let rec longer (xs, ys) match (xs, ys) with | (_, []) -> true (* line 1 *) | ([],_) -> false (* line 2 *) | (_::xs,_::ys) -> longer (xs, ys) (* line 3 *) (a) For each of the following alternate orders of the branches, indicate one of the following: (A) The function would still be correct. (B) The function would still type-check (no unreachable branch) but would no longer be correct. (C) The function would no longer type-check (due to an unreachable branch). i. line 2; then line 1; then line 3 ii. line 3; then line 1; then line 2 iii. line 3; then line 2; then line 1 iv. line 1; then line 3; then line 2

let rec longer (xs, ys) match (xs, ys) with | (_, []) -> true (* line 1 ) | ([],_) -> false ( line 2 ) | (_::xs,_::ys) -> longer (xs, ys) ( line 3 *) (a) For each of the following alternate orders of the branches, indicate one of the following: (A) The function would still be correct. (B) The function would still type-check (no unreachable branch) but would no longer be correct. (C) The function would no longer type-check (due to an unreachable branch). i. line 2; then line 1; then line 3 ii. line 3; then line 1; then line 2 iii. line 3; then line 2; then line 1 iv. line 1; then line 3; then line 2

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

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Given the following function, what happens if a[] contains just one element that doesn't match val? int binarySearch(int a[], int first, int last, int val){ if (first > last) return -1; int middle = (first + last) / 2; if (a[middle] == val) return middle; if (a[middle] < val) return binarySearch(a, middle+1, last, val); else return binarySearch(a, first, middle-1, val);} Group of answer choices binarySearch never calls itself again and terminates (recursion never happens) binarySearch calls itself once then terminates (recursion happens once) binarySearch calls itself twice then terminates (recursion happens twice) binarySearch calls itself 3 times then terminates (recursion happens 3 times)
Implement the following function, without using any data structure or #include <bits/stdc++.h> /* Given two vectors of chars, check if the two vectors are permutations of each other, i.e., they contains same values, in same or different order.e.g., V1=[‘a’,’b’,’a’] and V2=[‘b’,’a’,’a’] stores same multi-set of data points: i.e., both contains two ‘a’, and one ‘b’. e.g., V3=[‘a’,’c’,’t’,’a’] and V4=[‘a’,’c’,’t’] are not same multi-set. V3 contains two ‘a’s, while V4 has only one ‘a’. Note: when considering multiset, the number of occurrences matters. @param list1, list2: two vectors of chars @pre: list1, list2 have been initialized @post: return true if list1 and list2 stores same values (in same or different order); return false, if not. */  bool SameMultiSet (vector<char> list1, vector<char> list2) THIS IS NOT THE CORRECT SOLUTION / include headers #include <bits/stdc++.h> // deinfe the namespace using namespace std; // function fo rchecking same bool…
Describe each of the following sets by listing its elements:a. {x | x ∈ N and x2 − 5x + 6 = 0}b. {x | x ∈ R and x2 = 19}c. {x | x ∈ N and x2 − 2x − 8 = 0}}d. {x | x ∈ Z and x ∗ x = 4}
Implement the following function, without using any data structure. /* Given two vectors of chars, check if the two vectors are permutations of each other, i.e., they contains same values, in same or different order.e.g., V1=[‘a’,’b’,’a’] and V2=[‘b’,’a’,’a’] stores same multi-set of data points: i.e., both contains two ‘a’, and one ‘b’. e.g., V3=[‘a’,’c’,’t’,’a’] and V4=[‘a’,’c’,’t’] are not same multi-set. V3 contains two ‘a’s, while V4 has only one ‘a’. Note: when considering multiset, the number of occurrences matters. @param list1, list2: two vectors of chars @pre: list1, list2 have been initialized @post: return true if list1 and list2 stores same values (in same or different order); return false, if not. */  bool SameMultiSet (vector<char> list1, vector<char> list2)
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.
We have the following rectangle records in a list of dictionaries. Test Cases: rectangle_records = [{"ID": "Rect1", "Length": 40, "Breadth": 25, "Color": "red"}, {"ID": "Rect2", "Length": 30, "Breadth": 20, "Color": "blue"}, {"ID": "Rect3", "Length": 70, "Breadth": 45, "Color": "green"}, {"ID": "Rect4", "Length": 20, "Breadth": 10, "Color": "purple"}] Using Insertion Sort, Write a python function sort_rectangles that takes rectangle_records and record_title as a parameter and sort the rectangle_records in ascending order by record_title. You need to return the updated rectangle_records list. NOTE: The type of record_title input can be ID, Length, Breadth, or Color. >>> sort_rectangles(rectangle_records, "ID") [{"ID": "Rect1", "Length": 40, "Breadth": 25, "Color": "red"}, {"ID": "Rect2", "Length": 30, "Breadth": 20, "Color": "blue"}, {"ID": "Rect3", "Length": 70, "Breadth": 45, "Color": "green"}, {"ID": "Rect4", "Length": 20, "Breadth": 10, "Color": "purple"}] >>>…
Computer Science you are given a string X of length n and another string Y of length m ≤n. Say,the indexes p1, p2, p3, p4 and q1, q2, q3, q4 form two sub-sequences, i.e., 0 ≤ p1 < p2 < p3 < p4 < nand 0 ≤q1 < q2 < q3 < q4 < n; then, they are non-overlapping if p4 < q1.The task is to count the maximum number of non-overlapping sub-sequences of X that are thesame as Y . Thus, if X = GAXTYAWBGTAUGBTABGRGTAXB and Y = GTAB, then the answer is3 as shown by the red fonts. We cannot select the underlined GTAB as it overlaps with a red GTAB(i.e., among overlapping sub-sequences, you can select only one of them).Describe an O(m + n) time algorithm to obtain the count. Write a pseudo-code.
For the following C++ code find and write the recurrence relation. You need to model the runtime of function "Func" in terms of n. (only the recurrence relation in terms of n, No output of the code or final runtime analysis is required) s= array L[] start index e= array L[] end index void Func(int L[], int s, int e) { if (s < e) { i=s-1; for (int j = s; j <= e - 1; j++) { if (A[j] <= x) { i++; swap (&L[i], &L[j]); } } swap (&L[i + 1], &L[e]); int k = i+1; Func(L, s, k - 1); Func(L, k + 1, e); } } Please explain how you get the relation Thank you!
Write a template version of the iterative binary search algorithm from Display 13.8 which only searches an array of integers for an integer key. Specify requirements on the template parameter type. Discuss the requirements on the template parameter type.
Create a function that takes a grid of # and -, where each hash (#) represents amine and each dash (-) represents a mine-free spot.Return a grid, where each dash is replaced by a digit, indicating the number ofmines immediately adjacent to the spot i.e. (horizontally, vertically, anddiagonally).Example of an input:[ ["-", "-", "-", "#", "#"],["-", "#", "-", "-", "-"],["-", "-", "#", "-", "-"],["-", "#", "#", "-", "-"],["-", "-", "-", "-", "-"] ]Example of the expected output:[ ["1", "1", "2", "#", "#"],["1", "#", "3", "3", "2"],["2", "4", "#", "2", "0"],["1", "#", "#", "2", "0"],["1", "2", "2", "1", "0"] ]Here is a tip. When checking adjacent positions to a specific position in the grid,the following table might assist you in determining adjacent indexes:NW position =current_row - 1current_col - 1N position =current_row - 1current_colNE position =current_row - 1current_col + 1W position =current_rowcurrent_col - 1Current position =current_rowcurrent_colE position…
Consider there are random numbers and their size is not predictable initially. The requirement is to store them in a way that is efficient in the following means: the chosen data structure is memory efficient a lot of insertions and deletions are expected from the given positions a lot of insertions and deletions are expected based on noduplicate allowed rule a lot of insertions and deletions are expected from given positions’ successor/predecessor
Consider an input string TAM of letters ‘A’, ‘M’, and ‘T’. This string, which is given by the user, ends with ‘#’. It should be stored in a table (or array), called TAMUK. The number of each of these letters is unknown. We have a function, called SWAP(TAM,i,j), which places the ith letter in the jth entry of string TAM and the jth letter in the ith entry of TAM. Note that SWAP(TAM,i,j) is defined for all integers i and j between 0 and length(TAM)–1, where length(TAM) is the number of letters of TAM. 1. Using our algorithmic language, write an algorithm, called Sort_TAM, which sorts the letters in the array TAMUK in a way that all T’s appear first, followed by all A’s, and followed by all M’s. The algorithm Sort_TAM should have one parameter: The array TAMUK. Also, your solution is correct only if the following four constraints are satisfied: - Constraint 1: Each letter (‘A’, ‘M’, or ‘T’) is evaluated only once. - Constraint 2: The function SWAP(TAM,i,j) is used only when it is…