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Examples of Recursion

Instructor: Mainak Chaudhurimainakc@cse.iitk.ac.in

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Sum of natural numbersclass SumOfNaturalNumbers { public static void main (String arg[]) { int m = 3, n = 10; if (m > n) { System.out.println (“Invalid input!”); } else { System.out.println (“Sum of numbers

from ” + m + “ to ” + n + “ is ” + Sum(m, n));

} }

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Sum of natural numbers public static int Sum (int m, int n) { int x, y; if (m==n) return m; x = Sum(m, (m+n)/2); y = Sum((m+n)/2+1, n); return (x+y); }} // end class

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GCD revisited• Recall that gcd (a, b) = gcd (a-b, b)

assuming a > b.– Directly defines a recurrence

public static int gcd (int a, int b) { if ((a==1) || (b==1)) return 1; if (a==b) return a; if (a < b) return gcd (a, b-a); if (a > b) return gcd (a-b, b);}

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GCD revisited• Refinement: gcd (a, b) = gcd (a-nb, b)

for any positive integer n such that a >= nb, in particular n = a/b, assuming a > b

public static int gcd (int a, int b) { if (0==a) return b; if (0==b) return a; if (a < b) return gcd (a, b%a); if (a > b) return gcd (a%b, b);}

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Towers of Hanoi• Three pegs, one with n disks of

decreasing diameter; two other pegs are empty

• Task: move all disks to the third peg under the following constraints– Can move only the topmost disk from one

peg to another in one step– Cannot place a smaller disk below a

larger one• An example where recursion is much

easier to formulate than a loop-based solution

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Towers of Hanoi• We want to write a recursive method

THanoi (n, 1, 2, 3) which moves n disks from peg 1 to peg 3 using peg 2 for intermediate transfers

• The first step is to formulate the algorithm– Observation: THanoi (n, 1, 2, 3) Ξ THanoi

(n-1, 1, 3, 2) followed by transferring the largest disk to peg 3 from peg 1 and calling THanoi (n-1, 2, 1, 3)

– Stopping condition: n = 1

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Towers of Hanoiclass TowersOfHanoi { public static void main (String arg[]) { int n = 10; THanoi(n, 1, 2, 3); }

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Towers of Hanoi public static void THanoi (int n, int source, int

extra, int destination) { if (n > 1) { THanoi (n-1, source, destination, extra); } System.out.println (“Move disk ” + n + “

from peg ” + source + “ to peg ” + destination);

if (n > 1) { THanoi (n-1, extra, source, destination); } } // How many moves needed? 2n-1} // end class

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Towers of Hanoi• Total number of method calls

Let Tn be the number of method calls to solve for n disks

Tn = 2Tn-1 + 1 for n > 1; T1 = 1

– Use generating function to solve the recurrence (worked out in class on board)

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Fibonacci series• Number of method calls

– Refer to the program in the last lecture

– Let Tn be the number of method calls to find the nth Fibonacci number

Tn = Tn-1 + Tn-2 + 1 for n > 2; T1 = T2 = 1

– Use generating function to solve the recurrence

– Observe that Tn = Fn – 1 where Fn is the nth Fibonacci number

Tn = (21-n/√5)((1+√5)n – (1-√5)n) – 1

– The number (1+√5)/2 is called the golden ratio, which is lim n ∞￫ (Fn+1/Fn)