Simpler boolean truth table?

Question

I'm doing a CodingBat exercise and would like to learn to write code in the most efficient way. On this exercise, I was just wondering if there's a shorter way to write this code.

monkeyTrouble(true, true) → true
monkeyTrouble(false, false) → true
monkeyTrouble(true, false) → false

public boolean monkeyTrouble(boolean aSmile, boolean bSmile) {
 if (aSmile && bSmile) {
 return true;
 }
 if (!aSmile && !bSmile) {
 return true;
 }
 return false; 
}

• Nikita U.: there is "Show Solution" button that shows 3 solutions on the page you linked

Answer 1

Sometimes it is easy to forget that the simplest logical constructs like boolean are comparable with the == operator, and that, in Java, (false == false) is true.

With this in mind, your code could become:

public boolean monkeyTrouble(boolean aSmile, boolean bSmile) {
 return aSmile == bSmile;
}

---

It may be easier to see how to get there if you first transform your original code into

public boolean monkeyTrouble(boolean aSmile, boolean bSmile) {
 if ((aSmile && bSmile) || (!aSmile && !bSmile)) {
 return true;
 } else {
 return false; 
 }
}

... which could become

public boolean monkeyTrouble(boolean aSmile, boolean bSmile) {
 return (aSmile && bSmile) || (!aSmile && !bSmile);
}

From there, you may come to the realization that "both true or both false" is equivalent to "both the same".

---

Here is a verification of the output:

public static boolean monkeyTrouble(boolean aSmile, boolean bSmile) {
 return aSmile == bSmile;
}
private static void testTruth(boolean a, boolean b) {
 System.out.printf("monkeyTrouble(%s, %s) = %s\n", a, b, monkeyTrouble(a, b));
}
public static void main(String[] args) {
 testTruth(true, true);
 testTruth(true, false);
 testTruth(false, true);
 testTruth(false, false);
}

This produces:

monkeyTrouble(true, true) = true
monkeyTrouble(true, false) = false
monkeyTrouble(false, true) = false
monkeyTrouble(false, false) = true

• Warren van Rooyen: But there are two conditions there. The one states that if both monkeys are smiling or if neither of them are smiling, return 'true'. So wouldn't simply making aSmile == bSmile ignore those two conditions?
• rolfl: if both are smiling then aSmile == bSmile is true == true, and that is true. If neither are smiling then aSmile == bSmile is false == false, and that resolves to true. If only one is smiling, then aSmile == bSmile is false == true, and that resolves to false.
• rolfl: Trust me, I'm a monkey, and I know monkeyTrouble() ;-)
• Mathieu Guindon: truth of false and false is true - doesn't that sound like monkey business? (false && false) == false returns true, no? Then false and false is... false.
• rolfl: @retailcoder : big difference between false == false and false && false.... ;-)

Answer 2

This is the exclusive-or (or XOR) condition negated.

You can simply do this:

public boolean monkeyTrouble(boolean aSmile, boolean bSmile) {
 return !(aSmile ^ bSmile);
}

or, as it is so simply, you can use it in your code without the function.

---

Explanation of XOR operator ^:

a ^ b = c
1 0 1
0 1 1
0 0 0
1 1 0

• 200_success: The negation of XOR is also known as XNOR.
• user2357112: Wrong precedence there.
• Wayne Conrad: Shouldn't this be: !(aSmile ^ bSmile)?
• António Almeida: @WayneConrad, you are right, Its corrected! It also works the other way, but this is the correct one, thanks.