Homework 4
==========


Problem 1:

String s1 = "CSCI 161: Introduction to Programming 1";
s1 ==> "CSCI 161: Introduction to Programming 1"

String s2 = s1.substring (10, 20);
s2 ==> "Introducti"

String s3 = s2.toUpperCase();
s3 ==> "INTRODUCTI"

int indexOfO1 = s2.indexOf("o");
indexOfO1 ==> 4

int indexOfO2 = s3.indexOf("o");
indexOfO2 ==> -1

int indexOfT = s2.indexOf("t");
indexOfT ==> 2

String small = s3.substring(indexOfO1, indexOfT);
|  java.lang.StringIndexOutOfBoundsException thrown: begin 4, end 2, length 10
|        at String.checkBoundsBeginEnd (String.java:3116)
|        at String.substring (String.java:1885)
|        at (#7:1)

String large = s1.toUpperCase().substring(20);
large ==> "ON TO PROGRAMMING 1"


Problem 2:

if (grade >= 60) {
  System.out.println("Pass");
} else {
  System.out.println("Fail");
}


Problem 3:

Scanner input = new Scanner(System.in);
double number = input.nextDouble();
double answer = -1;
if (number > 0) {
  answer = Math.sqrt(number);
}


Problem 4:

if (gpa >= 3.95) {
  System.out.println ("Magna Cum Laude");
} else if (gpa >= 3.7) {
  System.out.println ("Summa Cum Laude");
} else if (gpa >= 3.5) {
  System.out.println ("Cum Laude");
}


Problem 5:

Scanner input = new Scanner(System.in);
int a = input.nextInt();
int b = input.nextInt();
double average = ((double)a + (double)b) / 2.0;
System.out.println(average);




Homework 5
==========


Problem 1:

z is odd:                      z % 2 == 1    OR    z % 2 != 0
y is positive:                 y > 0
Either x or y is even:         x % 2 != y % 2
y is a multiple of z:          y % z == 0
x and z are of opposite signs: x * z < 0    OR    x > 0 && z < 0 || x < 0 && z > 0


Problem 2:

ifElseMystery1(3, 20);

  z = z + 9; // because z (4) <= x (3)
  y++;       // because z (13) <= y (20)

  Output: "13 21"

ifElseMystery1(4, 5);

  z = z + 1; // because z (4) <= x (4)
  y++;       // because z (5) <= y (5)

  Output: "5 6"

ifElseMystery1(5, 5);

  z = z + 1; // because z (4) <= x (5)
  
  Output: "6 5"

ifElseMystery1(6, 10); 

  z = z + 1; // because z (4) <= x (5)
  y++;       // because z (5) <= y (10)

  Output: "7 11"


Problem 3:

public static int daysInMonth (int month) {
  if (month == 2) {
    return 28;
  } else if (month == 4 || month == 6 || month == 9 || month == 11) {
    return 30;
  } else {
    return 31;
  }
}


Homework 7
==========


Problem 1:

A). x starts off at one; loop increments x by 10; loop condition is x < 100
    1, 11, 21, 31, 41, 51, 61, 71, 81, 91, [101]
    10 times

B). max starts off at 10; loop condition is max < 10; loop never executes
    0 times

C). x starts off at 250; loop condition is x % 3 != 0; loop doesn't change x
    Since x never changes, and 250 % 3 is always != 0, the loop runs forever

D). x starts off at 2; loop condition is x < 200; we multiply x by itself each iteration
    2, 4, 16, [256]
    3 times

E). words is initially "a"; loop condition is length of word < 10; surround word by "b"s
    "a", "bab", "bbabb", "bbbabbb", "bbbbabbbb", ["bbbbbabbbbb"]
    5 times

F). x starts off at 100; loop condition is x > 0; we divide x by 2 each iteration
    100, 50, 25, 12, 6, 3, 1, [0]
    7 times


Problem 2:

x is our parameter; x = 1, z = 0;
For each iteration we increment z by one and multiply y by 2
we will evolve each iteration until 2 * y <= x
Finally, we print "y z"


mystery(1)

-> 2 * y <= x   | x = 1, y = 1, z = 0

Output: "1 0"


mystery(6)

-> 2 * y <= x   | x = 6, y = 1, z = 0
-> 2 * y <= x   | x = 6, y = 2, z = 1
-> 2 * y <= x   | x = 6, y = 4, z = 2

Output: "4 2"


mystery(19)

-> 2 * y <= x   | x = 6, y = 1, z = 0
-> 2 * y <= x   | x = 6, y = 2, z = 1
-> 2 * y <= x   | x = 6, y = 4, z = 2
-> 2 * y <= x   | x = 6, y = 8, z = 3
-> 2 * y <= x   | x = 6, y = 16, z = 4

Output: "16 4"


mystery(19)

-> 2 * y <= x   | x = 6, y = 1, z = 0
-> 2 * y <= x   | x = 6, y = 2, z = 1
-> 2 * y <= x   | x = 6, y = 4, z = 2
-> 2 * y <= x   | x = 6, y = 8, z = 3
-> 2 * y <= x   | x = 6, y = 16, z = 4
-> 2 * y <= x   | x = 6, y = 32, z = 5

Output: "32 5"


mystery(19)

-> 2 * y <= x   | x = 6, y = 1, z = 0
-> 2 * y <= x   | x = 6, y = 2, z = 1
-> 2 * y <= x   | x = 6, y = 4, z = 2
-> 2 * y <= x   | x = 6, y = 8, z = 3
-> 2 * y <= x   | x = 6, y = 16, z = 4
-> 2 * y <= x   | x = 6, y = 32, z = 5
-> 2 * y <= x   | x = 6, y = 64, z = 6

Output: "64 6"


Problem 2:

int x = 27;
int y = -1;
int z = 32;
boolean b = false;

a). !b

    true

b). b || true

    true

c). (x > y) && (y > z)

    (27 > -1) is false; therefore false

d). (x == y) || (x <= z)

    (27 == -1) || (27 <= 32); therefore true

e). !(x % 2 == 0)

    !(27 % 2 == 0); therefore true

f). (x % 2 != 0) && b

    False because b must be true but is false

g). b && !b

    False (tautology)

h). b || !b

    True (tautology)

i). (x < y) == b

    (27 < -1) is false; therefore, true

j). !(x / 2 == 13) || b || (z * 3 == 96)

    z * 3 is 96; therefore true

k). (z < x) == false

    z is greater than x; therefore true

l). !((x > 0) && (y < 0))

    DeMorgan's: x < 0 || y > 0
    x is greater than 0 and y is less than 0
    False