¬[[p → ¬q) v s] = [p Aq A -s] -[[p → ¬q) → s] = [¬(p^q) A¬s] [(pA¬q) → s] = [¬s → (-p v q)] [¬s → (-p v q)] = [s V q V ¬p]
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Please answer true or false
I think the first is true,second is true, third is false and fourth is false
but I want to check my answers
![-[(p → ¬q) V s] = [p^q A -s]
¬[(p → ¬9) → s] = [¬(p^q) A¬s]
[(pA¬q) → s] = [¬s → (-p v q)]
[-s
→ (-p v q)] = [s V q v -p]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F191441f0-66db-457d-b504-dd20fe1485fa%2F234d9368-cc0f-4940-a2c2-0f0cc9294db5%2Fy95z6n_processed.jpeg&w=3840&q=75)
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- Suppose you have a signature scheme S (which is correct and existentially unforgeable), and S can be used to sign any t-bit message. And you have a hash function H which outputs t bits and is collision-resistant. Consider a modified signature scheme S’ which can sign messages of unlimited length, where: S’.Sign(sk, m) = S.sign(sk, H(m)) Prove that this scheme is existentially unforgeable as long as S is existentially unforgeable and H is collision-resistant.Computer Science (a) Let h be a collision resistant hash function. Now consider the following function H that is defined for binary strings of even length. Let x = x1||x2 with x1 ∈ F2n and x2 ∈ Fn2 . Then H(x) = h(x1 ⊕ x2). Prove that this function is not collision resistant.Consider the set F = {1, −1, i, −i} with an operation ✕ defined by the table. ✕ 1 −1 i −i 1 1 −1 i −i −1 −1 1 −i i i i −i −1 1 −i −i i 1 −1 a ✕ b means find the entry in row a and column b; for example, −1 ✕ (−i) = i (the entry in row −1 and column −i). Find each of the following. (a) −1 ✕ i (b) i ✕ i (c) −i ✕ i (d) −i ✕ 1 (e) 1 ✕ i (f) −i ✕ −i
- Given g = {(1,c),(2,a),(3,d)}, a function from X = {1,2,3} to Y = {a,b,c,d}, and f = {(a,r),(b,p),(c,δ),(d,r)}, a function from Y to Z = {p, β, r, δ}, write f o g as a set of ordered pairs.Draw the portion of the state space tree generated by LCBB for the following instances. n = 4, m = 15, (P₁, ..., P) = (10, 10, 12, 18) (w₁,..... W 4) = (2, 4, 6, 9).link(a,b). link(a,c). link(b,c). link(b,d). link(c,d). link(d,e). link(d,f). link(e,f). link(f,g). Using The above Formulate the appropriate Prolog predicate "path(X,Y,N)" which is true if (and only if) there is a path of length "N" from node "X" to node "Y". For example, there is a path of length 2 from "a" to "d": "a->b->d", but also "a->c->d", and so "path(a,d,2)" gives "true" (two times). There is also a path of length 3 from "a" to "d": "a->b->c->d". Test this predicate out on the above network to verify whether or not it is working correctly. Once this is working correctly, note now, that e.g., "path(a,e,N)." will give multiple answers:
- Given the visited node for breadth first search , starting with s , given the following adjacency lists adj(s) = [a,c.d] adj(a) = [], adj(c) = [e.b] adj(b) = [d], adj(d) = [c], adj(e) = [s], a) sabdec b) sacdeb c) sadcbe d)sacebdlink(a,b). link(a,c). link(b,c). link(b,d). link(c,d). link(d,e). link(d,f). link(e,f). link(f,g). Using the above Formulate the appropriate Prolog predicate "path(X,Y,N)" which is true if (and only if) there is a path of length "N" from node "X" to node "Y". For example, there is a path of lengthConsider the (directed) network in the attached document We could represent this network with the following Prolog statements: link(a,b). link(a,c). link(b,c). link(b,d). link(c,d). link(d,e). link(d,f). link(e,f). link(f,g). Now, given this network, we say that there is a "connection" from a node "X" to a node "y" if we can get from "X" to "Y" via a series of links, for example, in this network, there is a connection from "a" to "d", and a connection from "c" to "f", etc.
- Prove the theorem Suppose that a hash function h is chosen randomly from a universal collection of hash functions and has been used to hash n keys into a table T of size m, using chaining to resolve collisions. If key k is not in the table, then the expected length E Œnh.k/ of the list that key k hashes to is at most the load factor ˛ D n=m. If key k is in the table, then the expected length E Œnh.k/ of the list containing key k is at most 1 C ˛.What is the time and space complexity of this function? It is a dfs function which goes through every possible path from node to another without cycles. Time complexity could be either O(n!) or O(2^n), or is there another answer? What is the space complexity? def dfs(currency_pairs, source, target): graph = defaultdict(dict) for s1, s2, rate_val1, rate_val2 in currency_pairs: graph[s1][s2] = rate_val1 graph[s2][s1] = rate_val2 def backtrack(current, seen): if current == target: return 1 product = 0 if currentingraph: neighbors = graph[current] for neighbor in neighbors: if neighbor not in seen: seen.add(neighbor) product = max(product, graph[current][neighbor] * backtrack(neighbor, seen)) seen.remove(neighbor) return product return backtrack(source,…Consider P, the set of palindromes over Σ, defined recursively as follows: Basis: λ ∈ P and ∀c ∈ Σ (c ∈ P) Recursive: ∀w ∈ P, c ∈ Σ (cwc ∈ P) Use structural induction to prove that r(w) = w for all w ∈ P.