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, then 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 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
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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, then 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
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
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- 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, then use a recursive method to create and output all possible orderings of those names, one ordering per line. When the input is: Julia Lucas MiaOnly in Python 3! 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, then 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 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 JuliaWrite 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 main.cpp #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 when -1 is read. Then call recursive method.return 0;}
- 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…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 Partially done code (C++): #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 when -1 is read. Then call recursive method.return 0;}python 3 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, then 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 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 note: the code should have 'for loops, 'if' statements for the base case, and 'else' statements for recursive case. provided code: def all_permutations(permList, nameList):# TODO: Implement method to create and output all permutations of the list of names. if __name__ == "__main__": nameList = input().split(' ')permList = []all_permutations(permList, nameList)
- python 3 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, then 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 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 question: is it any way that i can use ('if' statement as base case, and 'else' statement as recursive case) in the code below? thanks. code: def all_permutations(permList, nameList):# TODO: Implement method to create and output all permutations of the list of names.def createPermutationsList(nameList):f = len(nameList) if f == 0:return [] if f == 1:return [nameList] permList = [] for i in range(f):newList = nameList[i]remaining = nameList[:i] + nameList[i+1:]for p in createPermutationsList(remaining):permList.append([newList] + p)…In C++ 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 hen 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 #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 when -1 is read. Then call recursive method. return 0;}1. Write a recursive method expFive(n) to compute y=5^n. For instance, if n is 0, y is 1. If n is 3, then y is 125. If n is 4, then y is 625. The recursive method cannot have loops. Then write a testing program to call the recursive method. If you run your program, the results should look like this: > run RecExpTest Enter a number: 3 125 >run RecExpTest Enter a number: 3125 2. For two integers m and n, their GCD(Greatest Common Divisor) can be computed by a recursive function. Write a recursive method gcd(m,n) to find their Greatest Common Divisor. Once m is 0, the function returns n. Once n is 0, the function returns m. If neither is 0, the function can recursively calculate the Greatest Common Divisor with two smaller parameters: One is n, the second one is m mod n. Although there are other approaches to calculate Greatest Common Divisor, please follow the instructions in this question, otherwise you will not get the credit. Meaning your code needs to follow the given algorithm. Then…
- The word ladder game was invented by Lewis Carroll in 1877. The idea is to begin with a start word and then change one letter at a time until you arrive at an end word. Each word along the way must be an English word. For example, starting from FISH, you can arrive at MAST through the following word ladder:FISH, WISH, WASH, MASH, MAST Write a program that uses recursion to find the word ladder given a start word and an end word, or that determines no word ladder exists. Use the file words.txt that is available online with the source code for the book as your dictionary of valid words. This file contains 87,314 words. Your program does not need to find the shortest word ladder between words; any word ladder will do if one exists. list aalii aardvark aardvarks aardwolf aba abaca abaci abacist aback abacus abacuses abaft abalone abalones abamp abampere abandon abandoned abandonee abandoner abandonersExercise-3: Write a recursive and iterative methods to convert a decimal number to its binary equivalent string. The iterative algorithm (in pseudo-code) for converting a decimal integer into a binary integer as follows: 1. If the integer is 0 or 1, its binary equivalent is 0 or 1. 2. If the integer is greater than or equal to 2 do the following: 3. Divide the integer by 2. 4. Separate the result into a quotient and remainder. 5. Divide the quotient again and repeat the process until the quotient is zero. 6. Write down all remainders in reverse order as a string. 7. This string is the binary equivalent of the given integer. // Recursive decimal to binary method public static String dec2binRecursive(int n) { if (n<2) return n+ " ". else return dec2binRecursive(n/2) + n%2; } a) Write the Complete program for above Recursive decimal to binary method Algorit b) Iterative decimal to binary methodWrite a program that has a main() function that calls a recursive method repeatPrint(String s, .....) that prints out the strings as shown in the example below: repeat Print ("CHICAGO", ....); C CH CHI CHIC CHICA CHICAG CHICAGO CHICAGO HICAGO ICAGO CAGO AGO GO 0