Midterm (AY20/21)
Last updated
Last updated
This question assesses whether students can form a logical expression given a set of conditions.
For this kind of question about forming a logical expression given a set of conditions, if the conditions are not too many, we should try the Truth Table Method. That is list all the possible conditions and see the output's relationship between the different set of possibilities.
Y
Y
Y
Y
Y
Y
N
Y
Y
N
Y
Y
Y
N
N
Y
N
Y
Y
N
N
Y
N
N
N
N
Y
Y
N
N
N
N
Then, what we need to do is to find all the rows that make the output True/Y. We will find that when is_with_gf == true, output is always true. When !is_with_mom && is_raining, the output will be true. Otherwise, the output will be false. So, we can form our final logical expression as:
This question is another tricky question about the invariant. But before we start finding the invariant, let's find what the code is doing by substituting some numbers in.
After a series of trial and error, we may find that the function is basically doing res = n/3 + n/7. And we should find out that n = i-1. Using these two results, we can form our invariant that relates res and i as res = (i - 1)/3 + (i - 1)/7. And we start proving that the loop variant holds at Line A.
Proof:
Let's assume that the loop invariant holds at the start of the loop. After that, we may or may not enter the first if structure.
Enter the if structure, which means i % 3 == 0. Now, our res = (i - 1)/3 + (i - 1)/7 + 1. Since i % 3 == 0, we can find that i % 3 == (i - 1) % 3 + 1. So, our invariant will become res = i / 3 + (i - 1)/7.
Not enter the if structure, which means i % 3 != 0. Now, we can say the (i - 1) / 3 == i / 3, so we can also get res = i / 3 + (i - 1)/7.
Simiarly, we can do the analysis on the second if structure and finally we will get res = i/3 + i/7. Now, we execute i = i + 1. Our invariant will become res = (i - 1)/3 + (i - 1)/7, which is the same as when we enter the loop.
Sometimes if it is hard to prove the loop invariant from forward, we can use the answer and try to prove it backward (think of what we want at last and then think forwardly about how can we get that)
This is a recursion question, but it wants to test how to convert the if-else structure recursion to a single return statement including logical expressions only. But the basic idea is to write the if-else structure recursion first!!!
So, for this question, the normal are_coprime_upto_c() should be
First thing first, to make our thinking easier, let's consider when we will return true.
When c==1 or
when !(a % c == 0 && b % c == 0) && are_coprime_upto_c(a, b ,c - 1), assuming that are_coprime_upto_c() will return as expected.
Knowing this, we can form our single return statement with the logical expression only easily. That's
If the variables are not too much, always try the Truth Table Method since it will make your life and analysis much easier!
"Imprecisions using floating-point values might mean that certain conditions may be evaluated incorrectly", this occurred when you want to do the equality test between a floating number and an integer.
When you want to form a single return statement in recursion, write out the normal if-else recursion first, then consider under which cases we will get true output.
Before we try to form the invariant, do some examples and see what the code is doing!!!
The columns of the table are the bool variables given in the table, the output is bool. Since we have three variables, so the total possibilities will be , which means we will have 8 rows in total.
As discussed in . To prove a program is false, one counterexample is enough. But to prove an algorithm is true, we may use assertion or some other proof techniques. Usually try to find counterexample first!!!
