Complete the program shown in the 'Answer' box below by filling in the blank so that the program prints2 2We're dead, FredThere are multiple solutions; any one will do.Answer:def silly2(nums, indices, extra=None):print(len(nums), len(indices))new_indices = []for i in indices:if i >= 0 and i < len(nums):new_indices.append(i)if extra:new_indices.append (extra)try:for index in new_indices:if index < len(nums):print(nums [index])except IndexError:print ("We're dead, Fred")silly2

Complete the program shown in the 'Answer' box below by filling in the blank so that the program prints 22 We're dead, Fred There are multiple solutions; any one will do. Answer: def silly2(nums, indices, extra-None): print(len(nums), len(indices)) new_indices - (0 for i in indices: if i >- 0 and i < len(nums): new_indices.append(1) if extra: new_indices.append (extra) try: for index in new_indices: if index < len(nums): print(nums [index]) except IndexError: print ("We're dead, Fred") silly2

Complete the program shown in the 'Answer' box below by filling in the blank so that the program prints 22 We're dead, Fred There are multiple solutions; any one will do. Answer: def silly2(nums, indices, extra-None): print(len(nums), len(indices)) new_indices - (0 for i in indices: if i >- 0 and i < len(nums): new_indices.append(1) if extra: new_indices.append (extra) try: for index in new_indices: if index < len(nums): print(nums [index]) except IndexError: print ("We're dead, Fred") silly2

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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Similar questions

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
The goal is to rewrite the function, below, such that passes in a different list of parameters, particularly eliminating the need to pass low and high for each recursive call to binary_search. defbinary_search(nums,low,high,item): mid=(low+high)//2iflow>high:returnFalse #The item doesn't exist in the list!elifnums[mid]==item:returnTrue# The item exists in the list!elifitem<nums[mid]:returnbinary_search(nums,low,mid-1,item)else:returnbinary_search(nums,mid+1,high,item) The new function should be prototyped below. The number of changes between the given version, and the one requested is not significant. defbinary_search(nums,item):pass# Remove this and fill in with your code Tip: If you consider that high and low are used to create a smaller version of our problem to be processed recursively, the version requested will do the same thing, just through a different, more Pythonic technique.
Given is the code to print numbers from 1 to n in increasing order recursively. But it contains few bugs that you need to rectify such that all the test cases pass.Input Format :Integer nOutput Format :Numbers from 1 to n (separated by space)Constraints :1 <= n <= 10000Sample Input 1 : 6Sample Output 1 :1 2 3 4 5 6Sample Input 2 : 4Sample Output 2 :1 2 3 4.
Write a program called Recursive_fibonacci.java that implements a recursive function for computing the nth term of a Fibonacci Sequence. In the main method of your program accept the value of n from the user as a command-line argument and then call your function named Fibonacci with this value. The output should look exactly like what is pictured below.
The Fibonacci algorithm is a famous mathematical function that allows us to create a sequence of numbers by adding together the two previous values. For example, we have the sequence:1, 1, 2, 3, 5, 8, 13, 21…Write your own recursive code to calculate the nth term in the sequence. You should accept a positive integer as an input, and output the nth term of the sequence.Once you have created your code, add comments describing how the code works, and the complexity of any code you have created.
The task is to get input (n) and make a list from 2 to n and remove its multiples from the list by applying recursive functions. Example: Input: 6 Output: [2, 3, 4, 5, 6] How to apply this python code in the task? def createList(n): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all the TODO for this function return [] def removeMultiples(x, arr): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all the TODO for this function return []
Keeping data structures and algorithms in mind, can you help me to write these two functions in java? The factorial of a positive integer n —which we denote as n!—is the product of n and the factorial of n − 1. The factorial of 0 is 1. Write two different recursive methods that each return the factorial of n. Include the two methods in a file called Test.java What to submit: The Test.java
Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2; if (target == nums[mid]) return mid; else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);}For the statement below, indicate whether you think it is True or False. If you like, you can provide a description of your answer for partial credit in case you are incorrect.4) The code “else return binarySearch(nums, mid + 1, high, target);” is a base case for this function
Below is a recursive version of binarySearch:int binarySearch(int nums[], int low, int high, int target){ if (low > high) return -1; int mid = (low + high)/2; if (target == nums[mid]) return mid; else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target);}For the statement below, indicate whether you think it is True or False. If you like, you can provide a description of your answer for partial credit in case you are incorrect. 1) The code “if (low > high) return -1;” is a base case for this function
Write in JAVA When analyzing data sets, such as data for human heights or for human weights, a common step is to adjust the data. This adjustment can be done by normalizing to values between 0 and 1, or throwing away outliers. For this program, adjust the values by dividing all values by the largest value. The input begins with an integer indicating the number of floating-point values that follow. Assume that the list will always contain fewer than 20 floating-point values. Output each floating-point value with two digits after the decimal point, which can be achieved as follows:System.out.printf("%.2f", yourValue);
Write a program that lists all ways people can line up for a photo (all permutations of a list of strings). The program will read a list of one word names (until -1), and use a recursive method to create and output all possible orderings of those names, one ordering per line. When the input is: Julia Lucas Mia -1 then the output is (must match the below ordering): Julia Lucas Mia Julia Mia Lucas Lucas Julia Mia Lucas Mia Julia Mia Julia Lucas Mia Lucas Julia partial code below, only lines 9-11, and 18 can be added onto, the rest must stay the same. #include <vector>#include <string>#include <iostream> using namespace std; // TODO: Write method to create and output all permutations of the list of names.void AllPermutations(const vector<string> &permList, const vector<string> &nameList) { } int main(int argc, char* argv[]) { vector<string> nameList; vector<string> permList; string name; // TODO: Read in a list of names; stop…
How to solve the problem by FOLLOWING this python code format? def createList(n): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all the TODO for this function return [] def removeMultiples(x, arr): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all the TODO for this function return [] def Sieve_of_Eratosthenes(list): #Base Case/s if len(list) < 1 : return list #Recursive Case/s else: return [list[0]] +…