Transcript An Example - Cengage Learning

Programming and Problem Solving
With Java
Chapter 5
Loop Control
Statements
The for Statement
The while Statement
The do-while Statement
Numerical Accuracy
Numerical Analysis
Copyright 1999, James M. Slack
Loops: Introduction
Loop
Group of statements the computer
executes over and over, as long
as some criterion holds
Statement
Statement
Three loop statements in Java
for
while
do-while
Statement
Statement
Programming and Problem Solving With Java
2
Loops: Introduction
Many real-world activities
involve loops
Example: Library checkout
Clear up patron's overdue
materials and outstanding fines, if any.
Scan or type patron's ID number. Update ID if not current
and valid.
Scan material code for each item.
Stamp due date on each item.
Give materials and library card back to the patron.
Checkout librarian does these steps until library
closes
Programming and Problem Solving With Java
3
Loops: Kinds
Two kinds of loops
Counting
Event-controlled
Counting loop
Computer knows, when it begins the loop, how many
times to execute the body
Counting loop statement in Java: for
Event-controlled
Computer stops the loop when a condition is no longer
true
Event-controlled loop statements in Java: while, do-while
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4
Loops: Kinds
Counting loops are subset of event-controlled
"Event" of counting loop: counting variable reaches the
limit
Event-controlled loops
Counting loops
Programming and Problem Solving With Java
5
Flow of
control
The for Statement
initalization
part
The counting loop statement in Java
Example
Initialization
Condition
Is
condition
true?
Increment
No
for (int count = 1; count <= 5; count++)
{
System.out.println(count);
}
Output
1
2
3
4
5
Programming and Problem Solving With Java
Yes
for-body
increment part
Next statement in
program
6
The for Statement: Example
Sum the first 20 integers
Flow of
control
int number = 1
int total = 0;
for (int number = 1; number <= 20; number++)
{
total = total + number;
}
System.out.println("1 + 2 + ... + 20 is "
+ total);
Output
Is number
<= 20?
No
Yes
total = total +
number
1 + 2 + ... + 20 is 210
number++
Next statement in
program
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The for Statement: Example
Flow of
control
initalization
part
What does this display?
for (int count = 10; count <= 6; count++)
{
System.out.println(count);
}
Is
condition
true?
No
Yes
for-body
increment part
Next statement in
program
Programming and Problem Solving With Java
8
The for Statement: Syntax
Can leave out initialization, condition, or increment
int total = 0, number = 1;
for (; number <= 20;)
{
total = total + number;
number++;
}
System.out.println("1 + 2 + ... + 20 is " + total);
Don't need to use same variable throughout
int y = 10;
for (int x = 0; x < y; y--)
{
System.out.println(x + " " + y);
}
(Confusing!)
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9
The for Statement: Draw Circles
Example: drawing circles in turtle graphics
Can
simulate
circles with
drawPolygon()
Too many
sides makes
drawing too
slow
20 to 40 sides
looks ok
Programming and Problem Solving With Java
360 sides
4.2 seconds
180 sides
1.9 seconds
90 sides
0.9 seconds
45 sides
0.4 seconds
20 sides
0.2 seconds
10 sides
0.1 seconds
10
The for Statement: drawPolygon()
The drawPolygon() method (Chapter 2)
// drawPolygon: Draws a regular polygon with the given
//
number of sides, and all sides are of
//
length size
void drawPolygon(int numSides, int size)
throws TurtleException
{
for (int side = 1; side <= numSides; side++)
{
this.move(size);
this.turnRight(360 / numSides);
}
}
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11
The for Statement: drawPolygon()
Drawing circles with drawPolygon()
How to draw circle with specific radius?
Circumference of a circle
2 x  x radius
Distance around a polygon
sides x size
Set these equal to each other, solve for size
2 x  x radius = sides x size
size = (2 x  x radius) / sides
In Java
int size = (int) Math.round(2 * Math.PI * radius) / NUM_SIDES;
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12
The for Statement: drawCircle()
Drawing circles with drawPolygon()
First version of drawCircle()
// drawCircle: (First version) Draws a circle of the given
//
radius, to the right of the turtle. The turtle
//
finishes at the same position and direction as
//
before. The pen must be down beforehand, and is
//
down afterward.
static final int NUM_SIDES = 20; // Must divide into 360
public void drawCircle(int radius)
throws TurtleException
{
int size = (int) Math.round(2 * Math.PI * radius)
/ NUM_SIDES;
this.drawPolygon(NUM_SIDES, size);
}
Computes size of 20-sided polygon
to achieve desired radius
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The for Statement: drawCircle()
Test drawCircle()
// This program draws a circle inside a square,
// to see how accurate the circle drawing method is
import turtlegraphics.*;
import SmartTurtle;
public class TestCircle
{
static final int CIRCLE_SIZE = 300;
Oops!
public static void main[] (String args)
throws TurtleException
{
SmartTurtle myTurtle = new SmartTurtle();
}
}
myTurtle.goLeft(200);
myTurtle.turnAround();
myTurtle.move(CIRCLE_SIZE);
myTurtle.turnAround();
myTurtle.drawSquare(CIRCLE_SIZE * 2);
myTurtle.move(CIRCLE_SIZE);
myTurtle.drawCircle(CIRCLE_SIZE);
Programming and Problem Solving With Java
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The for Statement: drawCircle()
What's wrong with drawCircle()?
Turtle draws first side of polygon
straight up
Turtle draws last side of polygon
at an angle
Need to balance
these two angles
At beginning
First edge of
polygon
Turtle's starting
position
this.turnRight(360 / NUM_SIDES / 2);
At end
Last edge of
polygon
this.turnLeft(360 / NUM_SIDES / 2);
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The for Statement: drawCircle()
Final version of drawCircle()
// drawCircle: (Final version) Draws a circle of the given
//
radius, to the right of the turtle. The turtle
//
finishes at the same position and direction as
//
before. The pen must be down beforehand, and is
//
down afterward.
static final int NUM_SIDES = 20; // must divide into 360
// with an even quotient
public void drawCircle(int radius)
throws TurtleException
{
this.turnRight(360 / NUM_SIDES / 2);
int size = (int) Math.round(2 * Math.PI * radius)
/ NUM_SIDES;
this.drawPolygon(NUM_SIDES, size);
this.turnLeft(360 / NUM_SIDES / 2);
}
Ah...
 New restriction: number of sides must
divide into 360 with even quotient
(So can divide this by 2 for start
and end angles)
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The while Statement
Flow of
control
One of two event-controlled loop
statements
Is
condition
true?
Body keeps executing as
long as condition is true
Example
int i = 0;
while (i < 3)
{
System.out.println(i);
i++;
}
No
Yes
while-body
Output
0
1
2
Programming and Problem Solving With Java
Next statement in
program
17
The while Statement
while similar to if statement
while statement
if
statement
while (i < 3)
{
System.out.println(i);
i++;
}
while
statement
Yes
Flow of
control
Is
condition
true?
Is
condition
true?
No
if statement
Flow of
control
No
Yes
if-body
while-body
Next statement in
program
Next statement in
program
if (i < 3)
{
System.out.println(i);
i++;
}
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The while Statement: Sentinels
Sentinel loop
Sentinel: not data, but marks the end of data
Sentinel loop: reads data values until sentinel
Example
Sum series of numbers terminated
by zero
10, 20, 30, 0
Data values
Programming and Problem Solving With Java
Sentinel
19
The while Statement: Sentinels
Sentinel loop example
int sum = 0, number;
number = Keyboard.readInt("Enter first number: ");
while (number != 0)
{
sum = sum + number;
number = Keyboard.readInt("Enter next number: ");
}
System.out.println("The sum is " + sum);
Enter first number: 10
Enter next number: 20
Enter next number: 30
Enter next number: 0
The sum is 60
Trace
Statement
number
sum
int sum = 0, number;
number = Keyboard.readInt("Enter first number (0 to stop): )");
while (number != 0)
sum = sum + number;
number = Keyboard.readInt("Enter next number (0 to stop): )");
while (number != 0)
sum = sum + number;
number = Keyboard.readInt("Enter next number (0 to stop): )");
while (number != 0)
sum = sum + number;
number = Keyboard.readInt("Enter next number (0 to stop): )");
while (number != 0)
System.out.println("The sum is " + sum);
—
10
10
10
20
10
10
30
30
30
0
0
0
0
0
0
10
10
10
30
30
30
60
60
60
60
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The while Statement: Infinite Loop
Infinite loop
Loop that doesn't stop
Easy to write accidentally with while statement
Infinite loop example 1
while (x < 3)
{
y++;
}
Infinite loop example 2
sum = 0;
count = 0;
while (count < 100)
{
sum = sum + count;
}
System.out.println("Sum of numbers from 1 to 100 is " + sum);
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The while Statement: Infinite Loop
Infinite loop example 3
answer = Keyboard.readChar("Say hi? (y/n)");
while (answer == 'y')
System.out.println("Hi!");
answer = Keyboard.readChar("Say hi? (y/n)");
System.out.println("Bye!");
Infinite loop example 3
int x = 0;
while (x < 10);
{
System.out.println(x);
x++;
}
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The while Statement: Infinite Loop
Why doesn't compiler catch infinite loops?
Halting problem: Impossible for a computer program to
detect all possible infinite loops in another program
Up to the programmer to avoid infinite loops
When you write a while
statement
Make sure some statement in
the while-body makes the
condition false
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The while Statement: Menus
Example: Writing menu-based programs
Menu: List of selections that user can pick from
Writing a text-based menu system with while
Present menu
Get selection from user
As long as not quit
•Do the selection
•Present the menu
•Get next selection
Programming and Problem Solving With Java
File Edit View Flip Window Help
New
Open
Save
Save As
Exit
Drawing
24
The while Statement: Menus
Skeleton for text-based menu
// Present the menu the first time
// Get first selection from user
while (selection != quitOption)
{
// Do selection
// Present the menu again
// Get next selection from user
}
Menu
0.
1.
2.
3.
Quit
Enter beginning balance
Enter interest rate
Compute ending balance
Example: Compute bank balance
--- Bank Balance Menu --0. Quit
1. Enter beginning balance
2. Enter interest rate
3. Compute ending balance
Enter selection (0, 1, 2, or 3): _
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The while Statement: Menus
Bank Balance Program
Present menu
Get selection from user
As long as not quit
•Do the selection
•Present the menu
•Get next selection
int selection;
double balance = 0.00;
double rate = 0.00;
// Display program title
System.out.println("--- Compute End-of-Year Bank Balance ---");
System.out.println();
// Display menu, get first selection
System.out.println("--- Bank Balance Menu ---");
System.out.println("0. Quit");
System.out.println("1. Enter beginning balance");
System.out.println("2. Enter interest rate");
System.out.println("3. Compute ending balance");
selection = Keyboard.readInt("Enter selection (0, 1, 2, or 3): ", 0, 3);
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The while Statement: Menus
// Handle the first selection; keep handling selections until
// user picks the quit selection (0)
while (selection != 0)
{
switch (selection)
{
case 1:
balance = Keyboard.readDouble("Enter beginning balance: ");
break;
case 2:
rate = Keyboard.readDouble("Enter interest rate: ");
break;
case 3:
System.out.println("Ending balance is " + (balance + balance * rate));
break;
default:
System.out.println("Problem with switch statement");
break;
}
}
// Display menu, get next selection
System.out.println();
System.out.println("--- Bank Balance Menu ---");
System.out.println("0. Quit");
System.out.println("1. Enter beginning balance");
System.out.println("2. Enter interest rate");
System.out.println("3. Compute ending balance");
selection = Keyboard.readInt("Enter selection (0, 1, 2, or 3): ", 0, 3);
System.out.println("Bye!");
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The while Statement: Counting
Writing counting loops with while
Can convert for statement to while
for (initialization; condition; increment)
{
for-body
}
initialization;
while (condition)
{
for-body;
increment;
}
int total = 0;
int number = 1, total = 0;
for (int number = 1; number <= 20; number++)
{
total = total + number;
}
while (number <= 20)
{
total = total + number;
number++;
}
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Kinds of Loops
Condition
at the
top
Flow of
control
Condition
at the
bottom
Flow of
control
Condition
in the
middle
body
(begin)
body
Yes
condition
Flow of
control
condition
condition
Yes
Yes
No
No
No
body
(end)
statement
statement
body
statement
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Kinds of Loops
Condition at the top
Condition tested before the body
Java statements: for and while
Condition at the bottom
Condition tested after the body
Java statements: do-while
Condition in the middle
Condition tested inside the body
No built-in Java statement
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The do-while Statement
The "other" event-controlled loop
statement
Body keeps executing as
long as condition is true
Flow of
control
do-while-body
Example
int i = 0;
do
{
System.out.println(i);
i++;
} while (i < 3);
Output
0
1
2
Programming and Problem Solving With Java
Yes
Is
condition
true?
No
Next statement in
program
31
The do-while Statement
Difference from while statement
do-while executes body at least once
while statement may not execute body at all
Example
while
int i = 10;
while (i < 3)
{
System.out.println(i);
i++;
}
do-while
int i = 10;
do
{
System.out.println(i);
i++;
} while (i < 3);
Output
(nothing)
Programming and Problem Solving With Java
10
32
The do-while Statement: Format
Poor style
do
{
do-while-body
}
while (condition);
Looks like the beginning
of a while statement
Programming and Problem Solving With Java
Preferred style
do
{
do-while-body
} while (condition);
The brace before while
distinguishes it as part
of a do-while
33
The do-while Statement: Example
Program to help children learn multiplication
int guessNum = 0, response;
System.out.println("What is "
+ operand2
do
{
guessNum++;
response
= Keyboard.readInt("Enter
} while (response != operand1
+ operand1 + " times "
+ "?");
guess number " + guessNum + ": ");
* operand2);
System.out.println("Correct!");
Equivalent while statement requires
two copies of body
 One before loop starts
 One inside the loop
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34
Numerical Accuracy
// Add up 0.10 until the sum is 1.0
(Wrong)
public class AddTo1
{
public static void main(String[] args)
{
double total = 0.0;
}
}
while (total != 1.0)
{
total = total + 0.10;
System.out.println("Total so far: " + total);
}
System.out.println("Done");
Total
Total
Total
Total
Total
Total
Total
Total
Total
Total
Total
Total
Total
Total
Programming and Problem Solving With Java
so
so
so
so
so
so
so
so
so
so
so
so
so
so
far:
far:
far:
far:
far:
far:
far:
far:
far:
far:
far:
far:
far:
far:
0.1
0.2
0.30000000000000004
0.4
0.5
0.6
0.7
0.7999999999999999
0.8999999999999999
0.9999999999999999
1.0999999999999999
1.2
1.3
1.4000000000000001
35
Numerical Accuracy: Round-off
0.1+0.1+0.1+ 0.1+0.1+0.1+0.1+0.1+0.1+0.1  1.0?
Round-off error
Can happen only with floating-point arithmetic
Computer stores numbers in binary, not base 10
Representable number: can store in finite number of
digits
Unrepresentable in base 10:
Representable in base 10:
1/3 = 0.333333333...
1/4 = 0.25
Some representable numbers in base 10 are
unrepresentable in binary
0.10 (decimal) = 0.00011001100110011001100 . . . (binary)
Can only store approximation of 0.10
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Numerical Accuracy: Round-off
Round-off error can happen with representable
numbers
Example:
1.3707275390625 (decimal) = 1.0101111011101 (binary)
Suppose computer can only store 8 binary digits
Binary
Decimal
1.0101111011101
1.3707275390625
 1.0101111
 1.3671875
0.0000000011101
0.0035400390625
Computer's version of number is an approximation
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Numerical Accuracy: Comparisons
Don’t compare floating-point numbers with == or !=
double total = 0.0;
while (total != 1.0) ...
Instead:
// Avoid
Use integers
int total = 0;
while (total != 10) ...
// Use 1 for 1/10
Use <=, >=, <, or >
double total = 0.0;
while (total <= 1.0) ...
// Good approach
Check if numbers are very close
double total = 0.0;
// Can set exact tolerance
while (Math.abs(total - 1.0) > 0.00001) ...
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Numerical Accuracy: Arithmetic
Avoid arithmetic on floating-point numbers
that are very different in size
Simulate space probe from earth to Proxima Centauri
double centimetersToProximaCentauri = 4067815793829482158.0;
while (centimetersToProximaCentauri > 0.0)
{
// Simulate action of space probe here ...
centimetersToProximaCentauri = centimetersToProximaCentauri - 1.0;
}
This loop is infinite: computer stores
4067815793829482158.0 as 4067815793829482000.0
4067815793829482000.0
- 1.0
4067815793829482000.0
One solution: use kilometers instead
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Condition-in-the-Middle Loops
Hypothetical condition-in-the-middle loop (not Java)
loop
{
number = Keyboard.readInt("Number (0 to stop): ");
until (number == 0);
sum = sum + number;
}
Can use as while or do-while
Move until up for while, down for do-while
Equivalent while loop
number = Keyboard.readInt("Enter first number (0 to stop): ");
while (number != 0)
{
sum = sum + number;
number = Keyboard.readInt("Enter next number (0 to stop): ");
}
Must duplicate part of body before loop starts
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Condition-in-the-Middle Loops
Can use break statement to stop loop early
Put "if (condition) break" in infinite loop
while (true)
{
number = Keyboard.readInt("Number (0 to stop): ");
if (number == 0) break;
sum = sum + number;
}
System.out.println("Sum is " + sum);
Doesn't work if break is inside switch
Must label the loop statement
loop:
while (true)
{
switch (...)
{
case xx: break loop;
...
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The continue Statement
Syntax
continue;
or
continue label;
Example
for (i = 0; i < 100; i++)
{
if (i % 2 == 0) // Ignore even numbers
{
continue;
}
System.out.println(i);
}
Action
Control skips rest of loop iteration; starts next iteration
for loop: computer executes the increment part before
starting next iteration
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Numerical Analysis
Study of use of arithmetic in computer programs
Some issues in numerical analysis
Minimize round-off errors
Solve problems that don't have exact
solution (or exact solution is difficult to find)
Finding roots of mathematical functions
Root of function: Value of x that makes
function 0
Often difficult to find roots algebraically
Example: find all roots of f(x) = x5 - 5x + 1
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Numerical Analysis: Roots
Find roots of f(x) = x5 - 5x + 1
Plot the function
30
20
10
f(x)
0
-10
-20
-30
-2
-1
0
1
2
x
Three roots: one is between -1 and -2
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Numerical Analysis: Bisection
Use bisection to find root of f(x) = x5 - 5x + 1
Plot shows a root between -1 and -2
Pick 2 values of x, one on either side of root
x value
f(x) value
Above zero
-1 -1.5
5 0.90625
Zero value
???
0
Below zero
-2
-21
Use -1.5 as approximation of root, plug into function
f (1.5)  (1.5)5  5(1.5)  1
 7.59375  7.5  1
 0.90625
0.90625 > 0, so it replaces old "above 0" value
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Numerical Analysis: Bisection
Use bisection to find root of f(x) = x5 - 5x + 1
Next evaluate f(x) at -1.75 (halfway between -1.5 and -2)
x value
f(x) value
Above zero
-1.5
0.90625
Zero value
???
0
Below zero
-2 -1.75
-21-6.66301
f(-1.75) is -6.66301
-6.66301 < 0, so it replaces old "below 0" value
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Numerical Analysis: Bisection
Keep using bisection to get as close to 0 as desired
Each iteration cuts range by half
2
1
0
-1
Upper
-2
Lower
-3
Midpoint
-4
f(Midpoint)
-5
-6
-7
-8
0
1
2
3
4
5
6
7
8
9
10
11
12
Iterations
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Numerical Analysis: Bisection
Bisection advantages
Bisection disadvantages
 Simple and effective
 Little or no round-off error
 Slow
 Only finds one root at a time
 Can't tell if function has
other roots
 Appropriate for continuous
intervals only
2
1.5
1
0.5
f(x)
0
-0.5
Noncontinuous
function
-1
-1.5
-2
-1
-0.5
0
0.5
1
x
Programming and Problem Solving With Java
48
Numerical Analysis: Bisection
Bisection Program
//
//
//
//
//
//
//
//
Find zero of a function by bisection.
Displays bound1, bound2, midpoint between bound1 and
bound2, and f(midpoint). If f(midpoint) is not within
+/- ERROR, cuts area in half and repeats.
NOte: Required that f(bound2) < 0 < f(bound1), and there
exists f(x) such that f(x) == 0, x is between
bound1 and bound2, and f() is continuous between
f(bound1) and f(bound2)
import Keyboard;
import Format;
public class Bisection
{
static final double ERROR = 0.001; // Maximum difference between
// generated solution and zero
static final int WIDTH = 14;
static final int DECIMALS = 5;
// Width of output numbers
// Decimals of output numbers
// evaluateFunction: Returns f(x) (Put your function here)
static double evaluateFunction(double x)
{
return Math.pow(x, 5) - (5 * x) + 1;
}
Programming and Problem Solving With Java
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Numerical Analysis: Bisection
public static void main(String[] args)
throws java.io.IOException
{
double bound1, bound2;
System.out.println("--- Find Zeros of a Function "
+ "by Bisection ---");
System.out.println();
// Get initial bounds for the value of x from the user
System.out.println("Enter two values that surround the value");
System.out.println("of x that makes the function 0.");
bound1 = Keyboard.readDouble("Value of x that gives "
+ "positive value "
+ "from function: ");
bound2 = Keyboard.readDouble("Value of x that gives "
+ "negative value "
+ "from function: ");
System.out.println();
double midpoint;
System.out.println(Format.padRight("Bound1", WIDTH)
+ Format.padRight("Bound2", WIDTH)
+ Format.padRight("Midpoint", WIDTH)
+ Format.padRight("f(Midpoint)", WIDTH));
System.out.println();
Programming and Problem Solving With Java
50
Numerical Analysis: Bisection
// Do first cut before the loop
midpoint = (bound1 + bound2) / 2;
System.out.println(Format.pad(bound1, WIDTH, DECIMALS)
+ Format.pad(bound2, WIDTH, DECIMALS)
+ Format.pad(midpoint, WIDTH, DECIMALS)
+ Format.pad(evaluateFunction(midpoint), WIDTH, DECIMALS));
}
}
// Keep reducing the area between the bounds until the
// f(midpoint) is within ERROR tolerance of 0.0
while (Math.abs(evaluateFunction(midpoint)) > ERROR)
{
if (evaluateFunction(midpoint) > 0.0)
bound1 = midpoint;
else
bound2 = midpoint;
midpoint = (bound1 + bound2) / 2;
System.out.println(Format.pad(bound1, WIDTH, DECIMALS)
+ Format.pad(bound2, WIDTH, DECIMALS)
+ Format.pad(midpoint, WIDTH, DECIMALS)
+ Format.pad(evaluateFunction(midpoint), WIDTH, DECIMALS));
}
Program uses Format.pad() (from Chapter 7)
 Format.pad(floating-point-number, width, decimals)
 Format.pad(string, width)
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51
30
Num. Analysis: Bisection
20
10
f(x)
Bisection program run on
f(x) = x5 - 5x + 1
0
-10
-20
-30
-2
-1
0
1
x
--- Find Zeros of a Function by Bisection --Enter two values that surround the value
of x that makes the function 0.
Value of x that gives positive value from function: -1
Value of x that gives negative value from function: -2
Bound1
Bound2
Midpoint
f(Midpoint)
-1.00000
-1.50000
-1.50000
-1.50000
-1.50000
-1.53125
-1.53125
-1.53906
-1.53906
-1.54102
-1.54102
-1.54150
-1.54150
-2.00000
-2.00000
-1.75000
-1.62500
-1.56250
-1.56250
-1.54687
-1.54687
-1.54297
-1.54297
-1.54199
-1.54199
-1.54175
-1.50000
-1.75000
-1.62500
-1.56250
-1.53125
-1.54687
-1.53906
-1.54297
-1.54102
-1.54199
-1.54150
-1.54175
-1.54163
0.90625
-6.66309
-2.20596
-0.50073
0.23783
-0.12241
0.05994
-0.03068
0.01477
-0.00792
0.00343
-0.00224
0.00060
Programming and Problem Solving With Java
52
2
Num. Analysis: Bisection
1.5
1
0.5
f(x)
Bisection program run on
f(x) = (6x3 + 4x2 + 1) / 25x
0
-0.5
-1
-1.5
-2
--- Find Zeros of a Function by Bisection ---
-1
-0.5
0
0.5
x
Enter two values that surround the value
of x that makes the function 0.
Value of x that gives positive value from function: 1.0
Value of x that gives negative value from function: -0.4
Bound1
Bound2
Midpoint
f(Midpoint)
1.00000
0.30000
0.30000
0.12500
0.03750
0.03750
0.01562
0.00469
0.00469
0.00195
0.00059
0.00059
0.00024
0.00007
0.00007
0.00003
0.00001
-0.40000
-0.40000
-0.05000
-0.05000
-0.05000
-0.00625
-0.00625
-0.00625
-0.00078
-0.00078
-0.00078
-0.00010
-0.00010
-0.00010
-0.00001
-0.00001
-0.00001
0.30000
-0.05000
0.12500
0.03750
-0.00625
0.01562
0.00469
-0.00078
0.00195
0.00059
-0.00010
0.00024
0.00007
-0.00001
0.00003
0.00001
-0.00000
0.20293
-0.80740
0.34375
1.07300
-6.40099
2.56256
8.53409
-51.20012
20.48031
68.26676
-409.60002
163.84004
546.13335
-3276.80000
1310.72000
4369.06667
-26214.40000
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