# comparing boolean logic expression

Callie Jones

Greenhorn

Posts: 1

posted 4 years ago

- 0

Hi I am studying for an exam and after looking at a past paper I found out that we are being tested on something we haven't been taught in our course.

I don't have a clue how to tackle the question so I am wondering if anyone could explain what kind of method you should use? I don't know anything about maths/logic so idiot-level explanations would be appreciated! All we got taught was that Java has these boolean operators - but no information about how to work with them.

The question is:

I don't have a clue how to tackle the question so I am wondering if anyone could explain what kind of method you should use? I don't know anything about maths/logic so idiot-level explanations would be appreciated! All we got taught was that Java has these boolean operators - but no information about how to work with them.

The question is:

**Which two Java boolean expressions are equivalent to this expression?****Select two from the following:**

Ralph Cook

Ranch Hand

Posts: 479

posted 4 years ago

- 0

The "not" operator can be applied through the rest of the expression, reversing all the operators: || becomes &&, < becomes >=, etc.

If you reverse the expression

(count > 2) || (count <= 1)

You get

(count <= 2) && (count > 1)

which is D, since && does not depend on the order of its operands. The tricky part is reversing

You can check it by evaluating what integer(s) will fit both expressions -- in this case, only the integer 2 fits the expression, assuming they mean to be talking about integers.

rc

If you reverse the expression

(count > 2) || (count <= 1)

You get

(count <= 2) && (count > 1)

which is D, since && does not depend on the order of its operands. The tricky part is reversing

**everything**.You can check it by evaluating what integer(s) will fit both expressions -- in this case, only the integer 2 fits the expression, assuming they mean to be talking about integers.

rc

posted 4 years ago

- 0

You should check this out: De Morgan's laws

"Any fool can write code that a computer can understand. Good programmers write code that humans can understand." --- Martin Fowler

Please correct my English.

posted 4 years ago

- 0

The rule Ralph mentioned is called De Morgan's rule. (Which Wouter already mentioned...)

Now, not everybody knows that one. You could have checked all the possible solutions, limiting your possible values from 0 to 2; anything < 0 is handled by the 0 case, and anything >= 2 is handled by the 2 case. So the original expression would return true only if count == 2. You then check all of the other expressions and check out which other only returns true for count == 2. And that's two: D and E.

A is wrong because of the ||; it would return true for, well, any value of count. Likewise for B. C would return true for only count == 1. F is essentially the same as C.

E is right because it's the direct application of De Morgan's rule:

!((count > 2) || (count <= 1))

is the same as, De Morgan

!(count > 2) && !(count <= 1)

is the same as, && is associative

!(count <= 1) && !(count > 2) (answer E)

is the same as

(count > 1) && (count <= 2) (answer D)

Well actually, there is one flaw in this logic. && is a shortcut operator. That means that the right operand will not be evaluated if the left one is already true. Therefore, the associative swapping isn't always allowed. In this case there is no problem with it though.

Now, not everybody knows that one. You could have checked all the possible solutions, limiting your possible values from 0 to 2; anything < 0 is handled by the 0 case, and anything >= 2 is handled by the 2 case. So the original expression would return true only if count == 2. You then check all of the other expressions and check out which other only returns true for count == 2. And that's two: D and E.

A is wrong because of the ||; it would return true for, well, any value of count. Likewise for B. C would return true for only count == 1. F is essentially the same as C.

E is right because it's the direct application of De Morgan's rule:

!((count > 2) || (count <= 1))

is the same as, De Morgan

!(count > 2) && !(count <= 1)

is the same as, && is associative

!(count <= 1) && !(count > 2) (answer E)

is the same as

(count > 1) && (count <= 2) (answer D)

Well actually, there is one flaw in this logic. && is a shortcut operator. That means that the right operand will not be evaluated if the left one is already true. Therefore, the associative swapping isn't always allowed. In this case there is no problem with it though.

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