Oct8 - 131 slid

Oct 8th
Lab08.
Quiz review
Triangle and Stripes
http://www.slideshare.net/takyeon

Quiz review
No
i++
++i
Use and then increase
Increase and then use
int i = 3;
int a = i++; // a = 3, i = 4
int b = ++a; // b = 4, a = 4

Quiz review
str
maxCount 100
• Whenever a new variable is declared, it is
added to STACK.
• Primitive data types are stored in STACK
• byte, short, int, long, float, double, boolean, char
• Other data types are stored in HEAP.
• String, Integer, Scanner, …
"Hello"
"HELLO"
• Data in HEAP are not immediately
deleted but unlinked, and will be
garbage-collected.

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int size = sc.nextInt();
for(int row=1;row<=size;row++) {
for(int col=1;col<=size;col++) {
System.out.print(row*col + " ");
}
System.out.println();
}
}

Lab – 2D drawing
Two methods.
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
M1. Iterate pixels to paint
for (int i=0; i<size; i++) {
grid.setColor(i, i, Color.BLUE);
}
Intuitive and efficient 
M2. Iterate every pixel, use if conditionals to
check pixels to paint
for (int row=0; row<size; row++) {
for (int col=0; col<size; col++) {
if(row==col) {
grid.setColor(row, col, Color.BLUE);
}
}
}
Complex and inefficient 
BUT! More generalizable

Lab – 2D drawing
Two methods.
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
M1. Iterate pixels to paint
Very difficult 
4,0 4,1 4,2 4,3 4,4 M2. Iterate every pixel, use if conditionals to
check pixels to paint
for (int row=0; row<size; row++) {
for (int col=0; col<size; col++) {
if(row!=col) {
grid.setColor(row, col, Color.BLUE);
}
}
}
You can simply inverse the conditional logic
Now you want to
paint all the pixels
except the diagonal
line.

More examples.
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
row>2
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
row%2 == 1
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
col%2 == 1
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
row-col
0 -1 -2 -3 -4
1 0 -1 -2 -3
2 1 0 -1 -2
3 2 1 0 -1
4 3 2 1 0
(row-col)>=0
Diagonal shapes
require both row
and col
Linear shapes require either row
or col.

Transformation > Move
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
(row+1, col)
0,1 0,2 0,3 0,4 0,5
1,1 1,2 1,3 1,4 1,5
2,1 2,2 2,3 2,4 2,5
3,1 3,2 3,3 3,4 3,5
4,1 4,2 4,3 4,4 4,5
(row+1)-col
-1 -2 -3 -4 -5
0 -1 -2 -3 -4
1 0 -1 -2 -3
2 1 0 -1 -2
3 2 1 0 -1
(row+1)-col >= 0
To move a shape to left by 1
pixel,
replace "row" with "row+1"

Transformation > Horizontal Flip.
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
0,4 0,3 0,2 0,1 0,0
1,4 1,3 1,2 1,1 1,0
2,4 2,3 2,2 2,1 2,0
3,4 3,3 3,2 3,1 3,0
4,4 4,3 4,2 4,1 4,0
Horizontal
Flip
(row, 4-col)
row-(4-col)
-4 -3 -2 -1 0
-3 -2 -1 0 1
-2 -1 0 1 2
-1 0 1 2 3
0 1 2 3 4
(row-(4-col))>=0
To flip a shape, multiple row or
column by -1, and add size
col
-col
size-col
col
4 := size of the shape – 1
Why -1? Because our row and
col index started from 0.

Transformation > Vertical Flip.
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
Vertical
Flip
(4-row, col)
4,0 4,1 4,2 4,3 4,4
3,0 3,1 3,2 3,3 3,4
2,0 2,1 2,2 2,3 2,4
1,0 1,1 1,2 1,3 1,4
0,0 0,1 0,2 0,3 0,4
(4-row)-col
4 3 2 1 0
3 2 1 0 -1
2 1 0 -1 -2
1 0 -1 -2 -3
0 -1 -2 -3 -4
(4-row)-col >= 0

0 -1 -2 -3 -4
1 0 -1 -2 -3
2 1 0 -1 -2
3 2 1 0 -1
4 3 2 1 0
(row-col)>=0
-4 -3 -2 -1 0
-3 -2 -1 0 1
-2 -1 0 1 2
-1 0 1 2 3
0 1 2 3 4
(row-(4-col))>=0
Horizontal
Flip
Vertical flip
4 3 2 1 0
3 2 1 0 -1
2 1 0 -1 -2
1 0 -1 -2 -3
0 -1 -2 -3 -4
(4-row)-col >= 0
Vertical flip
0 1 2 3 4
-1 0 1 2 3
-2 -1 0 1 2
-3 -2 -1 0 1
-4 -3 -2 -1 0
(4-row)-(4-col) >= 0
Horizontal
Flip
col-row >= 0

Oct 6th
Lab07.
Loop applications

Finding common factors of two numbers
Common factors can divide both numbers.
E.g. Common factors of 9 and 12  1 and 3
public void commonFactor(int n1, int n2) {
for(int i=1; i<=min(n1,n2); i++) {
if(n1%i==0 && n2%i==0) {
System.out.println(i);
}
}
}
Common factors of 24 and 78  1, 2, 3, and 6

compareTo method
String s1 = "aaa";
String s2 = "aac";
int k = s1.compareTo(s2); // k => -2
Compares s1 and s2 lexicographically.
Negative if s1 precedes s2
Positive if s1 follows s2
Zero if s1 is equal to s2

Get multiple words, find the first and the last words
1) Using while loop, keep asking words until "STOP"
2) Using compareTo, update the first and the last words
3) Print out

SquareGrid.java
Drawing shapes on 2D grid
ExampleDriver.java
• Prompt a shape question
• Create an empty grid
• Draw the requested shape
OperatorMaker.java
drawOp (SquareGrid grid, int symbol)
minus, plus, divide, multiply (SquareGrid grid)
You will change only these methods 

Single loop for drawing a line
0 1) How can we get the middle row number?
3
size : 7
int size = grid.getHt();
int midRow = size / 2;
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
2) How to draw a line?
• Iterate over columns (0 – 4)
• Paint the middle cell
for (int iCol=0; iCol<size; iCol++) {
grid.setColor(midRow, iCol, Color.BLUE);
}

0,0 0,1 0,2 0,3 0,4 1) How can we get the middle column number?
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
int size = grid.getWd();
int midCol = size / 2;
• Iterate over rows (0 – 4)
• Paint the middle cell
for (int iRow=0; iRow<size; iRow++) {
grid.setColor(iRow, midCol, Color.BLUE);
}
Notice that drawing horizontal and vertical lines are quite similar.
We just switched row and column variables.

0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
• Iterate over rows or columns (0-4)
• Paint diagonal cells.
grid.setColor(iRow, iRow, Color.BLUE);
}

grid.setColor(midRow, iCol, Color.BLUE);
}
grid.setColor(iRow, midCol, Color.BLUE);
}
grid.setColor(iRow, iRow, Color.BLUE);
}
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
0,0 0,1 0,2 0,3 0,4
1,0 1,1 1,2 1,3 1,4
2,0 2,1 2,2 2,3 2,4
3,0 3,1 3,2 3,3 3,4
4,0 4,1 4,2 4,3 4,4
grid.setColor(iCol, iCol, Color.BLUE);
}
or

Iterating over
the columns,
paint the middle
cell.
Iterating over
the columns,
paint the middle
cell.
Iterating over
the rows,
paint the center
cell.
Iterating over
the columns,
paint i-th cell.
Draw Plus,
Divide,
and Divide (rotated).

Sep 29th
Lab05.
1. Recap the quiz #1
2. String Class
3. static vs. instance method

Recap quiz #1.
PRINT your names in
the grade server.
NO nicknames.

Recap quiz #1.
Use specific technical keywords
e.g. What does CVS “do” for us?
1. Check out / Download the starter files
2. Store / Save multiple versions of the source code
share, deliver, get, access, connected people, ...
Penalties for inaccurate extra info
e.g. CVS runs our code and give us grades.  -1 for incorrect extra info.

String Class
Create a new String object with an
initial value “hello”
String s = “hello”;
String objects have many convenient methods,
upperS = s.toUpperCase(); // will set upperS to “HELLO”
whereIsl= s.indexOf(‘l’); // will find the position of the first ‘l’
newS = s.replace(‘e’,’a’); // will set newS to “hallo”

int type vs. Integer Class
int i=0;
Primitive data type
Integer i = 17;
Wrapper Class
Faster A little slower
don’t have much method provide methods
- convert to string
- generate hash codes

Static vs. Instance method
Intance methods need a sheep as a subject.
bob.eat();
bob.smileTo(clara);
bob.getPenNumber();
Static methods are about all the sheeps.
Sheep.getTotalSheep();
Sheep.removeAll();
Sheep.addSheep(‘evan’);

Flow of Control
1. Top-to-bottom statements
2. Method calls
3. Conditional statements
4. Iteration (loop)
for, while, ...

Two goals of iteration
1. Automation
Reduce repetition of code
System.out.println(“****”);
How can we reduce?
for(int i=0;i<3;i++) {
}
2. Abstraction
Code for various situations
How can we print n-number of “*”?

From manual & concrete to automatic & abstract
Level 1. Draw 30 by 10 rectangle (hard-coded)
System.out.println(“**********”);
... 27 more lines
 Too many copy & paste. Hard to modify.
Level 2. Draw 30 by 10 rectangle (single-loop)
int row=0;
while(row<30) {
row++;
}
 A little more compact.
 Still too many * for each line.

From manual & concrete to automatic & abstract
Level 3. Draw 30 by 10 rectangle (nested-loop)
int row=0, col=0;
while(row<30) {
while(col<10) {
System.out.print(“*”);
}
}
 Much compact.
 Cannot change # of row and col
Level 4. Draw height by width (nested-loop, parameterized)
int row=0, col=0;
int height=30, width=10;
while(row<height) {
while(col<width) {
System.out.print(“*”);
}
}
 Compact
 Can draw any sized rectangle

iterartor
line 1
0
value
line 2
0
target
line 4

answer
line 1
1
i
line 1
1
j
line 1
0

Oct8 - 131 slid

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Oct8 - 131 slid