1. Lecture 8. How to Form Recursive relations 1

2. Recap Asymptotic analysis helps to highlight the order of growth of functions to compare algorithms Common asymptotic notation are of O, o, ,  and . Definitions of each notation uses the two constants k (or n0 ) and c. Constant k refer to the point where two function shows same value or fro where growth of one function start. While constant c refer to the value of multiple time growth of f(n) with respect to g(n). 2

3. Steps to solve and analysed problem (Without using Recursive approach) Step-1:- Take and understand problem. – Such as find the sum of elements of an array of size N Step-2:- Write algorithm or Pseudo code – Use C/C++ syntax to write a pseudo code (already discused) Step-3:- Analyse algorithm or pseudo code (Three ways) – Consider execution time (based on computer, not preferred) – Consider total steps (based on programmer style, not preferred) – Make a function f(n) with respect to input size n (Preferred, for this type, follow step 4, 5 and 6) Step-4:- Assign run time cost to each step of algorithm/pseudo code – c1, c2,... Cn to all steps 3

4. Steps to solve and analysed problem (Without using Recursive approach) Step-5:- Make a general function f(n) by considering input size n and run time cost of algorithm. – Such as f(n) = c1+c2*N + c3 etc Step-6:- Calculate rate/order of growth of f(n) (i.e. Use Big O because it refer to worst case) – Such as O(f(n)) = O(c1 +c2*N + c3 ) = O(N) – Apply rules of order of growth which are Remove constant, low order terms and coefficient with high order term) Step-7:- if you have two algorithms or pseudo codes then use asymptotic definition O, ,Θ for worst, best and average case analysis. – Asymptotic definition will help you show the rate of growth of a function (f(n))with respect to another function g(n) 4

5. 5 Recursive Algorithms Recursive algorithms solve the problem by solving smaller versions of the problem – If the smaller versions are only a little smaller, the algorithm can be called a reduce and conquer algorithm – If the smaller versions are about half the size of the original, the algorithm can be called a divide and conquer algorithm

6. Analyzing Recursive Algorithms Remember the difference between recursive algorithms and iterative algorithms Iterative – uses a loop to do the work Recursive – a function (method) calls itself - usually involves a base case - recursive call is on a “simpler” set of data 6

7. Analyzing Recursive Algorithms Iterative: Int power (int a, int p) {result = 1; For(i=1;i<=p;i++) result = result * a; Return result} 7

8. Analyzing Recursive Algorithms Recursive: int power( int a, int p) calling point:- Res=power(3,2) { If(p == 1) return a Else return a*power(a, p-1); } 8

9. Analyzing Recursive Algorithms  Check to see if the problem is small enough to solve directly  If not, then divide the data into smaller problems  Call the function on the smaller sets of data and form partial solutions  Combine the partial solutions to form the final solution 9

10. Analyzing Recursive Algorithms A recursive function to find the factorial of a number void Factorial(N) calling point:- Factorial(4) if N == 1 then return 1; else {smaller = N-1; answer = Factorial(smaller) return (N * answer) } 10

11. The Recursion Pattern Recursion: when a method calls itself Classic example--the factorial function: – n! = 1· 2· 3· ··· · (n-1)· n Recursive definition: As a C/C++ method: // recursive factorial function int recursiveFactorial(int n) { if (n == 0) return 1;// basis case else return n * recursiveFactorial(n- 1);// recursive case } 11

12. Visualizing Recursion Recursion trace A box for each recursive call An arrow from each calling point to called point An arrow from each called point to calling point showing return value Example recursion trace: recursiveFactorial(4) (3) (2) (1) (0) return1 call return1*1=1 2*1=2 3*2=6 4*6=24 final answer call 12

13. Recurrence Relations Two types:  Only a few simple cases: T(1) = 40 T(2) = 40 T(n) = 2T(n-2)-15 2. Several simple cases: T(n) = 4 if n<=4 T(n) = 4T(n/2) – 1 if n>4 13

14. 14 Recursive Algorithm Analysis The analysis depends on – the preparation work to divide the input – the size of the smaller pieces – the number of recursive calls – the concluding work to combine the results of the recursive calls

15. Content of a Recursive function/Method Base case(s). – Values of the input variables for which we perform no recursive calls are called base cases (there should be at least one base case). – Every possible chain of recursive calls must eventually reach a base case. Recursive calls. – Calls to the current method. – Each recursive call should be defined so that it makes progress towards a base case. 15

16. What is a recurrence relation? A recurrence relation, T(n), is a recursive function of an integer variable n. Like all recursive functions, it has one or more recursive cases and one or more base cases. Example: The portion of the definition that does not contain T is called the base case of the recurrence relation; the portion that contains T is called the recurrent or recursive case. 16

17. What is a recurrence relation? (Cont !!!) Recurrence relations are useful for expressing the running times (i.e., the umber of basic operations executed) of recursive algorithms The specific values of the constants such as a, b, and c (in the previous slide) are important in determining the exact solution to the recurrence. Often however we are only concerned with finding an asymptotic upper bound on the solution. We call such a bound an asymptotic solution to the recurrence. 17

18. Forming Recurrence Relations For a given recursive method, the base case and the recursive case of its recurrence relation correspond directly to the base case and the recursive case of the method. Example 1: Write the recurrence relation for the following method: The base case is reached when n = = 0. The method performs one comparison. Thus, the number of operations when n = = 0, T(0), is some constant a. public void f (int n) { if (n > 0) { cout<<n; f(n-1); } } 18

19. Forming Recurrence Relations When n > 0, the method performs two basic operations and then calls itself, using ONE recursive call, with a parameter n – 1. Therefore the recurrence relation is: T(0) = a for some constant a T(n) = b + T(n – 1) for some constant b In General, T(n) is usually a sum of various choices of T(m ), the cost of the recursive subproblems, plus the cost of the work done outside the recursive calls: T(n ) = aT(f(n)) + bT(g(n)) +... + c(n) where a and b are the number of subproblems, f(n) and g(n) are subproblem sizes, and c(n) is the cost of the work done outside the recursive calls [Note: c(n) may be a constant] 19

20. Example 2: Write the recurrence relation for the following method: The base case is reached when n == 1. The method performs one comparison and one return statement. Therefore, T(1), is some constant c. public int g(int n) { if (n == 1) return 2; else return 3 * g(n / 2) + g( n / 2) + 5; } Forming Recurrence Relations (Cont !!!) 20

21. Forming Recurrence Relations (Cont !!!) When n > 1, the method performs TWO recursive calls, each with the parameter n / 2, and some constant # of basic operations. Hence, the recurrence relation is: T(1) = c for some constant c T(n) = b + 2T(n / 2) for some constant b 21

22. Example 3: Write the recurrence relation for the following method: The base case is reached when n == 1 or n == 2. The method performs one comparison and one return statement. Therefore each of T(1) and T(2) is some constant c. long fibonacci (int n) { // Recursively calculates Fibonacci number if( n == 1 || n == 2) return 1; else return fibonacci(n – 1) + fibonacci(n – 2); } Forming Recurrence Relations (Cont !!!) 22

23. When n > 2, the method performs TWO recursive calls, one with the parameter n - 1, another with parameter n – 2, and some constant # of basic operations. Hence, the recurrence relation is: T(n) = c if n = 1 or n = 2 T(n) = T(n – 1) + T(n – 2) + b if n > 2 Forming Recurrence Relations (Cont !!!) 23

24. Example 4: Write the recurrence relation for the following method: The base case is reached when n == 0 or n == 1. The method performs one comparison and one return statement. ThereforeT(0) and T(1) is some constant c. long power (long x, long n) { if(n == 0) return 1; else if(n == 1) return x; else if ((n % 2) == 0) return power (x, n/2) * power (x, n/2); else return x * power (x, n/2) * power (x, n/2); } Forming Recurrence Relations (Cont !!!) 24

25. At every step the problem size reduces to half the size. When the power is an odd number, an additional multiplication is involved. To work out time complexity, let us consider the worst case, that is we assume that at every step an additional multiplication is needed. Thus total number of operations T(n) will reduce to number of operations for n/2, that is T(n/2) with seven additional basic operations (the odd power case) Hence, the recurrence relation is: T(n) = c if n = 0 or n = 1 T(n) = 2T(n /2) + b if n > 2 Forming Recurrence Relations (Cont !!!) 25

26. Summary We have to use seven steps from taking a problem scenario and to analyzing its run time complexity. These steps are for those problems which are attempted without recursive approach. A recursive function or method s that method which called itself from within body until its base condition fulfill. There are two components of a recursive method i.e. base and recursive part. A recursive relation is form by considering base condition and total recursive calls with some constants 26

27. Summary Constants for recursive relation are a, b and c or others. – One constant refer to cost of base condition (i.e a) – One constant refer to total recursive calling (i.e. b) – One constant refer to cost of those steps which are performed outside the recursive calls. Always consider worst case analysis for recursive relations. 27

28. In Next Lecturer In next lecture, We will discuss the arithmetic and geometric series and its sum and also mathematical induction. 28