Booleans and conditionals: Correct Solution


Part 1

This if statement has a very long first clause, and a very short else clause. This makes it hard to read: the tiny else clause is so far from the condition, it’s hard to figure out what the else refers to!

if (!spre() && cusm && (scri || fu) && trar) {
    ...
    ...
    // Pretend there is lots of code here
    ...
    ...
} else {
    spla();
}

Improve readability by refactoring this conditional so that its two clauses are swapped: what is now the second clause (the else clause) comes first, and the first clause comes second.

Solution

if (!trar || !fu && !scri || !cusm || spre()) {
    spla();
} else {
    ...
    ...
    // Pretend there is lots of code here
    ...
    ...
}

Things to double-check in your solution:


Part 2

Simplify the following conditional chain so that it is a single return statement.

if (criNodres() == 0 || hessja() && wuc) {
    if (cin && oswo()) {
        if (oswo()) {
            return true;
        }
        if (madash() != 1) {
            return true;
        }
    }
}
return false;

Solution

return (madash() != 1 || cin) && oswo() || criNodres() == 0 || hessja() && wuc;

Bonus challenge: rewrite the if/else chain above so that instead of consisting of many return true; statements with one return false; at the end, it has many return false; statements with one return true; at the end.

Solution

if (!hessja() && criNodres() != 0 && !oswo() || !cin && madash() == 1) {
    if (!cin && madash() == 1) {
        if (!oswo()) {
            return false;
        }
    }
    if (criNodres() != 0) {
        return false;
    }
    if (!wuc) {
        return false;
    }
}
return true;

Part 3

Simplify the following messy chain of conditionals:

if ((wo >= re) == true) {
    houSeou();
}
if (bi == true && (wo >= re) != true) {
    peatwi();
} else if ((pa != 1) == true && (wo >= re) != true && bi != true) {
    emfru();
}
if (frol && (wo >= re) != true && bi != true && (pa != 1) != true) {
    ecflon();
} else if (a == true && (wo >= re) != true && bi != true && (pa != 1) != true && !frol) {
    cordi();
}

Solution

{
    if (wo >= re) {
        houSeou();
    }
    if (bi) {
        peatwi();
    }
    if (pa != 1) {
        emfru();
    }
    if (frol) {
        ecflon();
    }
    if (a) {
        cordi();
    }
}

Things to double-check in your solution:


Related puzzles: