简化 if 条件布尔表达式

Question

I have this if-condition in my code:

if (a||a&&!b){
// do some stuff
}

with that intial values from my junit test :

boolean a=true, b = true;

as I recognized later the statement can be simplified to:

if (a&&b)

becomes green: Assert.assertTrue(a||a&&!b == a&&b);

Are there any further simplifications? How could I have recognized that this boolean expression could be simplified?

Answer 1

Just check the truth table (I've added parentheses)

  a || (a && (!b))

  a|b|result
  ----------
  T|T|T 
  T|F|T 
  F|T|F 
  F|F|F

As we can see, the formula doesn't depend on b , and can be simplified to only

  a

Finally

  if (a) {
    // do some stuff
  }

For formulae with many variables when truth table is too long for manual analysis you can use Karnaugh maps as Eomm proposed.

Answer 2

One common technique used to simplify complex Boolean expressions is Karnaugh Maps . It is relatively easy to learn, and it can help you produce shorter expressions, or even build Boolean expressions from a truth table.

Karnaugh Map for your expression is very simple - it looks like this:

It simplifies to a , not a && b .

Answer 3

You could use Karnaugh map for simplify boolean logic.

With 3 or more variables is the best way

Answer 4

a || a && !b 

is not equal to

a && b

It is equal to just a .

I suppose that in your JUnit test you used a specific combination of values for a and b where the results match, but that does not mean that the expressions are equivalent—and in fact they aren't. A quick way to convince yourself of that is checking the combination

a = true, b = false;

Your original expression clearly yields true for all cases where a == true , but your second expression will yield false whenever b == false .

As for a formal proof of equivalence to just a , take the expansion

a == a && (b || !b)
  == a && b || a && !b

Plugging into your original expression:

a || a && !b == a && b || a && !b || a && !b
             == a && b || a && !b
             == a

Answer 5

Wolfram Alpha actually does boolean algebra simplification. It might take a little converting the answer it gives you to make it compatible with your programming language, but the concept works and checks out fine. Here's a link to the page