1. RECURSION Go back, Jack, and do it again.

2. Recursion 2  Recursion occurs whenever "something" is defined in terms of itself.  Structural recursion: when the data itself is self- referential. A class contains an instance of that class.  Functional recursion: when a procedure is self- referential. A method calls itself.

3. Example 3  Problem: Write a method to compute the sum of the numbers 1 through N where N is a non-negative integer parameter. Iterative Solution 1 int sum(int n) { int sum = 0; for(int i=1; i<=n; i++) { sum = sum + i; } return sum; } Iterative Solution 2 int sum(int n) { int sum = 0; while(n > 0) { sum += n; n--; } return sum; }

4. Recursive Solution  The goal is to compute: 0 + 1 + 2 +... + (N-1) + N  We can define sum(n) as Recursive Solution int sum(int n) { if(n == 0) { return 0; } else { return sum(n-1)+n; } Base Case Reduction step (makes the input smaller)

5. Variable Scope and Lifetime 5  Class variables belong to Classes  Always exist  Instance variables belong to Objects  Are “born” when the object is created and “die” when the object is destroyed by the garbage collector.  Local variables belong to methods  Are “born” when the method is called and “die” when the method ends.  The time between the “birth” and “death” of a variable is called the “lifetime” of a variable

6. Basic Approach  Establish a base case for which the solution is trivial.  Execute a reduction step which makes the solution converge towards the base case.  The reduction step reduces the input size (makes the problem simpler)  The reduced problem is identical in nature to the larger problem.  The results of solving this reduced problem can then be used to solve the larger problem

7. Greatest Common Divisor 7  Goal: Find the largest number D that evenly divides M and N. otherwise if n = 0 gcd(n, m%n) m gcd(m,n) = Base Case Reduction step (makes the input smaller) gcd(1440, 408) = gcd(408, 216) gcd(408, 216) = gcd(216, 192) gcd(216,192) = gcd(192, 24) gcd(192, 24) = gcd(24, 0) gcd(24, 0) = 24 Example Mr. Euclid of Alexandria 325 BC to 265 BC

8. Greatest Common Divisor 8 otherwise if n = 0 gcd(n, m%n) m gcd(m,n) = Base Case Reduction step (makes the input smaller) Implementation int gcd(int m, int n) { if(n == 0) { return m; } else { return gcd(n, m%n); } Mr. Euclid

9. Greatest Common Divisor 9  When a method is invoked, an activation of that method is created.  Examine what occurs when executing “int z = gcd(1440, 408)” Activation 1 int gcd(int m=1440, int n=408) { if(n == 0) { return m; } else { return gcd(n, m%n); } Activation 2 int gcd(int m=408, int n=216) { if(n == 0) { return m; } else { return gcd(n, m%n); } Activation 3 int gcd(int m=216, int n=192) { if(n == 0) { return m; } else { return gcd(n, m%n); } Activation 4 int gcd(int m=192, int n=24) { if(n == 0) { return m; } else { return gcd(n, m%n); } Activation 5 int gcd(int m=24, int n=0) { if(n == 0) { return m; } else { return gcd(n, m%n); }

10. Conversion to Binary 10 Goal: Print the binary representation of the integer N. Implementation void printBinary(int n) { if(n == 0) { return; } printBinary(n/2); System.out.print(n%2); } Algorithm to print the binary representation of N Stop if N = 0 Write a ‘1’ if N is odd and a ‘0’ if N is even Move the pen to the left one digit Print the binary representation of the integer N/2 CallOutput print(43)1 print(21)11 print(10)011 print(5)1011 print(2)01011 print(1)101011 print(0)101011

11. Watch Out! 11 Mr. Fibonacci 1170 to 1250 Goal: Compute the Nth fibonacci number if n = 11 otherwise if n = 0 fib(n-1)+fib(n-2) 0 fib(n) = Recursive Implementation int fib(int n) { if(n == 0 || n == 1) { return n; } else { return fib(n-2) + fib(n-1); }

12. Watch Out! 12 Mr. Fibonacci Goal: Compute the Nth fibonacci number Recursive Implementation int fib(int n) { if(n == 0 || n == 1) { return n; } else { return fib(n-2) + fib(n-1); } Why is this so terribly slow?

13. Reverse Printing 13 Goal: Print a string backwards! Let S be a string consisting of the characters [C 0, C 1, C 2, C 3, …, C n-2, C n-1 ] Note that Reverse(S) = [C n-1, C n-2, …, C 3, C 2, C 1, C 0 ] Reverse(C 0, C n-2 ) Reverse(S) = [C n-1, Reverse(C 0, C n-2 )] Recursive Implementation void printBackwards(String s, int k) { System.out.println(s.charAt(k)); if(k > 0) { printBackwards(s, k-1); }

14. Reversing an Array 14 Goal: Reverse the elements in an array. Let A be an array consisting of the elements [X 0, X 1, X 2, X 3, …, X n-2, X n-1 ] Note that Reverse(A) = [X n-1, X n-2, …, X 3, X 2, X 1, X 0 ] Reverse(A, X 1, X n-2 ) Reverse(A, f, l) = [X l, Reverse(A, X f+1, X l-1 ), X f ] Recursive Implementation void arrayReverse(int[] data, int first, int last) { if(first < last) { int temp = data[first]; data[first] = data[last]; data[last] = temp; arrayReverse(data, first+1, last-1); }

15. Pop Quiz! 15  Write a single recursive method to print the numbers N to 0.  Make the signature be “void print(int n)”. Feeble Attempt 1 void print(int n) { System.out.println(n); print(n-1); } Feeble Attempt 2 void print(int n) { if(n > 0) return; System.out.println(n); print(n-1); } 3 rd Times the Charm void print(int n) { if(n < 0) return; System.out.println(n); print(n-1); }

16. MineSweeper Example 16 public void select(int x, int y) { if coordinate (x,y) is off of the game-board or coordinate (x,y) is already selected then return mark (x,y) as selected if coordinate (x,y) is not a bomb AND not adjacent to a bomb then select(x-1, y-1) select(x, y-1) select(x+1, y-1) select(x-1, y) select(x+1, y) select(x-1, y+1) select(x, y+1) select(x+1, y+1) } 1111B 1B2111 12B1 222 1B1 111 111 1B1 Initially – all cells are UNSELECTED Now, select cell (3,1)