The count_crossings_and_nestings function has a computational complexity of O(n²), where n is the number of arcs in the input. This complexity arises from the nested loops iterating over pairs of arcs to check for crossings and nestings. As the number of arcs (n) increases: • The runtime of the function will grow quadratically, making it inefficient for large values of n. • The memory requirement of the function grows linearly with the number of arcs since it doesn't store any additional data structures. For large inputs, the function may become impractical due to its quadratic time complexity. To improve efficiency for larger inputs, alternative algorithms or optimizations should be considered.

The count_crossings_and_nestings function has a computational complexity of O(n²), where n is the number of arcs in the input. This complexity arises from the nested loops iterating over pairs of arcs to check for crossings and nestings. As the number of arcs (n) increases: • The runtime of the function will grow quadratically, making it inefficient for large values of n. • The memory requirement of the function grows linearly with the number of arcs since it doesn't store any additional data structures. For large inputs, the function may become impractical due to its quadratic time complexity. To improve efficiency for larger inputs, alternative algorithms or optimizations should be considered.

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

This function uses a curious mix of iteration and recursion: function F(n) if n < 1 t <- O return 1 for i <- 0 to n for j <- i to n t<- t+j return t + F(n-1) The number of basic operations (additions and subtractions) performed is: ○ O(n) Ⓒ (n²) (n² log n) Ⓒ (n³) Ө Ⓒ (n4)
I have two functions in a function, one has a time complexity of O(n) and the other one has a time complexity of O(n^2). what is the overall time complexity of the function? def function(): def function1(): def function2():
Sudoku is a popular logic puzzle that uses a 9 by 9 array of squares that are organized into 3 by 3 subarrays. The puzzle solver must fill in the squares with the digits 1 to 9 such that no digit is repeated in any row, any column, or any of the nine 3 by 3 subgroups of squares. Initially, some squares are filled in already and cannot be changed. For example, the following might be a starting configuration for a Sudoku puzzle: Create a class SudokuPuzzle.java Download SudokuPuzzle.java that has the attributes • board—a 9 by 9 array of integers that represents the current state of the puzzle, where 0 indicates a blank square • start—a 9 by 9 array of boolean values that indicates which squares in board are given values that cannot be changed and the following methods: • SudokuPuzzle—a constructor that creates an empty puzzle • toString—returns a string representation of the puzzle that can be printed • addInitial(row, col, value)—sets the given square to the given value as an…
int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;
function recursion(B[0..n − 1], i) if n == 0 then return False if n == 1 then return (B[0] == i)x ← recursion(B[0........n/2 − 1], i) y ← recursion(B[n/2..........4 × n/2 − 1], i) z ← recursion(B[4 × n/2..........n − 1], i) return (x OR y OR z) input array is of length n = 2^p, p is a positive integer Write the recursive formula for above algorithm as of worst case inputs.
Please code in JAVA implement and test the GET-MEMORY algorithm This algorithm uses the Next-Fit(First-Fit-With-A-Roving-Pointer) technique. then implement and test the FREE-MEMORY algorithm. Implement the “GET_MEMORY” and “FREE_MEMORY” algorithms. Comprehensive testing must be done for each algorithm. Following are sample run results for each: GET_MEMORY IS RUNNING……… Initial FSB list FSB# Location Size 1 7 4 2 14 10 3 30 20 . . . . . . Rover is 14 ---------------------------------------------------------------------------- Allocation request for 5 words Allocation was successful Allocation was in location 14 FSB# Location Size…
Create an ABM function that takes the following parameters: n := number of paths to be simulated m := number of discretization points per path S0 := initial starting point dS=μdt+σdW Program the function by using two nested "for loops" def ABM(n,m,S0,mu,sigma,dt): np.random.seed(999) arr = # create 2D zeros array with the correct dimensions arr[,] = #initialize column 0 # fill in array entries for i in : for j in : arr[i,j] = return arr
Merge sort is an efficient sorting algorithm with a time complexity of O(n log n). This means that as the number of elements (chocolates or students) increases significantly, the efficiency of merge sort remains relatively stable compared to other sorting algorithms. Merge sort achieves this efficiency by recursively dividing the input array into smaller sub-arrays, sorting them individually, and then merging them back together. The efficiency of merge sort is primarily determined by its time complexity, which is , where n is the number of elements in the array. This time complexity indicates that the time taken by merge sort grows logarithmically with the size of the input array. Therefore, even as the number of chocolates or students increases significantly, merge sort maintains its relatively efficient performance. Regarding the distribution of a given set of x to y using iterative and recursive functions, the complexity analysis depends on the specific implementation of each…
Lee has discovered what he thinks is a clever recursive strategy for printing the elements in a sequence (string, tuple, or list). He reasons that he can get at the first element in a sequence using the 0 index, and he can obtain a sequence of the rest of the elements by slicing from index 1. This strategy is realized in a function that expects just the sequence as an argument. If the sequence is not empty, the first element in the sequence is printed and then a recursive call is executed. On each recursive call, the sequence argument is sliced using the range 1:. Here is Lee’s function definition: def printAll(seq): if seq: print(seq[0]) printAll(seq[1:]) Write a program that tests this function and add code to trace the argument on each call. Does this function work as expected? If so, are there any hidden costs in running it?
Lee has discovered what he thinks is a clever recursive strategy for printing the elements in a sequence (string, tuple, or list). He reasons that he can get at the first element in a sequence using the 0 index, and he can obtain a sequence of the rest of the elements by slicing from index 1. This strategy is realized in a function that expects just the sequence as an argument. If the sequence is not empty, the first element in the sequence is printed and then a recursive call is executed. On each recursive call, the sequence argument is sliced using the range 1:. Here is Lee’s function definition: Write a program that tests this function and add code to trace the argument on each call. Does this function work as expected? If so, are there any hidden costs in running it?
The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a c) def f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] d) def f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return a
In programming, a recursive function calls itself. The classical example is factorial(n), which can be defined recursively as n*factorial(n-1). Nonethessless, it is important to take note that a recursive function should have a terminating condition (or base case), in the case of factorial, factorial(0)=1. Hence, the full definition is: factorial(n) = 1, for n = 0 factorial(n) = n * factorial(n-1), for all n > 1 For example, suppose n = 5: // Recursive call factorial(5) = 5 * factorial(4) factorial(4) = 4 * factorial(3) factorial(3) = 3 * factorial(2) factorial(2) = 2 * factorial(1) factorial(1) = 1 * factorial(0) factorial(0) = 1 // Base case // Unwinding factorial(1) = 1 * 1 = 1 factorial(2) = 2 * 1 = 2 factorial(3) = 3 * 2 = 6 factorial(4) = 4 * 6 = 24 factorial(5) = 5 * 24 = 120 (DONE) Exercise (Factorial) (Recursive): Write a recursive method called factorial() to compute the factorial of the given integer. public static int factorial(int n) The recursive algorithm is:…