for loops

This weeks lesson will cover:

• break and continue

• for loops

• nested loops

⭐ - is difficulty of question

for loops

Simple for examples ⭐

הדפס מספרים 0 עד 3

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

        for (int i = 0; i < 4; i++) {
            System.out.println(i);
        } // end of for
    }
}

0
1
2
3

?מה יודפס כאן

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

        int i = 10;
        for ( ; i > 5; i--) 
            System.out.print(i + ", ");
    }
}

10, 9, 8, 7, 6,

?מה יודפס כאן

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

        for (int i = 10; i < 20; i++) {
            System.out.print(i + ", ");
            i++;
        } // end of for
    }
}

10, 12, 14, 16, 18,

?מה יודפס כאן

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

        for (int i = 0; i < 10; i++) {
            if (i==3)
                continue;
            else if (i==6)
                break;
            System.out.print(i + ", ");
        } // end of for
    }
}

0, 1, 2, 4, 5,

explanation

• the keyword continue will continue to the next number in the loop

• the keyword break will exit the loop altogether.

Factorial and Sum of values ⭐⭐

חשב עצרת וסכום עבור n

$$n! = 1 \cdot 2 \cdot3\cdot ... \cdot n \qquad and \qquad \sum _{i=0}^{n}i=1+2+..+n$$

example n=4

$$4! = 1 \cdot 2 \cdot3\cdot 4=24 \qquad and \qquad \sum _{i=0}^{4}i=1+2+3+4=10$$

public class Main {
    public static void main(String[] args) {
        int n = 5; // or enter from user
        int factorial = 1;

        for (int i = 1; i <= n; i++)
            factorial*=i;

        System.out.println(n + "! = " + factorial);

        int sum = 0;
        for (int i = 1; i <= n; i++)
            sum+=i;

        System.out.println("Summation of " + n + " = " + sum);
    }
}

5! = 120
Summation of 5 = 15

version 2

חשב את הערך של n עד שמקבלים -1

public class Main {
    public static void main(String[] args) {
        int n;


        while ((n = MyConsole.readInt("enter num: "))!=-1){
            int factorial = 1;
            for (int i = 1; i <= n; i++)
                factorial*=i;

            System.out.println(n + "! = " + factorial);

            int sum = 0;
            for (int i = 1; i <= n; i++)
                sum+=i;

            System.out.println("Summation of " + n + " = " + sum);
        }

    }
}

enter num:  5
5! = 120
Summation of 5 = 15
enter num:  10
10! = 3628800
Summation of 10 = 55
enter num:  -1

Question - Formula ⭐⭐

תקודדו את הנוסחה הבא:

$$(a+2 ^0\cdot b),(a+2 ^0\cdot b+2 ^1\cdot b),..,,(a+2 ^0\cdot b+2 ^1\cdot b+...+2 ^{n-1}\cdot b)$$

public class Main {
    public static void main(String[] args) {
        int a = MyConsole.readInt("enter a: ");
        int b = MyConsole.readInt("enter b: ");
        int n = MyConsole.readInt("enter n: ");

        int sum = a;

        for (int i = 0; i < n; i++) {
            sum+=Math.pow(2,i)*b;
            System.out.println(sum);
        }
    }
}

enter a:  3
enter b:  5
enter n:  6

הלוואה מבנק ⭐⭐

פלוני לקח הלוואה מהבנק ומחזיר 10 אחוז מהלוואה כל חודש הדפיסו

• כמה כסף הוא מחזיר כל חודש

• כמה הוא עדיין חייב אחרי 3 חודשים מאז שהוא התחיל לפרוע בחזרה

public class Main{
    public static void main(String[] args) {
        int amount = MyConsole.readInt("enter loan amount: ");
        int original_amount = amount;

        for (int i = 1; i < 4; i++){
            System.out.println("month " + i + " payback: " + amount*.10);
            amount-= (amount * .10);
        }

        System.out.println("He still owes " + amount + " from the " + original_amount + " loan");
    }
}

enter loan amount:  5000
month 1 payback: 500.0
month 2 payback: 450.0
month 3 payback: 405.0
He still owes 3645 from the 5000 loan

Catalan number ⭐⭐⭐

חשב את מספר קלטן n

$$C_{n}={\frac {(2n)!}{(n+1)!\,n!}}\qquad {\text{for }}n\geq 0.$$

for example

$$C_{4}={\frac {(2 \cdot4)!}{(4+1)!\,4!}}={\frac {(8)!}{(5)!\,4!}} = 14$$

?מה זה מספר קלטן

זהו מספר האפשריות לניתן להגיע מפינה השמאלית למטה לפינה הימנית למעלה בלי לחצות את הקו וללכת רק ימינה ולמעלה

public class Main{
    public static void main(String[] args) {
        int n = MyConsole.readInt("enter number: ");

        int numerator = 1;
        int denominator1 = 1;
        int denominator2 = 1;

        //numerator
        for (int i = 1; i <= n * 2; i++)
            numerator *= i;

        //denominator
        for (int i = 1; i <= n; i++)
            denominator1 *= i;

        denominator2 = denominator1*(n+1);

        int result = numerator/(denominator1*denominator2);
        System.out.println("C_"+ n + " = " + result);
    }
}

version 2

עכשיו תתדפיסו את כל המספרי קלטן עד n

public class Main{
    public static void main(String[] args) {
        int n = MyConsole.readInt("enter number: ");

        for (int each_c = 0; each_c <= n; each_c++) {
            int numerator = 1;
            int denominator1 = 1;
            int denominator2 = 1;

            //numerator
            for (int i = 1; i <= each_c * 2; i++)
                numerator *= i;

            //denominator
            for (int i = 1; i <= each_c; i++)
                denominator1 *= i;

            denominator2 = denominator1*(each_c+1);

            int result = numerator/(denominator1*denominator2);
            System.out.println("C_"+ each_c + " = " + result);
        }
    }
}

enter number:  6
C_0 = 1
C_1 = 1
C_2 = 2
C_3 = 5
C_4 = 14
C_5 = 42
C_6 = 132

Nested Loops

Print stars ⭐⭐

כתבו תוכנית אשר ידפיס משולש כוכבים עד גובה n

example:

enter number:  4
*
**
***
****      <--- The height of the triangle is 4 stars
***
**
*

code:

public class Main{
    public static void main(String[] args) {
        int n = MyConsole.readInt("enter number: ");

        //*
        //**
        //***
        //****
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <=i ; j++)
                System.out.print("*");
            System.out.println();
        }


        //***
        //**
        //*
        for (int i = n-1; i > 0; i--) {
            for (int j = i; j > 0 ; j--)
                System.out.print("*");
            System.out.println();
        }
    }
}

כתוב תוכנית אשר יציג את לוח הכפל (עד10)

1x1 = 1	2x1 = 2	3x1 = 3	4x1 = 4	5x1 = 5	6x1 = 6	7x1 = 7	8x1 = 8
1x2 = 2	2x2 = 4	3x2 = 3	4x2 = 8	5x2 = 10	6x2 = 12	7x2 = 14	8x2 = 16
1x10 = 10	2x10 = 20	3x10 = 30	4x10 = 40	5x10 = 50	6x10 = 60	7x10 = 70	8x10 = 80

public class Main{
    public static void main(String[] args) {
        for (int i = 1; i < 10; i++) {
            for (int j = 1; j < 10; j++)
                System.out.print(j + "x" + i +"=" +(i*j) +"\t");
            System.out.println();
        }
    }
}

1x1=1    2x1=2    3x1=3    4x1=4    5x1=5    6x1=6    7x1=7    8x1=8    9x1=9    
1x2=2    2x2=4    3x2=6    4x2=8    5x2=10    6x2=12    7x2=14    8x2=16    9x2=18    
1x3=3    2x3=6    3x3=9    4x3=12    5x3=15    6x3=18    7x3=21    8x3=24    9x3=27    
1x4=4    2x4=8    3x4=12    4x4=16    5x4=20    6x4=24    7x4=28    8x4=32    9x4=36    
1x5=5    2x5=10    3x5=15    4x5=20    5x5=25    6x5=30    7x5=35    8x5=40    9x5=45    
1x6=6    2x6=12    3x6=18    4x6=24    5x6=30    6x6=36    7x6=42    8x6=48    9x6=54    
1x7=7    2x7=14    3x7=21    4x7=28    5x7=35    6x7=42    7x7=49    8x7=56    9x7=63    
1x8=8    2x8=16    3x8=24    4x8=32    5x8=40    6x8=48    7x8=56    8x8=64    9x8=72    
1x9=9    2x9=18    3x9=27    4x9=36    5x9=45    6x9=54    7x9=63    8x9=72    9x9=81

Later on we will use more nested loops for accessing multi-dimension arrays

e^x ⭐⭐⭐⭐

compute e^x using the following formula:

{\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots }

example:

$$e^3 =20.0855369232$$

if accurray or n=12 we get

$$e^3 = 1 + 3 + \frac{3^2}{2!} + \frac{3^3}{3!} + \frac{3^4}{4!} + .. + \frac{3^{12}}{12!} =20.0855369232$$

public class Main{
    public static void main(String[] args) {
        double e = 1;
        double x = MyConsole.readDouble("enter exponent: ");
        double accuracy = MyConsole.readDouble("enter accuracy: ");

        for (int n = 1; n <= accuracy; n++) {
            int factorial = 1;

            //find factorial
            for (int i = 1; i <= n; i++) {
                factorial*=i;
            } // end of inner for
            e += Math.pow(x, n)/factorial;
        } // end of outer for
        System.out.println("e^"+ x +" = " + e);
    }
}

enter exponent:  3
enter accuracy:  12
e^3.0 = 20.08521256087662

in the outer loop

for (int n = 1; n <= accuracy; n++) {
            int factorial = 1;

each time we reset factorial to calculate the factorial

while the inner loop runs until n each time

for (int n = 1; n <= accuracy; n++) {
            ...
            for (int i = 1; i <= n; i++) {

version 2

display each accuracy until n

public class Main{
    public static void main(String[] args) {
        double e = 1;
        double x = MyConsole.readDouble("enter exponent: ");
        double accuracy = MyConsole.readDouble("enter accuracy: ");

        double real_value_of_e = 2.718281828459045;

        for (int i = 1; i <= accuracy; i++) {
            e = 1;

            for (int n = 1; n <= i; n++) {
                int factorial = 1;

                //factorial
                for (int j = 1; j <= n; j++) {
                    factorial*=j;
                } // end of inner for
                e += Math.pow(x, n)/factorial;
            } // end of middle for
            System.out.println("accuracy: "+ i + ", e^"+ (int)x +" = " + e + ", loss of: " + (real_value_of_e-e));
        } // end of outer for
    }
}

enter exponent:  1
enter accuracy:  10
accuracy: 1, e^1 = 2.0, loss of: 0.7182818284590451
accuracy: 2, e^1 = 2.5, loss of: 0.2182818284590451
accuracy: 3, e^1 = 2.6666666666666665, loss of: 0.05161516179237857
accuracy: 4, e^1 = 2.708333333333333, loss of: 0.009948495125712054
accuracy: 5, e^1 = 2.7166666666666663, loss of: 0.0016151617923787498
accuracy: 6, e^1 = 2.7180555555555554, loss of: 2.262729034896438E-4
accuracy: 7, e^1 = 2.7182539682539684, loss of: 2.7860205076724043E-5
accuracy: 8, e^1 = 2.71827876984127, loss of: 3.0586177750535626E-6
accuracy: 9, e^1 = 2.7182815255731922, loss of: 3.0288585284310443E-7
accuracy: 10, e^1 = 2.7182818011463845, loss of: 2.7312660577649694E-8

think how large 12! is its 479,001,600

Remember than e = 2.718281828459045 which isn't bad at all, each time we are accurate by one decimal

Graph ⭐⭐

תכתבו תוכנית אשר מקבלת 2 מספרים שלמים m וn

ומדפיסה בטרמינל את הישר שלהם על גרף בין 10 למינוס 10 בציר הx והציר הy

$$y = m \cdot x+n$$

img

public class Main{
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int m = MyConsole.readInt("m = ");
        int n = MyConsole.readInt("n = ");
        for (int y = 10; y >= -10; y--) {
            for (int x = -10; x <= 10; x++) {
                if (y == m*x + n) {
                    System.out.print(" * ");
                }
                else if (y == 0 && x == 0){
                    System.out.print(" + ");
                }
                else if (x == 0) {
                    System.out.print(" | ");
                }
                else if (y == 0) {
                    System.out.print(" - ");
                }
                else {
                    System.out.print("   ");
                }
            }
            System.out.println();
        }

    }
}

m =  2
n =  3
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