Recursion. Recursion is: Recursion splits a problem:

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Amy Lynch
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1 Recursion

2 Recursion Recursion is: A problem solving approach, that can... Generate simple solutions to... Certain kinds of problems that... Would be difficult to solve in other ways Recursion splits a problem: Into one or more simpler versions of itself Stop the break down process at a special case whose solution is obvious (Base Case)

3 Recursion: examples Process (e.g. count, paint) nested dolls: 1. if there is only one doll 2. do what it needed for it else 3. do what is needed for the outer doll 4. Process the inner nest in the same way

4 Recursion: examples Sum of 1 N(assume that N>=1): 1. if N=1 2. return 1 as the sum 3. else if 4. return the sum of 1 N 1 plus N

5 Recursion: the general case 1. if problem is small enough 2. solve it directly 3. else 4. break into one or more smaller subproblems 5. solve each subproblem recursively 6. combine results into solution to whole problem

6 Recursion Recursion can solve many programming problems that are difficult to conceptualize and solve linearly In the field of artificial intelligence, recursion often is used to write programs that exhibit intelligent behavior: playing games of chess proving mathematical theorems recognizing patterns, and so on Recursive algorithms can compute factorials compute a greatest common divisor process data structures (strings, arrays, linked lists, etc.) search efficiently using a binary search find a path through a maze, and more sort efficiently, e.g. quick sort define a computer language

7 Recursion At least one small case that you can solve directly A way of breaking a larger problem down into: One or more smaller subproblems Each of the same kind as the original A way of combining subproblem results into an overall solution to the larger problem

8 Factorial N! = N * (N-1) * (N-2) * * 3 * 2 * 1 int factorial (int n) { int i, result; result = 1; for (i = 1; i <= n; i++) { result = result * i; return ( result );

9 Factorial Recursive N! definition N! = 1 if N = 0 = N * (N-1)! Otherwise int fact ( int n) { if (n == 0) return (1); return (n * fact(n - 1) );

10 Factorial n = 3 A: fact(n 1)=? return? n = 2 A: fact(n 1)=? return? n = 1 return 1 n = 3 A: fact(n 1)=? 2 return? 6 n = 2 A: fact(n 1)=? 1 return? 2 n = 1 return 1

11 Factorial Iterative 1. 2 local variables 2. 3 statements 3. Saves solution in an intermediate variable Recursive 1. No local variables 2. One statement 3. Returns result in single expression How about efficiency?

12 Array Sum int SumArray (int a[ ], int size) { int j, sum = 0; for ( j = 0; j < size; j++ ) sum += a[ j ]; return sum;

13 Array Sum int SumArray ( int a [ ], int size) { if (size == 0) return 0; else return array[ size 1] + SumArray (a, size 1);

14 Binary search Search a sorted array: 1. if the array is empty 2. return false as the search result (not present) 3. else if the middle element == target 4. return true 5. else if target < middle element 6. recursively search elements before middle 7. else 8. recursively search elements after the middle

15 bool Search ( int a[], int size, int key){ if (size < 1) return false; if (size == 1) return (key==a[0]); middle=size/2; if (key < a[middle]) return Search(a, middle, key); else if (key == middle) return true; else return (a+middle+1, size-middle-1, key);

16 Multiplying Mice/Rabbits Month 1 Month 2 Quite difficult to keep track of the population explosion. Month 3 Month 4 Month 5 Month 6 Month 7 Babies Recursive relation: #babies in Month 7= #mice in Month 5 #adults in Month 7 = #mice in Month 6

17 Fibonacci numbers Definition F(n) = F(n 1) + F(n 2); 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,. Base cases?

18 Fibonacci numbers int fibonum(int n){ if (n <= 0) return 0; else if (n == 1) return 1; else return(fibonum(n-1)+ fibonum(n-2))

19 Fibonacci numbers int fibonum(int n){ if (n == 1) return 1; else if (n == 2) return 1; else return(fibonum(n-1)+ fibonum(n-2))

20 Fibonacci numbers 8 m(6) Return m(5)+m(4) 5 3 m(5) m(4) Return m(4)+m(3) 3 2 Return m(3)+m(2) 2 1 m(4) m(3) Return m(3)+m(2) 2 1 Return m(2)+m(1) 1 1 m(3) Return m(2)+m(1) 1 1 m(2) Return 1 m(3) Return m(2)+m(1) 1 1 m(2) Return 1 m(2) Return 1 m(1) Return 1 m(2) Return 1 m(1) Return 1 m(2) Return 1 m(1) Return 1

21 Is Palindrome? A palindrome is a string that is the same as its reverse radar and able was I ere I saw elba. How do we check if a string is a palindrome?

22 Is Palindrome? A palindrome is a string that is the same as its reverse radar and able was I ere I saw elba. r a d a r How do we check if a string is a palindrome? First char =?= last char? Yes? Check the rest of the string: recursion. Base case?

23 Is Palindrome? How do we check if a string is a palindrome? First char =?= last char? Yes? Check the rest of the string: recursion. size = size 2 Base case: String size = 2 String size = 1 String size = 0

24 Is Palindrome? int ispalindrome(char s[], int size){ if (size<=1) return 1; else return ((s[0]==s[size-1]) && ispalindrome(s+1, size-2)); ;

25 String reverse Print a string in reverse order Iterative vs. recursive solutions

26 The 8 queen puzzle What is the maximum number of queens that can be placed on an n x n chessboard such that no two attack one another? The answer is n Eight queens for the usual 8x8 board

27 The 8 queen puzzle

28 The 8 queen puzzle The 8 queen puzzle What is the maximum number of queens that can be placed on an 8 x 8 chessboard such that no two attack one another? Answer: 8 There are 92 distinct solutions There are 12 unique solutions discounting symmetrical answers (rotations/reflections)

29 8 queen puzzle It s a very popular problem in CS, especially in algorithm courses. N=1 N=2 N=3

30 8 queen puzzle Recursion We can place at most one queen in each row If we have placed d queens in the first d rows How do we place the next queen in row d+1? Check every cell in the row Any conflict with the existing d queens? Base case? We have successfully placed 8 queens and arrive at row 9

31 const int SIZE=8; const char BACKG='.'; const char FOREG='*'; void initilizesolution(char solution[][size], int s); void eightqueens(char solution[][size], int s, int depth); int isfeasible(int p1, int p2, char solution[][size], int s); void printsolution(char solution[][size], int s); int main(){ //init char solution[size][size]; initilizesolution(solution, SIZE); //search for solutions eightqueens(solution, SIZE, 0); return 0;

32 const int SIZE=8; const char BACKG='.'; const char FOREG='*'; void initilizesolution(char solution[][size], int s); void eightqueens(char solution[][size], int s, int depth); int isfeasible(int p1, int p2, char solution[][size], int s); void printsolution(char solution[][size], int s); int main(){ //init char solution[size][size]; initilizesolution(solution, SIZE); //search for solutions eightqueens(solution, SIZE, 0); void initilizesolution(char solution[][size], int s){ for (int i=0; i<s; i++) for (int j=0; j<s; j++) solution[i][j]=backg; return 0;

33 const int SIZE=8; const char BACKG='.'; const char FOREG='*'; void initilizesolution(char solution[][size], int s); void eightqueens(char solution[][size], int s, int depth); int isfeasible(int p1, int p2, char solution[][size], int s); void printsolution(char solution[][size], int s); int main(){ //init char solution[size][size]; initilizesolution(solution, SIZE); //search for solutions eightqueens(solution, SIZE, 0); return 0; void initilizesolution(char solution[][size], int s){ for (int i=0; i<s; i++) for (int j=0; j<s; j++) solution[i][j]=backg; void printsolution(char solution[][size], int s){ cout << "solution:" << endl; for (int i=0; i<s; i++){ for (int j=0; j<s; j++) cout << solution[i][j]; cout << endl; cout << endl;

34 void eightqueens(char solution[][size], int s, int depth){ //base case if(depth==s){ count++; cout << count << endl; printsolution(solution, s); return;

35 void eightqueens(char solution[][size], int s, int depth){ //base case if(depth==s){ count++; cout << count << endl; printsolution(solution, s); return;???????? //try to place one more queen for (int j=0; j<s; j++){ if (isfeasible(depth, j, solution, s)){ //recursive search solution[depth][j]=foreg; eightqueens(solution, s, depth+1); //reset solution[depth][j]=backg;

36 int isfeasible(int p1, int p2, char solution[][size], int s){ int i, j; for (i=0; i<p1; i++){ if(solution[i][p2]==foreg) return 0; for (i=p1 1, j=p2 1; i>=0 && j>=0; i, j ){ if(solution[i][j]==foreg) return 0;??? for (i=p1 1, j=p2+1; i>=0 && j<s; i, j++ ){ if(solution[i][j]==foreg) return 0;

37 int isfeasible(int p1, int p2, char solution[][size], int s){ int i, j; for (i=0; i<p1; i++){ if(solution[i][p2]==foreg) return 0; for (i=p1 1, j=p2 1; i>=0 && j>=0; i, j ){ if(solution[i][j]==foreg) return 0;??? for (i=p1 1, j=p2+1; i>=0 && j<s; i, j++ ){ if(solution[i][j]==foreg) return 0;

38 int isfeasible(int p1, int p2, char solution[][size], int s){ int i, j; for (i=0; i<p1; i++){ if(solution[i][p2]==foreg) return 0; for (i=p1 1, j=p2 1; i>=0 && j>=0; i, j ){ if(solution[i][j]==foreg) return 0;??? for (i=p1 1, j=p2+1; i>=0 && j<s; i, j++ ){ if(solution[i][j]==foreg) return 0;

39 8 queen puzzle Recursion We can place at most one queen in each row If we have placed d queens in the first d rows How do we place the next queen in row d+1? Check every cell in the row Any conflict with the existing d queens? Base case? We have successfully placed 8 queens and arrive at row 9

CMSC 132: Object-Oriented Programming II. Recursive Algorithms. Department of Computer Science University of Maryland, College Park

CMSC 132: Object-Oriented Programming II Recursive Algorithms Department of Computer Science University of Maryland, College Park Recursion Recursion is a strategy for solving problems A procedure that