, the question is asking how many edges need to be removed from ( K_n ) to get a spanning tree. I know that a complete graph ( K_n ) has ( frac{n(n - 1)}{2} ) edges. So, if we need a spanning tree with ( n - 1 ) edges, the number of edges to remove would be the total edges minus the edges in the spanning tree.Let me write that down:Total edges in ( K_n ): ( frac{n(n - 1)}{2} )Edges in a spanning tree: ( n - 1 )So, edges to remove: ( frac{n(n - 1)}{2} - (n - 1) )Simplify that:Factor out ( n - 1 ): ( (n - 1)left( frac{n}{2} - 1 right) )Which is ( (n - 1)left( frac{n - 2}{2} right) )So, ( frac{(n - 1)(n - 2)}{2} ) edges need to be removed.Wait, let me check that. If n is 3, for example, ( K_3 ) has 3 edges. A spanning tree would have 2 edges, so we remove 1 edge. Plugging into the formula: ( frac{(3 - 1)(3 - 2)}{2} = frac{2 * 1}{2} = 1 ). That works. For n=4, ( K_4 ) has 6 edges. A spanning tree has 3 edges, so we remove 3 edges. The formula gives ( frac{(4 - 1)(4 - 2)}{2} = frac{3 * 2}{2} = 3 ). Perfect. So, the formula seems correct.So, Sub-problem 1 answer is ( frac{(n - 1)(n - 2)}{2} ).Moving on to Sub-problem 2: Now, they want to implement a new protocol to reduce latency. Each edge has a weight between 1 and 10, uniformly distributed. They want the average latency of the spanning tree to be less than or equal to a threshold ( T ). We need to find the probability of that happening.Hmm. So, the spanning tree has ( n - 1 ) edges. Each edge's weight is uniformly distributed between 1 and 10. So, each weight ( w(e) ) is a random variable with a uniform distribution on [1,10].The average latency would be the average of these ( n - 1 ) weights. So, the average ( bar{w} = frac{1}{n - 1} sum_{i=1}^{n - 1} w(e_i) ).We need the probability that ( bar{w} leq T ).Since ( n ) is large, we can use the Central Limit Theorem. The sum of a large number of independent, identically distributed random variables will be approximately normally distributed.First, let's find the mean and variance of a single edge's weight.For a uniform distribution on [1,10], the mean ( mu ) is ( frac{1 + 10}{2} = 5.5 ).The variance ( sigma^2 ) is ( frac{(10 - 1)^2}{12} = frac{81}{12} = 6.75 ).So, each ( w(e_i) ) has mean 5.5 and variance 6.75.Now, the sum ( S = sum_{i=1}^{n - 1} w(e_i) ) will have mean ( (n - 1)mu = 5.5(n - 1) ) and variance ( (n - 1)sigma^2 = 6.75(n - 1) ).The average ( bar{w} = frac{S}{n - 1} ) will have mean ( mu = 5.5 ) and variance ( frac{sigma^2}{n - 1} = frac{6.75}{n - 1} ).Therefore, ( bar{w} ) is approximately normally distributed with mean 5.5 and standard deviation ( sqrt{frac{6.75}{n - 1}} ).We need ( P(bar{w} leq T) ).This is the probability that a normal variable with mean 5.5 and standard deviation ( sqrt{frac{6.75}{n - 1}} ) is less than or equal to ( T ).To find this probability, we can standardize it:( Z = frac{bar{w} - 5.5}{sqrt{frac{6.75}{n - 1}}} )So, ( P(bar{w} leq T) = Pleft( Z leq frac{T - 5.5}{sqrt{frac{6.75}{n - 1}}} right) )Which is ( Phileft( frac{T - 5.5}{sqrt{frac{6.75}{n - 1}}} right) ), where ( Phi ) is the standard normal cumulative distribution function.Alternatively, we can write it as:( Phileft( frac{(T - 5.5)sqrt{n - 1}}{sqrt{6.75}} right) )Simplify ( sqrt{6.75} ). Since 6.75 is 27/4, so ( sqrt{6.75} = sqrt{27/4} = (3sqrt{3})/2 approx 2.598 ).But maybe we can leave it as ( sqrt{6.75} ) for exactness.So, the expression is ( Phileft( frac{(T - 5.5)sqrt{n - 1}}{sqrt{6.75}} right) ).Alternatively, since ( 6.75 = 27/4 ), we can write ( sqrt{6.75} = sqrt{27}/2 = (3sqrt{3})/2 ), so:( Phileft( frac{2(T - 5.5)sqrt{n - 1}}{3sqrt{3}} right) )But perhaps it's better to just write it as ( Phileft( frac{(T - 5.5)sqrt{n - 1}}{sqrt{6.75}} right) ).So, that's the probability expression.Wait, but hold on. Is the spanning tree selected in a particular way? Because in a complete graph, there are many spanning trees, each with different sets of edges. But the problem says "the average weight of the edges in the spanning tree". So, does it matter which spanning tree is chosen? Or is it considering all possible spanning trees?Hmm, the problem says "the average weight of the edges in the spanning tree". It doesn't specify which spanning tree, but since the weights are assigned randomly, maybe we can assume that all spanning trees are equally likely? Or perhaps, since the weights are assigned uniformly, the average over all possible spanning trees would be the same as the average of a random spanning tree.But actually, in the problem statement, it's not clear whether the spanning tree is fixed or random. Wait, the initial network is a complete graph, and edges have weights assigned uniformly. Then, the Minister wants to ensure that the average latency does not exceed T. So, I think the spanning tree is being selected, but the weights are random variables.But in reality, the spanning tree is a specific set of edges, but since the weights are random, the average is a random variable. So, we need the probability that this random variable is less than or equal to T.But in the problem, it's about the average weight of the edges in the spanning tree. So, if the spanning tree is fixed, then it's the average of n-1 specific edges. But since the edges are randomly weighted, the average is a random variable.But in the problem, it's not specified whether the spanning tree is fixed or random. But since the weights are assigned uniformly, perhaps the spanning tree is also randomly selected? Or maybe it's a minimum spanning tree? Wait, the problem doesn't specify.Wait, the problem says: "the average weight of the edges in the spanning tree". It doesn't specify which spanning tree, so perhaps it's considering any spanning tree, but since the weights are random, the average is a random variable.But actually, if the spanning tree is fixed, then the average is the average of n-1 specific edges. But since the weights are assigned uniformly, the average would be a random variable with mean 5.5 and variance 6.75/(n-1). So, regardless of the spanning tree, as long as the edges are independent, the distribution of the average would be the same.But wait, in a complete graph, all edges are present, but when you choose a spanning tree, you're selecting n-1 edges. However, in reality, in a complete graph, the edges are not independent because selecting one edge affects the others? Wait, no, the weights are assigned independently. So, each edge has its own weight, independent of others.Therefore, regardless of which spanning tree is chosen, the average of its edges is the average of n-1 independent uniform variables on [1,10]. So, the distribution is the same for any spanning tree.Therefore, the probability that the average is less than or equal to T is the same as the probability that the average of n-1 independent uniform [1,10] variables is less than or equal to T.Hence, the expression I derived earlier applies.So, the probability is ( Phileft( frac{(T - 5.5)sqrt{n - 1}}{sqrt{6.75}} right) ).Alternatively, since ( sqrt{6.75} = sqrt{27/4} = (3sqrt{3})/2 ), we can write it as:( Phileft( frac{2(T - 5.5)sqrt{n - 1}}{3sqrt{3}} right) )But I think it's fine to leave it as ( sqrt{6.75} ).So, summarizing:The probability is the standard normal CDF evaluated at ( frac{(T - 5.5)sqrt{n - 1}}{sqrt{6.75}} ).Therefore, the expression is ( Phileft( frac{(T - 5.5)sqrt{n - 1}}{sqrt{6.75}} right) ).I think that's the answer for Sub-problem 2.Final AnswerSub-problem 1: The minimum number of edges to remove is boxed{frac{(n - 1)(n - 2)}{2}}.Sub-problem 2: The probability is boxed{Phileft( frac{(T - 5.5)sqrt{n - 1}}{sqrt{6.75}} right)}.</think>

🧠:<think>Alright, so I'm trying to tackle these two questions about designing an efficient dynamic array system using a balanced binary tree. Let me start by understanding each part step by step.Question 1: Proving Average Time Complexity with Self-Balancing BSTsOkay, the first question is about proving that using a self-balancing BST like AVL or Red-Black tree ensures that any sequence of n insertions and deletions has an average time complexity of O(log n). I remember that balanced BSTs maintain a height that's logarithmic in the number of nodes, which is why their operations are efficient. But I need to formalize this.So, a balanced BST ensures that the tree's height is always O(log n). For example, in an AVL tree, the height difference between any two subtrees is at most 1. This property is maintained through rotations during insertions and deletions. Similarly, Red-Black trees use coloring to ensure the tree remains roughly balanced, with a maximum height of 2 log n or something like that.Now, for insertion and deletion operations, each involves traversing from the root to a leaf, which takes time proportional to the height of the tree. Since the height is O(log n), each operation is O(log n). But the question mentions the average time complexity over n operations. I think this relates to amortized analysis, where even though some operations might take longer, the average over many operations remains logarithmic.Wait, but in self-balancing BSTs, each individual operation is O(log n), so the average should also be O(log n). Maybe I need to discuss the properties that ensure this. The key is that the tree remains balanced after each operation, preventing the worst-case scenario of a skewed tree which would have O(n) operations.Mathematically, the height h of a balanced BST with n nodes satisfies h = O(log n). Therefore, any operation that requires traversing the height will take O(log n) time. Since each insertion and deletion involves such a traversal, their time complexity is O(log n). Thus, over n operations, the average is also O(log n).I should also mention that these trees maintain their balance through specific mechanisms. For AVL trees, it's rotations, and for Red-Black trees, it's rotations and recoloring. These operations ensure that the tree doesn't become too unbalanced, keeping the height logarithmic.Question 2: Huffman Coding for Space EfficiencyThe second part is about compressing tree nodes using Huffman coding. Each node stores an integer, and we encode these based on their access frequency. The goal is to construct an optimal Huffman tree and find the average encoded length, then discuss how this affects space complexity.First, Huffman coding assigns variable-length codes to symbols (here, integers) based on their frequency. More frequent symbols get shorter codes. The Huffman tree is built by combining the two least frequent nodes until one tree remains.Given a sequence of access frequencies, I need to construct the Huffman tree. Let's say the frequencies are given as a list. I would start by creating a priority queue (min-heap) of nodes, each with its frequency. Then, repeatedly extract the two nodes with the smallest frequencies, combine them into a new parent node with frequency equal to the sum, and insert this back into the queue. This continues until one node remains, which becomes the root.The average encoded length is the sum over all nodes of (frequency * depth). Since Huffman coding is optimal, this average is minimized. Therefore, the space saved is maximized.Now, how does this impact the overall space complexity? Each node's integer is now represented by a variable-length code. If the integers are large, say 32-bit or 64-bit, replacing them with shorter codes can significantly reduce the space. However, the Huffman tree itself requires some storage. Each node in the Huffman tree needs to store the code length or the actual code, which might add overhead. But if the savings from shorter codes outweigh the overhead, the overall space complexity improves.I should also consider that the Huffman tree is built once based on access frequencies, so the construction time is a one-time cost. The space savings come from the more efficient encoding of frequently accessed nodes, which are likely to be more common, thus their shorter codes contribute more to the average.Wait, but in the context of a dynamic array managed by a BST, each node's value is part of the array. If we Huffman encode these values, we can store them more efficiently. However, when performing operations like insertion or deletion, we might need to decode these values, which could add some overhead. But since the time complexity is dominated by the tree operations, which are O(log n), the decoding time, which is likely O(1) or O(log k) where k is the code length, shouldn't affect the overall time complexity significantly.So, in summary, using Huffman coding can reduce the space needed to store the node values, especially if some values are accessed more frequently and thus encoded with shorter codes. This leads to an overall improvement in space efficiency without significantly impacting the time complexity.Putting It All TogetherFor the first part, the key points are the properties of self-balancing BSTs that maintain logarithmic height, ensuring each operation is O(log n). For the second part, constructing the Huffman tree optimally compresses the node values, reducing space usage based on access frequencies.I think I've covered the main ideas, but I should make sure to explain the mathematical principles clearly, like the height of the tree and how it relates to log n, and the optimality of Huffman coding in minimizing the average code length.Final Answer1. The use of a self-balancing BST ensures that each insertion and deletion operation has an average time complexity of (O(log n)) due to the tree's logarithmic height. This is proven by the properties of such trees maintaining balance through rotations and recoloring, ensuring operations traverse (O(log n)) nodes.2. Applying Huffman coding based on access frequencies constructs an optimal tree with minimal average encoded length, improving space efficiency. The average encoded length is calculated as the sum of (frequency × code length) for all nodes, leading to reduced space usage for frequently accessed values.The final answers are:1. boxed{O(log n)}2. The average encoded length is minimized, improving space efficiency, with the exact value depending on the frequencies but generally reducing the overall space complexity.</think>

🧠:<think>Alright, so I'm trying to help Alex with his baking optimization problem. He's a software developer who likes baking, and now he's into mathematical modeling to make his bread's texture perfect. Cool! The problem has two parts, so I'll tackle them one by one.Problem 1: Finding Critical Points of the Texture FunctionFirst, the texture function is given as:( T(x, y, z) = 100e^{-left( frac{(x-30)^2}{50} + frac{(y-60)^2}{100} + frac{(z-200)^2}{250} right)} )I need to find the critical points and determine if they are local maxima, minima, or saddle points.Okay, critical points occur where the gradient of T is zero. That is, where the partial derivatives with respect to x, y, and z are all zero.Let me compute the partial derivatives.First, let's denote the exponent as:( E = frac{(x-30)^2}{50} + frac{(y-60)^2}{100} + frac{(z-200)^2}{250} )So, ( T = 100e^{-E} )The partial derivative of T with respect to x is:( frac{partial T}{partial x} = 100e^{-E} cdot left( -frac{2(x-30)}{50} right) = -100e^{-E} cdot frac{(x-30)}{25} )Similarly, partial derivative with respect to y:( frac{partial T}{partial y} = 100e^{-E} cdot left( -frac{2(y-60)}{100} right) = -100e^{-E} cdot frac{(y-60)}{50} )And with respect to z:( frac{partial T}{partial z} = 100e^{-E} cdot left( -frac{2(z-200)}{250} right) = -100e^{-E} cdot frac{(z-200)}{125} )To find critical points, set each partial derivative to zero.So,1. ( -100e^{-E} cdot frac{(x-30)}{25} = 0 )2. ( -100e^{-E} cdot frac{(y-60)}{50} = 0 )3. ( -100e^{-E} cdot frac{(z-200)}{125} = 0 )Since ( e^{-E} ) is always positive, the terms multiplying it can't be zero unless the other factors are zero.So,1. ( (x - 30) = 0 ) => ( x = 30 )2. ( (y - 60) = 0 ) => ( y = 60 )3. ( (z - 200) = 0 ) => ( z = 200 )So, the only critical point is at (30, 60, 200).Now, we need to determine the nature of this critical point. Since the function T is a product of exponentials, which are all positive, and the exponent is a sum of squared terms, this is a Gaussian-like function. The maximum occurs at the mean of each variable, which is exactly at (30, 60, 200). So, this should be a local maximum.But to confirm, let's compute the second partial derivatives and use the second derivative test.Compute the Hessian matrix:( H = begin{bmatrix}frac{partial^2 T}{partial x^2} & frac{partial^2 T}{partial x partial y} & frac{partial^2 T}{partial x partial z} frac{partial^2 T}{partial y partial x} & frac{partial^2 T}{partial y^2} & frac{partial^2 T}{partial y partial z} frac{partial^2 T}{partial z partial x} & frac{partial^2 T}{partial z partial y} & frac{partial^2 T}{partial z^2}end{bmatrix} )First, compute the second partial derivatives.Starting with ( frac{partial^2 T}{partial x^2} ):We have ( frac{partial T}{partial x} = -100e^{-E} cdot frac{(x-30)}{25} )Differentiate again with respect to x:( frac{partial^2 T}{partial x^2} = -100e^{-E} cdot frac{1}{25} - 100e^{-E} cdot frac{(x-30)}{25} cdot left( -frac{2(x-30)}{50} right) )Wait, that seems a bit complicated. Maybe it's better to compute it step by step.Let me denote ( f = e^{-E} ), so ( T = 100f )Then, ( frac{partial T}{partial x} = 100 cdot frac{partial f}{partial x} )Similarly, ( frac{partial^2 T}{partial x^2} = 100 cdot frac{partial^2 f}{partial x^2} )Compute ( frac{partial f}{partial x} = -e^{-E} cdot frac{2(x - 30)}{50} )Then, ( frac{partial^2 f}{partial x^2} = -e^{-E} cdot frac{2}{50} + e^{-E} cdot left( frac{2(x - 30)}{50} right)^2 cdot 2 )Wait, no. Let me compute it correctly.Using the chain rule:( frac{partial f}{partial x} = -e^{-E} cdot frac{2(x - 30)}{50} )Then,( frac{partial^2 f}{partial x^2} = frac{partial}{partial x} left( -e^{-E} cdot frac{2(x - 30)}{50} right) )This is the derivative of a product: derivative of (-e^{-E}) times (2(x - 30)/50) plus (-e^{-E}) times derivative of (2(x - 30)/50).So,( frac{partial^2 f}{partial x^2} = e^{-E} cdot left( frac{2(x - 30)}{50} right)^2 cdot 2 + (-e^{-E}) cdot frac{2}{50} )Wait, let's compute it step by step.First term: derivative of (-e^{-E}) is e^{-E} * derivative of E with respect to x.Derivative of E with respect to x is (2(x - 30))/50.So,( frac{partial}{partial x} (-e^{-E}) = e^{-E} cdot frac{2(x - 30)}{50} )Then, multiply by (2(x - 30)/50):First part: ( e^{-E} cdot frac{2(x - 30)}{50} cdot frac{2(x - 30)}{50} = e^{-E} cdot frac{4(x - 30)^2}{2500} )Second part: (-e^{-E}) times derivative of (2(x - 30)/50) which is 2/50.So, second part: ( -e^{-E} cdot frac{2}{50} )Therefore, overall:( frac{partial^2 f}{partial x^2} = e^{-E} cdot left( frac{4(x - 30)^2}{2500} - frac{2}{50} right) )Similarly, compute ( frac{partial^2 f}{partial y^2} ) and ( frac{partial^2 f}{partial z^2} ).But since the function is a product of Gaussians in x, y, z, the Hessian should be diagonal, and the cross partials should be zero.Wait, actually, since E is a sum of squares, the function f is a product of exponentials, each depending on a single variable. Therefore, the mixed partial derivatives should be zero.Hence, the Hessian matrix will be diagonal.So, at the critical point (30, 60, 200), let's compute the second derivatives.At x = 30, (x - 30) = 0, so:( frac{partial^2 f}{partial x^2} = e^{-E} cdot left( 0 - frac{2}{50} right) = -e^{-E} cdot frac{2}{50} )Similarly, for y:( frac{partial^2 f}{partial y^2} = e^{-E} cdot left( frac{4(y - 60)^2}{10000} - frac{2}{100} right) )At y = 60, (y - 60) = 0, so:( frac{partial^2 f}{partial y^2} = -e^{-E} cdot frac{2}{100} )Similarly, for z:( frac{partial^2 f}{partial z^2} = e^{-E} cdot left( frac{4(z - 200)^2}{62500} - frac{2}{250} right) )At z = 200, (z - 200) = 0, so:( frac{partial^2 f}{partial z^2} = -e^{-E} cdot frac{2}{250} )Therefore, the Hessian matrix at (30, 60, 200) is:( H = 100 cdot e^{-E} cdot begin{bmatrix}- frac{2}{50} & 0 & 0 0 & - frac{2}{100} & 0 0 & 0 & - frac{2}{250}end{bmatrix} )Since E at (30,60,200) is zero, e^{-E} = 1.So,( H = 100 cdot begin{bmatrix}- frac{2}{50} & 0 & 0 0 & - frac{2}{100} & 0 0 & 0 & - frac{2}{250}end{bmatrix} = begin{bmatrix}-4 & 0 & 0 0 & -2 & 0 0 & 0 & -0.8end{bmatrix} )All the eigenvalues of this matrix are negative (since all diagonal elements are negative and it's diagonal), which means the Hessian is negative definite. Therefore, the critical point at (30,60,200) is a local maximum.So, that answers the first part.Problem 2: Solving the Heat Equation with Given Boundary ConditionsNow, the second part is about solving the heat equation in three dimensions with specific boundary conditions.The heat equation is:( frac{partial u}{partial t} = alpha left( frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} + frac{partial^2 u}{partial z^2} right) )Boundary conditions: u is zero on all faces of the cube of side length L. So, u(0,y,z,t) = u(L,y,z,t) = 0, similarly for y and z.Initial condition: u(x,y,z,0) = f(x,y,z)We need to derive the general solution.This is a standard problem in PDEs. The solution can be found using separation of variables.Assume a solution of the form:( u(x,y,z,t) = X(x)Y(y)Z(z)T(t) )Plugging into the heat equation:( X Y Z frac{dT}{dt} = alpha left( X'' Y Z + X Y'' Z + X Y Z'' right) T )Divide both sides by ( alpha X Y Z T ):( frac{1}{alpha T} frac{dT}{dt} = frac{X''}{X} + frac{Y''}{Y} + frac{Z''}{Z} )Since the left side depends only on t and the right side depends only on x, y, z, both sides must be equal to a constant, say -λ.So,( frac{1}{alpha T} frac{dT}{dt} = -lambda )and( frac{X''}{X} + frac{Y''}{Y} + frac{Z''}{Z} = -lambda )Let me denote:( frac{X''}{X} = -lambda_x )( frac{Y''}{Y} = -lambda_y )( frac{Z''}{Z} = -lambda_z )So,( -lambda_x - lambda_y - lambda_z = -lambda )Thus,( lambda = lambda_x + lambda_y + lambda_z )Now, each of X, Y, Z satisfies an ODE with boundary conditions.Given the boundary conditions, for X(x):X(0) = X(L) = 0Similarly for Y(y) and Z(z):Y(0) = Y(L) = 0Z(0) = Z(L) = 0So, the solutions for X, Y, Z are sine functions.For X(x):( X_n(x) = sinleft( frac{npi x}{L} right) ), with ( lambda_x = left( frac{npi}{L} right)^2 ), n = 1,2,3,...Similarly, Y_m(y) = sin(mπy/L), λ_y = (mπ/L)^2, m = 1,2,3,...Z_k(z) = sin(kπz/L), λ_z = (kπ/L)^2, k = 1,2,3,...Thus, the eigenvalues are:( lambda_{n,m,k} = left( frac{npi}{L} right)^2 + left( frac{mpi}{L} right)^2 + left( frac{kpi}{L} right)^2 )Now, the time-dependent part:( frac{dT}{dt} = -alpha lambda_{n,m,k} T )This is a first-order ODE, solution is:( T_{n,m,k}(t) = e^{-alpha lambda_{n,m,k} t} )Therefore, the general solution is a triple sum over n, m, k:( u(x,y,z,t) = sum_{n=1}^infty sum_{m=1}^infty sum_{k=1}^infty A_{n,m,k} sinleft( frac{npi x}{L} right) sinleft( frac{mpi y}{L} right) sinleft( frac{kpi z}{L} right) e^{-alpha lambda_{n,m,k} t} )Where ( A_{n,m,k} ) are constants determined by the initial condition.To find ( A_{n,m,k} ), use the initial condition:( u(x,y,z,0) = f(x,y,z) = sum_{n=1}^infty sum_{m=1}^infty sum_{k=1}^infty A_{n,m,k} sinleft( frac{npi x}{L} right) sinleft( frac{mpi y}{L} right) sinleft( frac{kpi z}{L} right) )This is a three-dimensional Fourier sine series. The coefficients ( A_{n,m,k} ) can be found by:( A_{n,m,k} = frac{8}{L^3} int_0^L int_0^L int_0^L f(x,y,z) sinleft( frac{npi x}{L} right) sinleft( frac{mpi y}{L} right) sinleft( frac{kpi z}{L} right) dx dy dz )So, putting it all together, the general solution is:( u(x,y,z,t) = sum_{n=1}^infty sum_{m=1}^infty sum_{k=1}^infty left[ frac{8}{L^3} int_0^L int_0^L int_0^L f(x',y',z') sinleft( frac{npi x'}{L} right) sinleft( frac{mpi y'}{L} right) sinleft( frac{kpi z'}{L} right) dx' dy' dz' right] times sinleft( frac{npi x}{L} right) sinleft( frac{mpi y}{L} right) sinleft( frac{kpi z}{L} right) e^{-alpha left( frac{n^2 + m^2 + k^2}{L^2} right) pi^2 t} )That's the general solution.Summary of Thoughts:For the first problem, I recognized that the function is a multivariate Gaussian, so the maximum occurs at the mean values. I computed the partial derivatives, found the critical point, and then used the second derivative test (Hessian) to confirm it's a local maximum.For the second problem, I recalled that the heat equation with Dirichlet boundary conditions can be solved using separation of variables. I broke it down into spatial and temporal parts, solved the ODEs, and expressed the solution as a triple Fourier sine series. The coefficients are determined by the initial condition through integration.I think I covered all the necessary steps, but I should double-check the Hessian computation to ensure I didn't make any sign errors. Also, in the heat equation solution, I need to make sure the eigenvalues and the separation constants are correctly accounted for.Final Answer1. The critical point is at ( boxed{(30, 60, 200)} ) and it is a local maximum.2. The general solution is:( u(x,y,z,t) = sum_{n=1}^infty sum_{m=1}^infty sum_{k=1}^infty left( frac{8}{L^3} int_0^L int_0^L int_0^L f(x',y',z') sinleft( frac{npi x'}{L} right) sinleft( frac{mpi y'}{L} right) sinleft( frac{kpi z'}{L} right) dx' dy' dz' right) sinleft( frac{npi x}{L} right) sinleft( frac{mpi y}{L} right) sinleft( frac{kpi z}{L} right) e^{-alpha left( frac{n^2 + m^2 + k^2}{L^2} right) pi^2 t} )So, the boxed answer for the critical point is:( boxed{(30, 60, 200)} )</think>

🧠:<think>Okay, so I have this problem where a conservative Louisianan wants to analyze voting patterns over four elections. He's concerned about political maneuvering and biased media, so he's looking into the data himself. The data includes the number of votes for Party A and Party B in each election, as well as the total voter turnout. There are two sub-problems here. The first one is about calculating the percentage change in votes for Party A and determining if there's a statistically significant trend using linear regression. The second sub-problem is a chi-square test to see if voter turnout is related to the votes for Party A and B.Starting with the first sub-problem. I need to calculate the percentage change in Party A's votes from one election to the next. So, for each subsequent election, I'll take the difference in votes between that election and the previous one, divide by the previous election's votes, and multiply by 100 to get a percentage. But wait, percentage change can be a bit tricky because it's relative. So, if Party A had, say, 100 votes in Election 1 and 150 in Election 2, that's a 50% increase. But if they had 200 votes in Election 3, that's only a 33% increase from Election 2. So, the percentage change can vary even if the absolute change is the same. Once I have these percentage changes, I need to see if there's a trend. That's where linear regression comes in. I'll model the percentage change as a function of time (election number). The regression equation will be something like Percentage Change = β0 + β1*(Election Number) + ε, where ε is the error term. The slope β1 will tell me if there's an increasing or decreasing trend. If β1 is positive and statistically significant, that means Party A's vote percentage is increasing over time. If it's negative and significant, the opposite. If it's not significant, there's no clear trend. To perform the linear regression, I need the percentage changes for each election. Let me denote them as PC1, PC2, PC3, where PC1 is the change from Election 1 to 2, PC2 from 2 to 3, and PC3 from 3 to 4. Then, I can set up my data points as (1, PC1), (2, PC2), (3, PC3). I can use software like Excel or R to compute the regression, but since I'm doing this manually, I'll recall the formula for the slope in simple linear regression. The slope β1 is calculated as the covariance of the two variables divided by the variance of the independent variable (election number). Covariance is the average of the product of the deviations of each variable from their respective means. Variance is the average of the squared deviations from the mean for the independent variable. So, first, I'll compute the mean of the election numbers (which are 1, 2, 3) and the mean of the percentage changes. Then, for each pair, I'll subtract the mean from the election number and the percentage change, multiply them together, sum those products, and divide by the sum of the squared deviations of the election numbers from their mean. This will give me the slope. Then, I can calculate the intercept β0 as the mean of the percentage changes minus the slope times the mean of the election numbers. After getting the regression equation, I need to interpret the slope. If it's positive, Party A's vote percentage is increasing each election; if negative, decreasing. The significance will depend on the p-value associated with the slope. If the p-value is less than 0.05, we can say the trend is statistically significant.Moving on to the second sub-problem, the chi-square test for independence. This test is used to determine if there's a significant association between two categorical variables. In this case, the variables are voter turnout and the votes for Party A and B. Wait, but voter turnout is a continuous variable, right? It's the total number of voters in each election. Votes for Party A and B are also continuous. So, a chi-square test might not be directly applicable because it typically works with categorical data, like counts in different categories. Hmm, maybe I need to reframe this. Perhaps, instead of using the actual vote counts, I can categorize the voter turnout into high and low, and similarly categorize the votes for Party A and B into high and low. Then, I can create a contingency table and perform the chi-square test. Alternatively, maybe the problem expects me to treat voter turnout as a categorical variable by grouping the elections into categories based on turnout. For example, if turnout was above a certain threshold or below. But without specific thresholds, it's hard to categorize. Alternatively, I could consider each election as a separate category, but that might not make sense because the chi-square test would then be testing the independence across elections, which isn't the intended use. Wait, perhaps the variables are voter turnout and the proportion of votes for Party A and B. But the chi-square test isn't suitable for proportions either. Maybe I need to use a different test, like a correlation test or a regression analysis. But the problem specifically mentions a chi-square test for independence. So, perhaps I need to think differently. Maybe the data is structured in a way that each election's voter turnout and the votes for each party can be considered as observed frequencies in a contingency table. Let me think. For each election, we have the total votes for A, total votes for B, and total turnout. So, for each election, the votes for A and B are counts, and the total turnout is the sum. So, if I create a 4x2 contingency table where each row is an election, and the columns are Party A and Party B votes, then I can perform a chi-square test to see if the distribution of votes is independent of the election. But that might not be the right approach because each election is a separate observation. Alternatively, if I consider voter turnout as a separate variable, maybe I can create a table where rows are high/low turnout and columns are high/low votes for Party A and B. But without clear categories, this is tricky. Alternatively, maybe the problem is expecting a chi-square goodness-of-fit test, comparing observed vote counts to expected counts under some null hypothesis. But the problem mentions a test for independence, which suggests it's about the relationship between two variables. Wait, perhaps the variables are voter turnout and party preference. So, for each election, we can consider the voter turnout as a factor and the votes for A and B as the outcomes. But again, since each election is a separate data point, it's unclear how to structure the contingency table. I might need to reconsider. Maybe the problem is expecting a different approach. Perhaps, instead of treating each election separately, we can aggregate the data. But with only four elections, that might not be feasible. Alternatively, maybe the problem is considering voter turnout as a categorical variable with four levels (each election) and the votes for A and B as counts. Then, a chi-square test can be used to see if the distribution of votes (A vs B) is independent of the election (turnout). Wait, that might make sense. So, we can set up a 4x2 contingency table where each row is an election, and the columns are Party A and Party B. The entries are the number of votes each party received in each election. Then, we can perform a chi-square test for independence to see if the distribution of votes (A vs B) is independent of the election (which is associated with voter turnout). If the chi-square statistic is significant, it would suggest that the distribution of votes is not independent of the election, implying that voter turnout is related to the votes for Party A and B. But I'm not entirely sure if this is the correct way to structure the test. Alternatively, maybe the problem expects a different approach, such as a chi-square test comparing observed and expected vote counts based on some model. I think the key here is that the chi-square test for independence is used when we have two categorical variables and want to see if they are related. In this case, the two variables could be "Election" (which is associated with voter turnout) and "Party" (A or B). So, each election has a certain voter turnout, and within each election, the votes are split between Party A and B. If we set up a contingency table with elections as rows and parties as columns, the chi-square test will tell us if the distribution of parties is independent of the election, which is influenced by voter turnout. Therefore, I'll proceed by creating a 4x2 contingency table with the votes for Party A and B in each election. Then, calculate the expected counts under the assumption of independence, compute the chi-square statistic, and compare it to the critical value at a 0.05 significance level. If the calculated chi-square statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a significant relationship between voter turnout and the votes for Party A and B. Otherwise, we fail to reject the null hypothesis.So, to summarize my approach for both sub-problems:1. For the percentage change and linear regression: - Calculate percentage change in Party A's votes from each election to the next. - Set up a simple linear regression model with election number as the independent variable and percentage change as the dependent variable. - Calculate the slope and intercept, then interpret the slope's significance.2. For the chi-square test: - Create a 4x2 contingency table with elections as rows and Party A/B as columns, filled with vote counts. - Perform the chi-square test for independence to see if voter turnout (through election differences) is related to the distribution of votes between the parties.I think this is the way to go. Now, I'll proceed to outline the steps in detail for each sub-problem.</think>