🧠:<think>First, I need to determine the number of Ancient, Medieval, and Modern history books in the store. The total number of books is 120, and there are two given ratios: Ancient to Medieval is 3:2, and Medieval to Modern is 4:5.To combine these ratios, I'll find a common multiple for the Medieval category. The Ancient to Medieval ratio can be scaled to 6:4, and the Medieval to Modern ratio remains 4:5. This gives a combined ratio of Ancient:Medieval:Modern as 6:4:5.Next, I'll calculate the total number of parts in the ratio: 6 + 4 + 5 = 15 parts. Each part represents 120 / 15 = 8 books. Therefore, the number of Ancient books is 6 * 8 = 48, Medieval books are 4 * 8 = 32, and Modern books are 5 * 8 = 40.Now, to calculate the total cost for the enthusiast's purchase. The enthusiast buys 5 Ancient books, 3 Medieval books, and 2 Modern books. The original prices are 30, 25, and 20 respectively. However, there's a 10% discount on Ancient books due to the membership.First, calculate the cost of Ancient books with the discount: 5 books * 30 * 0.90 = 135. Then, calculate the cost of Medieval books: 3 books * 25 = 75. Finally, calculate the cost of Modern books: 2 books * 20 = 40. Adding these together, the total cost is 135 + 75 + 40 = 250.</think>

🧠:<think>Alright, so I have these two problems to solve related to web crawlers and scoring functions. Let me try to tackle them one by one. Starting with the first problem: deriving the iterative method to approximate the PageRank vector. Hmm, I remember that PageRank is a way to measure the importance of web pages based on their links. The formula given is (mathbf{p} = alpha M mathbf{p} + (1 - alpha) mathbf{v}). I think this is the standard PageRank equation where (alpha) is the damping factor, usually around 0.85, (M) is the transition matrix, and (mathbf{v}) is the teleportation vector which is often a uniform distribution over all pages.Since the web graph is large and sparse, we can't compute this directly using matrix inversion or something like that. So, we need an iterative method. I recall that the power method is commonly used for this. The power method is an algorithm that can find the dominant eigenvector of a matrix, which in this case corresponds to the PageRank vector.So, how does the power method work? I think it starts with an initial guess for the eigenvector, which in this case could be a uniform distribution where each page has equal rank. Then, we repeatedly multiply this vector by the transition matrix (M) scaled by (alpha) and add the teleportation term scaled by (1 - alpha). Let me write that down. Let (mathbf{p}^{(k)}) be the PageRank vector at iteration (k). The iterative formula would be:[mathbf{p}^{(k+1)} = alpha M mathbf{p}^{(k)} + (1 - alpha) mathbf{v}]Yes, that makes sense. So, starting with (mathbf{p}^{(0)}), which is usually a vector of ones or a uniform distribution, we keep applying this update until the vector converges, meaning the changes between iterations are below a certain threshold.But wait, the transition matrix (M) is derived from the adjacency matrix of the graph, right? Each entry (M_{ij}) represents the probability of moving from page (j) to page (i). Since the graph is sparse, (M) will have a lot of zeros, which is good because it means each multiplication step isn't too computationally intensive.Also, the teleportation vector (mathbf{v}) is typically a uniform vector, so each entry is (1/n) where (n) is the number of pages. This accounts for the random jumps a user might make to any page, hence the name teleportation.So, putting it all together, the iterative method is just repeatedly applying this formula until convergence. That should give us the PageRank vector (mathbf{p}).Moving on to the second problem: proving that the scoring function (S_i = beta p_i + (1 - beta) R_i) is bounded between 0 and 1. Given that (R_i = mathbf{f}_i cdot mathbf{q}), and both (mathbf{f}_i) and (mathbf{q}) are unit vectors, the dot product (R_i) is the cosine of the angle between them. Since the cosine function ranges between -1 and 1, (R_i) is between -1 and 1. But wait, in the context of content relevance, I think (R_i) should be non-negative because a negative relevance doesn't make much sense. Maybe the feature vectors are such that the dot product is non-negative? Or perhaps the scoring function is designed to handle that.But the problem statement doesn't specify that (R_i) is non-negative, just that (|mathbf{f}_i| = 1) and (mathbf{q}) is fixed. So, technically, (R_i) could be between -1 and 1. However, in practice, relevance scores are often non-negative, but I can't assume that here.But the scoring function (S_i) is a weighted average of (p_i) and (R_i). Since (beta) is between 0 and 1, let's see:If (R_i) can be as low as -1, then (S_i) could be as low as (beta p_i - (1 - beta)). But (p_i) is a PageRank score, which is a probability distribution, so (p_i) is between 0 and 1. Similarly, if (R_i) is 1, then (S_i) is (beta p_i + (1 - beta)). But wait, the problem says to prove that (S_i) is bounded between 0 and 1. So, maybe I need to consider the possible ranges.Let me think. Since (p_i) is a probability, (0 leq p_i leq 1). (R_i) is a dot product of two unit vectors, so ( -1 leq R_i leq 1 ). So, (S_i = beta p_i + (1 - beta) R_i). Let's find the maximum and minimum possible values.The maximum occurs when (p_i) is maximum (1) and (R_i) is maximum (1). So, (S_i = beta * 1 + (1 - beta) * 1 = beta + 1 - beta = 1).The minimum occurs when (p_i) is minimum (0) and (R_i) is minimum (-1). So, (S_i = beta * 0 + (1 - beta) * (-1) = - (1 - beta)). But this would be negative, which contradicts the statement that (S_i) is bounded between 0 and 1.Hmm, so maybe I'm missing something. Perhaps (R_i) is non-negative? Let me check the problem statement again. It says "content relevance score (R_i)", which is a dot product. If the feature vectors are such that the dot product is non-negative, maybe because they are in the same space or something, but it's not specified.Alternatively, maybe the scoring function is designed to take the absolute value or something, but the problem doesn't say that. Alternatively, perhaps the teleportation vector (mathbf{v}) is such that it ensures non-negativity, but no, (mathbf{v}) is just a vector.Wait, maybe I need to consider that in the context of the scoring function, negative relevance doesn't make sense, so perhaps (R_i) is actually the absolute value of the dot product, or it's normalized to be non-negative. But the problem doesn't specify that.Alternatively, maybe the scoring function is designed such that even if (R_i) is negative, the combination with (p_i) ensures it's non-negative. Let's see.If (R_i) is negative, say (R_i = -c) where (c geq 0), then (S_i = beta p_i - (1 - beta)c). Since (p_i geq 0), and (c leq 1) because (R_i geq -1), then (S_i geq - (1 - beta)). But this can be negative if (1 - beta > 0), which it is since (beta in [0,1]).So, unless there's a constraint that (R_i) is non-negative, (S_i) can be negative. But the problem says to prove it's bounded between 0 and 1. So, perhaps I'm missing a step.Wait, maybe the content relevance score is defined differently. The problem says (R_i = mathbf{f}_i cdot mathbf{q}), but if (mathbf{f}_i) and (mathbf{q}) are unit vectors, their dot product is the cosine similarity, which is between -1 and 1. But in practice, relevance scores are often non-negative, so maybe the problem assumes that (R_i) is non-negative. Alternatively, maybe the scoring function is adjusted to ensure non-negativity.Alternatively, perhaps the scoring function is a convex combination, but since (beta) is between 0 and 1, it's a weighted average. But if (R_i) can be negative, then the lower bound isn't necessarily 0.Wait, maybe I need to think differently. Let's consider the maximum and minimum possible values of (S_i):Maximum: When (p_i = 1) and (R_i = 1), (S_i = 1).Minimum: When (p_i = 0) and (R_i = -1), (S_i = - (1 - beta)). But this is negative, which contradicts the statement.So, unless there's a constraint that (R_i geq 0), the scoring function isn't necessarily bounded below by 0. But the problem says to prove it's bounded between 0 and 1, so perhaps I need to consider that (R_i) is non-negative.Wait, maybe the feature vectors are such that the dot product is non-negative. For example, if all feature vectors and the query vector are in the same direction, but that's not necessarily the case.Alternatively, perhaps the scoring function is defined as (S_i = beta p_i + (1 - beta) max(R_i, 0)), but the problem doesn't say that.Wait, maybe I'm overcomplicating. Let's see: since (beta in [0,1]), and (p_i in [0,1]), and (R_i in [-1,1]), then:The maximum value of (S_i) is when (p_i = 1) and (R_i = 1), so (S_i = 1).The minimum value is when (p_i = 0) and (R_i = -1), so (S_i = - (1 - beta)). But since (beta in [0,1]), (1 - beta in [0,1]), so the minimum is (- (1 - beta)), which is between -1 and 0.But the problem says to prove it's bounded between 0 and 1. So, unless there's a constraint on (R_i) being non-negative, this isn't true. Therefore, perhaps the problem assumes that (R_i geq 0). Let me check the problem statement again.It says: "the content relevance score (R_i) for each page (i) is given by the dot product of the page's feature vector (mathbf{f}_i) and a query vector (mathbf{q}), i.e., (R_i = mathbf{f}_i cdot mathbf{q}). Assuming ( |mathbf{f}_i| = 1 ) for all (i) and (mathbf{q}) is fixed."So, it doesn't specify that (R_i) is non-negative. Hmm. Maybe I need to consider that in the context of the scoring function, negative relevance is treated as zero. Or perhaps the scoring function is designed to be non-negative by construction.Alternatively, maybe the scoring function is a convex combination, but since (beta) is between 0 and 1, it's a weighted average. But if (R_i) can be negative, then the lower bound isn't necessarily 0.Wait, perhaps the problem is considering that (R_i) is non-negative because it's a relevance score, so even though mathematically it can be negative, in practice, it's treated as non-negative. So, assuming (R_i geq 0), then (S_i) would be between 0 and 1.But the problem doesn't state that, so I'm not sure. Alternatively, maybe the scoring function is designed such that even if (R_i) is negative, the combination with (p_i) ensures it's non-negative. Let's see.Suppose (R_i) is negative, say (R_i = -c) where (c geq 0). Then (S_i = beta p_i - (1 - beta)c). Since (p_i geq 0), and (c leq 1), but without knowing the relationship between (p_i) and (c), we can't guarantee (S_i geq 0).Wait, maybe the problem assumes that (R_i) is non-negative because it's a relevance score, so even though the dot product can be negative, in practice, it's considered as zero or something. But the problem doesn't specify that.Alternatively, maybe the scoring function is defined as (S_i = beta p_i + (1 - beta) max(R_i, 0)), but again, the problem doesn't say that.Hmm, this is confusing. Let me think differently. Maybe I can express (S_i) in terms of its maximum and minimum possible values.Given that (p_i in [0,1]) and (R_i in [-1,1]), then:The maximum value of (S_i) is when (p_i = 1) and (R_i = 1), so (S_i = beta + (1 - beta) = 1).The minimum value is when (p_i = 0) and (R_i = -1), so (S_i = - (1 - beta)). But since (beta in [0,1]), (1 - beta in [0,1]), so the minimum is between -1 and 0.But the problem says to prove it's bounded between 0 and 1. So, unless there's a constraint that (R_i geq 0), this isn't true. Therefore, perhaps the problem assumes that (R_i geq 0), or that the scoring function is adjusted to ensure non-negativity.Alternatively, maybe the scoring function is a convex combination where both terms are non-negative, so (S_i) is non-negative. But since (R_i) can be negative, that's not necessarily the case.Wait, maybe I'm missing something. Let's consider that (R_i) is the dot product of two unit vectors, so it's the cosine of the angle between them. Cosine is between -1 and 1, but in the context of relevance, maybe we take the absolute value or something. But the problem doesn't specify that.Alternatively, perhaps the scoring function is designed such that even if (R_i) is negative, the combination with (p_i) ensures it's non-negative. Let's see:If (S_i = beta p_i + (1 - beta) R_i), and we want (S_i geq 0), then:(beta p_i + (1 - beta) R_i geq 0)But since (p_i geq 0) and (R_i geq -1), the worst case is when (R_i = -1):(beta p_i - (1 - beta) geq 0)(beta p_i geq (1 - beta))But since (p_i leq 1), this would require:(beta geq (1 - beta))Which implies (beta geq 0.5). But (beta) can be less than 0.5, so this isn't guaranteed.Therefore, unless there's a constraint on (R_i) being non-negative, the scoring function isn't necessarily bounded below by 0. So, perhaps the problem assumes that (R_i geq 0), or that the scoring function is adjusted to ensure non-negativity.Alternatively, maybe the scoring function is defined as (S_i = beta p_i + (1 - beta) max(R_i, 0)), but the problem doesn't say that.Wait, maybe I'm overcomplicating. Let's think about the problem again. It says to prove that (S_i) is bounded between 0 and 1. So, perhaps I need to consider that both (p_i) and (R_i) are non-negative, even though mathematically (R_i) can be negative.Alternatively, maybe the scoring function is designed such that even if (R_i) is negative, the combination with (p_i) ensures it's non-negative. Let's see:If (R_i) is negative, say (R_i = -c), then (S_i = beta p_i - (1 - beta)c). Since (p_i geq 0), and (c leq 1), but without knowing the relationship between (p_i) and (c), we can't guarantee (S_i geq 0).Wait, maybe the problem is considering that (R_i) is non-negative because it's a relevance score, so even though the dot product can be negative, in practice, it's treated as zero or something. But the problem doesn't specify that.Alternatively, maybe the scoring function is defined as (S_i = beta p_i + (1 - beta) max(R_i, 0)), but again, the problem doesn't say that.Hmm, I'm stuck here. Maybe I need to proceed under the assumption that (R_i) is non-negative, even though mathematically it can be negative. So, assuming (R_i geq 0), then:Since (p_i in [0,1]) and (R_i in [0,1]), then (S_i = beta p_i + (1 - beta) R_i) is a convex combination, so it's between 0 and 1.Yes, that makes sense. So, if (R_i) is non-negative, then (S_i) is a weighted average of two numbers between 0 and 1, hence it's also between 0 and 1.But the problem doesn't specify that (R_i) is non-negative, so I'm not sure. Maybe I need to consider that (R_i) can be negative, but the scoring function is designed to ensure it's non-negative. Alternatively, perhaps the problem assumes that (R_i) is non-negative.Given that the problem asks to prove (S_i) is between 0 and 1, I think the intended approach is to assume that (R_i) is non-negative, perhaps because it's a relevance score, so even though mathematically it can be negative, in practice, it's non-negative.So, assuming (R_i geq 0), then (S_i) is a convex combination of (p_i) and (R_i), both of which are between 0 and 1, hence (S_i) is also between 0 and 1.Alternatively, even if (R_i) can be negative, the scoring function is designed such that the weights ensure it's non-negative. But without more information, I think the intended approach is to assume (R_i) is non-negative.So, to summarize:1. For the first problem, the iterative method is the power method, starting with an initial guess and repeatedly applying (mathbf{p}^{(k+1)} = alpha M mathbf{p}^{(k)} + (1 - alpha) mathbf{v}) until convergence.2. For the second problem, assuming (R_i) is non-negative (as it's a relevance score), then (S_i) is a convex combination of (p_i) and (R_i), both between 0 and 1, hence (S_i) is also between 0 and 1.</think>

🧠:<think>Okay, so I have this problem about a creative writing major developing an interactive storytelling platform. It uses a directed graph where nodes are story events and edges are transitions. There are two parts to the problem.Starting with part 1: I need to derive a recursive formula for the number of nodes ( V_n ) at depth ( n ) in the story graph. Then, find the total number of nodes up to depth ( n ).Alright, let's parse the information given. The initial node ( v_0 ) has 3 outgoing edges. Each subsequent node ( v_i ) (where ( i geq 1 )) has 2 outgoing edges. So, the graph starts with one node at depth 0, which branches into 3 nodes at depth 1. Each of those 3 nodes then branches into 2 nodes each at depth 2, right?So, if I think about it, the number of nodes at each depth is growing exponentially. At depth 0, it's 1 node. At depth 1, it's 3 nodes. At depth 2, each of those 3 nodes branches into 2, so 3*2=6 nodes. At depth 3, each of those 6 nodes branches into 2, so 6*2=12 nodes, and so on.So, the number of nodes at depth ( n ) seems to follow a pattern where each subsequent depth is multiplied by 2. But wait, the initial depth 0 is 1, depth 1 is 3, depth 2 is 6, depth 3 is 12... Hmm, so actually, it's 3 multiplied by 2^{n-1} for n >=1, and 1 for n=0.Wait, let me check:- Depth 0: 1 node- Depth 1: 3 nodes (1*3)- Depth 2: 3*2 = 6 nodes- Depth 3: 6*2 = 12 nodes- Depth 4: 12*2 = 24 nodesSo, yes, it's 3*2^{n-1} for n >=1. So, the recursive formula would relate ( V_n ) to ( V_{n-1} ). Since each node at depth n-1 branches into 2 nodes, the number of nodes at depth n is 2 times the number of nodes at depth n-1.But wait, for n=1, it's 3, which is 3*2^{0}=3. For n=2, it's 6=3*2^{1}. So, the general formula is ( V_n = 3*2^{n-1} ) for n >=1, and ( V_0 = 1 ).But the question asks for a recursive formula. So, how can I express ( V_n ) in terms of ( V_{n-1} )?Well, since each node at depth n-1 branches into 2 nodes, then ( V_n = 2*V_{n-1} ) for n >=1, with ( V_0 =1 ).But wait, hold on. At n=1, ( V_1 = 3 ), but according to the recursive formula ( V_1 = 2*V_0 = 2*1=2 ), which is not correct. So, that suggests my initial thought is wrong.Hmm, maybe the recursion isn't as straightforward because the initial node has a different number of outgoing edges. So, perhaps the recursion isn't uniform across all depths.Wait, so maybe I need to adjust the recursion. Let's think about the structure.At depth 0: 1 node.Each node at depth 0 branches into 3 nodes at depth 1.Each node at depth 1 branches into 2 nodes at depth 2.Each node at depth 2 branches into 2 nodes at depth 3.And so on.So, the number of nodes at each depth is:- ( V_0 = 1 )- ( V_1 = 3 )- ( V_2 = 3*2 = 6 )- ( V_3 = 6*2 = 12 )- ( V_4 = 12*2 = 24 )- etc.So, the recursion is ( V_n = 2*V_{n-1} ) for n >=2, with ( V_0 =1 ) and ( V_1 =3 ).So, that's the recursive formula. It's piecewise: for n=0, it's 1; for n=1, it's 3; and for n>=2, it's 2*V_{n-1}.Alternatively, maybe we can write it as:( V_n = begin{cases}1 & text{if } n = 0 3 & text{if } n = 1 2 V_{n-1} & text{if } n geq 2end{cases} )But perhaps the problem expects a single recursive formula without piecewise definition. Hmm.Alternatively, maybe we can express it as ( V_n = 2 V_{n-1} ) for n >=1, but with an initial condition that ( V_0 =1 ) and ( V_1 =3 ). That might be acceptable.But let's see: if n=1, V1=3=2*V0=2*1=2, which is not correct. So, that suggests that the recursion isn't consistent for n=1. So, perhaps the recursion is different for n=1 and n>=2.Alternatively, maybe the initial node is a special case, and the recursion is Vn = 2 V_{n-1} for n >=1, but starting from V0=1, which would give V1=2, but that contradicts the given information that V1=3.So, perhaps the correct approach is to model the number of nodes at each depth as Vn = 3*2^{n-1} for n >=1, and V0=1.But the question asks for a recursive formula, so perhaps we can express Vn in terms of V_{n-1}.Given that Vn = 2*V_{n-1} for n >=2, but V1=3, which is 3=2*V0 + something? Wait, 3=2*1 +1.Hmm, maybe the recursion is Vn = 2*V_{n-1} + c, where c is some constant.But let's see:If n=1: V1=3=2*V0 + c => 3=2*1 +c => c=1.Then, check for n=2: V2=6=2*V1 +c=2*3 +1=7, which is not correct.So, that doesn't work.Alternatively, maybe it's a different kind of recursion.Wait, perhaps the number of nodes at depth n is equal to the number of nodes at depth n-1 multiplied by the number of outgoing edges per node at depth n-1.But in this case, for n=1, V1=3=1*3.For n=2, V2=3*2=6.For n=3, V3=6*2=12.So, in general, Vn = V_{n-1} * k, where k is the number of outgoing edges from nodes at depth n-1.But since for n>=1, the nodes have 2 outgoing edges, except for the initial node which has 3.So, the recursion is:V0=1V1=3Vn=2*V_{n-1} for n>=2So, that's a piecewise recursion.Alternatively, if we want a single formula, perhaps using indicator functions or something, but that might complicate.Alternatively, maybe express it as Vn = 3*2^{n-1} for n >=1, which is a closed-form formula, but the question asks for a recursive formula.So, in conclusion, the recursive formula is:V0 = 1V1 = 3Vn = 2*V_{n-1} for n >=2Now, the second part is to determine the total number of nodes |V| up to depth n.So, the total number of nodes is the sum of V0 + V1 + V2 + ... + Vn.Given that V0=1, V1=3, V2=6, V3=12, V4=24, etc.So, the sum S_n = 1 + 3 + 6 + 12 + 24 + ... + Vn.Looking at the pattern:S_0 =1S_1=1+3=4S_2=1+3+6=10S_3=1+3+6+12=22S_4=1+3+6+12+24=46Hmm, so the sum seems to be following a pattern where each term is doubling and adding something.Wait, let's see:S_n = 1 + 3 + 6 + 12 + ... + 3*2^{n-1} for n >=1.Wait, actually, for n=0, S0=1.For n>=1, S_n = 1 + 3*(1 + 2 + 4 + ... + 2^{n-1}).Because V1=3=3*2^0, V2=6=3*2^1, V3=12=3*2^2, etc.So, the sum S_n = 1 + 3*(sum_{k=0}^{n-1} 2^k) for n >=1.The sum inside is a geometric series: sum_{k=0}^{m} 2^k = 2^{m+1} -1.So, for n >=1, S_n =1 + 3*(2^{n} -1) =1 +3*2^{n} -3= 3*2^{n} -2.Wait, let's check:For n=1: S1=1+3=4. According to formula: 3*2^1 -2=6-2=4. Correct.For n=2: S2=1+3+6=10. Formula: 3*2^2 -2=12-2=10. Correct.For n=3: S3=1+3+6+12=22. Formula: 3*2^3 -2=24-2=22. Correct.For n=4: S4=1+3+6+12+24=46. Formula: 3*2^4 -2=48-2=46. Correct.So, the total number of nodes up to depth n is 3*2^n -2 for n>=1, and 1 for n=0.But since the problem says "up to depth n", and n is presumably >=0, we can write it as:|V| = 3*2^n -2 for n >=1, and |V|=1 for n=0.Alternatively, we can express it as |V| = 3*2^n -2 + [n=0]*(1 - (3*2^0 -2)).But perhaps it's better to write it as a piecewise function:|V| = begin{cases}1 & text{if } n =0 3*2^n -2 & text{if } n geq1end{cases}Alternatively, since for n=0, 3*2^0 -2=3-2=1, so actually, the formula 3*2^n -2 works for all n>=0.Wait, let's check n=0: 3*2^0 -2=3-2=1. Correct.n=1: 3*2^1 -2=6-2=4. Correct.So, actually, the formula |V|=3*2^n -2 works for all n>=0.Therefore, the total number of nodes up to depth n is 3*2^n -2.So, summarizing part 1:Recursive formula:V0=1V1=3Vn=2*V_{n-1} for n>=2Total nodes up to depth n: |V|=3*2^n -2.Now, moving on to part 2.The major uses a hash function h(x)=(a x + b) mod p, where p is prime. We need to prove that h(x) is a bijection over integers modulo p, and find a and b such that it's a valid bijection.First, recall that a function is a bijection if it's both injective and surjective. For a function from Z_p to Z_p, it's a bijection if it's a permutation of Z_p.For a linear function h(x)=a x +b mod p, it's a bijection if and only if a is invertible modulo p, i.e., gcd(a,p)=1.Since p is prime, any a not divisible by p will be invertible.So, to ensure h(x) is a bijection, we need a and p to be coprime, i.e., a ≠0 mod p.Additionally, b can be any integer, as adding a constant doesn't affect bijectivity, as long as a is invertible.So, the conditions are:- a must be an integer such that gcd(a,p)=1, i.e., a is not a multiple of p.- b can be any integer.Therefore, to ensure h(x) is a bijection, a must be invertible modulo p, which for prime p means a ≠0 mod p, and b can be any value.So, the values of a and b are:- a is any integer where a mod p ≠0.- b is any integer.But since we're working modulo p, we can represent a as any integer from 1 to p-1, and b as any integer from 0 to p-1.So, in conclusion, for h(x) to be a bijection, a must be coprime with p (i.e., a ≠0 mod p), and b can be any integer.Therefore, the values of a and b are:- a ∈ {1,2,...,p-1}- b ∈ {0,1,2,...,p-1}So, any a not divisible by p and any b will work.To prove that h(x) is a bijection:First, show it's injective: suppose h(x1)=h(x2) mod p.Then, a x1 + b ≡ a x2 + b mod pSubtracting b: a x1 ≡ a x2 mod pSince a is invertible, multiply both sides by a^{-1}: x1 ≡x2 mod p.Thus, injective.Since it's a function from Z_p to Z_p and injective, it's also surjective (by pigeonhole principle), hence bijective.Therefore, h(x) is a bijection if and only if a is invertible modulo p, i.e., a and p are coprime.So, the conditions are a ≠0 mod p, and b can be any integer.So, summarizing part 2:h(x) is a bijection over Z_p if and only if a is coprime with p (i.e., a ≠0 mod p), and b can be any integer.Therefore, the values of a and b are:- a is any integer such that a mod p ≠0.- b is any integer.But since we're working modulo p, a can be any integer from 1 to p-1, and b can be any integer from 0 to p-1.So, that's the conclusion.Final Answer1. The recursive formula for the number of nodes at depth ( n ) is ( V_n = 2V_{n-1} ) for ( n geq 2 ) with ( V_0 = 1 ) and ( V_1 = 3 ). The total number of nodes up to depth ( n ) is ( |V| = 3 times 2^n - 2 ).2. The hash function ( h(x) ) is a bijection if ( a ) is coprime with ( p ) (i.e., ( a notequiv 0 pmod{p} )) and ( b ) is any integer. Thus, ( a ) can be any integer not divisible by ( p ) and ( b ) can be any integer.The final answers are:1. Recursive formula: ( V_n = 2V_{n-1} ) for ( n geq 2 ) with ( V_0 = 1 ) and ( V_1 = 3 ). Total nodes: ( boxed{3 times 2^n - 2} ).2. ( a ) must be coprime with ( p ) and ( b ) can be any integer. Thus, ( a ) and ( b ) are ( boxed{a notequiv 0 pmod{p}} ) and ( b ) is any integer.</think>

🧠:<think>Okay, so I need to help John figure out the optimal portfolio allocation for his new client. The client wants a balanced approach, maximizing returns while keeping the portfolio's standard deviation below 10%. John has three assets: stocks, bonds, and real estate. Each has different expected returns, standard deviations, and correlations. First, let me try to understand the problem. We need to maximize the expected return of the portfolio, subject to the standard deviation constraint. The portfolio consists of three assets, so we have three weights: w_S for stocks, w_B for bonds, and w_R for real estate. The sum of these weights should be 1 because it's a fully invested portfolio.So, the first part is to formulate the optimization problem. I think this involves setting up the objective function and the constraints. The objective function is the expected return of the portfolio, which is a weighted average of the expected returns of each asset. The constraints are that the portfolio's standard deviation is less than or equal to 10%, and the sum of the weights equals 1.Let me write down the expected return formula. The expected return of the portfolio, E(R_p), is:E(R_p) = w_S * E(R_S) + w_B * E(R_B) + w_R * E(R_R)Plugging in the numbers:E(R_p) = 0.08 * w_S + 0.04 * w_B + 0.06 * w_RWe need to maximize this.Now, the standard deviation of the portfolio is a bit more complicated because it involves the variances and covariances between the assets. The formula for the variance of the portfolio, Var(R_p), is:Var(R_p) = (w_S)^2 * Var(R_S) + (w_B)^2 * Var(R_B) + (w_R)^2 * Var(R_R) + 2 * w_S * w_B * Cov(R_S, R_B) + 2 * w_S * w_R * Cov(R_S, R_R) + 2 * w_B * w_R * Cov(R_B, R_R)We know the standard deviations, so we can compute the variances by squaring them:Var(R_S) = (0.15)^2 = 0.0225Var(R_B) = (0.05)^2 = 0.0025Var(R_R) = (0.10)^2 = 0.01Next, the covariance terms can be calculated using the correlation coefficients. Covariance is correlation multiplied by the product of standard deviations.Cov(R_S, R_B) = ρ(S,B) * σ_S * σ_B = 0.2 * 0.15 * 0.05 = 0.0015Cov(R_S, R_R) = ρ(S,R) * σ_S * σ_R = 0.5 * 0.15 * 0.10 = 0.0075Cov(R_B, R_R) = ρ(B,R) * σ_B * σ_R = 0.3 * 0.05 * 0.10 = 0.0015So, plugging all these into the variance formula:Var(R_p) = (w_S)^2 * 0.0225 + (w_B)^2 * 0.0025 + (w_R)^2 * 0.01 + 2 * w_S * w_B * 0.0015 + 2 * w_S * w_R * 0.0075 + 2 * w_B * w_R * 0.0015We need this variance to be less than or equal to (0.10)^2 = 0.01, since the standard deviation is the square root of variance.So, the constraints are:1. Var(R_p) ≤ 0.012. w_S + w_B + w_R = 1Additionally, we might want to include non-negativity constraints, but since it's a balanced portfolio, maybe short selling isn't allowed, so weights should be ≥ 0. But the problem doesn't specify, so perhaps we can assume that weights can be any real numbers as long as they sum to 1.But I think in portfolio optimization, usually, weights are non-negative unless specified otherwise. So, maybe we should include w_S, w_B, w_R ≥ 0.But the problem doesn't mention it, so perhaps we can proceed without that for now.So, summarizing, the optimization problem is:Maximize E(R_p) = 0.08 w_S + 0.04 w_B + 0.06 w_RSubject to:Var(R_p) = 0.0225 (w_S)^2 + 0.0025 (w_B)^2 + 0.01 (w_R)^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R ≤ 0.01And:w_S + w_B + w_R = 1So, that's the formulation.Now, for part 2, we need to use the Lagrange multiplier method to derive the equations necessary to solve for the weights.The Lagrange multiplier method is used to find the extrema of a function subject to equality constraints. In this case, we have an inequality constraint on the variance, but since we are maximizing return, the optimal portfolio will lie on the boundary of the feasible region, meaning the variance will be exactly 0.01. So, we can convert the inequality constraint into an equality.So, our problem becomes:Maximize E(R_p) = 0.08 w_S + 0.04 w_B + 0.06 w_RSubject to:Var(R_p) = 0.0225 (w_S)^2 + 0.0025 (w_B)^2 + 0.01 (w_R)^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R = 0.01And:w_S + w_B + w_R = 1So, now we have two equality constraints. To use Lagrange multipliers, we can set up the Lagrangian function, which incorporates the objective function and the constraints with multipliers.Let me denote the Lagrangian as:L = 0.08 w_S + 0.04 w_B + 0.06 w_R - λ1 [0.0225 (w_S)^2 + 0.0025 (w_B)^2 + 0.01 (w_R)^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R - 0.01] - λ2 [w_S + w_B + w_R - 1]Where λ1 and λ2 are the Lagrange multipliers.To find the extrema, we take the partial derivatives of L with respect to each variable (w_S, w_B, w_R, λ1, λ2) and set them equal to zero.So, let's compute the partial derivatives.First, partial derivative with respect to w_S:∂L/∂w_S = 0.08 - λ1 [2*0.0225 w_S + 0.003 w_B + 0.015 w_R] - λ2 = 0Similarly, partial derivative with respect to w_B:∂L/∂w_B = 0.04 - λ1 [2*0.0025 w_B + 0.003 w_S + 0.003 w_R] - λ2 = 0Partial derivative with respect to w_R:∂L/∂w_R = 0.06 - λ1 [2*0.01 w_R + 0.015 w_S + 0.003 w_B] - λ2 = 0Partial derivative with respect to λ1:∂L/∂λ1 = - [0.0225 (w_S)^2 + 0.0025 (w_B)^2 + 0.01 (w_R)^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R - 0.01] = 0Which simplifies to:0.0225 (w_S)^2 + 0.0025 (w_B)^2 + 0.01 (w_R)^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R = 0.01And partial derivative with respect to λ2:∂L/∂λ2 = - [w_S + w_B + w_R - 1] = 0Which simplifies to:w_S + w_B + w_R = 1So, now we have five equations:1. 0.08 - λ1 [0.045 w_S + 0.003 w_B + 0.015 w_R] - λ2 = 02. 0.04 - λ1 [0.005 w_B + 0.003 w_S + 0.003 w_R] - λ2 = 03. 0.06 - λ1 [0.02 w_R + 0.015 w_S + 0.003 w_B] - λ2 = 04. 0.0225 (w_S)^2 + 0.0025 (w_B)^2 + 0.01 (w_R)^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R = 0.015. w_S + w_B + w_R = 1So, these are the equations we need to solve for w_S, w_B, w_R, λ1, and λ2.This seems a bit complex because it's a system of nonlinear equations. Solving this analytically might be challenging, so in practice, we might use numerical methods or optimization software. But since the problem only asks to derive the equations, not solve them, I think we're done here.Wait, but let me double-check the partial derivatives to make sure I didn't make a mistake.For ∂L/∂w_S:The derivative of the objective function is 0.08.Then, the derivative of the variance constraint with respect to w_S is:2*0.0225 w_S + 0.003 w_B + 0.015 w_RSo, multiplied by λ1, it's λ1*(0.045 w_S + 0.003 w_B + 0.015 w_R)Then, the derivative of the weight sum constraint is 1, so multiplied by λ2, it's λ2.So, the equation is 0.08 - λ1*(0.045 w_S + 0.003 w_B + 0.015 w_R) - λ2 = 0Similarly for the others.Yes, that seems correct.So, the equations are correctly derived.Therefore, the optimization problem is set up, and the Lagrangian equations are as above.I think that's the answer they're looking for.Final AnswerThe optimization problem is formulated to maximize the expected return subject to the standard deviation constraint, and the Lagrangian equations are derived as follows:1. The expected return is maximized: boxed{E(R_p) = 0.08 w_S + 0.04 w_B + 0.06 w_R}2. Subject to the variance constraint: boxed{0.0225 w_S^2 + 0.0025 w_B^2 + 0.01 w_R^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R leq 0.01}3. And the weight sum constraint: boxed{w_S + w_B + w_R = 1}The Lagrangian equations are:1. (0.08 - lambda_1 (0.045 w_S + 0.003 w_B + 0.015 w_R) - lambda_2 = 0)2. (0.04 - lambda_1 (0.005 w_B + 0.003 w_S + 0.003 w_R) - lambda_2 = 0)3. (0.06 - lambda_1 (0.02 w_R + 0.015 w_S + 0.003 w_B) - lambda_2 = 0)4. (0.0225 w_S^2 + 0.0025 w_B^2 + 0.01 w_R^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R = 0.01)5. (w_S + w_B + w_R = 1)These equations are represented as:boxed{begin{aligned}&0.08 - lambda_1 (0.045 w_S + 0.003 w_B + 0.015 w_R) - lambda_2 = 0 &0.04 - lambda_1 (0.005 w_B + 0.003 w_S + 0.003 w_R) - lambda_2 = 0 &0.06 - lambda_1 (0.02 w_R + 0.015 w_S + 0.003 w_B) - lambda_2 = 0 &0.0225 w_S^2 + 0.0025 w_B^2 + 0.01 w_R^2 + 0.003 w_S w_B + 0.015 w_S w_R + 0.003 w_B w_R = 0.01 &w_S + w_B + w_R = 1end{aligned}}</think>