CS101 Students for CS101 Students
Possible CS101 Exam Questions & Programs
Last Update
December 15, 2015 - 3:29 pm
Note: I asked a number of students prepare possible midterm/final questions and their answers.
By possible midterm/exam question what I mean is questions that are difficult and easy enough for our exams (definition of easy and difficult is according to the judgment of the submitter it does not reflect my judgment about difficulty level of the questions). I also include recommendations by your peers/classmates.
I publish them with no editing. So if you study them please be critical since they may have some typos/bugs.
I will update this document irregularly as the suggestions come and add the newest one to the top.
F. Can
Bilkent - CS101 Sec. 4 & 5
December 15, 2015 - 3:29 pm
Rummy Garden & Playing Cards
by Alihan Hüyük
Problem Description: The Rummy Garden: Helping your father!
/**
* The Rummy Garden - Line Class
* Alihan Hüyük
* Please refer to the question description "The Rummy Garden"
*/
public class Line
{
//I chose to define a line with a slope value and a point
private double m;
private Point p;
public Line(Point p1, Point p2)
{
p = new Point(p1);
//I do not want p to change values if I edit p1 in elsewhere so I copy p1.
m = (p2.getY() - p1.getY()) / (p2.getX() - p1.getX());
//m will end up being infinite if the line is perfectly vertical
}
public Boolean onTop(Point p)
{
if(!Double.isFinite(m)) return null;
//question does not make sense
return p.getY() < this.p.getY() + m * (p.getX() - this.p.getX());
//it is just some algebra you can easily verify it
//used smaller signed as garden's y-axis is inverted
}
public Boolean onLeft(Point p)
{
if(m == 0) return null;
//question does not make sense
if(!Double.isFinite(m)) return p.getX() < this.p.getX();
//the case where the line is perfectly vertical
return p.getX() < this.p.getX() + (1/m) * (p.getY() - this.p.getY());
//yet again some basic algebra
}
}
//==========================================================================
/**
* The Rummy Garden - Point Class
* Alihan Hüyük
* Please refer to the question description "The Rummy Garden"
*/
public class Point
{
private double x;
private double y;
public double getX() { return x; }
public double getY() { return y; }
public Point(double x, double y)
{
this.x = x;
this.y = y;
}
public Point()
{
//calls the first constructor
this(0, 0);
}
public Point(Point p)
{
this(p.x, p.y);
}
}
//==========================================================================
import java.util.Scanner;
/**
* The Rummy Garden - Main Class
* Alihan Hüyük
* Please refer to the question description "The Rummy Garden"
*/
public class RummyMain
{
public static void main(String[] args)
{
Scanner scanner = new Scanner(System.in);
Line left, right, top, bottom;
System.out.println("Enter the left most vertical line:");
left = scanLine(scanner);
System.out.println();
System.out.println("Enter the right most vertical line:");
right = scanLine(scanner);
System.out.println();
System.out.println("Enter the top most vertical line:");
top = scanLine(scanner);
System.out.println();
System.out.println("Enter the bottom most vertical line:");
bottom = scanLine(scanner);
System.out.println();
while(true)
{
double x, y;
System.out.print("x: ");
x = scanner.nextDouble();
if(x < 0) break;
System.out.print("y: ");
y = scanner.nextDouble();
if(y < 0) break;
Point p = new Point(x, y);
int xCoordinate = 2;
if(right.onLeft(p))
{
xCoordinate = 1;
if(left.onLeft(p)) xCoordinate = 0;
}
int yCoordinate = 2;
if(bottom.onTop(p))
{
yCoordinate = 1;
if(top.onTop(p)) yCoordinate = 0;
}
int region = yCoordinate * 3 + xCoordinate + 1;
System.out.println("This point is in the region number " + region + ".");
System.out.println();
}
}
public static Line scanLine(Scanner scanner)
{
double x, y;
Point p1, p2;
System.out.print("\tx1: ");
x = scanner.nextDouble();
System.out.print("\ty1: ");
y = scanner.nextDouble();
p1 = new Point(x, y);
System.out.print("\tx2: ");
x = scanner.nextDouble();
System.out.print("\ty2: ");
y = scanner.nextDouble();
p2 = new Point(x, y);
return new Line(p1, p2);
}
}
Problem Description: Playıng Cards
/**
* Playing Cards - Card Class
* Alihan Hüyük
* Please refer to the question description "Playing Cards"
*/
public class Card
{
private int suit;
private int value;
// -1-
public Card(int suit, int value)
{
this.suit = suit;
this.value = value;
}
// -2-
public boolean smallerThan(Card other)
{
if(suit < other.suit) return true;
return value < other.value;
}
// -3-
public String toString()
{
String string = "";
if(value == 1) string += "Ace";
else if(value == 2) string += "Two";
else if(value == 3) string += "Three";
else if(value == 4) string += "Four";
else if(value == 5) string += "Five";
else if(value == 6) string += "Six";
else if(value == 7) string += "Seven";
else if(value == 8) string += "Eight";
else if(value == 9) string += "Nine";
else if(value == 10) string += "Ten";
else if(value == 11) string += "Jack";
else if(value == 12) string += "Queen";
else if(value == 13) string += "King";
string += " of ";
if(suit == 0) string += "Clubs";
else if(suit == 1) string += "Diamonds";
else if(suit == 2) string += "Spades";
else if(suit == 3) string += "Hearths";
return string;
}
// -4-
public boolean equals(Card other)
{
return suit == other.suit && value == other.value;
}
}
//==========================================================================
import java.util.ArrayList;
/**
* Playing Cards - Deck Class
* Alihan Hüyük
* Please refer to the question description "Playing Cards"
*/
public class Deck
{
ArrayList<Card> deck;
// -1-
public Deck()
{
deck = new ArrayList<Card>();
}
// -2-
public void addRandom()
{
int suit = (int) (Math.random() * 4);
int value = (int) (Math.random() * 13) + 1;
Card newCard = new Card(suit, value);
//checking if we can add this new card
boolean canAdd = true;
for(Card card : deck)
{
canAdd = !card.equals(newCard);
}
if(canAdd) deck.add(newCard);
else addRandom(); //try with a different random card
//A better way to implement this would be to make a list of cards which are not
//in the deck and choose a random card among them.
}
// -3-
public void sort()
{
//I will use buble sort for the implementation of this method.
//You can refer to some external sources if explanation is needed.
int lastIndex = deck.size() - 2;
while(lastIndex >= 0)
{
for(int i = 0; i <= lastIndex; i++)
{
if(deck.get(i + 1).smallerThan(deck.get(i)))
{
//swapping i and i + 1
Card temp = deck.get(i);
deck.set(i, deck.get(i + 1));
deck.set(i + 1, temp);
}
}
lastIndex--;
}
}
// -4-
public void shuffle()
{
//I will shuffle the deck by performing a set number of random swaps
//As there are maximum 52 cards, I chose that number to be 52.
for(int i = 0; i < 52; i++)
{
//choosing cards to swap
int i1 = (int) (Math.random() * deck.size());
int i2 = (int) (Math.random() * deck.size());
//swapping
Card temp = deck.get(i1);
deck.set(i1, deck.get(i2));
deck.set(i2, temp);
}
}
// -5-
public void cut()
{
ArrayList<Card> newDeck = new ArrayList<Card>();
int cutPosition = (int) (Math.random() * deck.size());
for(int i = cutPosition; i < deck.size(); i++)
{
newDeck.add(deck.get(i));
//first adding the cards after the cutPosition (in order)
}
for(int i = 0; i < cutPosition; i++)
{
newDeck.add(deck.get(i));
//then adding the cards before the cutPosition (in order)
}
deck = newDeck;
}
// -6-
public String toString()
{
String string = "";
for(Card card : deck)
string += card + "\n";
//removing the excess "\n" left from the last iteration of for loop
string.substring(0, string.length() - 1);
return string;
}
}
December 13, 2015 - 12:49 pm
Point - Vector Classes & Some Recommendations
by Kadir Can Çelik
CS101 – Questions About Classes
One question that comes to mind when learning to write programs is about how it will be useful in the future. As seen in lab07, we can simulate the behavior of a die and build a game upon it. In fact, what you can do with programming is probably only limited by your own imagination. In the following questions, I wanted to show briefly how useful would programming be if you want to do stuff about maths or physics. We will implement a Point and a Vector class. Actually, sciences is one of the areas that need programming as they need simulations and numerical computations. We will not do any fancy stuff for now but these baby steps may take us to that place, in future.
Some recommendations:
- Use the keyword “this”, get used to it, even if you don't use it, you might see some code involving it.
- Use anonymous objects.
- Not just implement the features that are mentioned, implement other features that seem reasonable to you.
- Use the methods that you write in a class, in another classes. You know what they say, don't invent the wheel again and again.
- Always write testers, as everyone is likely to make a mistake, find your mistake before it causes serious stuff.
- Observe, some strange things may happen. Try to understand these strange things, Google and the textbook are my best friends and highly suggest you to use them, too. I may not promise for anybody else but if you think I may help you, feel free to contact me via kcancelik [at] gmail.com. If there is a problem with the codes, please let me know.
1) Write a class that represents a point in two dimensions using Cartesian coordinates. Obviously it will have two instance variables, corresponding to x coordinate and y coordinate.
To make it more interesting, assign a name automatically to every Point object like “P1”, “P2” etc…
Write two constructor methods, one without any parameters and one with two parameters which are coordinates.
- Write accessors and mutators for the instance variables except for the name.
- Implement important methods like toString, equals, and compareTo.
- toString method should return the name and the coordinates.
- equals method should compare the coordinates and return a boolean value.
- compareTo should compare two points with respect to their distance from the origin, (0, 0). (I suggest writing a separate method for getting the distance from another point, as it looks useful for many reasons. )
- I highly recommend implementing a clone method to avoid problems about references. Actually, there is more into the clone method if you are interested enough to check the textbook, Cloneable interface although it is not necessary, at least for now.
- This should seem easy to you after doing the lab07 and lab08 assignments. However, there are some new stuff. For example, how can you name every object according to their construction order? “P1”, “P2”, “P3”… This looks like a counter, and in fact this counter cannot be inside an object because it goes up whenever a new object gets created. It looks like it is a property of a class rather than an object. We have seen such things in the past.
- If you were able to implement the naming thing, what happens if you create anonymous objects? Does the count go up or is an assignment necessary? What would you expect and want to happen in such a case?
- Note that what we wrote as the Point class takes place in two dimensions. How would you implement it in three dimensions? Four dimensions? Maybe you are some kind of a mad scientist and you have to work in N dimensions. Three and four may seem reasonably easy as they just require new variables for coordinates. For now, N dimensions require more work, but it is doable as well. Just think about it. Hopefully, there is an easier way to do it in N dimensions, and hopefully it will be covered soon, at least the name of lab08 suggests it. (That was a clue maybe!) When ArrayLists are covered in the class, try to implement a Point class covering coordinates in N-d.
2) Use the Point class as an instance variable to implement a Vector class. Although it may seem a good idea to have two instance variables for the end point and the start point, I suggest you to implement it only using one instance variable as vectors have equivalence relationship. This is just a technical detail, do not bother with it too much. Also, in Vector class, don't bother with the naming thing about the Point class, it was just to see how to implement such a counter into a class, and in fact it was unnecessary for the Point class in my opinion.
- Again create two constructors, one without any parameters and one with two Point parameters. Yes, you will take two parameters and you will find a way to fit this information into one Point object.
- One important property of vectors is their length, find a way to calculate it.
- Note that a clone method is almost every time useful, also toString, equals and compareTo.
- One another property of vectors is that they may undergo some operation which is called “scalar product”. Implement it.
- You can use the scalar production method to find out the angle between two vectors, implement it. You may need some library methods, use the API documentation to find them. In fact, using API documentation is an important skill.
public class Point {
private static int lastId = 0;
private String name;
private double x;
private double y;
// Default constructor w/o any parameters
public Point () {
x = 0.0;
y = 0.0;
lastId = lastId + 1;
name = "P" + lastId;
}
// A constructor with two parameters for the two instance variables
public Point (double x, double y) {
this.x = x;
this.y = y;
lastId = lastId + 1;
name = "P" + lastId;
}
// Accessor methods
public double getX () {
return x;
}
public double getY () {
return y;
}
// Mutator methods
public void setX (double x) {
this.x = x;
}
public void setY (double y) {
this.y = y;
}
public double getDistance (Point other) {
double result;
result = Math.sqrt (Math.pow (x - other.x, 2) + Math.pow (y - other.y, 2));
return result;
}
public Point clone() {
Point result;
result = new Point();
result.setX (this.x);
result.setY (this.y);
return result;
}
public String toString () {
String result;
result = name + " x: " + x + " y: " + y;
return result;
}
public boolean equals (Point other) {
boolean result;
result = (this.x == other.x && this.y == other.y);
return result;
}
public int compareTo (Point other) {
int result;
double distance1;
double distance2;
// Ouch, here we increment the lastId by two by creating two anonymous objects
distance1 = this.getDistance(new Point ());
distance2 = other.getDistance(new Point());
lastId = lastId - 2;
if ( distance1 < distance2 ) {
result = -1;
} else if ( distance1 == distance2 ) {
result = 0;
} else {
result = 1;
}
return result;
}
} // End of Point class
public class Vector {
private Point positionVector;
// Default constructor
public Vector() {
positionVector = new Point();
}
// A constructor with two Point parameters
public Vector (Point initialPoint, Point finalPoint) {
positionVector = new Point();
positionVector.setX (finalPoint.getX() - initialPoint.getX());
positionVector.setY (finalPoint.getY() - initialPoint.getY());
}
public Point getPositionVector () {
return (positionVector.clone());
}
public void setPositionVector (Point positionVector) {
this.positionVector = positionVector.clone();
}
public double getLength() {
return positionVector.getDistance();
}
public Vector clone() {
Vector result;
result = new Vector ();
result.setPositionVector (positionVector);
return result;
}
public String toString () {
String result;
result = positionVector.toString();
return result;
}
public boolean equals (Vector other) {
boolean result;
result = (this.getPositionVector().equals (other.getPositionVector()));
return result;
}
public int compareTo (Vector other) {
int result;
double length1;
double length2;
length1 = this.getLength();
length2 = other.getLength();
if ( length1 < length2 ) {
result = -1;
} else if ( length1 == length2 ) {
result = 0;
} else {
result = 1;
}
return result;
}
public double getScalarProduct (Vector other) {
double result;
result = this.positionVector.getX() * other.positionVector.getX();
result += this.positionVector.getY() * other.positionVector.getY();
return result;
}
public double getAngleBetween (Vector other) {
double result;
result = Math.acos (this.getScalarProduct (other) / (this.getLength() * other.getLength()));
return result;
}
} // End of Vector class
Check TC Kimlik No. Validity
by Ünal Ege Gaznepoğlu
/**
* @author Unal Ege Gaznepoglu
* @date 2015/12/12
*
* As you know, there are some certain restrictions which a number should
* comply if that number is an Turkish ID Number. This class will store a
* long integer and when requested it will check if this number is a valid
* ID Number or not.
*
* Checking function will return number of errors. If it returns 0, then
* provided number is a Turkish Citizenship Number probably.
*
* For purposes of this program, we will only care about 3 of the
* rules.
*
* After creating a checker class, we can create another method to create
* random numbers which comply with all of the conditions.
* (Although this is not a P = NP problem (my opinion at first glance),
* I will first create a random number then check whether it is a
* Citizenship number or not. If it is not, then I will decrease the value by one.
* This will keep happening until it complies with all of the rules.
*
* Another approach might be solving solution set for all equations, then plugging in
* required amount of random digits. If requested, I can code it as well.
*/
public class TcKimlikNo
{
private long _number;
public TcKimlikNo(long number)
{
_number = number;
}
public long generate()
{
_number = Math.random()*900000000000L + 100000000000L;
while(this.check() != 0)
{
_number --;
}
return _number;
}
public int check()
{
//keeping up each result in this ArrayList.
ArrayList<Boolean> conditions = new ArrayList<Boolean>();
//to ease up the process, let's keep digits seperated in another ArrayList.
//There is no possibility of having a zero in first digit, so there is no
//specially coded situation in while loop.
ArrayList<Integer> digits = new ArrayList<Integer>();
long tempNumber = _number;
while(tempNumber > 0)
{
digits.add((int)(tempNumber % 10));
tempNumber = tempNumber /10;
}
ArrayList<Integer> temp = new ArrayList<Integer>();
for(int i =0;i<11;i++) //to revert wrong order.
{
temp.add(digits.get(10-i));
}
digits = temp;
//lets start checking some of the conditions.
//first is summing up all 10 digits, modulus of 10 which should be equal to 11th digit.
int sumOfDigits = 0;
for(int i = 0;i<10;i++)
{
sumOfDigits += digits.get(i);
}
conditions.add(digits.get(10) == (sumOfDigits % 10)); //first condition is finished.
//now lets check the second condition. 7*(1th,3th,5th,7th and 9th) - (2th,4th,6th,8th) % 10 should be 10th digit.
sumOfDigits = 0;
for(int i =0;i<9;i++)
{
if(i%2 == 0)
sumOfDigits += (7*digits.get(i));
else
sumOfDigits += (-1*digits.get(i));
}
conditions.add(digits.get(9) == (sumOfDigits%10));
//and the third condition: 8 times 1,3,5,7,9 = 11.
sumOfDigits = 0;
for(int i = 0;i<5;i++)
{
sumOfDigits += 8*digits.get(2*i);
}
conditions.add(digits.get(10) == (sumOfDigits%10));
//lets print out the conditions.
System.out.println(conditions);
//now lets count numbers of false conditions.
int sum=0;
for(int i = 0; i<conditions.size(); i++)
{
sum += (1-boolToInt(conditions.get(i)));
}
return sum;
}
private static int boolToInt(Boolean x) //although there are much easier ways to do this, I loved this alternative at stackoverflow and wanted to share it.
{
return -("false".indexOf(""+x));
}
}
December 2, 2015 - 3:20 pm
Three programs easiest to hardest
by Alihan Hüyük
A PROGRAMMER'S BIRTHDAY
import java.util.Scanner;
/**
* Alihan Hüyük
* Please refer to the question description "A Programmer's
Birthday"
*/
public class Date
{
// 1 : properties
private int day;
private int month;
private int year;
// 2 : default constructor
public Date()
{
day = 7;
month = 10;
year = 1996;
}
// 3 : a constructor which takes three parameter for each of the
properties
public Date(int day, int month, int year)
{
// !! TRICKY PART
while(day <= 0) //while not if, because the day can be
-40 than we should add 30 to the day two times
{
day += 30;
month--;
}
while(day > 30)
{
day -= 30;
month++;
}
while(month <= 0)
{
month += 12;
year--;
}
while(month > 12)
{
month -= 12;
year++;
}
this.day = day;
this.month = month;
this.year = year;
}
// 4 : a constructor which takes only the number of days
public Date(int day)
{
this(31 + day, 12, 1999);
//the constructor takes care of the limits
}
// 5 : the static add and sub functions
public static Date add(Date d1, Date d2)
{
return new Date(d1.day + d2.day, d1.month + d2.month,
d1.year + d2.year);
//the constructor takes care of the limits
}
public static Date sub(Date d1, Date d2)
{
return new Date(d1.day - d2.day, d1.month - d2.month,
d1.year - d2.year);
//the constructor takes care of the limits
}
// 6 : toString function with leading zeros
public String toString()
{
String result = "";
if(day < 10) result += "0"; //adds leading
zeros if the day is an one-digit number
result += day + ".";
if(month < 10) result += "0";
result += month + ".";
result += year;
return result;
}
// * : calculating how much time a person lived
public static void main(String[] args)
{
Scanner scanner = new Scanner(System.in);
Date birth;
Date today;
int a, b, c;
System.out.println("Please enter your birthday. (as
day, month, year; one by one");
a = scanner.nextInt();
b = scanner.nextInt();
c = scanner.nextInt();
birth = new Date(a, b, c);
System.out.println();
System.out.println("Please enter the current date. (as
day, month, year; one by one");
a = scanner.nextInt();
b = scanner.nextInt();
c = scanner.nextInt();
today = new Date(a, b, c);
System.out.println();
System.out.println("You lived exactly this much: "
+ Date.sub(today, birth));
}
}
THINKING RATIONAL
/**
* Alihan Hüyük
* Please refer to the question description "Thinking Rational"
*/
public class Rational
{
// 1 : properties
private int n;
private int dn;
// 2: default constructor
public Rational()
{
n = 1;
n = 1;
}
// 3: a constructor with two parameters for each property
public Rational(int nominator, int denominator)
{
n = nominator;
dn = denominator;
}
// 4: a constructor which takes a double number
public Rational(double d)
{
// !! TRICKY PART
dn = 1;
while(d != (int)d) //this means d still has decimal points
{
d *= 10; //we multiply d with 10 to obtain an
integer, ex.: 1.25 -> 125
dn *= 10; //we count how many times we should
divide the result to get back the original number
}
//d has not decimal points now
n = (int) d;
}
// 5 : simplify function
public void simplify()
{
// !! TRICKY PART
int gcd = 1; //greatest common divisor
for(int i = 2; i <= Math.min(n, dn); i++)
{
if(n % i == 0 && dn % i == 0)
gcd = i;
}
n /= gcd;
dn /= gcd;
}
// 6 : arithmetic operation functions
public static Rational add(Rational q1, Rational q2)
{
Rational result = new Rational(q1.n * q2.dn + q1.dn * q2.n,
q1.dn * q2.dn);
result.simplify();
return result;
}
public static Rational sub(Rational q1, Rational q2)
{
Rational result = new Rational(q1.n * q2.dn - q1.dn * q2.n,
q1.dn * q2.dn);
result.simplify();
return result;
}
public static Rational mul(Rational q1, Rational q2)
{
Rational result = new Rational(q1.n * q2.n, q1.dn * q2.dn);
result.simplify();
return result;
}
public static Rational div(Rational q1, Rational q2)
{
Rational result = new Rational(q1.n * q2.dn, q1.dn * q2.n);
result.simplify();
return result;
}
// 7 : equals function
public boolean equals(Rational other)
{
//if a:b = c:d then ad = bc
return n * other.dn == dn * other.n;
}
// 8 : biggerThan and smallerThan
public boolean biggerThan(Rational other)
{
//we are checking if a:b > c:d
if(Math.signum(dn) == Math.signum(other.dn))
{
//if b and d has the same sign then a*d > b*c
return n * other.dn > dn * other.n;
}
else
{
//if b and d has different signs then a*d <
b*c
return n * other.dn < dn * other.n;
}
}
public boolean smallerThan(Rational other)
{
// !! SLIGHTLY TRICKY PART
return !equals(other) && !biggerThan(other);
}
// 9 : this was a really bad idea
// 10 : isInteger function
public boolean isInteger()
{
//if the denominator divides the nominator, the number is an
integer
return n % dn == 0;
}
// 11 : toString function
public String toString()
{
return n + " over " + dn;
}
}
DIGITAL ATTENDANCE
/**
* Alihan Hüyük
* Please refer to the question description "Digital Attendance"
*/
public class Student
{
//properties
private String name;
public boolean attendance;
private Student nextStudent;
//functionality 1 : initialize list with the first student’s name
public Student(String name)
{
this.name = name;
attendance = false;
nextStudent = null; //there is not a next student yet
//your secret "feature"
if(name.equals("Alihan Hüyük")) attendance = true;
}
//functionality 2 : add students to the list one by one
public void add(String name)
{
if(nextStudent == null)
{
//this is the end of the list so we should add
the new student
nextStudent = new Student(name);
}
else
{
// !! TRICKY PART
//this is not the end of the list
//we should continue to look for the end of the
list
nextStudent.add(name);
}
}
//functionality 3 : how long the list is
public int size()
{
if(nextStudent == null)
{
//this is the end of the list
//there is exactly one student after this point
return 1;
}
else
{
// !! TRICKY PART
//this is not the end of the list
//the number of students after this point is:
return 1 + nextStudent.size();
}
}
//functionality 4 : get the student with a specific name
public Student find(String name)
{
if(this.name.equals(name))
{
//we found the student
return this;
}
else if(nextStudent == null)
{
//this is the end of the list and we could not
find the student
return null;
}
else
{
// !! TRICKY PART
//we need to continue searching
//maybe the next student is who we are searching
for
return nextStudent.find(name);
}
}
}
Nov. 20, 2015 - 11:20 am
CS101-3
Mert MERMERCİ
EXERCISES FOR MIDTERM
Let’s consider a triangle which is made up with numbers.
1 The one which is
on the left can be a good example for that. Take a value from the
user
121 (the number
for the example is 3) and print out a triangle.
12321
If you do that, let’s try a different one.
1 Take a value from
the user and print out a triangle like that (the number in the
example
212 is 3).
32123
If you do the previous ones, let’s try something strange.
********* Take a height and a
thickness value from the user and print out a diamond shape.
*********** (the height value is 1 and thickness
value is 11 for the example)
*********
*******
*****
***
*
I know that these exercises are not tough but I think these can be good exercise for ‘for’ loops.( First two are working well for one digit numbers, I’m a beginner so if you have any comment or corrections please contact with me on mermertci@gmail.com .)
SOLUTION-1
// NUMBER TRIANGLE
int number;
number = scan.nextInt();
//First loop is for the lines of the triangle
for(int r = 1; r <= number; r++)
{
//Second loop is for the blanks.
for(int s = 0; s < number - r; s++)
{
System.out.print(" ");
}
/*
* Third loop is for printing out the
values.
* I divide the value part two pieces.
* First piece is printing out the numbers
till the value like 123.
*/
for(int t = 1; t <= r; t++)
{
System.out.print(t);
}
/*
* It's the second part of the value
printing.
* It prints out the reamining part 21.
* When we combine with the upper one 12321
is printing out.
*/
for(int u = r - 1;u > 0; u--)
{
System.out.print(u);
}
/*Another loop for blanks.
* This for loop is not necessary
especially.
* But if it is expected to print out
values different from blank, it's necessary.
*/
for(int v = 0; v < number - r; v++)
{
System.out.print(" ");
}
System.out.println();
}
SOLUTION-2
//NUMBER TRIANGLE 2
int number2;
number2 = scan.nextInt();
for(int a = 1; a <= number2; a++)
{
for(int b = 0; b < number2 - a; b++)
{
System.out.print(" ");
}
for(int c = a; c > 0; c--)
{
System.out.print(c);
}
for(int d = 2; d <= a ; d++)
{
System.out.print(d);
}
for(int e = 0; e < number2 - a; e++)
{
System.out.print(" ");
}
System.out.println();
}
SOLUTION - 3
//REAL DIAMOND
int height2;
int thickness2;
System.out.println("Enter the height: ");
height2 = scan.nextInt();
System.out.println("Enter the thickness: ");
thickness2 = scan.nextInt();
/*
* Thickness should be an odd number.
* If it's not, the printing shape would not be a
diamond.
*/
if(thickness2 % 2 == 1)
{
// This loop is for the head of the diamond.
for(int f = height2; f > 1; f--)
{
for(int g = 0; g < f; g++)
{
System.out.print("
");
}
for(int h = 0; h < thickness2 -
2*f; h++)
{
System.out.print("*");
}
for(int i = 0; i < f; i++)
{
System.out.print("
");
}
System.out.println();
}
// This loop is for the triangle in the
diamond.
for(int m = thickness2 ; m >= 1; m = m
- 2)
{
for(int n = 0; n < (thickness2 -
m)/2; n++)
{
System.out.print("
");
}
for(int o = 0; o < m; o++)
{
System.out.print("*");
}
for(int p = 0; p < (thickness2 -
m)/2; p++)
{
System.out.print("
");
}
System.out.println();
}
}
else
{
System.out.print("Invalid value");
}
CS101 – Midterm1 Question by Gökberk DAĞAŞAN & Ali Semi YENİMOL
Write a full program that draws a diamond with asterisks ( “*” ) but only the edges of the diamond will be visible. Side length will be given by user. For example, when user enter 4, it will print:
*
* *
* *
* *
* *
* *
*
Good luck in midterm1 everyone
SOLUTION
import java.util.Scanner;
public class DiamondWithSpace{
public static void main(String [] args){
//Variables
int sideLength;
int tmp;
Scanner scan = new
Scanner(System.in);
//Prompting the user
System.out.print( "Enter the
side length of the diamond : ");
sideLength = scan.nextInt();
for(int a = 0 ; a < sideLength ;
a++){
tmp =
sideLength;
sideLength =
a;
//Drawing
the top of the diamond
for ( int i
= 0 ; i < sideLength ; i++){
for( int k = 0 ; k < sideLength*2 ; k++){
if( k ==
sideLength - i || k == sideLength + i ){
System.out.print("*");
}
else{
System.out.print(" ");
}
}
System.out.println("");
}
//Drawing
the bottom of the diamond
for(int i =
sideLength - 2; i >=0 ; i--){
for(int k = 0 ; k < sideLength*2 ; k++){
if(k == sideLength - i || k == sideLength + i){
System.out.print("*");
}
else{
System.out.print(" ");
}
}
System.out.println("");
}
System.out.println("---------------------------------------------------");
sideLength =
tmp;
}
}
}
ALTERNATIVE SOLUTION
import java.util.Scanner ;
public class aa {
public static void main(String[] args) {
// Variables
int height;
// Code
Scanner sc = new Scanner(System.in);
height = sc.nextInt();
// Upper part being printed
for ( int i = 0 ; i < height ;
i++ )
{
for ( int j
= 0; j < height - i - 1; j++ )
{
System.out.print( " " );
}
System.out.print( "*" );
for ( int k
= 0; k < 2*i - 1; k++ )
{
System.out.print( " " );
}
if ( i != 0
)
{
System.out.print( "*" );
}
System.out.println();
}
// Bottom part being printed
for ( int m = height - 2 ; m >= 0
; m--)
{
for ( int b
= 0; b < height - 1 - m; b++ )
{
System.out.print( " " );
}
System.out.print( "*" );
for ( int z
= 0; z < 2*m -1; z++ )
{
System.out.print( " " );
}
if ( m != 0
)
{
System.out.print ( "*" );
}
System.out.println();
}
}
}
CS101 JAVA QUESTIONS BY COBAN, CAGAN SELIM
- Assume that you
are a compiler. Right now, you’re examining a basic factorial calculator
program which examines the divisions by 7. “factorialFunction” is your
method’s name which calculates factorials. A piece of code is going to
execute by the user:
import java.util.Scanner;
public class Question1
{
Scanner sc = new Scanner(System.in);
public static void main( String[] args){
//Initializing the variables and scanner
int val= sc.nextint();
//Entering the loop
for(i=1; i <= val ;
i++){
int factorialVal = factorialFunction.(i);
double result = factorialVal/7;
}
System.out.print.ln(“Factorial of ”val + “is” +
result);
}
- Show and correct
all syntax errors. Remember that Java is a case sensitive language!
- After you have
corrected all the syntax errors, what is the logical problem would be?
Explain briefly.
Answer A:
import java.util.Scanner;
public class Question1
{
public static void main( String[] args){
//Initializing the variables and scanner
Scanner sc = new Scanner(System.in);
int val= sc.nextInt();
//Entering the loop
for(int i=1; i <=
val; i++){
int factorialVal = factorialFunction(i);
double result = factorialVal/7;
System.out.println(result);
}
}
Answer B:
The logical problem in this code segment is an
uninteded integer division. When the all inputs are integers, “/” denotes does
not mean division in a mathematical sense, it means integer division because
both factorialVal and 7 are integers. As a consqeunce, the output will be
integers instead of doubles. If we enter 5, we’ll see this output:
0.0
0.0
0.0
3.0
17.0
- Today is the
first day of your new job at a software company. Your supervisor has
asked you to design a basic ASCII Generator Program. The program should
generate as follows:
- Get an integer
input from user.
- Generate square
including a left side diagonal consisted of o’s, and fill up the remaining
spaces by x.
If the user has entered 4, your generated figure should be like this:
oxxx
xoxx
xxox
xxxo
Answer:
import java.util.Scanner;
public class Question2
{
public static void main( String[] args){
Scanner sc = new Scanner(System.in);
int input = sc.nextInt();
String leftString="";
String rightString="";
int l=-1;
for(int c = input; c != 0; c--){
l++;
rightString =
"";
leftString =
"";
for(int i=0; i
< l; i++){
leftString += "x";
}
for(int i=0; i
< c-1 ; i++){
rightString += "x";
}
System.out.println(leftString + "o" + rightString);
}
}
}
- Your Calculus
Professor wants to round the nearest integer and get absolute values of a
set of grades which includes large number of rational numbers. Then in
order to help her, you have decided to write a method. Write the code
segment below. Kind request: Do not use abs() method!
public static int gradeMethod(double n) {
double result=n;
if(n >= 0){
// Does
nothing
}
else{
result *=
-1;
}
//Once the method gets the absolute
value of n, continues to round
int bufferRound = (int)result;
double dif = result - bufferRound;
if(dif >= 0.5){
bufferRound++;
}
else{
//Does nothing
}
return bufferRound;
}
Nov. 16, 2015 - 12:02 pm
CS101 – Midterm 1 Questions by K. Can Çelik
1. A positive integer is said to be “perfect” if the sum of all positive divisors except for the number itself is equal to that number. Write a static method that would take one formal parameter “number” of approriate type, that would return “true” if the number is perfect and “false” if the number is not perfect.
SOLUTION:
public static boolean isWonderful
(int number) {
int sum;
sum = 0;
for ( int i = 1; i < number; ++i ) {
if ( number % i == 0 ) {
sum = sum + i;
}
}
return ( sum == number );
}
2. You are given a method called mixString, and it is guaranteed that the parameter s will have at least two characters in it.
public static String mixString
(String s) {
String result;
int strLen;
int lastIndex;
result = “”;
strLen = s.length();
lastIndex = strLen / 2 - 1;
for ( int i = 0; i <= lastIndex; ++i ) {
result = result.concat (String.valueOf (s.charAt (i)));
result = result.concat (String.valueOf (s.charAt (strLen - i - 1)));
}
if ( strLen % 2 == 1 ) {
result = result.concat (String.valueOf (s.charAt (lastIndex + 1)));
}
return (result);
}
It appears that a programmer used that method as follows:
String mixed = mixString (“wneflurdo”);
What would be the value of the variable mixed after the program returns from the function? Find the answer by tracing the program.
SOLUTION: “wonderful”
3. Now that you traced the mixString method,
you probably understood the purpose of that method. Write a method inverseMix
that would inverse the effect of mixString, that is, this statement should be
true for all appropriate String s:
s ==
inverseMix (mixString (s));
Your
method's signature should be : “public static String inverseMix (String s)”.
SOLUTION :
public static String inverseMix
(String s) {
String result;
int strLen;
int counter;
result = "";
strLen = s.length();
counter = 0;
while ( counter < strLen ) {
result = result.concat (String.valueOf (s.charAt (counter)));
counter = counter + 2;
}
if ( strLen % 2 == 1 ) {
counter = strLen - 2;
} else {
counter = strLen - 1;
}
while ( counter > 0 ) {
result = result.concat (String.valueOf (s.charAt (counter)));
counter = counter - 2;
}
return (result);
}
CS Programming Exercise
Hüseyin Ziya İmamoğlu Section 5
Although, I checked the solutions with Dr. Java, there may be errors. If
there’s any, please mail me through h.ziya.im@gmail.com.
I am a beginner, so if you have simpler or more efficient ways of solving the
problems, again please mail me. Hope that, questions and answers will be
helpful. Best wishes and good luck in the exam!
- 13 is an unlucky
number. Given a string, write a method that will delete the number 13 from
the string and return it. The data may or may not include thirteen or the
data could be empty (“13thyt” should return as “thyt”).
- I love prime
numbers; they’re unique in their own sense. Write a full program that will
print the prime numbers starting from 1 to integer n inclusive where n is
provided by the user.
- So many people
love diamonds. Write a full program that will print a diamond with using
the symbol and the number of rows provided by the user (you don’t need to
check the user input).
Answers
- public static
String without13(String str)
{
String result = "";
int length = str.length();
for(int i = 1; i <= length ; i++)
{
String letter2 = "";
String letter1 =
str.substring(i-1,i);
if(length > i)
{
letter2 =
str.substring(i,i+1);
if(letter1.equals("1") && letter2.equals("3"))
{
i = i + 1;
letter1 =
"";
}
}
result = result + letter1;
}
return result;
}
- import
java.util.Scanner;
public class PrimeNumberPrinter
{
public static void main(String[] args)
{
//Variables
int n;
Scanner in = new Scanner(System.in);
boolean isPrime = true;
// Get the user input
System.out.println("Enter a
positive integer please");
n = in.nextInt();
//Initialize the remainder
int remainder = 0;
//Do appropriate data evaluation
if(n <= 1)
{
System.out.println("No prime number");
}
else
{
//If the n is more than 1, the
numbers will include 2
System.out.println("2");
/* For each integer value bigger or
equal to 3 ( since we included 2 we have to start counting from 3) and
less or equal to n, create an innet loop which will check whether the integer
value is divisible by another integer value ranging from 2 two the integer
value itself by checking the remainder. If it's equal to zero, the condition
will turn false and it’s not going to be printed. If it's not, print it after
completing the inner loop. The i will denote the integer value at a given
instant. */
for(int i = 3; i <= n; i++)
{
isPrime = true;
for(int k = 2; k < i; k++)
{
remainder = i % k;
if(remainder == 0)
{
isPrime = false;
}
}
if(isPrime)
{
System.out.println(i);
}
}
}
}
}
- import
java.util.Scanner;
public class DiamondPrinter
{
public static void main(String[] args)
{
//Variables
int noOfRows;
int noOfSymbols;
int noOfSpaces;
char symbol;
Scanner in = new Scanner(System.in);
Scanner scan = new
Scanner(System.in);
//Get the user input
System.out.println("Please
enter an odd integer value bigger than 1");
noOfRows = in.nextInt();
System.out.println("Please
enter a symbol");
symbol = scan.nextLine().charAt(0);
//Initialize the varibles for the
loop
noOfSymbols = 1;
noOfSpaces = noOfRows/2;
//Create a loop for the diamond
for(int i = 1; i <= noOfRows;
i++)
{
for(int k =
1; k <= noOfSpaces; k++)
{
System.out.print("
");
}
for(int j =
1; j <= noOfSymbols; j++)
{
System.out.print(symbol);
}
if(i <
noOfRows/2 +1)
{
noOfSpaces --;
noOfSymbols = noOfSymbols + 2;
}
else
{
noOfSpaces ++;
noOfSymbols = noOfSymbols - 2;
}
System.out.println();
}
}
}
Nov. 13, 2015
CS 101 Midterm Preparation Questions
Note: Assume that all needed headers are imported. Although I have checked multiple times, sometimes autocorrect “fixes” the capital letters at beginning of the line. Remember that Java is a case-sensitive language and assume that I have written the correct one.
Written by Ünal Ege Gaznepoğlu
Easier:
1. Write a program which will print out the following pattern with two different ways.
* * * * * * * *
* * * * * * * *
* * * * * * * *
* * * * * * * *
* * * * * * * *
* * * * * * * *
* * * * * * * *
* * * * * * * *
a. Using the minimum amount of System.out.println, strings are free to use.
b. Using the minimum amount of System.out.print, strings are forbidden to use.
Hint: In both cases, up to some extent, using for loops are highly recommended.
2. Write a program which will detect whether entered string is a palindrome or not. A palindrome is a word or number which remains same if it is reversed. For example “12321” is a palindrome and “44332211” is not a palindrome.
Hint: There are two different approaches to this problem. Apply both of them.
a. Revert the string and check if equal.
b. Check characters two by two.
Medium:
1. Your company is working on a graphing solution. Program should print out a graph of the given numbers and names to rows.
Example input:
CS,3
EE,5
ME,6
Output:
CS: ***
EE: *****
ME: ******
Your task is to code a method which will get only one input as a string and return the line for that input. For example, your method will only get “CS,3” string and return “CS: ***” string. You can use \t to tidy up the line.
2. Correct the given code segment so it does the intended function. If it already performs the intended function, then write TRUE, if not, simply write if the problem is a compile time error or a logic error. There might be more than one problem with the given code segment. If so, write for all of the errors.
a. The factorial function. This method was supposed to calculate the factorial of the given number.
public static int factorial(int x)
{
int returnValue = 0;
for(int i=1; i<=x; i++)
{
returnValue = returnValue * i;
}
}
b. The root finder for second-order terms (quadratic equations). This method was supposed to find the roots for a given second order equation. The input is in terms of coefficents and output is in terms of a printout string which contains the estimations for the values.
public static String findQuadraticRoots(int a,int b,int c)
{
double delta = Math.pow(b,2) – 4*a*c;
if(delta > 0)
return ("The roots are " + (-b-Math.sqrt(delta))/(2*a) + “and “ (-b+Math.sqrt(delta))/(2*a));
if(delta =0)
return("There is only one root for this equation and that is" + new Integer(-b/2*a));
if(delta<0)
return("Unfortunately, there are no real roots for given equation. ");
}
Challenging:
1. Write a method which calculates the Greatest Common Divisor and Least Common Multiplication of given numbers a and b without using arrays or recursion. Hint: You may interpret Euler’s algorithm or use some for’s to check common primes. Hint 2: You don’t need to check entire interval [2,min(a,b) -1], checking until Math.floor(sqrt[min(a,b)]) is enough. The proof is the graph of y = k/x which has the line y=x as a symmetry axis.
2. Think about a rectangle which has the dimensions AxB. Write a program which
a. Finds the smallest amount of equal squares needed to coat this rectangle.
b. Finds the smallest amount of squares needed to coat this rectangle.
Hint: Use the functions from question 1.
Solutions For Problems
Easier:
1. There are more than one solution for these problems. These are the ones that I have came up with.
a.
String printOut = " ";
for(int i=0;i<15;i++)
{
if(i%2 == 0)
printOut += "*";
else
printOut +=
" ";
}
for(int i=0;i<8;i++)
{
if(i %2 == 0)
System.out.println(printOut);
else
System.out.println("" + printOut);
}
b.
for(int i=0;i<8;i++)
{
for(int j=0;j<16;j++)
{
if((i%2==0 && j%2==0) ||
(i%2==1 && j%2==1) )
System.out.print("*");
else
System.out.print("
");
}
System.out.println();
}
2.
a. Using this method dramatically reduces the total lines of code. You can use your reverse method from Lab06 or use this one.
public static String reverse(String inputStr)
{
String returnStr = "";
For(int i =0;i<inputStr.length();i++)
{
returnStr += inputStr.charAt(inputStr.length()-i);
}
return returnStr;
}
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
String userInput = scan.nextLine();
if(reverse(userInput).equals(userInput))
System.out.println("This is a palindrome!");
else
System.out.println("This is not a palindrome.");
}
b. You can also solve this problem with the approach of checking characters at i’th and length-ith position.
public static void main(String[] Args)
{
Scanner scan = new Scanner(System.in);
String userInput = scan.nextLine();
Boolean allSame = true;
for(int i=0; i<(userInput.length()/2 +1) && allSame;i++)
{
if(userInput.charAt(i) != userInput.charAt(userInput.length()-i))
allSame = false;
}
if(allSame)
System.out.println("Yes, this is a palindrome. ");
else
System.out.println("No,it is not a palindrome. ")
}
Medium:
1. First, get the input, obviously. Next move is parsing (identify and split to the functional parts) the input. This can be done with the methods “stringVariableName.indexAt()” and “stringVariableName.substring()”. After parsing the input into the row name and the numbers of stars, we should first send off the row name and then the needed number of stars.
The reason why I prefer nextLine above all other input receiving methods is that it consumes the END OF LINE character. It is a good programming habit not to leave any wild EOL in System.in buffer because it may later confuse your program.
public static String prepareGraphRow()
{
Scanner scan = new Scanner(System.in);
String graphLine,userInput;
int value;
graphLine= "";
userInput = scan.nextLine();
value = userInput.substring(userinput.indexOf(","),userInput.length());
graphLine += userInput.substring(0,userinput.indexOf(","));
graphLine += "\t”;
for(int i = 1;i<=value;i++)
{
graphLine += "*";
}
return graphLine;
}
2. As you know, there are logic errors (fallacies) and syntax errors.
a. There were two different errors on this one.
i. The initialization of returnValue. As it is zero, after multiplying with any variable the result is still zero. To fix this up, one must initialize it with 1. This is a logic error which does not alert the compiler.
ii. Our method has a return type of "integer " but there is no return statement. This is a syntax error and compiler warns you if you do so.
b. There were three different errors on this one.
i. The syntax error at the first return statement. After "and ”, we should add a “+” (without quotes) sign to correct this one. The missing + sign creates two different strings instead of one. This is a syntax error as well.
ii. This is a bit tricky without really running the program to see this mistake. In the second return line, the expression is written as b/2*a, however this is not equal to the b/(2*a) due to the operation priorities. This is a logic error and can take hours from your miserable life until you find it.
iii. Although for every case you have returned some variable, but compiler does not understand this one because it does not know only and only one of the three conditionals can be true. So, writing up the method with if/else-if/else would be a better approach. Other solution is to mimic at the end of method as there is another return statement there, but this is not the best approach because your prior if/if/if statements might have some undefined situations.
Challenging:
1. To calculate greatest common divisor, we can use Euclid’s algorithm. Euclid’s algorithm works in following way:
public static int euclidGCD(int a,int b){
int larger = Math.max(a,b);
int smaller = Math.min(a,b);
int temporaryValue;
while(larger % smaller != 0)
{
temporaryValue = larger%smaller;
larger = smaller;
smaller = temporaryValue;
}
return larger; }
This method directly calculates the greatest common divisor for two integers. To calculate the least common multiplication, we can use the property AxB = GCD(A,B)*LCM(A,B)
Other approach to calculate the greatest common divisor is to use some for loops to find common prime factors of both numbers A and B. Be careful not to forget the multiple instances of the same prime factor. Example: Let A be 12 and B be 16. If we didn’t have the inner while condition then we would have gcd as 2 after this method but it is 4.
int gcd = 1;
for(int i=2;i<=Math.floor(Math.sqrt(Math.min(a,b)));i++)
{
if( a%i == 0 && b%i == 0)
while(a%i ==0 && b%i ==0)
{
gcd *= i;
a /= i;
b/= i;
}
}
2.
a. To calculate the smallest amount of equal squares to coat a surface AxB, we can use greatest common divisor method . After finding the dimensions for one side of a square, then calculating A*B/(dimension^2) would give us the right answer.
b. Without the word “equal”, there is a bit more bookkeeping to handle because we can use first larger squares and then get them smaller. Details for this problem can be found at Wikipedia within the title “Euclid’s Algorithm”. As I remember, there was a graphic animation of how this algorithm works on that page.
public static int findLeastSquare(int a, int b)
{
int amountOfSquares, greater,smaller,division,remainder;
greater = Math.max(a,b);
smaller = Math.min(a,b);
while(greater % smaller != 0)
{
division = greater/smaller;
remainder = greater % smaller;
greater = smaller;
smaller = remainder;
amountOfSquares += division;
}
return amountOfSquares;
}
I wish success in the first CS101 midterm which will take place in 21.11.2015.
CS101 – Midterm 1 Questions by K.
Can Çelik
1)
A positive integer is said to be “perfect” if the sum of all positive divisors
except for the number itself is equal to that number. Write a static method that
would take one formal parameter “number” of approriate type, that would return
“true” if the number is perfect and “false” if the number is not
perfect.SOLUTION:
public static boolean isWonderful (int number) {
int sum;
sum = 0;
for ( int i = 1; i < number;
++i ) {
if ( number % i == 0 ) {
sum = sum + i;
}
}
return ( sum == number );
}
2)
You are given a method called mixString, and it is guaranteed that the
parameter s will have at least two characters in it.
public static String mixString (String s) {
String result;
int strLen;
int lastIndex;
result = “”;
strLen = s.length();
lastIndex = strLen / 2 -
1;
for ( int i = 0; i <=
lastIndex; ++i ) {
result =
result.concat (String.valueOf (s.charAt (i)));
result =
result.concat (String.valueOf (s.charAt (strLen - i - 1)));
}
if ( strLen % 2 == 1 ) {
result = result.concat (String.valueOf (s.charAt
(lastIndex + 1)));
}
return (result);
}
It
appears that a programmer used that method as follows:
String
mixed = mixString (“wneflurdo”);
What
would be the value of the variable mixed after the program returns from the
function? Find the answer by tracing the program.
SOLUTION:
“wonderful”
3)
Now that you traced the mixString method, you probably understood the purpose
of that method. Write a method inverseMix that would inverse the effect of
mixString, that is, this statement should be true for all appropriate String s:
s
== inverseMix (mixString (s));
Your
method's signature should be : “public static String inverseMix (String s)”.
SOLUTION
:
public static String inverseMix (String s) {
String result;
int strLen;
int counter;
result = "";
strLen = s.length();
counter = 0;
while ( counter <
strLen ) {
result =
result.concat (String.valueOf (s.charAt (counter)));
counter =
counter + 2;
}
if ( strLen % 2 == 1 ) {
counter =
strLen - 2;
} else {
counter =
strLen - 1;
}
while ( counter > 0 )
{
result =
result.concat (String.valueOf (s.charAt (counter)));
counter =
counter - 2;
}
return (result);
}