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## RECURSION

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**RECURSION**• Self referential functions are called recursive (i.e. functions calling themselves) • Recursive functions are very useful for many mathematical operations**Factorial: Recursive Functions**• 1! = 1 • 2! = 2*1 = 2*1! So: n! = n*(n-1)! For N>1 • 3! = 3*2*1 = 3*2! 1! = 1 • 4! = 4*3*2*1 = 4 * 3! • Properties of recursive functions: • 1) what is the first case (terminal condition) • 2) how is the nth case related to the (n-1)th case Or more general: how is nth case related to <nth case**Recursive procedure**• A recursive task is one that calls itself. With each invocation, the problem is reduced to a smaller task (reducing case) until the task arrives at a terminal or base case, which stops the process. The condition that must be true to achieve the terminal case is the terminal condition.**#include <iostream.h>**long factorial(int); // function prototype int main() { int n; long result; cout << "Enter a number: "; cin >> n; result = factorial(n); cout << "\nThe factorial of " << n << " is " << result << endl; return 0; } long factorial(int n) { if (n == 1) return n; else return n * factorial(n-1); } Terminal condition to end recursion**long factorial(int n)**{ if (n == 1) return n; else /* L#6 */ return n * factorial(n-1); } n=1 Return value Addr of calling statement L#6: return 2 * factorial(1) n=2 Return value Addr of calling statement L#6: return 3 * factorial(2) n=3 Reserved for return value Main program: result = factorial(3) Addr of calling statement 1 2*1 3*2 ********Iterative solution for factorial:**int factorial (int n) { int fact; for (fact = 1;n>0; n--) fact=fact*n; return fact } Recursive functions can always be “unwound” and written iteratively**Example 2: Power function**• xn = 1 if n = 0 • = x * xn-1if n>0 • 53 = 5 * 52 • 52 = 5 * 51 • 51 = 5 * 50 • 50 = 1**Return 1**float power(float x, int n) { if (n == 0) return 1; else return x * power (x, n-1); } X=5, n=0 5 * power(5,0) X=5, n=1 5 * power(5,1) X=5, n=2 5 * power(5,2) X=5, n=3 Power(5,3) ********Power function xn**• Thus for n= 3, the recursive function is called 4 times, with n=3, n=2, n=1, and n=0 • in general for a power of n, the function is • called n+1 times. • Can we do better?**x7 = x (x3)2**int (7/2) = 3 We know that : x14 = (x7)2in other words: (x7 * x7) x3 = x (x1)2 xn = 1if n= 0 = x(xn/2)2if n is odd = (xn/2)2if n is even x1 = x (x0)2 x0 = 1 Floor of n/2**float power (float x, int n)**{ float HalfPower; //terminal condition if (n == 0) return 1; //can also stop at n=1 //if n Is odd if (n%2 == 1) { HalfPower = power(x,n/2); return (x * HalfPower *HalfPower): } else { HalfPower = power(x,n/2); return (HalfPower * HalfPower); } ******if (n == 0) return 1;**if (n%2 == 1) { HalfPower = power(x,n/2); return (x * HalfPower *HalfPower): } else { HalfPower = power(x,n/2); return (HalfPower * HalfPower); } Can also use the call: return power(x,n/2) * power (x,n/2) But that is much less efficient!**Recursive Power Function: divide & conquer**• How many calls to the power function does the second algorithm make for xn? Log2(n)!!!! ****Parts of a Recursive Function:**recursive_function (….N….) { //terminal condition if (n == terminal_condition) return XXX; else { … recursive_function(….<N….); } 1. 2.**int main(void)**{ int z; z= f ( f(2) + f(5)); cout << z; } int f(int x) { int returnvalue; if (x==1) || (x== 3) return x; else return (x * f(x-1)) } Recursion Example What gets printed? (2 + 60)! / 2 ****Recursion: Example 4**Write a recursive boolean function to return True (1) if parameter x is a member of elements 0 through n of an array.**Int inarray(int a[], int n, int x)**{ if (n<0) return FALSE; else if (a[n] == x) return TRUE; else return inarray(a,n-1,x); }**What is a better name for this function? (i.e., what does it**do?) • int main() • { • int RandyJackson[] = {1,2,3,4,5,6,7,8,9,99}; • int *paula; • paula = RandyJackson; • AmericanIdol(paula); • return 0; • } • int AmericanIdol(int *simon) • { • if (*simon == 99) return 0; • AmericanIdol(simon+1); • cout <<*simon<<endl; • return 0; • }**What gets printed?**• int r(int); • r(840); • return 0; • } • int r(int n) • { • cout << n <<endl; • if (n <= 4242) • r(2*n); • cout << n << endl; • return n; • }**840**• 1680 • 3360 • 6720 • 6720 • 3360 • 1680 • 840**Algorithm Analysis & Complexity**We saw that a linear search used n comparisons in the worst case (for an array of size n) and binary search had logn comparisons. Similarly for the power function -- the first one took n multiplications, the second logn. Is one more efficient than the other? How do we quantify this measure?**Efficiency**• CPU (time) usage • memory usage • disk usage • network usage • 1.Performance: how much time/memory/disk/... is actually used when a program is run. This depends on the machine, compiler, etc. as well as the code. • 2.Complexity: how do the resource requirements of a program or algorithm scale, i.e., what happens as the size of the problem being solved gets larger. • Complexity affects performance but not the other way around.**The time required by a method is proportional to the number**of "basic operations" that it performs. Here are some examples of basic operations: • one arithmetic operation (e.g., +, *). • one assignment • one test (e.g., x == 0) • one read • one write (of a primitive type)**Constant Time vs. Input Size**• Some algorithms will take constant time -- the number of operations is independent of the input size. • Others perform a different number of operations depending upon the input size • For algorithm analysis we are not interested in the EXACT number of operations but how the number of operations relates to the problem size in the worst case.**Big Oh Notation**• The measure of the amount of work an algorithm performs of the space requirements of an implementation is referred to as the complexity or order of magnitude and is a function of the number of data items. We use big oh notation to quantify complexity, e.g. O(n), or O(logn)**Big Oh notation**• O notation is an approximate measure and is used to quantify the dominant term in a function. • For example, if f(n) = n3 + n2 + n + 5 • then f(n) = O(n3) • (since for very large n, the n3 term • dominates)**Big Oh notation**For n = 5, the values in the array get printed (25 gets printed). After each row a new line gets printed (5 of them) Total work = n2 +n = O(n2) • for (I = 0; I<n; I++) • { • for (j=0; j<n; j++) • cout << a[I][j] << “ “; • cout << endl; • } For n = 1000, a[I][j] gets printed 1000000 times, endl only 1000 times.**Big Oh Definition**• Function f is of complexity or order at mostg, written with big-oh notation as f = O(g), if there exists a positive constant c and a positive integer n0such that • |f(n)| <= c|g(n)| for all n>n0 • We also say that f has complexity O(g)**|f(n)| <= c|g(n)| for all n>n0**Let f(n) = n2 + 5 Let g(n) = n2 so is f(n) = O(g(n)) or O(n2)? Yes, since there exists a constant c and a positive integer n0 to make the above statement true. For example, if c=2, n0 = 3 n2 + 5 <= 2n2for all n>3 This statement is always true for n>3**F(N) = 3 * N2 + 5. We can show that F(N) is O(N2) by**choosing c = 4 and n0 = 2. This is because for all values of N greater than 2: 3 * N2 + 5 <= 4 * N2 F(N) != O(N) because one can always find a value of N greater than any n0 so that 3 * N2 + 5 is greater than c*N. I.e. even if c = 1000, if N== 1M 3 * N2 + 5 > 1000 * N N>n0**Running**time O(n2) O(n) N Constants can make o(n) perform worse for low values**Time n=1 n=2 n=4 n=8 n=16 n=32**1 1 1 1 1 1 1 logn 0 1 2 3 4 5 n 1 2 4 8 16 32 nln 0 2 8 24 64 160 n2 1 4 16 64 256 1024 n3 1 8 64 512 4096 32768 2^n 2 4 16 256 65536 4294967296 n! 1 2 24 40326 2.1x1013 2613x1033**Determining Complexity in a Program:**1. Sequence of statements: statement 1; statement 2; ... statement k; total time = time(statemnt 1) + time(statemnt 2) + ...time(statemnt k) 2. If-then-else statements: total time = max(time(sequence 1),time(sequence 2)). For example, if sequence 1 is O(N) and sequence 2 is O(1) the worst-case time for the whole if-then-else statement would be O(N). 3. Loops for (i = 0; i < N; i++) { sequence of statements } The loop executes N times, so the sequence of statements also executes N times. Since we assume the statements are O(1), the total time for the for loop is N * O(1), which is O(N) overall.**Nested loops**for (i = 0; i < N; i++) { for (j = 0; j < M; j++) { sequence of statements } } The outer loop executes N times. Every time the outer loop executes, the inner loop executes M times. As a result, the statements in the inner loop execute a total of N * M times. Thus, the complexity is O(N * M). In a common special case where the stopping condition of the inner loop is j < N instead of j < M (i.e., the inner loop also executes N times), the total complexity for the two loops is O(N2).**Determining Complexity**• look for some clues and do some deduction to • arrive at the answer. Some obvious things— • Break the algorithm down into steps and analyze the complexity of each. For example, analyze the body of a loop first and then see how many times that loop is executed. • Look for for loops. These are the easiest statements to analyze! They give a clear upper bound, so they’re usually dead giveaways.—sometimes other things are going on in the loop which change the behavior of the algorithms. • Look for loops that operate over an entire data structure. If you know the size of the data structure, then you have some ideas about the running time of the loop. • Loops, loops. Algorithms are usually nothing but loops, so it is imperative to be able to analyze a loop!**General Rules for determining O**1. Ignoring constant factors: O(c f(N)) = O(f(N)), where c is a constant; e.g. O(20 N3) = O(N3) 2. Ignoring smaller terms: If a<b then O(a+b) = O(b), for example O(N2+N) = O(N2) 3. Upper bound only: If a<b then an O(a) algorithm is also an O(b) algorithm. For example, an O(N) algorithm is also an O(N2) algorithm (but not vice versa). 4. N and log N are "bigger" than any constant, from an asymptotic view (that means for large enough N). So if k is a constant, an O(N + k) algorithm is also O(N), by ignoring smaller terms. Similarly, an O(log N + k) algorithm is also O(log N). 5. Another consequence of the last item is that an O(N logN+N) algorithm, which is O(N(log N + 1)), can be simplified to O(NlogN).**Bubble sort -- analysis**• void bubble_sort(int array[ ], int length) • { • int j, k, flag=1, temp; • for(j=1; j<=length && flag; j++) • { flag=0; • for(k=0; k < (length-j); k++) • { if (array[k+1] > array[k]) • { • temp=array[k+1]; • array[k+1]= array[k]; • array[k]=temp; • flag=1; • } } } } N(N-1) = O(N2)**Review of Log properties**• logb x = p if and only if bp = x (definition) • logb x*y = logb x + logb y • logb x/y = logb x - logb y • logb xp = p logb x which implies that (xp)q = x(pq) • logb x = loga x * logb a log to the base b and the log to the base a are related by a constant factor. Therefore, O(N logb N), is the same as O(N loga N) because the big-O bound hides the constant factor between the logs. The base is usually left out of big-O bounds, I.e. O(N log N).**// this function returns the location of key in the list**// a -1 is returned if the value is not found int binarySearch(int list[], int size, int key) { int left, right, midpt; left = 0; right = size - 1; while (left <= right) { midpt = (int) ((left + right) / 2); if (key == list[midpt]) { return midpt; } else if (key > list[midpt]) left = midpt + 1; else right = midpt - 1; } O(logn)**When do constants matter?**When the problem size is “small” N 100*N N2/100 102 104 102 103 105 104 104 106 106 105 107 108 107 109 1012**Running Time**• Also interested in Best Case and Average Case Mission critical -- worst case important Merely inconvenient -- may be able to get away with Avg/Best case Avg case must consider all possible inputs