1

MT-201 Unit 6 : Advanced control structures & squares

A) Complex logical expressions And [&&], Or [ || ] , Not [ ! ] Short-circuit evaluation of logical expressions Bitwise operators [&], [ | ], [ ^ ]

B) Advanced branching statements Nested [ if / else ] statement [ break ] in [ switch / case ] statements

C) Advanced looping statements

D) Sorting (Insertion sort)

E) Multidimensional squares

F) Recursion — an advanced control structure

http://learn.ouhk.edu.hk/~t441063

2

How to express the following formula in Java ?

How about ?

Wrong !? How about adding more brackets ?

Still wrong. Depending on the content in x , it is :

Correct syntax :

1000 x

If ( 0 < x < 100 ) {. . . . . .

}

If ( (0 < x ) < 100 ) {. . . . . .

}

If ( true < 100 ) {. . . . . .

}

If ( false < 100 ) {. . . . . .

}

If ( (0 < x) && ( x < 100 )) {. . . . . .

}

If ( (x > 0) && ( x < 100 )) {. . . . . .

}

A) Complex logical expressions

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The [ && ] operator

in Table 6.1, conditions a and b denotes two conditions with values of either true or false

Each row in the table is a possible combination of values of conditions a and b.

rightmost column is result of conditions a && b.

Condition Result

a b a && b

true true true

true false false

false true false

false false falseTable 6.1 Truth table of the [ && ] operator

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Case-1 : We can treat sub-condition (0 < x ) as a and (x < 100 ) as b The given expression can be visualized in Figure 6.1, where

All possible result values of the given expression for variable x are partitioned in three regions Only the blue region can give a value true The other 2 red regions will give a value false

(0 < x) && ( x < 100 )

Figure 6.1 A number line with the two values 0 and 100 indicated

-50 -1 0 1 50 99 100 101 150

(x < 100)(0 < x)

(0 < x ) is true and(x < 100 ) is true So,(0 < x) && ( x < 100 ) will return true

(0 < x) is false So,(0 < x) && ( x < 100 ) will return false

( x < 100 ) is falseSo,(0 < x) && ( x < 100 ) will return false

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Value of x

Result of

x > 0

Result of

x < 100

Result of

( (0<x) && (x<100) )-50 (-50 >0)

false(-50 <100)

truefalse

0 (0 >0) false ( 0 <100) true false

50 (50 >0) true ( 50 <100) true true

100 (100 >0) true

( 100 <100) false

false

150 (150 >0) true

( 150 <100) false

false

-50 -1 0 1 50 99 100 101 150

(x < 100)(0 < x)

from Table 6.1, we can make Table 6.2 with different values of x Table 6.2 helps you investigate the relationship between the results of

sub-conditions and the overall result It helps you verify whether a condition is properly constructed

Table 6.2 Truth table for condition (x > 0) && (x < 100) for various values of x

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Value of x

Result of

x < 0

Result of

x > 100

Result of

( (x<0) && (x>100) )-50 (-50 <0) true (-50>100) false false

0 (0 <0) false (0 >100) false false

50 (50 <0) false (50 >100) false false

100 (100 <0) false (100>100) false false

150 (150 <0) false (150>100) true false

-50 -1 0 1 50 99 100 101 150

(x > 100)(x < 0)

Case-2 : Now, the condition is modified. Table 6.3 with different values of variable x is made as follows

Table 6.3 Truth table for condition (x < 0) && (x > 100) for various values of x

(x < 0) && ( x > 100 )

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the overall result of (x < 0) && (x > 100) is always false this condition has problems

as there is no negative value which can greater than 100 so, if the condition (x < 0) && (x > 100) is used as the

condition in a while loop or a for loop, the loop will terminate at once and the loop body will not be executed at all

However, if a condition that is always true is used as the condition for a loop construct the loop will repeat forever and it is an infinite loop (you’ll see an example soon)

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Value of x

Result of

x < 0

Result of

x < 100

Result of

( (x<0) && (x<100) )-50 (-50<0) true (-50<100) true true

0 (0<0) false (0<100) true false

50 (50<0) false (50<100) true false

100 (100<0) false (100<100) false false

150 (150<0) false (150<100) false false

-50 -1 0 1 50 99 100 101 150

(x < 100)(x < 0)

Case-3 : Now, the condition is modified to be (x < 0) && (x <100), From Table 6.4, we note that (x <100) seems redundant, and the given

expression can be simplified as (x < 0)

Table 6.4 Truth table for condition (x < 0) && (x < 100) for various values of x

(x < 0) && ( x < 100 )

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The [ || ] operator In English, we may have "x < 0" or "x > 100". In Java, it is written as:

(x < 0) || (x > 100) The || operator is the logical OR operator. The boolean

expressions on both sides of the || operator can be true or false,

Table 6.5 lists the result of the || operator

Condition Result

a b a || b

true true true

true false true

false true true

false false falseTable 6.5 Truth table of the [ || ] operator

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Value of x

Result of

x < 0

Result of

x > 100

Result of

( (x<0) || (x>100) )-50 (-50 <0) true (-50>100) false true

0 (0 <0) false (0 >100) false false

50 (50 <0) false (50 >100) false false

100 (100 <0) false (100>100) false false

150 (150 <0) false (150>100) true true

-50 -1 0 1 50 99 100 101 150

(x > 100)(x < 0)

Case-1 : using Table 6.5, different values of variable x is listed in Table 6.6

Table 6.6 Truth table for condition (x < 0) || (x > 100) for various values of x

(x < 0) || ( x > 100 )

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Equivalents codes

Expression Equivalent to

x >= 0 (x > 0) || (x == 0)

x <= 100 (x < 100) || (x == 100)

Two relational operators, >= and <=, can be written as two subconditions with an || operator.

For example:

The expressions in the left column are simpler and more readable than those in the right column.

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Value of x

Result of

x > 0

Result of

x < 100

Result of

( (x>0) || (x<100) )-50 (-50 >0)

false(-50<100) true true

0 (0 >0) false (0 <100) true true

50 (50 >0) true (50 <100) true true

100 (100 >0) true (100<100) false true

150 (150 >0) true (150<100) false true

-50 -1 0 1 50 99 100 101 150

(x < 100)(x > 0)

Case-2 : in Table 6.7, the result is always true. If this expression were used as the condition for an if loop, the

action part is always executed

Figure 6.7 Truth table for condition (x > 0) || (x < 100) for various values of x

(x > 0) || ( x < 100 )

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Operator precedence

relational operators such as { <, >, <=, >= } take precedence over the && operator and the || operator

so, parentheses, that enclose the sub-conditions, are not necessary. eg.

((0 < x) && (x < 100)) is equivalent to :

(0 < x && x < 100) It is not necessary to enclose the sub-expressions in

parentheses, but it is a good programming practice to enclose them because it can make the expression clearer, eg.

if ( (0 < x) && (x < 100) ) {......}

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The exclusive-OR, XOR, operator [ ^ ]

In Java, we write : (a ^ b) in Mathematics : (a and ~b) or (~a and b) 有你 無我 , 大家 相反 就 true

Condition Result

a b a ^ b

true true false

true false true

false true true

false false false

Table 6.5 Truth table of the short-circuited-OR operator [ || ]

相 反 就 true

相 同 就 false

相 同 就 false

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The negation, NOT, operator [ ! ] we can negate a boolean

expression result using a negation operator [ ! ]

The type of expression that follows the ! operator must be boolean

eg.

value of

xresult of

x < 0result of

!(x < 0 )

-5 true false

0 false true

5 false true

boolean expression

xresult of

!x

true false

false true

Note : the expression !(x < 0) is equivalent to x >= 0

value of

xresult of

x < 0result of

(x >= 0 )

-5 true false

0 false true

5 false true

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Self-Test 6.1

1. For variables a, b and c of type int, returns true if c is between the value of a and b. It returns false otherwise, provided that the value of a is less than or equal to b. a <= c && c <= b

2. For variables a and b of type int, returns true if the sign of the product of a and b is non-negative. It returns false otherwise. a * b >= 0, or !(a >= 0 ^ b >= 0)

Note: XOR works much faster than multiplication

3. For variables a, b, c and d of type int, returns true if the values of all variables are equal. Otherwise return false a == b && b == c && c == d

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Short-circuit evaluation of logical expressions Evaluations of [ && ] and [ || ] operators in Java are short-circuit sub-conditions on both sides of the && and || operators are

evaluated from left to right If sub-condition on left-hand side (the first operand) determines

the result of the entire condition, the sub-condition on its right-hand side (the second operand) is

not evaluated otherwise, evaluate the rest sub-conditions as usual

Condition Result

a b a && b

true true true

true false false

false true false

false false false

Condition Result

a b a && b

true true true

true false false

false don't care false

Table 6.1

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The || operator features the same short-circuit evaluation if the first operand (the condition a) of the || operator is

true, the overall result must be true. So, no need to evaluate the second operand (condition b)

&& short-circuits on false, while || does when true 如果考試唔記得 , 就 derive them from table 6.1 & 6.5

Condition Result

a b a || b

true true true

true false true

false true true

false false falseTable 6.5

Condition Result

a b a || b

true don't care true

false true true

false false false

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How short-circuiting works ? Consider

the [ && ] and [ || ] operator are left associated, so the above condition can be read as :

if the content of month is 4 , then we have

by short-circuit, month == 6 is skipped, so we have

similarly, month == 9 is skipped,

again, month == 11 is skipped, so the result is true

if (month==4 || month==6 || month==9 || month==11) {

day = 30;

}

( ( ( month==4 || month==6 ) || month==9 ) || month==11 )

( ( ( true || month==6 ) || month==9 ) || month==11)

( ( true || month==9 ) || month==11)

( true || month==11)

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Examples

if ((a>0) && (b>0)) If a’s value is –1, then the left operand (a > 0) is

false. The overall result must be false regardless of (b > 0)

With && operator, Java will not evaluate the expression (b>0) if (a>0) is false. This can improve performance by avioding unnecessary evaluations

if ((a>0) || (b>0)) If a’s value 1, then the left operand (a>0) is true.

The overall result must be true regardless of (b > 0) With || operator, Java will not evaluate the expression

(b>0) if (a>0) is true

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Precautions

Be careful when we construct a complex Boolean expression with short-circuit operators. It is dangerous to import the program logic to other

programming languages. Not all programming languages support short-circuit evaluations

Sometimes, the two operands of the operator have to be evaluated, especially the Boolean sub-expressions with side effects. eg. ((++i >0 ) && (++j>0))

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Non Short-Circuit Evaluation

the OR operator [ | ] similar to the || operator Unlike the || operator, even the result of the first

operand of the | operator is evaluated to be true, the second operand must be evaluated

the AND operator [ & ] similar to the && operator Unlike the && operator, even the result of the first

operand of the & operator is evaluated to be false, the second operand must be evaluated

23

Associativity & precedenceR

elat

iona

lB

oole

an

Precedence Associativity

>= > < <= Left

== != Left

& Left

^ Left

| Left

&& Left

|| Left

Incr

easi

ng

Pri

ori

ty

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Bitwise Operators

&, | and ^ operators are also known as bitwise operators when operands are integers. eg

0 1 0 0 1 0 0 17310 =

9810 =

7310 | 9810 = = 10710

0 1 1 0 0 0 1 0

0 1 1 0 1 0 1 1

pp-18

0 1 0 0 1 0 0 17310 =

9810 =

7310 & 9810 = = 6410

0 1 1 0 0 0 1 0

0 1 0 0 0 0 0 0

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B) Advanced branching statements Ternary conditional operator (? : )

eg

In this example, 0.9 is assigned to discount

Condition ? Value-for-true : value-for-false

if (isMmember) { discount = 0.8; }else { discount=0.9; }

public class Ternary {public static void main ( String args[] ) {

boolean isMember=false;double discount;discount= isMember ? 0.8 :

0.9 System.out.println(discount);

}}

pp-29

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Nested if

General format

or

consider the following codes

If the 1st statement is evaluated to be true, Java still continues the rest 3 statements, because these 4 statements are independent of each other. This will waste processing time

if ( (age<18) && (gender ==‘M’) ) { boyCount++; } // 1st statementif ( (age<18) && (gender==‘F’) ) { girlCount++; } // <---+\if ( (age>18) && (gender==‘M’) ) { manCount++;} // > restif ( (age>18) && (gender==‘F’) ) { ladyCount++;} // <---+/

if ( condition ) {- - - // if part

}else {

- - - // else part} // suggested style

if ( condition ) {- - - // if part

} else {- - - // else part

} // my current style

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We can use nested if/else statements to reduce

unnecessary condition evaluation :

Now, Java only evaluates the complex boolean expression two times, instead of four times better performance & easier maintenance

If age is changed to 21, then the prior version have to change all 4 statements it is error-prone if you forget to update any 1 of the 4

if (age<18) { // outer IFif (gender ==‘M’) { boyCount++;} //

inner IFelse {girlCount++;} //

inner ELSE} else {

if (gender==‘M’) {manCount++;} // inner IFelse {ladyCount++;} // inner

ELSE}

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In all cases, only 1 of the 4 blocks of statements will be evaluated

condition-1

condition-2 condition-3

statement(s) forcondition-1 is truecondition-2 is true

true

statement(s) forcondition-1 is truecondition-2 is false

true

statement(s) forcondition-1 is falsecondition-3 is true

statement(s) forcondition-1 is falsecondition-3 is false

Fig 6.5 A nested if / else construct pp-32

false

falsetruefalse

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Please use "if … else if … else if … else" to

make the program more readable

a better version ( the suggested version is on slide-26 )

if (mark>=90) { grade='A';} // suggestions :else { if (mark>=80) {grade='B';} // 1). use curly braces else { if (mark>=70) {grade='C';} // 2). indent each blocks

else { if (mark>=60) {grade='D';}else { grade='F';}

}}

}

if (mark>=90) { grade='A';} // grade is set by only 1 of else if (mark>=80) {grade='B';} // these 5 statement blocks

else if (mark>=70) {grade='C';} // ( Fig 6.5 has 4 blocks)else if (mark>=60) {grade='D';}

else { grade='F';}

pp-34~37

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Dangling-else problem

Please note that the else part always associates with the nearest if statement.

Curly braces can avoid the dangling-else problem eg. codes-1 will not return true for children

because Java treats codes-1 as codes-2

// codes-1boolean isAccepted = false;if (age>=18)

if ( gender == 'F' )isAccepted = true;

else isAccepted = true;

// codes-2boolean isAccepted = false;if ( age>=18 ) {

if ( gender == 'F' )isAccepted = true;

else isAccepted = true;

}

pp-38

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break in switch/case

public class DaysInMonth2 { public static void main (String args[ ]) {

int month = Integer.parseInt(args[0]);int days;if (month == 2) { days = 28; }else if (month == 4 || month == 6 || month == 9 || month == 11) {

days = 30;} else { days = 31; }System.out.println(days);

}}

public class DaysInMonth { public static void main (String args[ ]) { int month = Integer.parseInt(args[0]);

int days;switch (month) { case 2: days = 28; break; case 4: case 6: case 9: case 11: days = 30; break; default: days = 31; break;}System.out.println(days); // 1st stmt

}}

break statement is usually added to the end of each case statements causes control to break out of the switch and jump to the 1st statement

following the switch statement Without an explicit break statement, control will flow through subsequent

case statements ( eg. case 4, 6 & 9 )

following is equivalent using nested if/else

pp-40

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placing or not to place break

in a switch/case structure gives you greater flexibility

8 boolean variables are declared & initialized as false

If form is 4 or 5, all 8 variables are updated to true

if form is 3, then 7 variables will be updated to true

Otherwise, only the last 4 variables will be updated to true eg, when form is 1 or 2

Note : switch/case is more elegant & easier to read, even though it can be rewritten using if/else

boolean takeComputer = false;boolean takePhysics = false;boolean takeChemistry = false;boolean takeBiology = false;boolean takeChinese = false;boolean takeEnglish = false;boolean takeMaths = false;boolean takeMusic = false;switch ( form ) {

case 5: case 4: takeComputer = true;case 3: takePhysics = true;

takeChemistry = true;

takeBiology = true;default: takeChinese = true;

takeEnglish = true;takeMaths = true;takeMusic = true;break;

}

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Any number of statements can be placed in the loop body of a loop

these statements can include both branching ( eg. if/else , switch/case ) and looping ( eg. do/while , while , for )

the number of levels of nested loops is unrestricted

while ( . . . ) { // outer loop- - - -

- - - -}

C) Advanced looping statements

for ( . . . ) { // inner loop A

}

pp-42

do { - - - -

} while ( . . . )

34

Example:

Screen output :

when i=0, inner loop iterates once; so, print 1 "*" when i=1, inner loop iterates twice; so, print 2 "*"s when i=2, inner loop iterates 3 times; so, print 3 "*"s when i=3, inner loop iterates 4 times; so, print 4 "*"s when i=4, inner loop iterates 5 times; so, print 5 "*"s

public class Pattern { // Self-test 6.6 pp-110public static void main(String args[]) {

for (int i=0;i<5;i++) {for (int j=0; j<=i; j++)

{System.out.print("*");

}System.out.println(" i="+ i "); // start a new

line}

}}

* i=0** i=1*** i=2**** i=3***** i=4

ensures the next output starts at a new line

keep printing on the same line UNLESS it exceed the screen width

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refer to codes-1 inner for loop can access

both i and j outer for loop can only

access i but can't access j

refer to codes-2 declaring j outside the inner

loops, both loops can access both i and j

it is not preferable, especially when j is changed in both outer and inner loops,

because logic is complicated and the program is more difficult to read & maintain.

public class Pattern { // codes-1 (self test 6.6)

public static void main(String args[]) {for ( int i=0; i<5; i++) { // outer for ( int j=0; j<=i; j++) {// inner

System.out.print("*"); } System.out.println(" i="+ i"); }

}}public class Pattern1 { // codes-2 public static void main(String args[]) { for (int i=0; i<5; i++) {

int j;for (j=0; j<=i; j++) { j *= 2; System.out.print("*");}

System.out.println(" i="+i+", j="+j); } }}

* i=0 j=1** i=1 j=3** i=2 j=3*** i=3 j=7*** i=4 j=7

Screen output

declare i for outer loop

declare j for inner loop

declare j for inner loop

j is accessible by outer loop

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Revisiting Class & Driver conceptpublic class ObjectWithNestedLoop { public void showVariablesInNestedLoop ( int outerLimit, int innerLimit ) {

int count = 0;for (int i=0; i < outerLimit; i++) {

for (int j=0; j < innerLimit; j++) {count++;System.out.println( count + " : i=" + i + " & j="

+ j); }

}System.out.println( "Total number of iterations = "+ count );

}}

public class TestObjectWithNestedLoop { public static void main(String args[]) {

if (args.length < 2) { System.out.println( "Usage: java TestObjectWithNestedLoop" + " <outer limit> <inner limit>");}else { ObjectWithNestedLoop objectWithNL

= new ObjectWithNestedLoop(); int outerLimit=Integer.parseInt(args[0]); int innerLimit=Integer.parseInt(args[1]); objectWithNL.showVariablesInNestedLoop

(outerLimit, innerLimit);}

}}

pp 45~46

37

N:\unit-6>java TestObjectWithNestedLoop 3Usage: java TestObjectWithNestedLoop <outer limit> <inner limit>

java TestObjectWithNestedLoop 2 31 : i=0 & j=02 : i=0 & j=13 : i=0 & j=24 : i=1 & j=05 : i=1 & j=16 : i=1 & j=2Total number of iterations = 6

java TestObjectWithNestedLoop 3 21 : i=0 & j=02 : i=0 & j=13 : i=1 & j=04 : i=1 & j=15 : i=2 & j=06 : i=2 & j=1Total number of iterations = 6

java TestObjectWithNestedLoop 3 0Total number of iterations = 0

java TestObjectWithNestedLoop 0 3Total number of iterations = 0

java TestObjectWithNestedLoop 1 11 : i=0 & j=0Total number of iterations = 1

java TestObjectWithNestedLoop 1 21 : i=0 & j=02 : i=0 & j=1Total number of iterations = 2

java TestObjectWithNestedLoop 2 11 : i=0 & j=02 : i=1 & j=0Total number of iterations = 2

Testing on " TestObjectWithNestedLoop "

Inner loop was called twice; each time iterated once

Inner loop was called once & it iterated twice

ensure each cases are encountered

38

Break Vs Continue

The break statement, is usually used in switch/case statements can be used in the loop body When break statement is executed, the program

control jumps to the statement after the loop body, i.e. break the loop

The continue statement the program control

jumps to the end of the loop and then go back to the condition checking and continues

pp 52

39

Table 6.14 Positions of continue statement

do/while loop for loop while loopdo {

..if (condition){

..continue;

}else {

..continue;

}..

} while ( .. ) ;

for ( .. ; .. ; .. ) {..if (condition){

..continue;

}else {

..continue;

}..

}

while ( .. ) {..if (condition){

..continue;

}else {

..continue;

}..

}

pp 53

40

Table 6.15 break vs continue

break continue

users Can be used in all looping structures & switch/case statements

Can be used in looping structures only

effects quit the loop or switch/case statement Immediately & unconditionally

1. Skips the remaining statements in loop body

2. execute the update part of the for loop, &

3. check loop condition to determine if the loop continues or not

pp 54

41

Exampleclass ContinueAndBreak { public static void main (String args[]) {

int i, max=7;for (i=1; i<=max; i++) {

if (i==4) { break; }System.out.println("i = "+i);

}System.out.println("break => ends & don't print 4 & others");for (i=1; i<=max; i++) {

if (i==4) { continue; }System.out.println("i = "+i);

}System.out.println("continue => skip 4 & resume job");

}}// adopted from Kwan Ying's example

i = 1i = 2i = 3break => ends & don't print 4 & othersi = 1i = 2i = 3i = 5i = 6i = 7continue => skip 4 & resume job

Screen Output

42

D) Sorting

Binary search requires the elements must be ordered such as ascending order of numeric values Such an ordering operation is known as sorting For example, if you have an square object with element

type of int as follows,

it is expected after sorting, the contents of the square elements have become

12 32 4 87 22

4 12 22 32 87

pp 59

43

Insertion sortStep1: Separate the sorting scope

into 2 regions :sorted, unsorted

Step 2: Copy the first element, such as A, of the unsorted region to a variable storage

Step 3: Start from the last number of the sorted region, compare with the number in the variable storage . If it is greater than A, then it moves downward.

Step 4: Insert the number A to the empty space in the sorted region

12

32

4

87

22

sorted

unsorted32

storage

0

412

32

4

87

22

sorted

unsorted 4

12

12

32

87

22

4

4

12

32

87

22

sorted

unsorted

87

4

12

32

87

22

22

unsorted

sorted

4

12

22

32

87

22

12

32

4

87

22

32

4

12

32

32

87

22

pp 61~65

44

public class InsertionSorter2 {

public static void sort(int[] numbers) {for (int sorted = 0; sorted < numbers.length - 1; sorted++) {

System.out.print("DEBUG: sorted region = ");for (int i=0; i <= sorted; i++) { System.out.print(numbers[i] + "\t"); }System.out.print("\nDEBUG: unsorted region = ");for (int i=sorted + 1; i < numbers.length; i++) { System.out.print(numbers[i] + "\

t");}System.out.println();int storage = numbers[sorted + 1];System.out.println("DEBUG: Number to be moved = " + storage);int forCompare = sorted;while (forCompare >= 0 && numbers[forCompare] > storage) {

numbers[forCompare + 1] = numbers[forCompare];System.out.println("DEBUG: " + numbers[forCompare]

+ " is copied to the next element");forCompare--;

}numbers[forCompare + 1] = storage;System.out.println("DEBUG: " + storage + " is stored in numbers["

+ (forCompare + 1) + "]\n");}

}}

pp 71

45

public class TestInsertionSorter2 {

public static void main(String args[]) {

if (args.length == 0) {

System.out.println( "Usage: java TestInsertionSorter num1 num2 ..."); }

else {

int[] numbers = new int[args.length];

for (int i=0; i < args.length; i++) { numbers[i] = Integer.parseInt(args[i]) }

InsertionSorter2 sorter = new InsertionSorter2();

sorter.sort(numbers); // Sort numbers using square object

System.out.println("Sorted numbers:"); // Show the sorted numbers on screen

for (int i=0; i < numbers.length; i++) { System.out.print(numbers[i] + "\t"); }

System.out.println();

}

}

}

46

java TestInsertionSorter2 12 32 4 87 22DEBUG: sorted region = 12DEBUG: unsorted region = 32 4 87 22DEBUG: Number to be moved = 32DEBUG: 32 is stored in numbers[1]

DEBUG: sorted region = 12 32DEBUG: unsorted region = 4 87 22DEBUG: Number to be moved = 4DEBUG: 32 is copied to the next elementDEBUG: 12 is copied to the next elementDEBUG: 4 is stored in numbers[0]

DEBUG: sorted region = 4 12 32DEBUG: unsorted region = 87 22DEBUG: Number to be moved = 87DEBUG: 87 is stored in numbers[3]

DEBUG: sorted region = 4 12 32 87DEBUG: unsorted region = 22DEBUG: Number to be moved = 22DEBUG: 87 is copied to the next elementDEBUG: 32 is copied to the next elementDEBUG: 22 is stored in numbers[2]

Sorted numbers:4 12 22 32 87

Screen output pp 72

47

E) Multidimensional Arrays

It is an array of array Objects Declaration Syntax:

eg.

Alternatively

int [] [] a = new int [5] [10];

int [] [] a = new int[5][];a[0] = new int[10];a[1] = new int[10];a[2] = new int[10];a[3] = new int[10];a[4] = new int[10];

pp 77

type[][] <variable> //

variation 1

type <variable>[][] //

variation 2

48

Example-1

double[][] threeByFourTable;threeByFourTable = new double[3][4];

threeByFourTable

[0]

[1]

[2]

3

49

int [] [] square = { {8,3,4}, {1,5,9}, {6,7,2} };

square[0]

8

3

4

1

5

9

6

7

2

[1]

[2]

3

3

3

3

square[0]

[0]

[1]

[2]

square[0].length

square[0] [0]

square[1]

square[1] [1]

square[2]

square[2] [0]

pp 80~83

Example2

50

22222121

12121111

2221

1211

2221

1211

*

baba

baba

bb

bb

aa

aaBA

2222122121221121

2212121121121111

2221

1211

2221

1211

****

****

**

babababa

babababa

bb

bb

aa

aaBA

Matrix Manipulations ( out syllabus ) Given : two 2x2 matrix A and B

2221

1211

aa

aaA

pp 82

2221

1211

bb

bbB

10

01I

Addition of A and B

Identify Matrix, diagonal

elements are all '1's; others are '0'

Product of MatrixA and B

51

public class MatrixHandler {

public void showMatrix(double[][] matrix) {for (int row=0; row < matrix.length; row++) {for (int col=0; col < matrix[row].length; col++){ System.out.print(matrix[row][col] + "\t"); }System.out.println();

}}public double sumElements(double[][] matrix){

double total = 0.0;

// Add the value in each row and columnfor (int row=0; row < matrix.length; row++){for (int col=0; col < matrix[row].length; col++)

{ total += matrix[row][col]; }}return total;

}

// cont'd on the right

public double[][] createIdentityMatrix(int rank){double[][] matrix = new double[rank][rank];for (int i=0; i < rank; i++) {

matrix[i][i] = 1.0;}return matrix;

}public double[][] multiply(

double[][] a, double[][] b) {double[][] c = new

double[a.length][b[0].length];for (int row = 0; row < a.length; row++) {

for (int col = 0; col < b[0].length; col++) {c[row][col] = 0.0;for (int i=0; i < b.length; i++) {

c[row][col] += a[row][i] * b[i][col];}

}}return c;

}}

pp 83

52

8*46*37*45*3

8*26*17*25*1

87

65*

43

21* BApublic class TestMatrixHandler2 {

public static void main(String args[]) {double[][] a = { { 1, 2}, { 3, 4} };double[][] b = { {5, 6}, { 7, 8} };MatrixHandler handler = new MatrixHandler();System.out.println("The matrix A:");handler.showMatrix(a);System.out.println("\nThe matrix B:");handler.showMatrix(b);double[][] c = handler.multiply(a, b);System.out.println(

"\nThe product matrix of matrices A and B:");handler.showMatrix(c);System.out.println(

"\nSum of all elements in matrix product="+ handler.sumElements(c));

double[][] d = handler.createIdentityMatrix(4);System.out.println("\nIdentity matrix of rank 4");handler.showMatrix(d);

}}

The matrix A:1.0 2.03.0 4.0

The matrix B:5.0 6.07.0 8.0

The product matrix of matrices A and B:19.0 22.043.0 50.0

Sum of all elements in matrix product=134.0

Identity matrix of rank 41.0 0.0 0.0 0.00.0 1.0 0.0 0.00.0 0.0 1.0 0.00.0 0.0 0.0 1.0

Screen Output

53

Ragged ( 不平的 ) Arrays Java implements 2-dimensional arrays as 2 levels of 1-dimension

arrays as the 2nd level arrays objects are mutually independent, so they can

have different size so it is called ragged arrays Example int[][] ragged = new int[3][];

int[0] = new int[5];int[1] = new int[2];int[2] = new int[4];

ragged

[0]

[1]

[2]

3

pp 87

54

F) Recursion — an advanced control structure

Recursion is a problem-solving technique that, if given a problem, is resolved to another identical problem but with a smaller size.

Such an approach is repeated until a well-defined answer is obtained.

Usually, a method is recursive if it calls itself either directly or indirectly.

Recursion is considered as a more elegant and simpler solutions.

But it also incurs larger memory and time overhead.

pp 90

55

Self-test 6.9 : summation of N numbers

given

public class Summation {public int sum ( int n ) {

if ( n == 1 ) return 1 ; else return ( n + sum ( n-1 )

) ;}

}public class TestSummation {

public static void main(String args[]) {int n = Integer.parseInt(args[0]);Summation mySum = new Summation( );System.out.println( mySum.sum (n) );

}}

1or 1

13211

1

1

nsumnnnsum

nx

nnxnsumn

n

base condition is sum (1 ) = 1

56

public int sum ( int n ) {if ( n == 1 ) return 1 ; else return ( n + sum ( n-1 ) ) ;

}

public int sum ( int n ) {if ( n == 1 ) return 1 ; else return ( n + sum ( n-1 ) ) ;

}

public int sum ( int n ) {if ( n == 1 ) return 1 ; else return ( n + sum ( n-1 ) ) ;

}

Java TestSummation

3

2

3

1

6

3

2

1

The number 6 is returned to the driver program