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 =
// 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); } }
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Each call sets up a new instance of all the
When the method completes, control returns to
pow(4, 3) = 4 * pow(4, 2) = 4 * 4 * pow(4, 1) = 4 * 4 * 4 * pow(4, 0) = 4 * 4 * 4 * 1 = 64
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A method with a missing or badly written base
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!
Notice the following mathematical property:
= (9)6
= (81)3 = 81*(81)2
How does this "trick" work? Do you recognize it? How can we incorporate this optimization into our
What is the benefit of this trick? Go write it.
// 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); } }
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activation record: memory that Java allocates
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
What is the formula for the factorial of a
How would you write that recursively?
What is the base case? What is the recursive case?
Suppose you’re looking up a word in the
You probably won’t scan linearly thru the
What would be your strategy?
binarySearch(dictionary, word){ if (dictionary has one page) {// base case scan the page for word } else {// recursive case
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) } }
Write a method binarySearch that accepts
If the target value is not found, return -1
int index = binarySearch(data, 42); // 10 int index2 = binarySearch(data, 66); // -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
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
Thereafter each pair produces another pair each month
Rabbits never die.
How many pairs will there be after n
months?
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image from: http://www.jimloy.com/algebra/fibo.htm
The Fibonacci numbers are a sequence of
Write a method that, when given an integer i,
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recursive Fibonacci was expensive because
fibonacci(n) recomputed fibonacci(n-1, ... ,1) many
this is a common case of "overlapping
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 F3 F2 F0 F1 F4 F1 F3 F2 F0 F1 F1 F2 F0 F1
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Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one.
One disk is easy Two disks... Three disks... Four disk...
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Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one.
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Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one.
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Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one.
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Example: Tower of Hanoi, move all disks to third peg without ever placing a larger disk on a smaller one.
A parade consists of a set of bands and floats
To keep from drowning each other out, bands
Given the parade is of length n, how many
Let P(n) = the number of ways the parade
Parades can either end in a band or a float Let F(n) = the number of parades of length n
Let B(n) = the number of parades of length n
So:
P(n) = F(n) + B(n)
Consider F(n)
Since a float can be placed at next to anything,
F(n) = P(n-1)
Consider B(n)
The only way a band can end a parade is if the
B(n) = F(n-1) By substitution, B(n) = P(n-2)
So:
P(n) = P(n-1) + P(n-2)
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
We’ve reached the end of the 5 year mission,
The Enterprise enters a new solar system –
Unfortunately, we only have time to visit k of
How many different choices are there?
Let c(n,k) = the number of choices Let us consider a planet X C(n,k) = the number of choices that include
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)
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: