Logical Expression Logical expressionLet a, b, c € {0, 1} be bits. That is, cach of a, b, and c is either 0 or 1. For this problem, we equate false with 0and true with 1. Let ø be a logical expression$ = (? V ?) ^ (c V ?) ^ (6 V ?).We also write o as o(a, b, c) since it is a function of the values assigned to the three variables a, b, and c. Theformula o contains six terms in total. Two of the six terms are pre-filled. The third term is c, and the fifthterm is 6, the negation of b. The four remaining terms are cach one of {a,ā, b, b, c, c}. We want the formula Øto take the values listed in the following truth table.1111110| 01111111 1 101. Find a way to set the four remaining terms in ø so that the formula o takes the values in the above truthtable.At the beginning of your submission, give a table as shown here, and put one of {a, ā, b, b, c, t} in cachbox, indicating your answer.124Succinctly explain your methodology and justify your answer.2. Let 0 = (a → b) ^ (c→ b) ^ (ūÐ c). Give the truth table for 0. Show your calculations and justify. Givean interpretation of 0 in plain English. What does 0 compute? Explain.

Answered: Image /qna-images/answer/05e36e7b-ccc3-4a17-a911-36cea1695b3e.jpg

Logical expression Let a, b, c E {0, 1} be bits. That is, cach of a, b, and c is either 0 or 1. For this problem, we equate false with 0 and true with 1. Let ø be a logical expression $ = (? V ?) ^ (c V ?) ^ (b v ?). We also write o as ø(a, b, c) since it is a function of the values assigned to the three variables a, b, and c. The formula o contains six terms in total. Two of the six terms are prc-filled. The third term is c, and the fifth term is b, the negation of b. The four remaining terms are cach one of {a,ā,b, b, c,c}. We want the formula o to take the values listed in the following truth table. a 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1. Find a way to set the four remaining terms in ø so that the formula o takes the values in the above truth table.

Logical expression Let a, b, c E {0, 1} be bits. That is, cach of a, b, and c is either 0 or 1. For this problem, we equate false with 0 and true with 1. Let ø be a logical expression $ = (? V ?) ^ (c V ?) ^ (b v ?). We also write o as ø(a, b, c) since it is a function of the values assigned to the three variables a, b, and c. The formula o contains six terms in total. Two of the six terms are prc-filled. The third term is c, and the fifth term is b, the negation of b. The four remaining terms are cach one of {a,ā,b, b, c,c}. We want the formula o to take the values listed in the following truth table. a 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1. Find a way to set the four remaining terms in ø so that the formula o takes the values in the above truth table.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

Let Σn denote the set of all binary strings with lengths less than n. For example, Σ2 = {λ, 0, 1} and Σ3 = {λ, 0, 1, 00, 01, 10, 11} where λ is the empty string, i.e. the string of length 0. For x ∈ Σ we denote the length of x by |x|. a) How many functions f : Σn → Σn are there? b) How many one-to-one functions f : Σn → Σn are there?
3. Write a program to evaluate a polynomial \[ p(x)=c_{n} x^{n}+c_{n-1} x^{n-1}+\cdots+c_{2} x^{2}+c_{1} x+c_{0} \] and its derivative at a given pointt. The input is the row vectorcof coefficients of the polynomial arranged from highest to lowest power, and the numbert, and the two outputs arep(t)andp′(t). Computep(t)by first computing the row véctor \[ \left[t^{n}, t^{n-1}, \ldots, t^{2}, t, 1\right] \] and combine that with the vectorcto get the result. Computep′(t)by computing the coefficients of the derivative polynomial from the vectorc, and then evaluating that polynomial in the same way. Include some tests that provide evidence that your code produces the correct results. Please give proper explanation and typed answer only.
Let l be a line in the x-yplane. If l is a vertical line, its equation is x = a for some real number a. Suppose l is not a vertical line and its slope is m. Then the equation of l is y = mx + b, where b is the y-intercept. If l passes through the point (x₀, y₀), the equation of l can be written as y - y₀ = m(x - x₀). If (x₁, y₁) and (x₂, y₂) are two points in the x-y plane and x₁ ≠ x₂, the slope of line passing through these points is m = (y₂ - y₁)/(x₂ - x₁). Instructions Write a program that prompts the user for two points in the x-y plane. Input should be entered in the following order: Input x₁ Input y₁ Input x₂
Modulo arithmeticFor two integers ‘a’ and ‘b’ and a positive integer ‘n’, let + represent the operation a + b = (a+b)mod(n), that is, the result of a + b is the remainder of the usual sum a + b after dividing by n (using a natural representation). Similarly, let * represent the operation a * b = (a*b)mod(n), that is, the result of a * b is the remainder of the product a*b after dividing by n. For example using the set S = {2, 5, 8} and n = 9, 5 + 8 = 4, since 4 is the remainder after dividing 13 (5 + 8) by 9; and 2 * 8 = 7, since 7 is the remainder after dividing 16 (2*8) by 9.a. For each set S and number n specified below, create two “operation” tables (these are called Cayley tables), one showing the results for the operation +, and one showing the results of the operation * for each pair of elements from S (see the example tables in the notes).i. S1 = {0, 1} and n = 2ii. S2 = {1, 2} and n = 3iii. S3 = {0, 2, 4, 6} and n = 8iv. S4 = {1, 3, 5, 7} and n = 8v. S5 = {1, 2, 3, 4} and n =…
Computer Science Let a M be a finite set of messages, and let S(M) denote the set of all permutations of M (all bijective functions f : M → M). We’ll assume that if given a description of σ ∈ S(M), both σ and σ −1 are efficiently computable. Suppose P ⊂ S(M) is such that ∀x, y ∈ M, ∃σ ∈ P such that σ(x) = y. (a) Show that |P| ≥ |M|. (This is easy, but makes sure you’ve parsed the definition.) (b) Show that if |P| = |M|, then the following encryption scheme is perfectly secure, provided you only use it once: 1 Key generation: select a random σ ∈ P; Encryption: m 7→ σ(m) Decryption: c 7→ σ −1 (c) (c) Show that the above is false if |M| < |P| < 2|M|. (d) Observe that for any finite group G and any g ∈ G, the map x 7→ gx is a permutation of G. By viewing G itself as a set of permutations of G in this way, show that the above property is satisfied (with M = P = G). (e) The traditional xor one time pad is a special case of the above. What is the finite group in this case?
The sequence of values: n, c(n), c(c(n)), c(c(c(n))), c(c(c(c(n)))),... is known as the hailstone sequence starting at n. Implement c(n) as a function. Use predicates to test the parity of each input n.
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.
For the definition of x ∈ EXPR , na(x) denotes the number of a’s in the string, and we will use nop(x) to stand for the number of operators in x (the number of occurrences of + or ∗). Show that for every x ∈ EXPR, na(x)= 1 + nop(x).
Given a set of n positive integers, C = {c1,c2, ..., cn} and a positive integer K, is there a subset of C whose elements sum to K? A dynamic program for solving this problem uses a 2-dimensional Boolean table T, with n rows and k + 1 columns. T[i,j] 1≤ i ≤ n, 0 ≤ j ≤ K, is TRUE if and only if there is a subset of C = {c1,c2, ..., ci} whose elements sum to j. Which of the following is valid for 2 ≤ i ≤ n, ci ≤ j ≤ K? a) T[i, j] = ( T[i − 1, j] or T[i, j − ci]) b) T[i, j] = ( T[i − 1, j] and T[i, j − ci ]) c) T[i, j] = ( T[i − 1, j] or T[i − 1, j − ci ]) d) T[i, j] = ( T[i − 1, j] and T[i − 1, j − cj ]) In the above problem, which entry of the table T, if TRUE, implies that there is a subset whose elements sum to K? a) T[1, K + 1] b) T[n, K] c) T[n, 0] d) T[n, K + 1]
Show L = { w | w is an element of {0, 1, 2}* and (the number of 0s in w ≥ the number of 1s in w) and ((the number of 0s + the number of 1s in w) divides the number of 2s in w) } is not context free. You should assume that p is the pumping length and for any string s in L with |s| ≥ p, s can be broken into 5 pieces, s = uvxyz.
1. Consider the following dynamic programming implementation of the Knapsack problem: #include int find_max(int a, int b) { if(a > b) return a; return b; } int knapsack(int W, int *wt, int *val,int n) { int ans[n + 1][W + 1]; int itm,w; for(itm = 0; itm <= n; itm++) ans[itm][0] = 0; for(w = 0;w <= W; w++) ans[0][w] = 0; for(itm = 1; itm <= n; itm++) { for(w = 1; w <= W; w++) { if(wt[itm - 1] <= w) ans[itm][w] = ______________; else ans[itm][w] = ans[itm - 1][w]; } } return ans[n][W]; } int main() { int w[] = {10,20,30}, v[] = {60, 100, 120}, W = 50; int ans = knapsack(W, w, v, 3); printf("%d",ans); return 0; } Which of the following lines completes the above code? A. find_max(ans[itm – 1][w – wt[itm – 1]] + val[itm – 1], ans[itm – 1][w]) B. find_max(ans[itm – 1][w –…
Write a function that, given a list of non-overlapping intervals of items, takes an item as an input and identifies which, if any, interval that item lies in. If the items are integers and the intervals are 1643-2033, 5532-7643, 8999-10332, 5666653-5669321, then the query point 9122 is in the third interval while 8122 is not.