To implement the maze, I used a two-dimensional array, where the element of the array shows which side of the maze is open. The element of the array is a string of length four and contains only zero and one. Zero means the side is open and 1 means it is blocked. The first, second, third and fourth digits shows the state of the top, left, bottom and right side of one cell respectively. Please see the following example that shows how the maze is created. The left picture shows the maze, while the right shows the content of the array that represent the maze.

Make a code using Recursion The countSubstring function will take two strings as parameters and will return an integer that is the count of how many times the substring (the second parameter) appears in the first string without overlapping with itself. This method will be case insensitive. For example: countSubstring("catwoman loves cats", "cat") would return 2 countSubstring("aaa nice", "aa") would return 1 because "aa" only appears once without overlapping itself. public static int countSubstring(String s, String x) { if (s.length() == 0 || x.length() == 0) return 1; if (s.length() == 1 || x.length() == 1){ if (s.substring(0,1).equals(x.substring(0,1))){ s.replaceFirst((x), " "); return 1 + countSubstring(s.substring(1), x); } else { return 0 + countSubstring(s.substring(1), x); } } return countSubstring(s.substring(0,1), x) + countSubstring(s.substring(1), x); } public class Main { public static void main(String[] args) {...

CS211 Non-recursive solution for Towers of Hanoi Using the algorithm discussed in class, write an iterative program to solve the Towers of Hanoi problem. The problem: You are given three towers a, b, and c. We start with n rings on tower a and we need to transfer them to tower b subject to the following restrictions: 1. We can only move one ring at a time, and 2. We may never put a larger numbered ring on top of a smaller numbered one. There are always 3 towers. Your program will prompt the user for the number of rings. Here is the algorithm. Definition: A ring is "available" if it is on the top of one of the towers. Definition: The "candidate" is the smallest available ring that has not been moved on the most recent move. The first candidate is ring 1. The Algorithm: 1. Find the candidate. 2. Move the candidate (right or left, depending if the number of rings is odd or even) to the closest tower on which it can be placed. Move "around the circle" if necessary. 3. If not done, go back...

Write a recursive function (Java) called Fac which takes one positive integer argument (n) and returns n! You may not use a built-in factorial method or function.

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

T/F 11. A recursion will still be replaced by an iteration and the other way around.

When recursion is used to solve a problem, why must the recursive method call itself to solve a smaller version of the original problem?

There are two important parts to every simple recursive function: the base case, and the recursive call that makes progress towards the base case. Something that can go wrong with recursion when it is used incorrectly is a stack overflow. Explain two different ways that a recursive function could be written incorrectly that could lead to stack overflow. Hint: one has something to do with the base case, and the other with the recursive call. 1. Enter your answer here 2. Enter your answer here

The following recursive method get Number Equal searches the array x of 'n integers for occurrences of the integer val. It returns the number of integers in x that are equal to val. For example, if x contains the 9 integers 1, 2, 4, 4, 5, 6, 7, 8, and 9, then getNumberEqual(x, 9, 4) returns the value 2 because 4 occurs twice in x. public static int getNumberEqual(int x[], int n, int val) { if (n< 0) ( return 0; } else { if (x[n-1) == val) { return getNumberEqual(x, n-1, val) +1; } else { return getNumber Equal(x, n-1, val); } // end if ) // end if } // end get Number Equal Demonstrate that this method is recursive by listing the criteria of a recursive solution and stating how the method meets each criterion.

Identify the base case in this recursive function. Assume i> 0 when the function is invoked. (Line numbers are not part of the code.) 1. def add(i, j): 2. if i == 0: 3. return j 4. else: 5. return add(i - 1, j + 1) This function has no base case O line 5 line 1 line 4 O line 2 Question 6 The following recursive function is supposed to return a list that is the reverse of the list it was given. But it has a bug. What is the bug? (Line numbers are not part of the code.) MooP eok Dr

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in kotlin Write two versions of a recursive palindrome check function. One version should have a block body, and the other should have an expression body. Review the lecture slide about the substring function first. Use this algorithm: if the length of the string is less than 2, return true else if the first and last characters are not equal (!=) return false else return the result of a recursive call with the substring from the second character (the one at index 1) to the second to last character. Use a main that uses a loop to test the palindrome function using this list of strings: val ss = listOf("", "A", "AA", "AB", "AAA", "ABA", "ABB", "AAAA", "AABA", "ABBA", "ABCBA", "ABCAB")