评估Python中比较运算符的行为

Question

I'm TAing an introductory course on Python this semester (using 3.4) and recently came across an exercise about operator precedence and using parentheses to make a statement evaluate to true.

The exact question is:

Add a single pair of parentheses to the expression so that it evaluates to true.
1 < -1 == 3 > 4

I assumed that the correct answer would be:

1 < -1 == (3 > 4)

Given that the comparison operators are all on the same level of precedence, they should evaluate from left to right, so it should evaluate as such:

1 < -1 == (3 > 4) 
1 < -1 == False
False == False
True

But when I run the code it still returns false. I saw this question comparison operators' priority in Python vs C/C++ and the result of that expression makes sense to me; but in this case I've forced the evaluation of the latter statement before evaluating the rest of the expression, so I don't get why I'm still getting the wrong answer.

I've been staring at it for the past hour and I feel like I might be overlooking something obvious; if anyone could provide some insight as to what the correct solution might be it'd be greatly appreciated.

Answer 1

The task is provably impossible. Consider these three cases:

1. There is an open paren immediately before -1 .

2. There is an open paren between - and 1 .

3. There is an open paren anywhere else.

These three cases represent every possible location for the parentheses.

In the first case, we have 1 < ( ... ) , where the ellipsis is a boolean expression. Since 1 is not less than either True or False , the entire expression is False .

In the second case, we have 1 < -( ...) , where the ellipsis is a boolean expression. Since 1 is not less than either -True nor -False , the entire expression is False .

In the third case, we have 1 < -1 == ... . Since all legs of a expression of chained operators must be True, and and since 1 < -1 is False, the entire expression is False .

So, in every possible case, the result is False.

Answer 2

You are on the right track, but due to comparison chaining you will have:

1 < -1 == (3 > 4) 
1 < -1 == False
1 < -1 and -1 == False
False and False
False

Notice that we don't evaluate the second line left to right, rather we chain the two comparisons together with an and.

There are only a few valid ways to add one pair of parentheses to this, so we can check all the others easily:

(1 < -1 == 3 > 4)
False #Trivially

(1 < -1) == 3 > 4
False == 3 > 4
False == 3 and 3 > 4
False and False
False

1 < (-1 == 3) > 4
1 < False > 4
1 < False and False > 4
False and False
False

It looks like there is no answer to this problem!

Edit:

Whoops! Rob points out we forgot:

1 < -(1 == 3) > 4
1 < -False > 4
1 < 0 > 4
1 < 0 and 0 > 4
False and False
False

Answer 3

3>4 gives False . -1 == False gives False . 1 < False gives False . Hence, 1 < -1 == 3 > 4 becomes False .

Give proper brackets to make the statement semantically correct.

(1 < -1) == (3 > 4)

Answer 4

(1 < -1) == (3 > 4)

False == False

True