1. Sound Check

2. Some Final Words on Methods

3. RECALL:
Three major
paradigms for
programming

4. Java Methods Quick Review { System.out.println(“Hello World!”);
There exists in Java a single construct, the method, for both procedures and functions:
when a procedure is called for, specify the return type “void” before method name
public void printHelloWorld( ) {
System.out.println(“Hello World!”);
} // of printHelloWorld
Note: All methods must have parentheses for parameters even if no parameters!
Quick
Review

5. Java Methods Quick Review
Single construct for both procedures and functions:
when a function is called for, specify the
appropriate return type before method name
public float average (float fNum1, float fNum2, float fNum3) {
float fReturnVal;
fReturnVal =
(fNum1 + fNum2 + fNum3)/ 3;
return (fReturnVal);
} // of average
Quick
Review

6. Method Signatures Quick Review
“The signature of a method consists of the name of the method and the number and types of formal parameters to the method. A class may not declare two methods with the same signature, or a compile time error occurs.”
--Java Language Specification s.8.4.2
Method overloading occurs when identically named methods have different signatures.
public int getCube(int iNum){
return iNum*iNum*iNum; }
public int getCube(float fNum){
return (int)(fNum*fNum*fNum);
public int getCube(double dNum){
return (int) (dNum*dNum*dNum);
Quick
Review

7. Parameters Unlike Pseudocode, Java only has “in” parameters.
Thus no need to declare in, in/out or out ‘cause they’re all IN
Parameters do need to be declared just like variables:
public void demo(int i, float x)
More about parameters when we discuss objects

8. Where do Methods Go? Quick Review
Methods (and variables) are contained in Classes, as either class or instance members. More on creating classes and objects shortly . . .
Quick
Review
class zippy {
int pin;
public String yow() {
// some code }

9. Questions?

10. Iteration

11. Java Basics: Iteration Constructs
In Pseudocode, we had a single iteration construct, flexible enough to be used in all iteration contexts.
Actually (if you were paying attention last semester) we talked about three kinds of loops:
Sentinal Loops
Test-Last Loop
N-and-a-half Loops

12. Java Basics: Iteration Constructs
In Pseudocode, we had a single iteration construct, flexible enough to be used in all iteration contexts.
Java, like most programming languages, does not provide a single flexible construct.
Instead, Java offers three special case loop constructs, each good for a particular context.
Do not get accustomed to only one of them and try to use it for all situations.
To use them wisely, you should consider not only how they work, but also when each is best used…

13. Java Iteration Constructs: “While Loops”
When repeating steps, people naturally want to follow the pattern:
get a value, then process that value
The while loop construct calls for the unnatural pattern:
obtain the first loop control value before entering the loop itself;
then, within the loop body,
first do the process steps,
then do the get next steps

14. Unnatural? while(You haven’t gotten to my street) {
Keep going straight }
turn right

15. Java Iteration Constructs: “While Loops”
Pseudocode:
<get first value>
loop
exitif NOT(condition)
<process value>
<get next value>
endloop
Java example:
<get first value>
while (condition) {
<process value>
<get next value> }
Pseudocode flavor: Sentinal

16. Java Iteration Constructs: “Do While Loops”
Java example: do {
statement 1;
...
statement N;
} while (condition);
Pseudocode:
loop
statement 1
...
statement N
exitif (NOT(condition))
endloop
Pseudocode flavor: Test Last

17. Java Iteration Constructs: “For Loops”
Java syntax:
for (<initialization>; <continue if>;<increment>)
Java example:
int i;
for (i=0; i<10; i++) {
<some statements> }
Pseudocode:
i isoftype Num
i <- 0
loop
exitif (i =>10)
<some statements>
i <- i + 1
endloop

18. Secret! A for Loop can be exactly written as a while loop
while (i < 10) {
System.out.println(i);
i++; }
for(i = 0; i < 10; i++) {
System.out.println(i); }
This will help you understand the sequence of operations of a for loop

19. Java Iteration Constructs: “For Loops”
Common Problems with For Loops include:
--Spins on ending semicolon; code in braces executed once!
for (i=0; i<N; i++);
{ … }
--variable declared inside for loop signature; caution:
--the variable may be needed outside the for loop structure
for (int i=0; i<N; i++){…}

20. Potential Confusion public class Forex { static int i = 42;
public static void main(String args[])
for(int i=0; i < 5; i++)
System.out.println(i); }

21. Output 1 2 3 4 42

22. Java Iteration Constructs: When to Use
ASK:
Is it simply a count of the number of iterations?
Is it a value that the loop itself must compute?
Is it a value that already exists somewhere, and the loop only obtains it from elsewhere?
--The term “control variable” refers to the variable whose value is tested to determine if the loop should continue for another iteration or halt.
--For example, variable thisVar, below:
while (thisVar < = SOME_CONSTANT)
--To determine which loop construct is appropriate for a given situation, ask yourself “where does the control variable’s value come from?”

23. Java Iteration Constructs: When to Use
The for loop: used when the control variable is a simple count of the number of iterations,
e.g.: “create a loop that reads and processes the next 100 numbers.”
The while loop: used when the control variable has a value that already exists and is simply obtained by the loop.
e.g.: “create a loop that reads in numbers and processes them until it reads in a 100.”
The do-while loop: used when the control variable’s value must be calculated by the loop itself.
e.g.: “create a loop that reads in numbers until their sum is greater than 100.”

24. Which loop construct would you use if...
Java Iteration Constructs: Review
Which loop construct would you use if...
You need to perform a series of steps exactly N times?
You need to traverse a linked list of unknown size, and stop when you find a certain value?
You need to perform a series of steps at least once, and continue performing these steps for an an unknown number of times

25. Iteration Reiteration
Three kinds of iteration
for (...;...;...) {...} loops
Fixed number of iterations
while (...) {...} loops
Iteration condition evaluated in advance
do {...} while (...) loops
Iteration condition evaluated in the loop

26. Questions?

27. Arrays of Primitives The Idea: Same concepts you know from Pseudocode
A few differences in implementation
Java array declaration:
<elemType>[ ] <arrID> = new <elemType>[<size>];
e.g.: to declare an array of ten ints for storing numerical grades...
int[ ] iGradeArray = new int[10];
Array declaration error:
using parentheses, not brackets, e.g.,
int[ ] iGradeArray = new int(10);

28. Details
The odd looking syntax is because arrays (even arrays of primitives) are objects.
We'll explain in detail once we get to objects...
int[ ] iGradeArray = new int[10];
int iGradeArray[ ] = new int[10];

29. Arrays int[ ] iGradeArray = new int[10];
Example:
declare iGradeArray of 10 ints
initialize all 10 values to 0
int[ ] iGradeArray = new int[10];
int i; // when declaring and manipulating arrays,
// you may use the single-letter IDers
// i, j, k for indices due to convention
// (everybody knows what it is)
for (i=0; i < iGradeArray.length; i++) {
iGradeArray[i] = 0;
} // for loop
Great idea!
if you change
the array size,
you need only
change the
instantiation.
Notes:
Arrays know their own length
length is a field, not a method
Arrays are statically sized: you cannot change the length after declaration.
All arrays are objects, thus you must declare a reference, and instantiate it, and initialize it

30. Arrays int[ ] iGradeArray = new int[10]; int i;
More Notes:
Array indices begin at 0, not at 1
So, length is one greater than iMAX_INDEX
Thus, an error if you do:
Arrays
int[ ] iGradeArray = new int[10];
int i;
for (i=1; i <= iGradeArray.length; i++) {
iGradeArray[i] = 0;
} // for loop
Code above attempts to access elements 1..10
But... you have indices 0..9
So: it misses the 1st element (which is at index 0) it tries to go past 10th element (which is at index 9)

31. Questions?

32. Sample Quiz Problem
Given an array of ints, return the index of the largest element. Use the code below.
public int getLargestIndex(int[ ] iArray){
/* YOUR CODE GOES HERE */ }

33. Sample Quiz Problem
Given an array of ints, return the index of the largest element. Use the code below.
public int getLargestIndex(int[ ] iArray) {
int iLargest = -9999;
int iCounter;
for (iCounter = 0; iCounter < iArray.length;
iCounter++) {
if (iLargest < iArray[iCounter])
iLargest = iArray[iCounter]; }
return iLargest;
}// getLargestIndex
Does this work?

34. What if all the values are less than -9999
Sample Quiz Problem
Given an array of ints, return the index of the largest element. Use the code below.
public int getLargestIndex(int[ ] iArray) {
int iLargest = -9999;
int iCounter;
for (iCounter = 0; iCounter < iArray.length;
iCounter++) {
if (iLargest < iArray[iCounter])
iLargest = iArray[iCounter]; }
return iLargest;
}// getLargestIndex
What if all the values are less than -9999

35. Sample Quiz Problem
Given an array of ints, return the index of the largest element. Use the code below.
public int getLargestIndex(int[ ] iArray) {
int iLargest = iArray[0];
int iCounter;
for (iCounter = 0; iCounter < iArray.length;
iCounter++) {
if (iLargest < iArray[iCounter])
iLargest = iArray[iCounter]; }
return iLargest;
}// getLargestIndex
How about this?

36. Sample Quiz Problem
Given an array of ints, return the index of the largest element. Use the code below.
public int getLargestIndex(int[ ] iArray) {
int iLargest = iArray[0];
int iCounter;
for (iCounter = 1; iCounter < iArray.length;
iCounter++) {
if (iLargest < iArray[iCounter])
iLargest = iArray[iCounter]; }
return iLargest;
}// getLargestIndex
Why not start out
iCounter at 1, and not
0 like most for loops?

37. Now it should be perfect!
Sample Quiz Problem
Given an array of ints, return the index of the largest element. Use the code below.
public int getLargestIndex(int[ ] iArray) {
int iLargest = iArray[0];
int iCounter;
for (iCounter = 1; iCounter < iArray.length;
iCounter++) {
if (iLargest < iArray[iCounter])
iLargest = iArray[iCounter]; }
return iLargest;
}// getLargestIndex
Now it should be perfect!

38. READING THE DIRECTIONS
Sample Quiz Problem
Given an array of ints, return the index of the largest element. Use the code below.
NOTE THIS!
public int getLargestIndex(int[ ] iArray) {
int iLargest = iArray[0];
int iCounter;
for (iCounter = 1; iCounter < iArray.length;
iCounter++) {
if (iLargest < iArray[iCounter])
iLargest = iArray[iCounter]; }
return iLargest;
}// getLargestIndex
Test taking is about
READING THE DIRECTIONS

39. Sample Quiz Problem
What’s the solution? This is a good problem to work
out on your own. We don’t want the largest VALUE, we instead want the INDEX of the largest value.

40. Lessons Learned Read the question carefully.
Don’t fall into habits (e.g., all for loops must
start with the counter at 0).
Don’t make assumptions about input data (e.g.,
is NOT a good initial value).
Trace your code; watch for simple >, < errors.

41. Questions?

42. The Road Ahead Recursion Classes and Objects Constructors
Today
Recursion
Classes and Objects
Constructors
Object References
Parameters

43. A Question How many people here don’t like recursion? Why not?
A Promise: By the end of this lecture, you will say:
“Recursion Rocks My World!”

44. Recursion Let’s say you place two rabbits in a hutch.
What’s going to happen?

45. Two Months Later...
original rabbits
new rabbits
If it takes two months for rabbits to reach maturity, in two months you’ll have one productive pair and one (brand new) non-productive pair. (This assumes all rabbits live up to their reputation.)

46. The rabbits keep at it.
1 month old
0 months old
The next month, you get another pair . . .

47. Suppose 1 2 3 What if: The rabbits always had two offspring, always
male and female;
Rabbits always reached maturity in two months;
No rabbit dies.
How many pairs of rabbits do you have in a year? 1 2 3

48. Hare Raising Story KEY = one m/f pair Start End Month 1 End Month 2

49. Pairs of Rabbits 1 2 3 4 5 6 7 8 9 10 11 12
Month
Productive
Non-Productive
Total 13 21 34 55 89
144
See a
pattern
yet?

50. Let’s Take Another Example
Instead of rabbits, let’s use geometry.
Draw a square of size 1.
Rotating 90 degrees, add to it a square of size 1.
Rotating 90 degrees again, add a square of size 2.
Again, rotate and add a square of size 3, and so on.
Keep this up for the sequence we noted in the table:
1, 1, 2, 3, 5, 8, 13, 21, ,
What do you see?

51. 1

52. 1

53. 2

54. 3

55. 5

56. 8

57. 13

58. 21

59. 1 2 3 5 8 13 21

60. Does this look familiar?

61. It’s not just about rabbits.

62. The Truth
is Out There

63. See the pattern? It’s the Fibonacci Sequence.1 2 3 4 5 6 7 8 9 10 11 12
Month
Productive
Non-Productive
Total 13 21 34 55 89
144
It’s the Fibonacci
Sequence.
We used brute force to find the progression:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... ,
It turns out this pattern is repeated in many places: sea shells, sun flowers, pine cones, the stock market, bee hives, etc.

64. We’re not as smart as him. But that’s OK. We can code.
Writing the Formula
Given:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... ,
Can we write this as a formula for any number, n?
This guy could:
Jacques Binet
( ) 1+ 1- 5 2 - n
Fib(n) =
But let’s be honest.
We’re not as smart as him.
But that’s OK. We can code.

65. What If You Can’t Find the Formula?
Which one
would your
rather code
and debug?
For some
problems,
there might
not exist a
formula, and
recursion is
your only
option.
Suppose you didn’t know: 1+ 1- 5 2 - n
Fib(n) =
You could take Math 3012 or you could
instead manage with:
Fib(n) = Fib(n-1) + Fib(n-2),
Fib(0) = 0;
Fib(1) = 1;
(The value at any given place is the sum of the two prior values.)

66. Recursive Fibonacci We have our general rule:
Fib(n) = Fib(n-1) + Fib(n-2),
Fib(0) = 0;
Fib(1) = 1;
We can say a few things about it:
It’s defined recursively (duh).
It has a terminal condition (AHA!)
It can be determined and calculated (addition). 1 2 3

67. that gives and gets an int
Coding Fibonacci
public class FibTest {
public static int fib (int num) {
}// FibTest
What do we know to
start with? We know that
we need a method that
return the Fibonacci
value for a number at a
given position.
This suggests a method
that gives and gets an int

68. Coding Fibonacci What’s the FIRST thing we do with a recursive method?
public class FibTest {
public static int fib (int num) {
if (num == 0)
return 0;
}// FibTest
What’s the FIRST thing we do
with a recursive method?
We plan on how it will terminate!
We know one special case
for the Fibonacci sequence:
F(0) = 0

69. Coding Fibonacci We also know a second special case
public class FibTest {
public static int fib (int num) {
if (num == 0)
return 0;
else if (num == 1)
return 1;
}// FibTest
We also know a
second special case
that could terminate
our recursion:
F(1) = 1.

70. Coding Fibonacci The last part of our formula is merely:
public class FibTest {
public static int fib (int num) {
if (num == 0)
return 0;
else if (num == 1)
return 1;
else
return fib(num-1) + fib(num-2); }
}// FibTest
The last part of our
formula is merely:
F(n) = F(n-1) + F(n-2)

71. Is this safe? What if someone
Coding Fibonacci
public class FibTest {
public static int fib (int num) {
if (num == 0)
return 0;
else if (num == 1)
return 1;
else
return fib(num-1) + fib(num-2); }
}// FibTest
Is this safe? What if someone
passed in 0 to our method?
What happens?
What if they passed in 1?

72. It is our responsibility
Coding Fibonacci
public class FibTest {
public static int fib (int num) {
if (num == 0)
return 0;
else if (num == 1)
return 1;
else
return fib(num-1) + fib(num-2); }
public static void main(String[] args) {
for (int i=0; i < 10; i++)
System.out.println (fib(i));
}// FibTest
It is our responsibility
to write a main to test
this method.

73. numbers? More work is needed
Coding Fibonacci
public class FibTest {
public static int fib (int num) {
if (num == 0)
return 0;
else if (num == 1)
return 1;
else
return fib(num-1) + fib(num-2); }
public static void main(String[] args) {
for (int i=0; i < 10; i++)
System.out.println (fib(i));
}// FibTest
Are we done?
What about negative
numbers? More work is needed

74. Recursion Review So far, we’ve seen that for recursive behavior:
1) Recursion exists in all of nature.
2) It’s easier than memorizing a formula. Not every problem has a formula, but every problem can be expressed as a series of small, repeated steps.
3) Each step in a recursive process should be small, calculable, etc.
4) You absolutely need a terminating condition.

75. Honesty in Computer Science
1. To make life easy the typical examples given for recursion are factorial and the Fibonacci numbers.
2. Truth is the Fibonacci is a horror when calculated using “normal” recursion and there’s not really any big advantage for factorial.
3. So why all the fuss about recursion?
4. Recursion is absolutely great when used to write algorithms for recursively defined data structures like binary trees. Much easier than iteration!
5. Recursion is excellent for any divide & conquer algorithm like...

76. One More Example Suppose we wanted to create a method that solve:
Pow(x, y) = xy
In other words, the method returned the value of one number raised to the power of another:
public static double pow
(double value, int exponent);

77. Planning the Method
Unlike the Fibonacci example, our mathematical formula is not the complete answer.
Pow(x, y) = xy
We’re missing some termination conditions.
But we know:
x1 = x;
x0 = 1;
So we could use these as our terminating condition.

78. Attempt #1 Always, always start with some sort of terminating
public static double pow(double value, int exponent){
if (exponent == 0)
return 1D;
Always, always start with
some sort of terminating
condition. We know any number
raised to the zero power is one.

79. ... and any number raised to the
Attempt #1
public static double pow(double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
... and any number raised to the
power of one is itself.

80. Attempt #1 For all other values, we can return the number times the
public static double pow(double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else
return value * pow (value, exponent--); }
For all other values, we can
return the number times the
recursive call, using our exponent
as a counter. Thus, we calculate:
26 = 2*2*2*2*2*2

81. Attempt #1 When we run this, however, bad things happen. The program
public static double pow(double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else
return value * pow (value, exponent--); }
When we run this, however, bad
things happen. The program
crashes, having caused a stack
overflow. How can we solve this?

82. Attempt #1 public static double pow(double value, int exponent){
if (bDEBUG) System.out.println
(“Entering with: “ + value +
“, and exp: “ + exponent);
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else {
return value * pow (value, exponent--); }

83. Attempt #1 DOH! public static double pow(double value, int exponent){
if (bDEBUG) System.out.println
(“Entering with: “ + value +
“, and exp: “ + exponent);
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else {
return value * pow (value, exponent--); }
DOH!
Our debug statement tells us that the
exponent is never being decreased.
Evidently, the “exponent--” line is
not being evaluated before the recursive
call takes place. As it turns out, the
post-decrement operator -- is the problem.

84. Attempt #1 We decide that typing one extra
public static double pow(double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else {
exponent = exponent - 1;
return value * pow (value, exponent); }
We decide that typing one extra
line takes less time than debugging
such a subtle error. Things are
working now.

85. “Do I Have to Use Recursion?”
public static double pow(double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else {
exponent = exponent - 1;
return value * pow (value, exponent); }
How many would have preferred to do this with a “for loop” structure or some other iterative solution?
How many think we can make our recursive method even faster than iteration?

86. Nota Bene Our power function works through brute force recursion.
28 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2
But we can rewrite this brute force solution into two equal halves:
28 = 24 * 24
and
24 = 22 * 22
22 = 21 * 21
anything to the power 1 is itself!

87. Since these are the same we don't have to calculate them both!
And here's the cool part...
28 = 24 * 24
Since these are the same we don't have to calculate them both!

88. AHA! So only THREE multiplication operations have to take place:
28 = 24 * 24
24 = 22 * 22
22 = 21 * 21
So the trick is knowing that 28 can be solved by dividing the problem in half and using the result twice!

89. "But wait," I hear you say!
You picked an even power of 2. What about our friends the odd numbers?
Okay we can do odds like this:
2odd = 2 * 2 (odd-1)

90. "But wait," I hear you say!
You picked a power of 2. That's a no brainer!
Okay how about 221
221 = 2 * 220 (The odd number trick)
220 = 210 * 210
210 = 25 * 25
25 = 2 * 24
24 = 22 * 22
22 = 21 * 21

91. "But wait," I hear you say!
You picked a power of 2. That's a no brainer!
Okay how about 221
221 = 2 * 220 (The odd number trick)
220 = 210 * 210
210 = 25 * 25
25 = 2 * 24
24 = 22 * 22
22 = 21 * 21
That's 6 multiplications
instead of 20 and it gets
more dramatic as the
exponent increases

92. The Recursive Insight
If the exponent is even, we can divide and conquer so it can be solved in halves.
If the exponent is odd, we can subtract one, remembering to multiply the end result one last time.
We begin to develop a formula:
Pow(x, e) = 1, where e == 0
Pow(x, e) = x, where e == 1
Pow(x, e) = Pow(x, e/2) * Pow(x,e/2), where e is even
Pow(x, e) = x * Pow(x, e-1), where e > 1, and is odd

93. termination conditions
Solution #2
public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value; }
We have the same base
termination conditions
as before, right?

94. Solution #2 public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) { }
This little gem determines if
a number is odd or even.

95. Solution #2 public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) { }
Don’t like the mod operator?
Get used to it. The alternative is something mutually recursive and wacky, like this:
public static int isOdd(int num){
if (num==0)
return 1;
else
return isEven(num-1); }
public static int isEven(int num){
return 0;
return isOdd(num-1); }

96. Solution #2 public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) { }
That mod operator is starting
to look a lot better, eh? It merely tests if the number is evenly divided by two.
public static int isOdd(int num){
if (num==0)
return 1;
else
return isEven(num-1); }
public static int isEven(int num){
return 0;
return isOdd(num-1); }

97. Solution #2 We next divide the exponent in half.
public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) {
exponent = exponent / 2; }
We next divide the
exponent in half.

98. Solution #2 We recurse to find that half of the brute force
public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) {
exponent = exponent / 2;
double half = pow (value, exponent); }
We recurse to find that
half of the brute force
multiplication.

99. multiplied by themselves
Solution #2
public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) {
exponent = exponent / 2;
double half = pow (value, exponent);
return half * half; }
And return the two
halves of the equation
multiplied by themselves

100. Solution #2 If the exponent is odd, we have to reduce it by one . . .
public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) {
exponent = exponent / 2;
double half = pow (value, exponent);
return half * half; }
else {
exponent = exponent - 1;
If the exponent is odd,
we have to reduce it
by one . . .

101. recurse to solve that portion
Solution #2
public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) {
exponent = exponent / 2;
int half = pow (value, exponent);
return half * half; }
else {
exponent = exponent - 1;
double oneless = pow (value, exponent);
And now the exponent is
even, so we can just
recurse to solve that portion
of the equation.

102. Solution #2 public static double pow (double value, int exponent){
if (exponent == 0)
return 1D;
else if (exponent == 1)
return value;
else if (exponent % 2 == 0) {
exponent = exponent / 2;
int half = pow (value, exponent);
return half * half; }
else {
exponent = exponent - 1;
double oneless = pow (value, exponent);
return oneless * value;
We remember to multiply the value
returned by the original value, since
we reduced the exponent by one.

103. Recursion vs. Iteration:
Those of you who voted for an iterative solution are likely going to produce:
O(N)
In a Dickensian world, you would be fired for this.
While those of you who stuck it out with recursion are now looking at:
O(log2n)
For that, you deserve a raise.

104. STOP
Absolutely CRITICAL point coming up

105. Questions?