Recursion continued Exercise solution // Returns base ^ exponent. - PowerPoint PPT Presentation

Recursion continued

Exercise solution // Returns base ^ exponent. // Precondition: exponent >= 0 public static int pow(int x, int n) { if (n == 0) { // base case; any number to 0th power is 1 return 1; } else { // recursive case: x^n = x * x^(n-1) return x * pow(x, n-1); } }

How recursion works  Each call sets up a new instance of all the parameters and the local variables  When the method completes, control returns to the method that invoked it (which might be another invocation of the same method) pow(4, 3) = 4 * pow(4, 2) = 4 * 4 * pow(4, 1) = 4 * 4 * 4 * pow(4, 0) = 4 * 4 * 4 * 1 = 64 3

Infinite recursion  A method with a missing or badly written base case can causes infinite recursion public static int pow(int x, int y) { return x * pow(x, y - 1); // Oops! Forgot base case } pow(4, 3) = 4 * pow(4, 2) = 4 * 4 * pow(4, 1) = 4 * 4 * 4 * pow(4, 0) = 4 * 4 * 4 * 4 * pow(4, -1) = 4 * 4 * 4 * 4 * 4 * pow(4, -2) = ... crashes: Stack Overflow Error! 4

An optimization  Notice the following mathematical property: 3 12 = (3 2 ) 6 = (9) 6 = (81) 3 = 81*(81) 2  How does this "trick" work?  Do you recognize it?  How can we incorporate this optimization into our pow method?  What is the benefit of this trick?  Go write it.

Exercise solution 2 // Returns base ^ exponent. // Precondition: exponent >= 0 public static int pow(int base, int exponent) { if (exponent == 0) { // base case; any number to 0th power is 1 return 1; } else if (exponent % 2 == 0) { // recursive case 1: x^y = (x^2)^(y/2) return pow(base * base, exponent / 2); } else { // recursive case 2: x^y = x * x^(y-1) return base * pow(base, exponent - 1); } }

Activation records  activation record : memory that Java allocates to store information about each running method  return point ("RP"), argument values, local variable values  Java stacks up the records as methods are called; a method's activation record exists until it returns  Eclipse debug draws the act. records and helps us trace the behavior of a recursive method _ | x = [ 4 ] n = [ 0 ] | pow(4, 0) | RP = [pow(4,1)] | | x = [ 4 ] n = [ 1 ] | pow(4, 1) | RP = [pow(4,2)] | | x = [ 4 ] n = [ 2 ] | pow(4, 2) | RP = [pow(4,3)] | | x = [ 4 ] n = [ 3 ] | pow(4, 3) | RP = [main] | | | main 7

Factorial  What is the formula for the factorial of a number?  How would you write that recursively?  What is the base case?  What is the recursive case?

Dictionary lookup  Suppose you’re looking up a word in the dictionary (paper one, not online!)  You probably won’t scan linearly thru the pages – inefficient.  What would be your strategy?

Binary search binarySearch(dictionary, word){ if (dictionary has one page) { // base case scan the page for word } else { // recursive case open the dictionary to a point near the middle determine which half of the dictionary contains word if (word is in first half of the dictionary) { binarySearch(first half of dictionary, word) } else { binarySearch(second half of dictionary, word) } }

Binary search  Write a method binarySearch that accepts a sorted array of integers and a target integer and returns the index of an occurrence of that value in the array.  If the target value is not found, return -1 index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 value -4 2 7 10 15 20 22 25 30 36 42 50 56 68 85 92 103 int index = binarySearch(data, 42); // 10 int index2 = binarySearch(data, 66); // -1

Fibonacci’s Rabbits  Suppose a newly-born pair of rabbits, one male, one female, are put on an island. A pair of rabbits doesn’t breed until 2 months  old. Thereafter each pair produces another pair  each month Rabbits never die.   How many pairs will there be after n months? image from: http://www.jimloy.com/algebra/fibo.htm 12

Do some cases, see a pattern? m0: 1 young 1 m1: 1 mature 1 m2: 1 mature 1 young 2 m3: 2 mature 1 young 3 m4: 3 mature 2 young 5 m5: 5 mature 3 young 8 m6?

The pattern... m0: 1 young 1 m1: 1 mature 1 m2: 1 mature 1 young 2 m3: 2 mature 1 young 3 m4: 3 mature 2 young 5 m n = m n-1 (rabbits never die) + m n-2 (newborn pairs) How fast does this rabbit population grow?

Fibonacci numbers  The Fibonacci numbers are a sequence of numbers F 0 , F 1 , ... F n defined by: F 0 = F 1 = 1 F i = F i -1 + F i -2 for any i > 1  Write a method that, when given an integer i , computes the nth Fibonacci number.

Fibonacci numbers  recursive Fibonacci was expensive because it made many, many recursive calls  fibonacci(n) recomputed fibonacci(n-1, ... ,1) many times in finding its answer!  this is a common case of "overlapping subproblems ”, where the subtasks handled by the recursion are redundant with each other and get recomputed 16

Fibonacci code  Let's run it for n = 1,2,3,... 10, ... , 20,...  What happens if n = 5, 6, 7, 8, ...  Every time n increments with 2, the call tree more than doubles.. F5 F4 F3 F2 F1 F2 F3 F0 F1 F1 F2 F1 F0 F0 F1

Growth of rabbit population 1 1 2 3 5 8 13 21 34 ... every 2 months the population at least DOUBLES

Recursive Algorithms Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one. 19

Try to find the pattern by cases  One disk is easy  Two disks...  Three disks...  Four disk...

Recursive Algorithms Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one. 21

Recursive Algorithms Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one. 22

Recursive Algorithms Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one. 23

Recursive Algorithms Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one. 24

Parade  A parade consists of a set of bands and floats in a single line.  To keep from drowning each other out, bands cannot be placed next to another band  Given the parade is of length n, how many ways can it be organized

Counting ways  Let P(n) = the number of ways the parade can be organized.  Parades can either end in a band or a float  Let F(n) = the number of parades of length n ending in a float  Let B(n) = the number of parades of length n ending in a band  So:  P(n) = F(n) + B(n)

Recursive case  Consider F(n)  Since a float can be placed at next to anything, the number of parades ending in a float is equal to  F(n) = P(n-1)  Consider B(n)  The only way a band can end a parade is if the next to last unit is a float.  B(n) = F(n-1)  By substitution, B(n) = P(n-2)  So:  P(n) = P(n-1) + P(n-2)

Base case  How many parades configs can there be for:  n=1  2 – float or band  How many parade configs can there be for :  n=2  3  Float/float  Band/float  Float/band

Spock’s dilemma  We’ve reached the end of the 5 year mission, just a few days to go  The Enterprise enters a new solar system – which has n planets  Unfortunately, we only have time to visit k of those planets  How many different choices are there?

The algorithm  Let c(n,k) = the number of choices  Let us consider a planet X  C(n,k) = the number of choices that include choosing planet X + the number of choices that do not include planet X

The recurring cases  Including planet X  C(n-1, k-1)  Not including planet X  C(n-1, k)  So:  C(n,k) = C(n-1,k-1) + C(n-1, k)

The base cases  If k=0  We have chosen all the planets we can choose  What’s left is a single case  If k = n  We must choose all the remaining planets  What’s left is a single case  So: If (k == 0) || (k == n) return 1;