Loops
EECS2030: Advanced Object Oriented Programming Fall 2018
CHEN-WEI WANG
Learning Outcomes
Understand about Loops :
● Motivation: Repetition of similar actions ● Two common loops: for and while
● Primitive vs. Compound Statements
● Nesting loops within if statements
● Nesting if statements within loops
● Common Errors and Pitfalls
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Motivation of Loops
● We may want to repeat the similar action(s) for a (bounded) number of times.
e.g., Print the “Hello World” message for 100 times
e.g., To find out the maximum value in a list of numbers
● We may want to repeat the similar action(s) under certain circumstances.
e.g., Keep letting users enter new input values for calculating
the BMI until they enter “quit”
● Loops allow us to repeat similar actions either
○ a specified number of times; or
○ a specified condition holds true.
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for
while
The for Loop (1)
for (int i = 0; i < 100; i ++) { System.out.println("Welcome to Java!");
}
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The for Loop (2)
for (int i = 0; i < 100; i ++) { System.out.println("Welcome to Java!");
}
i < 100
Enter/Stay Loop?
Iteration
0 < 100
True
1
1 < 100
True
2
2 < 100
True
3
i
Actions
0 print,i ++ 1 print,i ++ 2 print,i ++
...
99 print,i ++
99 < 100
True
100
100 < 100
False
–
100
–
● The number of iterations (i.e., 100) corresponds to the number of times the loop body is executed.
● # of times that we check the stay condition (SC) (i.e., 101) is #
of iterations (i.e., 100) plus 1. [ True × 100; False × 1 ] 5 of 81
The for Loop (3)
for ( ; i < 100; ) { System.out.println("Welcome to Java!");
}
int i = 0
i ++
○ The “initial-action” is executed only once, so it may be moved right before the for loop.
○ The “action-after-each-iteration” is executed repetitively to make progress, so it may be moved to the end of the for loop body.
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int i = 0;
for (; i < 100; ) { System.out.println("Welcome to Java!");
i ++;
}
The for Loop: Exercise (1) Compare the behaviour of this program
for (int count = 0; count < 100; count ++) { System.out.println("Welcome to Java!");
}
and this program
for (int count = 1; count < 201; count += 2) { System.out.println("Welcome to Java!");
}
○ Are the outputs same or different?
○ It is similar to asking if the two intervals
[0,1,2,...,100) and [1,3,5,...,201)
contain the same number of integers.
○ Same, both loop bodies run exactly 100 times and do not depend
on the value of count. 7 of 81
The for Loop: Exercise (2) Compare the behaviour of this program
int count = 0;
for (; count < 100; ) {
System.out.println("Welcome to Java " + count + "!");
count ++; /* count = count + 1; */ }
and this program
int count = 1;
for (; count <= 100; ) {
System.out.println("Welcome to Java " + count + "!");
count ++; /* count = count + 1; */ }
Are the outputs same or different? Different, both loop body run exactly 100 times and depend on the value of count.
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The for Loop: Exercise (3)
Compare the behaviour of the following three programs:
Output: 12345
for (int i = 1; i <= 5 ; i ++) { System.out.print(i); }
int i = 1;
for ( ; i <= 5 ; ) {
System.out.print(i); i ++; }
Output: 12345
int i = 1;
for ( ; i <= 5 ; ) {
i ++; System.out.print(i); }
Output: 23456 9 of 81
The while Loop (1)
int count = 0;
while (count < 100) {
System.out.println("Welcome to Java!");
count ++; /* count = count + 1; */ }
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The while Loop (2)
int j = 3;
while (j < 103) {
System.out.println("Welcome to Java!"); j ++; /* j = j + 1; */ }
j
3 4 5 ... 102 103
Actions
print,j ++ print,j ++ print,j ++
print, j ++ –
j < 103
3 < 103
Enter/Stay Loop?
Iteration
4 < 103
True
1
5 < 103
True
2
102 < 103
103 < 103
True
True
3
100
False
● The number of
● # of times that we check the stay condition (SC) (i.e., 101) is #
–
(i.e., 100) corresponds to the number of times the loop body is executed.
iterations
of iterations (i.e., 100) plus 1. [ True × 100; False × 1 ] 11 of 81
The while Loop: Exercise (1) Compare the behaviour of this program
int count = 0;
while (count < 100) {
System.out.println("Welcome to Java!");
count ++; /* count = count + 1; */ }
and this program
int count = 1;
while (count <= 100) {
System.out.println("Welcome to Java!");
count ++; /* count = count + 1; */ }
Are the outputs same or different? Same, both loop bodies run exactly 100 times and do not depend on the value of count.
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The while Loop: Exercise (2) Compare the behaviour of this program
int count = 0;
while (count < 100) {
System.out.println("Welcome to Java " + count + "!");
count ++; /* count = count + 1; */ }
and this program
int count = 1;
while (count <= 100) {
System.out.println("Welcome to Java " + count + "!");
count ++; /* count = count + 1; */ }
Are the outputs same or different? Different, both loop body run exactly 100 times and depend on the value of count.
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Primitive Statement vs. Compound Statement
● A statement is a block of Java code that modifies value(s) of some variable(s).
● An assignment (=) statement is a primitive statement: it only modifies its left-hand-side (LHS) variable.
● An for or while loop statement is a compound statement: the loop body may modify more than one variables via other statements (e.g., assignments, if statements, and for or while statements).
○ e.g., a loop statement may contain as its body if statements
○ e.g., a loop statement may contain as its body loop statements ○ e.g., an if statement may contain as its body loop statements
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Compound Loop: Exercise (1.1)
How do you extend the following program
System.out.println("Enter a radius value:"); double radius = input.nextDouble();
double area = radius * radius * 3.14; System.out.println("Area is " + area);
with the ability to repeatedly prompt the user for a radius value, until they explicitly enter a negative radius value to terminate the program (in which case an error message is also printed)? System.out.println("Enter a radius value:");
double radius = input.nextDouble(); while (radius >= 0) {
double area = radius * radius * 3.14; System.out.println(“Area is ” + area); System.out.println(“Enter a radius value:”);
radius = input.nextDouble(); } System.out.println(“Error: negative radius value.”);
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Compound Loop: Exercise (1.2)
Another alternative: Use a boolean variable isPositive
1 System.out.println(“Enter a radius value:”); 2 double radius = input.nextDouble();
3
4
boolean isPositive = radius >= 0;
while (isPositive)
{
5 6 7 8
double area = radius * radius * 3.14; System.out.println(“Area is ” + area); System.out.println(“Enter a radius value:”); radius = input.nextDouble();
isPositive = radius >= 0; }
9
10 System.out.println(“Error: negative radius value.”);
● In L2: What if user enters 2? What if user enters -2? ● Say in L2 user entered 2, then in L8:
What if user enters 3? What if user enters -3?
● What if isPositive = radius >= 0 in L9 is missing? 16 of 81
Compound Loop: Exercise (1.3)
Another alternative: Use a boolean variable isNegative
1 System.out.println(“Enter a radius value:”); 2 double radius = input.nextDouble();
3
boolean isNegative = radius < 0;
4
5 6 7 8
double area = radius * radius * 3.14; System.out.println("Area is " + area); System.out.println("Enter a radius value:"); radius = input.nextDouble();
while (!isNegative)
{
isNegative = radius < 0; }
9
10 System.out.println("Error: negative radius value.");
● In L2: What if user enters 2? What if user enters -2? ● Say in L2 user entered 2, then in L8:
What if user enters 3? What if user enters -3?
● What if isNegative = radius < 0 in L9 is missing? 17 of 81
Converting between for and while Loops (1) ● To convert a while loop to a for loop, leave the initialization
and update parts of the for loop empty.
while(B) {
/* Actions */
}
is equivalent to:
for( ; B ; ) { /* Actions */
}
where B is any valid Boolean expression.
● However, when there is not a loop counter (i.e., i, count, etc.)
that you intend to explicitly maintain, stick to a while loop. 18 of 81
Converting between for and while Loops (2)
● To convert a for loop to a while loop, move the initialization part immediately before the while loop and place the update part at the end of the while loop body.
for(int i = 0 ; B ; i ++ ) { /* Actions */
}
is equivalent to:
int i = 0; while(B) {
/* Actions */
i ++; }
where B is any valid Boolean expression.
● However, when there is a loop counter (i.e., i, count, etc.) that
you intend to explicitly maintain, stick to a for loop. 19 of 81
Stay Condition vs. Exit Condition (1)
● A for loop or a while loop
○ stays to repeat its body when its stay condition (SC) evaluates to
true.
○ exits when its SC evaluates to false.
● Say we have two Boolean variables: boolean iP7sInStock, iP7sPlusInStock;
● When does the following loop exit?
!(iP7sInStock && iP7sPlusInStock)
this is equivalent to !iP7sInStock || !iP7sPlusInStock ● When does the following loop exit?
!(iP7sInStock || iP7sPlusInStock)
this is equivalent to !iP7sInStock && !iP7sPlusInStock 20 of 81
while(iP7sInStock && iP7sPlusInStock) { ... }
while(iP7sInStock || iP7sPlusInStock) { ... }
Stay Condition vs. Exit Condition (2)
Consider the following loop:
int x = input.nextInt(); while(10 <= x || x <= 20) {
/* body of while loop */
}
● It compiles, but has a logical error. Why?
[∵ negation of stay condition] [∵ law of disjunction] [∵lawofnegation]
● 10 > x && x > 20isequivalenttofalse,sincethereisno number smaller than 10 and larger than 20 at the same time.
● An exit condition being false means that there is no way to exit
from the loop! [infinite loops are BAD!] 21 of 81
● Think about the exit condition : ○ !(10 <= x || x <= 20)
○ !(10 <= x) && !(x <= 20) ○ 10 > x && x > 20
Problems, Data Structures, and Algorithms
● A well-specified computational problem precisely describes the desired input/output relationship.
○ Input: A sequence of n numbers ⟨a1, a2, . . . , an⟩
○ Output: The maximum number max in the input array, such that
max ≥ ai , where 1 ≤ i ≤ n
○ An instance of the problem: ⟨3, 1, 2, 5, 4⟩
● A
data in order to facilitate access and modifications.
data structure is a systematic way to store and organize ● An algorithm is:
○ A solution to a well-specified computational problem
○ A sequence of computational steps that takes value(s) as input
and produces value(s) as output
● Steps in an algorithm manipulate well-chosen data structure(s).
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Arrays: A Simple Data Structure
● An array is a linear sequence of elements. High
scores
● Types of elements in an array are the same. ○ an array of integers
○ an array of doubles
○ an array of characters
○ an array of strings
[int[]] [double[]] [char[]] [String[]] [boolean[]]
indices
○ an array of booleans
● Each element in an array is associated with an integer index. ● Range of valid indices of an array is constrained by its size.
○ The 1st element of an array has the index 0.
○ The 2nd has index 1.
○ The ith element has index i − 1.
○ The last element of an array has the index value that is equal to
the size of the array minus one. 23 of 81
Arrays: Initialization and Indexing
● Initialize a new array object with a fixed size: String[] names = new String[10];
● Alternatively, initialize a new array explicitly with its contents: String[] names = {“Alan”, “Mark”, “Tom”};
● Access elements in an array through indexing: String first = names[0];
String last = names[names.length – 1];
An illegal index triggers an ArrayInexOutOfBoundsException.
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Arrays: Iterations
● Iterate through an array using a for-loop:
for (int i = 0; i < names.length; i ++) { System.out.println (names[i]);
}
● Iterate through an array using a while-loop:
int i = 0;
while (i < names.length) {
System.out.println (names[i]);
i ++; }
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The for Loop: Exercise (3)
Problem: Given an array numbers of integers, how do you
print its average?
e.g., Given array {1, 2, 6, 8}, print 4.25.
int sum = 0;
for(int i = 0; i < numbers.length; i ++) {
sum += numbers[i]; }
double average = (double) sum / numbers.length; System.out.println("Average is " + average);
Q: What’s the printout when the array is empty
(e.g., int[] numbers = {};)?
A: Division by zero (i.e., numbers.length is 0). Fix?
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The for Loop: Exercise (4)
Problem: Given an array numbers of integers, how do you
print its contents backwards?
e.g.,Givenarray{1,2,3,4},print4 3 2 1.
Solution 1: Change bounds and updates of loop counter.
for(int i = numbers.length - 1; i >= 0; i –) { System.out.println(numbers[i]);
}
Solution 2: Change indexing.
for(int i = 0; i < names.length; i ++) { System.out.println(numbers[ names.length - i - 1 ]);
}
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The for Loop: Exercise (5)
Problem: Given an array names of strings, how do you print its
contents separated by commas and ended with a period?
e.g., Given array {”Alan”, ”Mark ”, ”Tom”}, print ”Names: Alan, Mark, Tom.”
System.out.print("Names:")
for(int i = 0; i < names.length; i ++) {
System.out.print(names[i]); if (i < names.length - 1) {
System.out.print(", "); }
} System.out.println(".");
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Array Iterations: Translating for to while (1) ● Use either when you intend to iterate through the entire array.
● int[] a = new int[100];
for(int i = 0; i < a.length; i ++) {
/* Actions to repeat. */
}
In a for loop, the initialization and update of the loop counter i
are specified as part of the loop header.
● int i = 0;
while(i < a.length) {
/* Actions to repeat. */
i ++; }
In a while loop, the loop counter i
○ Is initialized outside and before the loop header ○ Is updated at the end of the loop body
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Array Iterations: Translating for to while (2)
● In both the for and while loops:
○ The stay/continuation conditions are identical.
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○ The loop counter i is initialized only once before first entrance.
○ In each iteration, the loop counter i is executed at the end of the
loop body.
Compound Loop: Exercise (2)
Given an integer array:
int[] a = {2, 1, 3, 4, -4, 10}
How do you print out positive numbers only?
Hint: Use a for loop to iterate over the array. In the loop body, conditionally print out positive numbers only.
1 2 3 4 5
for(int i = 0; i < a.length; i ++) { if (a[i] > 0) {
System.out.println(a[i]); }
}
Exercise: Write the equivalent using a while loop. 31 of 81
Compound Loop: Exercise (3.1)
● Problem: Given an array of numbers, determine if it contains
all positive number.
1 2 3 4 5 6 7 8 9
● Change Line 5 to soFarOnlyPosNums = numbers[i] > 0;? ● Hints: Run both versions on the following three arrays:
int[] numbers = {2, 3, -1, 4, 5, 6, 8, 9, 100}; boolean soFarOnlyPosNums = true;
int i = 0;
while (i < numbers.length) {
soFarOnlyPosNums = soFarOnlyPosNums && (numbers[i] > 0);
i = i + 1; }
if (soFarOnlyPosNums) { /* print a msg. */ } else { /* print another msg. */ }
1. {2, 3, 1, 4, 5, 6, 8, 9, 100} 2. {2, 3, 100, 4, 5, 6, 8, 9, -1} 3. {2, 3, -1, 4, 5, 6, 8, 9, 100}
[all positive] [negative at the end] [negative in the middle]
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Compound Loop: Exercise (3.1) Demo (1)
1 2 3 4 5 6 7
int[] ns = {2, 3, -1, 4, 5}; boolean soFarOnlyPosNums = true; int i = 0;
while (i < ns.length) {
soFarOnlyPosNums = soFarOnlyPosNums && (ns[i] > 0);
i = i + 1; }
i
ns[i] > 0
soFarOnlyPosNums
true
i < ns.length
stay?
ns[i]
0 true 1 true 2 false 3 true 4 true
true
true
2
true
true
true
3
true
true
true
-1
false
true
true
4
false
true
true
5
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5
false
false
false
–
–
Compound Loop: Exercise (3.1) Demo (2)
1 2 3 4 5 6 7
int[] ns = {2, 3, -1, 4, 5}; boolean soFarOnlyPosNums = true; int i = 0;
while (i < ns.length) {
soFarOnlyPosNums = ns[i] > 0; /* wrong */
i = i + 1; }
i
ns[i] > 0
soFarOnlyPosNums
i < ns.length
stay?
ns[i]
0 true 1 true 2 false 3 true 4 true
true
true
true
2
true
true
true
3
true
false
true
true
-1
true
true
4
true
true
true
5
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5
false
false
–
–
true
Compound Loop: Exercise (3.2)
Problem: Given an array of numbers, determine if it contains all positive number. Also, for efficiency, exit from the loop as soon as you find a negative number.
1 2 3 4 5 6 7 8 9
int[] numbers = {2, 3, -1, 4, 5, 6, 8, 9, 100}; boolean soFarOnlyPosNums = true;
int i = 0;
while (soFarOnlyPosNums && i < numbers.length) {
soFarOnlyPosNums = numbers[i] > 0;
i = i + 1; }
if (soFarOnlyPosNums) { /* print a msg. */ } else { /* print another msg. */ }
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Compound Loop: Exercise (3.2) Demo
1 2 3 4 5 6 7
int[] ns = {2, 3, -1, 4, 5, 6, 8, 9, 100}; boolean soFarOnlyPosNums = true;
int i = 0;
while (soFarOnlyPosNums && i < ns.length) {
soFarOnlyPosNums = ns[i] > 0;
i = i + 1; }
i
ns[i] > 0
i < ns.length
ns[i]
soFarOnlyPosNums
stay?
0 true 1 true 2 false
true
true
true
2
true
true
true
3
true
true
true
-1
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3
false
true
false
–
–
Compound Loop: Exercise (3.3) Summary
Four possible solutions (posNumsSoFar is initialized as true): 1. Scan the entire array and accumulate the result.
2. Scan the entire array but the result is not accumulative.
3. The result is accumulative until the early exit point.
4. The result is not accumulative until the early exit point.
for (int i = 0; i < ns.length; i ++) { posNumsSoFar = posNumsSoFar ns[i] > 0; }
for (int i = 0; i < ns.length; i ++) { posNumsSoFar = ns[i] > 0; }
&&
for (int i = 0; posNumsSoFar posNumsSoFar = posNumsSoFar
&&
i < ns.length; i ++) { ns[i] > 0; }
&&
for (int i = 0; posNumsSoFar i < ns.length; i ++) { posNumsSoFar = ns[i] > 0; }
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&&
/* Not working. Why? */
Compound Loop: Exercise (4)
Given a non-empty integer array, e.g., int[] a = {2, 1, 3, 4, -4, 10}, find out its maximum element.
Hint: Iterate over the array. In the loop body, maintain the maximum found so far and update it when necessary.
1 2 3 4
int max = a[0];
for(int i = 0; i < a.length; i ++) {
if(a[i]>max){ max=a[i];}} System.out.println(“Maximum is ” + max);
Q: Will the program still work if we change the initialization in L1 toint max = 0?
A: NO ∵ Contents of a may be all smaller than this initial value (e.g., all negatives).
Q: Will the program still work if we change the initialization in L2
toint i = 1?
A: YES ∵ a[0] > a[0] is always false anyway. 38 of 81
Compound Loop: Exercise (4) Demo
1 2 3 4 5 6
int[] a = {2, 1, 3, 4, -4, 10}
int max = a[0];
for(int i = 0; i < a.length; i ++) {
if (a[i] > max) { max = a[i]; } }
System.out.println(“Maximum is ” + max);
i a[i] a[i] > max
update max? max
–
–
–
02
0 2 false N 2
1 1 false N 2 2 3 true Y 3 3 4 true Y 4 4-4false N 4 510 true Y 10
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Compound Loop: Exercise (5)
Problem: Given an array a of integers, how do determine if it is
sorted in a non-decreasing order?
e.g., Given {1, 2, 2, 4}, print true; given {2, 4, 3, 3} print false.
1 2
3 4
1 2
3 4
boolean isSorted = true; for(inti=0;i< a.length-1;i++){
isSorted = isSorted && (a[i] <= a[i + 1]); }
Alternatively (with early exit):
boolean isSorted = true;
for(int i = 0; isSorted && i < a.length - 1 ; i ++) {
isSorted = a[i] <= a[i + 1]; }
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Compound Loop: Exercise (5) Demo [A]
1 2 3 4 5
int[] a = {1, 2, 2, 4}
boolean isSorted = true; for(inti=0;i< a.length-1;i++){
isSorted = isSorted && (a[i] <= a[i + 1]); }
i a[i] a[i + 1] a[i] <= a[i + 1] isSorted
exit?
–
–
–
true
0N
0 1 2 true true N
1 2 2 true true N 2 2 4 true true Y
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Compound Loop: Exercise (5) Demo [B]
1 2 3 4 5
int[] a = {2, 4, 3, 3}
boolean isSorted = true; for(inti=0;i< a.length-1;i++){
isSorted = isSorted && (a[i] <= a[i + 1]); }
i a[i] a[i + 1] a[i] <= a[i + 1] isSorted
exit?
–
–
–
true
0N
0 2 4 true true N
1 4 3 false false N 23 3 true false Y
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Compound Loop: Exercise (5) Demo [C]
1 2 3 4 5
int[] a = {2, 4, 3, 3}
boolean isSorted = true;
for(int i = 0; isSorted && i < a.length - 1 ; i ++) {
isSorted = a[i] <= a[i + 1]; }
i a[i] a[i + 1] a[i] <= a[i + 1] isSorted
exit?
–
–
–
true
0N
0 2 4 true true N
1 4 3 false false Y
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Checking Properties of Arrays (1)
● Determine if all elements satisfy a property.
● We need to repeatedly apply the logical conjunction .
● As soon as we find an element that does not satisfy a property, then we exit from the loop.
e.g., Determine if all elements in array a are positive.
1 2 3 4
1 2 3 4
boolean allPos = true;
for(int i = 0; i < a.length; i ++) {
allPos = allPos && (a[i] > 0); }
Alternatively (with early exit):
boolean allPos = true;
for(int i = 0; allPos && i < a.length; i ++) {
allPos = a[i] > 0; }
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Checking Properties of Arrays (1): Demo
1 2 3 4 5
int[] a = {2, 3, -1, 4, 5, 6, 8, 9, 100}; boolean allPos = true;
for(int i = 0; allPos && i < a.length; i ++) {
allPos = a[i] > 0; }
i a[i]
a[i] > 0
allPos
exit?
–
–
true
0N
0 2 true true N
1 3 true true N
2 -1 false false Y
● Question: Why do we initialize allPos as true in Line 2?
● Question: What if we change the stay condition in Line 3 to only i < a.length?
Intermediate values of allPos will be overwritten! 45 of 81
Checking Properties of Arrays (2)
● ●
Determine if at least one element satisfies a property.
we find an element that satisfies a property, then we exit from the loop.
e.g., Is there at lease one negative element in array a? Version 1: Scanner the Entire Array
1 2
3 4
1 2 3 4
Version 2: Possible Early Exit
boolean foundNegative = false;
for(int i = 0; ! foundNegative && i < a.length; i ++) {
foundNegative = a[i] < 0; }
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As soon as
boolean foundNegative = false; for(int i = 0; i < a.length; i ++) {
foundNegative = foundNegative || a[i] < 0; }
Checking Properties of Arrays (2) Demo
1 2 3 4 5
int[] a = {2, 3, -1, 4, 5, 6, 8, 9, 100};
boolean foundNegative = false;
for(int i = 0; ! foundNegative && i < a.length; i ++) {
foundNegative = a[i] < 0; }
i a[i] 0N
a[i] < 0 false
foundNegative
!foundNegative
true N
true N false Y
exit?
–
–
false
true
0 2 1 3 2 -1
false
true
false false true
● Question: Why do we initialize foundNegative as false in Line 2?
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Observations
● In some cases, you must iterate through the entire array in order to obtain the result.
e.g., max, min, total, etc.
● In other cases, you exit from the loop as soon as you obtain the result.
e.g., to know if all numbers positive, it is certainly false as soon as you find the first negative number
e.g., to know if there is at least one negative number, it is certainly true as soon as you find the first negative number
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Compound Loop: Exercise (6.1)
Problem: Read an integer from the user, indicating how many strings exactly they want to enter, prompt them for that many strings, and then print them out.
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How many strings?
3
Enter a string:
Alan
Enter a string:
Mark
Enter a string:
Tom
You entered:
Alan Mark Tom
Compound Loop: Exercise (6.2)
1 2 3 4 5 6 7 8 9
10 11
12 13
Scanner input = new Scanner(System.in); System.out.println("How many strings?"); int howMany = input.nextInt();
String[] strings = new String[howMany]; for(int i = 0; i < howMany; i ++) {
System.out.println("Enter a string:"); String s = input.nextLine(); strings[i] = s;
}
System.out.println("You entered: "); for(int i = 0; i < strings.length ; i ++) {
System.out.print(strings[i] + " "); }
● In L10, the following three integer expressions have the same value: howMany, i, and strings.length.
● Scope of loop counter i declared in L5 is limited to L6 to L8.
● Scope of loop counter i declared in L11 is limited to L12. 50 of 81
Compound Loop: Exercise (7.1)
Problem: Read an integer from the user, indicating how many strings at most they want to enter, prompt them for up to that many strings, and then print them out as soon as exit is read.
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How many strings?
4
Enter a string:
Alan
Enter a string:
Mark
Enter a string:
exit
You entered:
Alan Mark
Compound Loop: Exercise (7.2)
1 2 3 4 5 6 7 8 9
10
11
12
13
14
15
Scanner input = new Scanner(System.in); System.out.println("How many strings?");
int howMany = input.nextInt();
boolean userWantsToContinue = true;
String[] strings = new String[howMany];
for(int i = 0; i < howMany && userWantsToContinue; i ++) {
System.out.println("Enter a string:"); String s = input.nextLine(); userWantsToContinue = !s.equals("exit"); if(userWantsToContinue) { strings[i] = s; }
}
System.out.println("You entered: "); for(int i = 0; i < strings.length ; i ++) {
System.out.print(strings[i] + " "); }
In L12, i may be that less than howMany slots are occupied.
Is this correct? [ NO ] 52 of 81
Compound Loop: Exercise (7.3)
How many strings?
4
Enter a string:
Alan
Enter a string:
Mark
Enter a string:
exit
You entered:
Alan Mark null null
● In L5, contents of array strings are all unknown:
{null, null, null, null}
● In L12, after exiting the loop, only two slots are filled: {"Alan", "Mark", null, null}
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Compound Loop: Exercise (7.4)
1 2 3 4 5 6 7 8 9
10
11
12
13
14
15
16
Scanner input = new Scanner(System.in); System.out.println("How many strings?"); int howMany = input.nextInt();
boolean userWantsToContinue = true; String[] strings = new String[howMany]; int numberOfStringsRead = 0;
for(int i = 0; i < howMany && userWantsToContinue; i ++) { System.out.println("Enter a string:");
String s = input.nextLine();
userWantsToContinue = !s.equals("exit"); if(userWantsToContinue) {
strings[i] = s;
numberOfStringsRead ++; } } System.out.println("You entered: ");
for(int i = 0; i < numberOfStringsRead ; i ++) {
System.out.print(strings[i] + " "); }
Is this correct?
[ YES ]
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Compound Loop: Exercise (7.5)
This version also works:
1 Scanner input = new Scanner(System.in); 2 System.out.println("How many strings?"); 3 int howMany = input.nextInt();
4
5 String[] strings = new String[howMany]; 6 int numberOfStringsRead = 0;
7 for(inti=0;i
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[ not always! ] [ ArrayIndexOutOfBoundException ]
Arrays: Indexing and Short-Circuit Logic (3.1)
1 2 3 4 5 6 7 8 9
10 11 12
Scanner input = new Scanner(System.in); System.out.println(“How many integers?”); int howMany = input.nextInt();
int[] ns = new int[howMany];
for(int i = 0; i < howMany; i ++) { System.out.println("Enter an integer"); ns[i] = input.nextInt(); }
System.out.println("Enter an index:"); int = input.nextInt();
i
if(
ns[i] % 2 == 0) {
0 <= i && i < ns.length &&
println(ns[i] + " at index " + i + " is even."); } else { /* Error: invalid index or odd ns[i] */ }
● Does the above code work?
ns[i] % 2 == 0 is evaluated only when the guard
[ always! ] ● Short-circuit effect of conjunction has L-to-R evaluations:
(i.e., 0 <= i && i < ns.length) evaluates to true. 58 of 81
Arrays: Indexing and Short-Circuit Logic (3.2)
1 2 3 4 5 6 7 8 9
10
11 12
Scanner input = new Scanner(System.in); System.out.println("How many integers?"); int howMany = input.nextInt();
int[] ns = new int[howMany];
for(int i = 0; i < howMany; i ++) { System.out.println("Enter an integer"); ns[i] = input.nextInt(); }
System.out.println("Enter an index:"); int = input.nextInt();
i
if(
ns[i] % 2 == 1) {
i < 0 || i >= ns.length ||
/* Error: invalid index or odd ns[i] */ }
else { println(ns[i] + ” at index ” + i + ” is even.”); }
● Does the above code work?
[ always! ] ● Short-circuit effect of disjunction has L-to-R evaluations:
ns[i] % 2 == 1 is evaluated only when the guard
(i.e., i < 0 || i >= ns.length) evaluates to false. 59 of 81
Arrays: Indexing and Short-Circuit Logic (4)
● ∵ Short-circuit evaluations go from left to right.
∴ Order in which the operands are placed matters!
● Consider the following changes to L10:
○ ns[i]%2==0 &&0<=i&&i
○0<=i&& ns[i]%2==0 &&i
● When does each change to L10 work and crash?
○ || i < 0 || i >= ns.length
ns[i] % 2 == 1
○ i < 0 || || i >= ns.length
○ i >= ns.length || || i < 0 60 of 81
ns[i] % 2 == 1
ns[i] % 2 == 1
Parallel Loops vs. Nested Loops
●
Parallel Loops :
Each loop completes an independent phase of work. e.g., Print an array from left to right, then right to left. System.out.println("Left to right:");
for(int i = 0; i < a.length; i ++) { System.out.println(a[i]); }
System.out.println("Right to left:"); for(int i = 0; i < a.length; i ++) {
System.out.println(a[a.length - i - 1]); }
Nested Loops :
Loop counters form all combinations of indices.
for(int i = 0; i < a.length; i ++) { for(int j = 0; j < a.length; j ++) {
System.out.println("(" + i + ", " + j + ")"); }}
●
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Nested Loops: Finding Duplicates (1)
● Given an integer array a, determine if it contains any duplicates. e.g., Print false for {1, 2, 3, 4}. Print true for {1, 4, 2, 4}.
● Hint: When can you conclude that there are duplicates? As soon as we find that two elements at difference indices happen to be the same
1 2 3 4 5 6
boolean hasDup = false;
for(int i = 0; i < a.length; i ++) {
for(int j = 0; j < a.length; j ++) {
hasDup = hasDup || (i != j && a[i] == a[j]);
} /* end inner for */ } /* end outer for */ System.out.println(hasDup);
● Question: How do you modify the code, so that we exit from the loops as soon as the array is found containing duplicates?
○ L2: for(...; && i < a.length; ...) ○ L3: for(...; && j < a.length; ...)
!hasDup
!hasDup
○ L4: hasDup = (i != j && a[i] == a[j]); 62 of 81
Nested Loops: Finding Duplicates (2)
1 2 3 4 5 6 7 8
/* Version 1 with redundant scan */ int[] a = {1, 2, 3}; /* no duplicates */ boolean hasDup = false;
for(int i = 0; i < a.length; i ++) {
for(int j = 0; j < a.length; j ++) {
hasDup = hasDup || (i != j && a[i] == a[j]);
} /* end inner for */ } /* end outer for */ System.out.println(hasDup);
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i
0
0
0
1
1
1
20true31 false 21true32 false 22false33 true
hasDup
false false false false false false false false false
j
0 1 2
i != j
false
true true
a[i]
1 1 1
a[j]
1 2 3
a[i] == a[j]
true
false false
0 1 2
true
false
true
2 2 2
1 2 3
false
true
false
Nested Loops: Finding Duplicates (3)
1 2 3 4 5 6 7 8
/* Version 1 with redundant scan and no early exit */ int[] a = {4, 2, 4}; /* duplicates: a[0] and a[2] */ boolean hasDup = false;
for(int i = 0; i < a.length; i ++) {
for(int j = 0; j < a.length; j ++) {
hasDup = hasDup || (i != j && a[i] == a[j]);
} /* end inner for */ } /* end outer for */ System.out.println(hasDup);
i
hasDup
j
0 false 0 false 0 true 1 true 1 true 1 true 20 true 4 4 true true 21true42 false true 22false44 true true
0 1 2
i != j
false
true true
a[i]
4 4 4
a[j]
4 2 4
a[i] == a[j]
true
false
true
0 1 2
true
false
true
2 2 2
4 2 4
false
true
false
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Nested Loops: Finding Duplicates (4)
1 2 3 4
5 6
7 8
/* Version 2 with redundant scan */ int[] a = {1, 2, 3}; /* no duplicates */ boolean hasDup = false;
for(int i = 0; i < a.length &&
i != j && a[i] == a[j]
i
0
0
0
1
1
1
20true31 false 21true32 false 22false33 true
hasDup
false false false false false false false false false
j
i != j
a[i]
a[j]
!hasDup
for(int j = 0; j < a.length && hasDup =
; i ++) {
; j ++) {
;
!hasDup
} /* end inner for */ } /* end outer for */ System.out.println(hasDup);
a[i] == a[j]
0 1 2
false
true true
1 1 1
1 2 3
true
false false
0 1 2
true
false
true
2 2 2
1 2 3
false
true
false
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Nested Loops: Finding Duplicates (5)
1 2 3 4
5 6
7 8
/* Version 2 with redundant scan and early exit */ int[] a = {4, 2, 4}; /* duplicates: a[0] and a[2] */ boolean hasDup = false;
for(int i = 0; i < a.length &&
!hasDup
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for(int j = 0; j < a.length && hasDup =
; i ++) {
; j ++) {
i j i != j a[i] a[j] a[i] == a[j]
hasDup
!hasDup
i != j && a[i] == a[j]
;
} /* end inner for */ } /* end outer for */ System.out.println(hasDup);
00false44 true false 01true42 false false 02 true 4 4 true true
Nested Loops: Finding Duplicates (6)
The previous two versions scan all pairs of array slots, but with redundancy: e.g., a[0] == a[2] and a[2] == a[0].
1 2 3 4
5 6
7 8
/* Version 3 with no redundant scan */
int[] a = {1, 2, 3, 4}; /* no duplicates */ boolean hasDup = false;
for(int i = 0; i < a.length && ; i ++) {
a[i] == a[j]
i hasDup
0 false 0 false 0 false 1 false 1 false 2334 false false
j
a[i]
a[j]
!hasDup
for(int ; j < a.length && hasDup = ;
} /* end inner for */ } /* end outer for */ System.out.println(hasDup);
; j ++) {
j=i+1
a[i] == a[j]
!hasDup
1 2 3
1 1 1
2 3 4
false false false
2 3
2 2
3 4
false false
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Nested Loops: Finding Duplicates (7)
1 2 3 4 5 6
7 8 9
10
/* Version 3 with no redundant scan:
* array with duplicates causes early exit
*/
int[] a = {1, 2, 3, 2}; /* duplicates: a[1] and a[3] */ boolean hasDup = false;
for(int i = 0; i < a.length && ; i ++) {
for(int ; j < a.length && hasDup = ;
} /* end inner for */ } /* end outer for */ System.out.println(hasDup);
; j ++) {
!hasDup
j=i+1
!hasDup
a[i] == a[j]
i j a[i] a[j] a[i] == a[j] hasDup
1 2 3
1 1 1
2 3 2
false false false
0
0
0
1223 false false 1322 true true
false false false
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Common Error (1):
Improper Initialization of Loop Counter
boolean userWantsToContinue; while (userWantsToContinue) {
/* some computations here */
String answer = input.nextLine();
userWantsToContinue = answer.equals("Y"); }
The default value for an initialized boolean variable is false. Fix?
boolean userWantsToContinue = true; while (userWantsToContinue) {
/* some computations here */
String answer = input.nextLine();
userWantsToContinue = answer.equals("Y"); }
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Common Error (2): Improper Stay Condition
for (int i = 0; i <= a.length; i ++) { System.out.println(a[i]);
}
The maximum index for array a is a.length - 1
Fix?
for (int i = 0; i < a.length; i ++) { System.out.println(a[i]);
}
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Common Error (3):
Improper Update to Loop Counter
Does the following loop print all slots of array a?
int i = 0;
while (i < a.length) {
i ++;
System.out.println(a[i]); }
The indices used to print will be: 1, 2, 3, . . . , a.length Fix?
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int i = 0;
while (i < a.length) {
System.out.println(a[i]);
i ++;
}}
int i = 0;
while (i < a.length) {
i ++;
System.out.println(a[i - 1]);
Common Error (4):
Improper Update of Stay Condition
1 2 3 4 5 6
String answer = input.nextLine();
boolean userWantsToContinue = answer.equals("Y");
while (userWantsToContinue) { /* stay condition (SC) */
/* some computations here */
answer = input.nextLine(); }
What if the user’s answer in L1 is simply Y?
An infinite loop!! ∵ SC never gets updated when a new answer is read. Fix?
String answer = input.nextLine();
boolean userWantsToContinue = answer.equals("Y"); while (userWantsToContinue) {
/* some computations here */
answer = input.nextLine(); userWantsToContinue = answer.equals("Y");
}
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Common Error (5):
Improper Start Value of Loop Counter
int i = a.length - 1; while (i >= 0) {
System.out.println(a[i]); i –; } while (i < a.length) {
System.out.println(a[i]); i ++; }
The value of loop counter i after the first while loop is −1! Fix?
int i = a.length - 1; while (i >= 0) {
System.out.println(a[i]); i –; } i = 0;
while (i < a.length) { System.out.println(a[i]); i ++; }
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Common Error (6): Wrong Syntax
How about this?
You meant:
How about this?
You meant:
or
while(int i = 0; i < 10; i ++) { ... }
for(int i = 0; i < 10; i ++) { ... }
for(i < 10) { ... }
while(i < 10) { ... }
for( ; i < 10 ; ) { ... }
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Common Error (7): Misplaced Semicolon
Semicolon (;) in Java marks the end of a statement (e.g., assignment, if statement, for, while).
int[] ia = {1, 2, 3, 4};
for (int i = 0; i < 10; i ++); {
System.out.println("Hello!"); }
Output?
Fix?
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Hello!
for (int i = 0; i < 10; i ++) { System.out.println("Hello!");
}
Index (1)
Learning Outcomes Motivation of Loops
The for Loop (1)
The for Loop (2)
The for Loop (3)
The for Loop: Exercise (1) The for Loop: Exercise (2) The for Loop: Exercise (3) The while Loop (1)
The while Loop (2)
The while Loop: Exercise (1)
The while Loop: Exercise (2)
Primitive Statement vs. Compound Statement
Compound Loop: Exercise (1.1)
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Index (2)
Compound Loop: Exercise (1.2)
Compound Loop: Exercise (1.3)
Converting between for and while Loops (1) Converting between for and while Loops (2) Stay Condition vs. Exit Condition (1)
Stay Condition vs. Exit Condition (2) Problems, Data Structures, and Algorithms Arrays: A Simple Data Structure
Arrays: Initialization and Indexing
Arrays: Iterations
The for Loop: Exercise (3)
The for Loop: Exercise (4)
The for Loop: Exercise (5)
Array Iterations: Translating for to while (1) 77 of 81
Index (3)
Array Iterations: Translating for to while (2) Compound Loop: Exercise (2)
Compound Loop: Exercise (3.1)
Compound Loop: Exercise (3.1) Demo (1) Compound Loop: Exercise (3.1) Demo (2) Compound Loop: Exercise (3.2)
Compound Loop: Exercise (3.2) Demo Compound Loop: Exercise (3.3) Summary Compound Loop: Exercise (4)
Compound Loop: Exercise (4) Demo Compound Loop: Exercise (5)
Compound Loop: Exercise (5) Demo [A]
Compound Loop: Exercise (5) Demo [B]
Compound Loop: Exercise (5) Demo [C]
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Index (4)
Checking Properties of Arrays (1)
Checking Properties of Arrays (1): Demo Checking Properties of Arrays (2)
Checking Properties of Arrays (2) Demo Observations
Compound Loop: Exercise (6.1)
Compound Loop: Exercise (6.2)
Compound Loop: Exercise (7.1)
Compound Loop: Exercise (7.2)
Compound Loop: Exercise (7.3)
Compound Loop: Exercise (7.4)
Compound Loop: Exercise (7.5)
Arrays: Indexing and Short-Circuit Logic (1)
Arrays: Indexing and Short-Circuit Logic (2)
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Index (5)
Arrays: Indexing and Short-Circuit Logic (3.1) Arrays: Indexing and Short-Circuit Logic (3.2) Arrays: Indexing and Short-Circuit Logic (4) Parallel Loops vs. Nested Loops
Nested Loops: Finding Duplicates (1)
Nested Loops: Finding Duplicates (2)
Nested Loops: Finding Duplicates (3)
Nested Loops: Finding Duplicates (4)
Nested Loops: Finding Duplicates (5)
Nested Loops: Finding Duplicates (6)
Nested Loops: Finding Duplicates (7)
Common Error (1):
Improper Initialization of Loop Counter
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Index (6) Common Error (2):
Improper Stay Condition
Common Error (3):
Improper Update to Loop Counter
Common Error (4):
Improper Update of Stay Condition
Common Error (5):
Improper Start Value of Loop Counter
Common Error (6): Wrong Syntax
Common Error (7): Misplaced Semicolon
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