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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
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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) =
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3
4 *
4 *
4 *
4 *
64
pow(4, 2)
4 * pow(4, 1)
4 * 4 * pow(4, 0)
4 * 4 * 1
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) =
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4
4 *
4 *
4 *
4 *
4 *
...
pow(4, 2)
4 * pow(4, 1)
4 * 4 * pow(4, 0)
4 * 4 * 4 * pow(4, -1)
4 * 4 * 4 * 4 * pow(4, -2)
crashes: Stack Overflow Error!
An optimization
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Notice the following mathematical property:
312 = (32)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?
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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
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activation record: memory that Java allocates
to store information about each running method
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x
RP
x
RP
x
RP
x
RP
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
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[ 4 ]
[pow(4,1)]
[ 4 ]
[pow(4,2)]
[ 4 ]
[pow(4,3)]
[ 4 ]
[main]
n = [ 0 ]
n = [ 1 ]
n = [ 2 ]
n = [ 3 ]
_
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pow(4, 0)
pow(4, 1)
pow(4, 2)
pow(4, 3)
main
Factorial
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What is the formula for the factorial of a
number?
How would you write that recursively?
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What is the base case?
What is the recursive case?
Dictionary lookup
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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
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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.
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If the target value is not found, return -1
index 0
1
value -4 2
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16
7 10 15 20 22 25 30 36 42 50 56 68 85 92 103
int index = binarySearch(data, 42);
int index2 = binarySearch(data, 66);
// 10
// -1
Fibonacci’s Rabbits
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Suppose a newly-born pair of
rabbits, one male, one female, are
put on an island.
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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
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Do some cases, see a pattern?
m0: 1 young
m1: 1 mature
m2: 1 mature
m3: 2 mature
m4: 3 mature
m5: 5 mature
m6?
1 young
1 young
2 young
3 young
1
1
2
3
5
8
The pattern...
m0:
m1:
m2:
m3:
m4:
1 young
1 mature
1 mature 1 young
2 mature 1 young
3 mature 2 young
1
1
2
3
5
mn = mn-1 (rabbits never die) +
mn-2 (newborn pairs)
How fast does this rabbit population grow?
Fibonacci numbers
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The Fibonacci numbers are a sequence of
numbers F0, F1, ... Fn defined by:
F0 = F 1 = 1
Fi = Fi-1 + Fi-2 for any i > 1
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Write a method that, when given an integer i,
computes the nth Fibonacci number.
Fibonacci numbers
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recursive Fibonacci was expensive because
it made many, many recursive calls
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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
Fibonacci code
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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
F1
F1 F1
F0
F2
F2
F3
F2
F3
F0 F1
F1
F0
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.
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Try to find the pattern by cases
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One disk is easy
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Two disks...
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Three disks...
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Four disk...
Recursive Algorithms
Example: Tower of Hanoi, move all disks to third peg without
ever placing a larger disk on a smaller one.
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Recursive Algorithms
Example: Tower of Hanoi, move all disks to third peg without
ever placing a larger disk on a smaller one.
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Recursive Algorithms
Example: Tower of Hanoi, move all disks to third peg without
ever placing a larger disk on a smaller one.
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Recursive Algorithms
Example: Tower of Hanoi, move all disks to third peg without
ever placing a larger disk on a smaller one.
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Parade
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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
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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:
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P(n) = F(n) + B(n)
Recursive case
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Consider F(n)
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Since a float can be placed at next to anything,
the number of parades ending in a float is equal to
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Consider B(n)
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The only way a band can end a parade is if the
next to last unit is a float.
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F(n) = P(n-1)
B(n) = F(n-1)
By substitution, B(n) = P(n-2)
So:
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P(n) = P(n-1) + P(n-2)
Base case
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How many parades configs can there be for:
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n=1
2 – float or band
How many parade configs can there be for :
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n=2
3
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Float/float
Band/float
Float/band
Spock’s dilemma
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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
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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
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Including planet X
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Not including planet X
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C(n-1, k-1)
C(n-1, k)
So:
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C(n,k) = C(n-1,k-1) + C(n-1, k)
The base cases
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If k=0
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If k = n
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We have chosen all the planets we can choose
What’s left is a single case
We must choose all the remaining planets
What’s left is a single case
So:
If (k == 0) || (k == n)
return 1;