3. Approximate the value of sum after the following code fragment, in terms of variable n in Big-Oh notation. int sum = 0; for (int i = 1; i <= n; i++) { sum++; } for (int j = 1; j <= n / 2; j++) { for (int k = 1; k <= n / 2; k++) { sum++; } }

3. Approximate the value of sum after the following code fragment, in terms of variable n in Big-Oh notation. int sum = 0; for (int i = 1; i <= n; i++) { sum++; } for (int j = 1; j <= n / 2; j++) { for (int k = 1; k <= n / 2; k++) { sum++; } }

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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Question 1 Work your way through the following code fragments. What would be printed? a) int i = -23; int * p = &i; printf("*p = %i\n", *p); b) int i; int * p = &i; printf("*p = %i\n", *p); c) int i = 48; int * p; printf("*p = %i\n", *p); d) int i = 10;int * p = &i;int j; j = ++*p; printf("j = %i\n", j);printf("i = %i\n", i); e) int i = 10, j = 20;int * p = &i;int * q = &j; *p = *q;printf("i = %i, j = %i\n", i, j);printf("*p = %i, *q = %i\n", *p, *q); i = 10; j = 20; p = q;printf("i = %i, j = %i\n", i, j);printf("*p = %i, *q = %i\n", *p, *q);
Answer the given question with a proper explanation and step-by-step solution. In Java, please. Consider the following problem. Input: an array of integers A[] Output: rearranges the array to have the following property: Suppose the first element in the original array A[], A[0] = x. In the rearranged array, suppose that x is at index i, that is A[i] = x. Then, we want the rearranged array to be such that A[j] <= x for all j < i and A[j] > x for all j > i. The rearranging should place all the values less than (or equal to) x to the "left" of x and all values larger than x to the right of x. Here's an example. Suppose the array has the elements in this initial order: 4 3 9 2 7 6 5 A[0] = 4. After rearranging elements, we get: 3 2 4 5 9 7 6 That is, A[0] = 4, is positioned in the resulting array so that all elements less than 4 (that is, 2, 3) are to its left (in no particular order), and all elements larger than 4 are to its right (again, in no particular order).
How can the following function be simplified so that it has a time complexity of O(n) or faster?For your information, the specifications of the functions are as follows: Using numbers ranging from 0 to 15 (inclusive), create all possible lists which sum up to K and have a length of N. Duplicated numbers are allowed as long as it fulfills the conditions above (this means [0,0,1], [0,1,0] and [1,0,0] are all correct outputs if K=1 and N=3). When instantiated with the list function, the list size of the function should be the number of all lists. For example, given K=23 and N=2, the expected list size is 8. The function must be able to accept N=10 and be finished before 9 seconds. Do not use itertools or external libraries.
Java problem implement this specification. In addition, while Scanner input and System.out output will take place only in the main method, all solitaire game simulation and computation of the first player win probability estimate should take place in a method getAdvantage that takes an integer parameter for the number of games to simulate, and returns a double-precision floating-point first player win probability. The main method will read the desired number of simulations, call method getAdvantage with that number, and report the returned number in the specified format. Specification: First-Player Advantage Pig is a folk jeopardy dice game with simple rules: Two players race to reach 100 points. Each turn, a player repeatedly rolls a die until either a 1 ("pig") is rolled or the player holds and scores the sum of the rolls (i.e. the turn total). At any time during a player's turn, the player is faced with two decisions: roll - If the player rolls a 1: the player scores nothing…
Which is the base case of the following recursion function: def mult3(n): if n == 1: return 3 else: return mult3(n-1) + 3 else n == 1 mult3(n) return mult3(n-1) + 3
Do not use static variables to implement recursive methods. USING JAVA Indicate the order of the following growth functions using < (e.g. n < n2) n2 log n n n2/ log n n! (1.5)n 3n
1: |R|←|P| Reserve space for |P| = 13 values.2: x ← n3: for i ← 0 ...(|P| − 1) do4: c ← x div Pi Number of multiplicands Pi in x.5: Ri ← c6: x ← x − c · Pi7: end for8: return RA Java programmer could implement Algorithm by first modelling the primitivenumbers with the enumeration type RomanNumeral. Each enum constant (I, IV, ..., M) is declared with its decimal value, which can be accessed with the function getValue().
JAVA Write a function to implement bottom-up minimal exchange algorithm for generating permutation of {1, 2, 3, 4, … , n}. Call your function from within a main program. Submit code and a sample output for the case n = 4. You may also submit a hand calculated version for (n = 4) if you rather not write a program. Please explain step-by-step.
Please explain Crime Wave more clearly for me please there's a picture of the decription and also please comment on each line if possible on the code below to for me to have a better understanding of what each line is doing. It is in C++. Please also give the space and time complexity and the reason why. #include <stdio.h>#include <string.h>#include <math.h>#define eps 1e-6double W[105][105];int N, M;int mx[105], my[105]; // match arrdouble lx[105], ly[105]; // label arrint x[105], y[105]; // used arrint hungary(int nd) { int i; x[nd] = 1; for(i = 1; i <= M; i++) { if(y[i] == 0 && fabs(W[nd][i]-lx[nd]-ly[i]) < eps) { y[i] = 1; if(my[i] == 0 || hungary(my[i])) { my[i] = nd; return 1; } } } return 0;}double KM() { int i, j, k; double d; memset(mx, 0, sizeof(mx)); memset(my, 0, sizeof(my)); memset(lx, 0, sizeof(lx)); memset(ly, 0, sizeof(ly));…
solve the problem in java:Consider the are n=2 subjects and needed =[4,5]answered questions, to pass.The student has answered =[2,4] questions in the two subjects so far, and can answer another q=1 questions across all subjects combined. The best outcome is to answer an additional question in the second subject on order to pass it, as 2 more answers required to pass the first subject. the max number of subject s that can be passed is 1. the function must return an integer that can represent the max number of subjects that can be passed. public static int maxNumsub(List <Integer> answered, List<Integer> needed, int q){//write code here}
Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…
Write a recursive function that, given a sequence of comparable values, returns the count of elements where the current element is less than the following ( next ) element in the given sequence. See the examples given below. def count_ordered ( seq ) : """ Input : A sequence of comparable elements Output : The number of elements that are less than the following element in the sequence Example : >>> count_ordered ( [ 1 , 2 , 3 , 4 , 5 , 6 ] ) 5 >>> count_ordered ( ( 1 , 12, 7.3 , -2,4 ) ) 2 >>> count_ordered ( 'Python' ) 2 >>> count_ordered ( [ 6 ] ) 0 >>> count_ordered ( [ ] ) 0 """ In the first example above , count_ordered ( [ 1,2,3,4,5,6 ] )the returned answer is 5 because for all the first 5 numbers the current number is less than the next number. In the second example above, count_ordered ( ( 1,12,7.3 , -2,4 ) )the…