2. Chapter 5: Recursion as a Problem-Solving Technique BACKTRACKING RECURSIVE GRAMMARS MATRIX OPERATIONS MATHEMATICAL INDUCTION RECURRENCE RELATIONS CS 240 19

3. CS 240 20 Recursion can be a very effective technique for solving what would otherwise be extremely elaborate problems. Sample recursive applications include:  Backtracking algorithms.  Programming language definition.  Matrix operations. In addition, a solid understanding of recursion is helpful in analyzing the time complexity of algorithms.  Mathematical induction.  Recurrence relations. Recursion-related analysis techniques include:

4. CS 240 21 Recursive Backtracking Example: Flood Fill Algorithm Starting with a “seed” pixel that’s inside a polygonal region, recursively visit the four adjacent pixels, coloring any that haven’t been colored, and that aren’t on the polygon’s boundary. Reaching the boundary is the recursion’s termination condition. Seed pixel Boundary pixel LRUD-visited pixel void floodfill(int x, int y) { if (!filled(x,y)) { color(x,y); floodfill(x-1,y); // Left floodfill(x+1,y); // Right floodfill(x,y-1); // Up floodfill(x,y+1); // Down }

5. CS 240 22 Recursive Grammar Example: A Calculator Grammar Languages (e.g., programming languages) are often defined by their grammars, sets of recursive rules which provide syntax. program:END expr_list END expr_list:expression ; expression ; expr_list expression:term + expression term – expression term term:primary / term primary * term primary primary:NUMBER NAME NAME = expression - primary ( expression ) Using this grammar, the following is a syntactically correct calculator program: pi = 3.1416; rad = 2.5; ht = 10; area = pi * rad * rad; surfacearea = 2 * (area + pi * rad * ht); END Recursive grammars are also prominent in more sophisticated languages, making the following language features possible: nested loops, conditionals, function calls nested loops, conditionals, function calls cascaded operators ( >, =, etc.) cascaded operators ( >, =, etc.) multiple cases in a switch statement multiple cases in a switch statement

6. CS 240 23 Many matrix operations may be defined recursively, with combinations of submatrix operations used to implement the large matrix operation.                                               Note that a matrix’s determinant is used to determine, among other things, whether the matrix is invertible. Recursive Matrix Operation Example: Determinants

7. CS 240 24 //////////////////////////////////////////// // Class definition file: Matrix.h // // The matrix class has two data members: // // a square array of integer values (all // // between 0 and 9), and an integer indi- // // cating the array size. Its member // // functions include constructors, func- // // tions to set and access specific array // // elements, a function to access a sub- // // array, a recursive determinant func- // // tion, and an output operator. // //////////////////////////////////////////// #ifndef MATRIX_H #include using namespace std; typedef int elementType; const int MAX_GRID_SIZE = 7; class matrix { public: // Class constructors matrix() {size = 0;} matrix(const matrix &m); matrix(int sz); //////////////////////////////////////////// // Class definition file: Matrix.h // // The matrix class has two data members: // // a square array of integer values (all // // between 0 and 9), and an integer indi- // // cating the array size. Its member // // functions include constructors, func- // // tions to set and access specific array // // elements, a function to access a sub- // // array, a recursive determinant func- // // tion, and an output operator. // //////////////////////////////////////////// #ifndef MATRIX_H #include using namespace std; typedef int elementType; const int MAX_GRID_SIZE = 7; class matrix { public: // Class constructors matrix() {size = 0;} matrix(const matrix &m); matrix(int sz); // Member functions int getSize() const { return size; } void setElement(int i, int j, elementType item); elementType getElement(int i, int j) { return table[i][j]; } elementType determinant(); friend ostream& operator << (ostream &os, const matrix &m); protected: // Data members elementType table[MAX_GRID_SIZE] [MAX_GRID_SIZE]; int size; // Member function matrix minor(int i, int j); }; #define MATRIX_H #endif // Member functions int getSize() const { return size; } void setElement(int i, int j, elementType item); elementType getElement(int i, int j) { return table[i][j]; } elementType determinant(); friend ostream& operator << (ostream &os, const matrix &m); protected: // Data members elementType table[MAX_GRID_SIZE] [MAX_GRID_SIZE]; int size; // Member function matrix minor(int i, int j); }; #define MATRIX_H #endif Recursive Determinant Program Example

8. CS 240 25 //////////////////////////////////////////// // Class implementation file: Matrix.cpp // // The implementation of the copy and // // initializing constructors, the setEle- // // ment, determinant, and minor member // // functions, and the output operator. // //////////////////////////////////////////// #include #include "Matrix.h" using namespace std; // Copy constructor: Copies existing mat.// matrix::matrix(const matrix &m) { size = m.size; for (int row = 0; row < m.size; row++) for (int col = 0; col < m.size; col++) table[row][col] = m.table[row][col]; } // Initializing constructor: Sets *this // // up as a sz x sz matrix of zeros. // matrix::matrix(int sz) { size = sz; for (int row = 0; row < sz; row++) for (int col = 0; col < sz; col++) table[row][col] = 0; } //////////////////////////////////////////// // Class implementation file: Matrix.cpp // // The implementation of the copy and // // initializing constructors, the setEle- // // ment, determinant, and minor member // // functions, and the output operator. // //////////////////////////////////////////// #include #include "Matrix.h" using namespace std; // Copy constructor: Copies existing mat.// matrix::matrix(const matrix &m) { size = m.size; for (int row = 0; row < m.size; row++) for (int col = 0; col < m.size; col++) table[row][col] = m.table[row][col]; } // Initializing constructor: Sets *this // // up as a sz x sz matrix of zeros. // matrix::matrix(int sz) { size = sz; for (int row = 0; row < sz; row++) for (int col = 0; col < sz; col++) table[row][col] = 0; } // SetElement Member Function: Sets the // // (i,j) element of the matrix to item. // void matrix::setElement(int i, int j, elementType e) { assert ((0<=i) && (i<size) && (0<=j) && (j<size)); table[i][j] = e; } // Determinant Member Function: Calculates // // & returns the determinant of the matrix. // elementType matrix::determinant() { elementType value = 0; if (size == 1) return table[0][0]; for (int col = 0; col < size; col++) { value += (elementType)pow(-1, 0+col) * table[0][col] * minor(0, col).determinant(); } return value; } // SetElement Member Function: Sets the // // (i,j) element of the matrix to item. // void matrix::setElement(int i, int j, elementType e) { assert ((0<=i) && (i<size) && (0<=j) && (j<size)); table[i][j] = e; } // Determinant Member Function: Calculates // // & returns the determinant of the matrix. // elementType matrix::determinant() { elementType value = 0; if (size == 1) return table[0][0]; for (int col = 0; col < size; col++) { value += (elementType)pow(-1, 0+col) * table[0][col] * minor(0, col).determinant(); } return value; } Notice how the size×size matrix is being evaluated recursively by using the top row to expand into size (size- 1)×(size-1) submatrices.

9. CS 240 26 // Output Operator: Outputs matrix // // as a grid of size rows with // // size columns in each row. // ostream& operator << (ostream &os, const matrix &m) { for (int row = 0; row < m.getSize(); row++) { for (int col = 0; col < m.getSize(); col++) os << setw(4) << m.table[row][col]; os << endl; } return os; } // Output Operator: Outputs matrix // // as a grid of size rows with // // size columns in each row. // ostream& operator << (ostream &os, const matrix &m) { for (int row = 0; row < m.getSize(); row++) { for (int col = 0; col < m.getSize(); col++) os << setw(4) << m.table[row][col]; os << endl; } return os; } // Minor Member Function: If *this is // // an nXn matrix, then this returns // // the (n-1)X(n-1) matrix that is // // *this w/row i & column j removed. // matrix matrix::minor(int i, int j) { int subrow, subcol; assert (size > 1); matrix submat(size-1); subrow = 0; for (int row = 0; row < size; row++) { subcol = 0; if (row != i) { for (int col = 0; col < size; col++) if (col != j) { submat.setElement(subrow, subcol, table[row][col]); subcol++; } subrow++; } return submat; } // Minor Member Function: If *this is // // an nXn matrix, then this returns // // the (n-1)X(n-1) matrix that is // // *this w/row i & column j removed. // matrix matrix::minor(int i, int j) { int subrow, subcol; assert (size > 1); matrix submat(size-1); subrow = 0; for (int row = 0; row < size; row++) { subcol = 0; if (row != i) { for (int col = 0; col < size; col++) if (col != j) { submat.setElement(subrow, subcol, table[row][col]); subcol++; } subrow++; } return submat; }

10. CS 240 27 //////////////////////////////////////////////////// // Program file: matrixDriver.cpp // // This program tests the matrix class by // // creating a random matrix of a user-specified // // size, outputting it, & taking its determinant. // //////////////////////////////////////////////////// #include #include "Matrix.h" using namespace std; int generateRandomNumber(int lowerBound, int upperBound); // The main function randomly generates & outputs // // a square matrix,, & determines its determinant.// void main() { int gridSize; cout << "SPECIFY THE MATRIX SIZE (a positive " << "integer less than " << MAX_GRID_SIZE+1 << "): "; cin >> gridSize; while ((gridSize MAX_GRID_SIZE)) { cout << "SORRY, ONLY VALUES BETWEEN 1 AND " \n "; cout << "SPECIFY THE MATRIX SIZE (a positive " << "integer less than " << MAX_GRID_SIZE+1 << "): "; cin >> gridSize; } //////////////////////////////////////////////////// // Program file: matrixDriver.cpp // // This program tests the matrix class by // // creating a random matrix of a user-specified // // size, outputting it, & taking its determinant. // //////////////////////////////////////////////////// #include #include "Matrix.h" using namespace std; int generateRandomNumber(int lowerBound, int upperBound); // The main function randomly generates & outputs // // a square matrix,, & determines its determinant.// void main() { int gridSize; cout << "SPECIFY THE MATRIX SIZE (a positive " << "integer less than " << MAX_GRID_SIZE+1 << "): "; cin >> gridSize; while ((gridSize MAX_GRID_SIZE)) { cout << "SORRY, ONLY VALUES BETWEEN 1 AND " \n "; cout << "SPECIFY THE MATRIX SIZE (a positive " << "integer less than " << MAX_GRID_SIZE+1 << "): "; cin >> gridSize; } matrix grid(gridSize); for (int row=0; row<gridSize; row++) for (int col=0; col<gridSize; col++) grid.setElement(row, col, generateRandomNumber(0,9)); cout << endl << "MATRIX:" << endl << grid << endl << endl; cout << "DETERMINANT: " << grid.determinant() << endl << endl; } // The generateRandomNumber function // // randomly generates an integer in the // // range between the parameterized low- // // erBound & upperBound values (inclu- // // sive). The first time it is called, // // it seeds the rand() random number // // generation function. // int generateRandomNumber(int lowerBound, int upperBound) { static bool firstTime = true; time_t randomNumberSeed; if (firstTime) { time(&randomNumberSeed); srand(randomNumberSeed); firstTime = false; } return (lowerBound + int((upperBound - lowerBound) * (float(rand()) / RAND_MAX))); } matrix grid(gridSize); for (int row=0; row<gridSize; row++) for (int col=0; col<gridSize; col++) grid.setElement(row, col, generateRandomNumber(0,9)); cout << endl << "MATRIX:" << endl << grid << endl << endl; cout << "DETERMINANT: " << grid.determinant() << endl << endl; } // The generateRandomNumber function // // randomly generates an integer in the // // range between the parameterized low- // // erBound & upperBound values (inclu- // // sive). The first time it is called, // // it seeds the rand() random number // // generation function. // int generateRandomNumber(int lowerBound, int upperBound) { static bool firstTime = true; time_t randomNumberSeed; if (firstTime) { time(&randomNumberSeed); srand(randomNumberSeed); firstTime = false; } return (lowerBound + int((upperBound - lowerBound) * (float(rand()) / RAND_MAX))); }

11. CS 240 28

12. CS 240 29 A mathematical “cousin” to recursion is the concept of induction. Step One: Prove The Base Case Formally demonstrate that it’s true for that smallest value, n 0. Step Two: Assume For Some General Case Assume that the it’s been proven true for all values through k, where k is at least n 0. Step Three: Prove For The Next Case Use the assumption that it’s been proven true for smaller cases to prove that it’s also true for k+1. When you want to prove that something is true for all integer values, beginning at a specific value n 0, perform the following steps:

13. CS 240 30 Induction Example Theorem A:  i = 1,n i = ½n(n + 1) for all n  1. Proof (by induction): Step One (Prove for the base case): For n = 1,  i = 1,1 i = 1 = ½(1)(1 + 1). Step Two (Assume for some general case): Assume for n = k:  i = 1,k i = ½k(k + 1). Step Three (Prove for the next case): Prove for n = k + 1:  i = 1,k+1 i = (k + 1) +  i = 1,k i = (k + 1) + ½k(k + 1) (by the assumption for case k) = ½(2)(k + 1) + ½k(k + 1) = ½(k + 1)(k + 2) = ½(k + 1)((k + 1) + 1).

14. CS 240 31 To prove that  i = 1,n i = ½n(n + 1) for all n  1, we started by proving that it was true for n = 1. Once we accomplished that, we assumed that it was true for some arbitrary value k and then proved that that made it true for the next value: k + 1. In essence, this last proof causes the truth of the theorem to “cascade” through all remaining values. Why Does Induction Work? TRUEFOR n = 1: 1 = ½(1)(2)TRUEFOR n = 1: 1 = ½(1)(2) Letting k = 1 TRUEFOR n = 2: 1 + 2 = ½(2)(3)TRUEFOR n = 2: 1 + 2 = ½(2)(3) Letting k = 2 TRUEFOR n = 3: 1+2+3 = ½(3)(4)TRUEFOR n = 3: 1+2+3 = ½(3)(4) Letting k = 3 Proof that if it’s true for n = k, then it’s also true for n = k + 1 TRUEFOR n = 4: 1+2+3 +4 = ½(4)(5)TRUEFOR n = 4: 1+2+3 +4 = ½(4)(5) Letting k = 4 TRUEFOR n = 5 : 1+2+3 +4+5 = ½(5)(6)TRUEFOR n = 5 : 1+2+3 +4+5 = ½(5)(6) And so on...

15. CS 240 32 Theorem Z: For any group of n people, all n have the same height. What’s Wrong With This Induction? Proof (by induction): Step One (Prove for the base case): For n = 1, the group consists of a single person, so the entire group obviously has the same height. Step Two (Assume for some general case): Assume for n = k: Any group of k people have the same height. Step Three (Prove for the next case): Prove for n = k + 1: Given a group of k + 1 people, remove one person. The resulting group of k people must, by the inductive hypothesis, have the same height. Reinsert the person that was removed and then remove a different person. The resulting group of k people must also have the same height. Thus, all k + 1 people must have the same height!

16. CS 240 33 int powerOf2(const int &n) { if (n == 0) return 1; return powerOf2(n-1) + powerOf2(n-1); } int powerOf2(const int &n) { if (n == 0) return 1; return powerOf2(n-1) + powerOf2(n-1); } One of the bridges between recursion and induction is the recurrence relation, which can be used to determine the execution time of a recursive function. Assuming that the execution time for arithmetic operations, condition checking, and returning are all the same, let’s also assume that there is a function T(n) such that it takes T(k) time to execute powerOf2(k). Examination of the code above allows us to conclude the following two facts: T(0) = 2 T(k) = 5 + 2T(k-1) for all k > 0 Using mathematical induction, we can prove that: T(k) = 7(2 k )-5 for all k  0 Step One (Prove for the base case): For n = 0, we already know that T(0) = 2, and 7(2 0 )-5 also evaluates to 2. Step One (Prove for the base case): For n = 0, we already know that T(0) = 2, and 7(2 0 )-5 also evaluates to 2. Step Two (Assume for some general case): Assume for n = k: T(k) = 7(2 k )-5. Step Two (Assume for some general case): Assume for n = k: T(k) = 7(2 k )-5. Step Three (Prove for the next case): We need to prove it for n = k + 1: We know that T(k+1) = 5 + 2T(k), and we’re assuming that T(k) = 7(2 k )-5, so we can conclude that T(k+1) = 5 + 2(7(2 k )-5) = 7(2 k+1 )-5, which is what we wanted. Step Three (Prove for the next case): We need to prove it for n = k + 1: We know that T(k+1) = 5 + 2T(k), and we’re assuming that T(k) = 7(2 k )-5, so we can conclude that T(k+1) = 5 + 2(7(2 k )-5) = 7(2 k+1 )-5, which is what we wanted.

17. CS 240 34 int powerOf2(const int &n) { if (n == 0) return 1; return 2*powerOf2(n-1); } int powerOf2(const int &n) { if (n == 0) return 1; return 2*powerOf2(n-1); } Let’s try the same approach on this alternate form of the function. Using the same assumptions as before, we get the following recurrence relation: T(0) = 2 T(k) = 4 + T(k-1) for all k > 0 This time, however, mathematical induction tells us that: T(k) = 4k+2 for all k  0