Except where otherwise noted, the contents of this document are Copyright 2019 Stuart Reges and Marty Stepp.
Goals for this problem set:
boolean expressions and variables to represent logical true/false expressionsboolean type
		The boolean type represents logical values of true or false.  Combine boolean expressions with logical operators && (and), || (or), and ! (not).
	
Example:
boolean test1 = 7 < 10; // true boolean test2 = (1 == 2); // false if ((test1 || test2) && 2 + 2 != 5) { System.out.print("hello"); // output: hello }
String methods with boolean results| Method name | Description | 
|---|---|
| string .equals(string) | whether the two strings are identical | 
| string .equalsIgnoreCase(string) | whether the two strings are identical, ignoring capitalization | 
| string .startsWith(string) | whether this string begins with the characters of the given string | 
| string .endsWith(string) | whether this string ends with the characters of the given string | 
| string .contains(string) | whether the characters of the given string occur within this string | 
String name = "Professor Smith";
if (name.startsWith("Prof")) {
    System.out.println("When are your office hours?");
}
		Write the result of each expression as either true
		or false, given the following variables.
	
int x = 12; int y = 7; int z = 28; String s = "mid term";
| x < 14 | true | |
| !(x % 2 < 1) | false | |
| x < y || x < z | true | |
| z / x < x / y * x | true | |
| s.length() == y | false | |
| s.toUpperCase().equals("MID TERM") | true | |
| !s.equals("mid term") || x * y != z | true | |
| s.substring(z / x).length() > y | false | 
 
    
  
    Write a method named allDigitsOdd that returns whether every
    digit of a positive integer is odd.  Your method should
    return true if the number consists entirely of odd digits
    and false if any of its digits are even.  0, 2, 4, 6, and 8
    are even digits, and 1, 3, 5, 7, 9 are odd digits.
  
    For example, allDigitsOdd(135319) returns true
    but allDigitsOdd(9145293)
    returns false.
  
    Hint: You can pull apart a number into its digits using /
    10 and % 10.
  
 
    
  
    Write a method hasMidpoint that accepts three
    integers as parameters, and returns true if one of the numbers
    is the midpoint of the other two and returns false otherwise.
  
    For example, the call hasMidpoint(3, 7, 5) would
    return true because one of the parameters (5) is the midpoint
    of the other two (3 and 7).
  
 
    
  
    Write a method before that takes as parameters two month/day
    combinations and that returns whether or not the first date comes before
    the second date (true if the first month/day comes before the
    second month/day, false if it does not).  The method will take
    four integers as parameters that represent the two month/day combinations.
  
The first integer in each pair represents the month and will be a value between 1 and 12 (1 for January, 2 for February, etc, up to 12 for December). The second integer in each pair represents the day of the month (a value between 1 and 31). One date is considered to come before another if it comes earlier in the year.
 
    
  
    Write a method sameDashes that takes two strings as parameters
    and that returns whether or not they have dashes in the same places
    (returning true if they do and returning false
    otherwise). For example, below are four pairs of strings of equal length
    that have the same pattern of dashes. Notice that the last pair has no
    dashes at all.
    
string 1: "hi--there-you." "-15-389" "criminal-plan" "abc" string 2: "12--(134)-7539" "-xy-zzy" "(206)555-1384" "9.8"To be considered a match, the strings must have exactly the same number of dashes in exactly the same positions. The Strings might be of different length.
 
    
  
    This attempted solution to Self-Check 5.15 (isVowel) has
    several problems:
// Returns whether the given string represents a vowel:
// a, e, i, o, or u, case insensitively.
public static boolean isVowel(String s) {
    if (s == "a") {
        return true;
    } else if (s == "e") {
        return true;
    } else if (s == "i") {
        return true;
    } else if (s == "o") {
        return true;
    } else if (s == "u") {
        return true;
    } else {
        return false;
    }
}
  
Fix the following aspects of the code:
public static boolean isVowel(String s) {
    s = s.toLowerCase();
    if (s.equals("a") || s.equals("e") || s.equals("i")
            || s.equals("o") || s.equals("u")) {
        return true;
    } else {
        return false;
    }
}
  
  The above can be improved. "Boolean Zen" version:
public static boolean isVowel(String s) {
    s = s.toLowerCase();
    return s.equals("a") || s.equals("e") || s.equals("i")
           || s.equals("o") || s.equals("u");
}
  
 
		
	
		Identify whether each assertion is always/never/sometimes true at each point.
	
| x > y | z == 0 | x == y | |
|---|---|---|---|
| A | SOMETIMES | ALWAYS | SOMETIMES | 
| B | SOMETIMES | SOMETIMES | NEVER | 
| C | ALWAYS | NEVER | NEVER | 
| D | NEVER | NEVER | NEVER | 
| E | NEVER | SOMETIMES | ALWAYS | 
public static void mystery(int x, int y) {
    int z = 0;
    // Point A
    while (x != y) {
        // Point B
        z++;
        if (x > y) {
            // Point C
            x = x / 10;
        } else {
            // Point D
            y = y / 10;
        }
    }
    // Point E
    System.out.println(x + " " + y + " " + z);
}