Due by 10:00am on Tuesday, 8/2
Complete the quiz and submit it before 10:00am on Tuesday, 8/2. You must work alone, but you may talk to the course staff (see Asking Questions below). You may use any course materials, including an interpreter, course videos, slides, and readings. Please do not discuss these specific questions with your classmates, and do not scour the web for answers or post your answers online.
Your submission will be graded automatically for correctness. Your implementations do not need to be efficient, as long as they are correct. We will apply additional correctness tests as well as the ones provided. Though we will not release these hidden tests to you, you will receive autograder feedback specifying whether or not you have passed all tests.
Asking Questions: If you believe you need clarification on a question, make a private post on Piazza. Please do not post publicly about the quiz contents. If the staff discovers a problem with the quiz or needs to clarify a question, we will email the class via Piazza. You can also come to office hours to ask questions about the quiz or any other course material, but no answers or hints will be provided in office hours.
Submission: When you are done, submit with
python3 ok --submit. You may submit more than once before
the deadline; only the final submission will be scored.
ok program helps you test your code and track your progress.
The first time you run the autograder, you will be asked to log in with your
@berkeley.edu account using your web browser. Please do so. Each time you run
ok, it will back up your work and progress on our servers.
You can run all the doctests with the following command:
To test a specific question, use the
-q option with the
name of the function:
python3 ok -q <function>
By default, only tests that fail will appear. If you
want to see how you did on all tests, you can use the
python3 ok -v
If you do not want to send your progress to our server or you have any
problems logging in, add the
--local flag to block all
python3 ok --local
When you are ready to submit, run
ok with the
python3 ok --submit
Readings: You might find the following references useful:
cond form to
As we saw in Lecture 20, Scheme source code is data. Every non-primitive expression is a list, and we can write procedures that manipulate other programs just as we write procedures that manipulate lists.
In Problem 19 of the project, you'll write a procedure
let special forms within a piece of Scheme code into
lambda procedure calls.
In this problem, we'll look at how we can convert a
cond special form into
an equivalent structure of nested
For example, the following
cond form from the
(cond ((null? s) '()) ((f (car s)) (cons (car s) (filter f (cdr s)))) (else (filter f (cdr s))))
could instead be rewritten as a series of nested
(if (null? s) '() (if (f (car s)) (cons (car s) (filter f (cdr s))) (filter f (cdr s))))
Note that the logic for evaluating both expressions is the same. You first
(null? s) and if it's true, you return
'(). If not, you then
(f (car s)) and if it's true, you return
(cons (car s) (filter f (cdr s))). If that's not true, then you've reached
the else case and can just return
(filter f (cdr s)).
Before we convert an entire
cond form, we'll create a few procedures to
extract different parts of the form that we can use in our solution.
cond form consists of a series of clauses, which are lists containing two
items, a predicate and a consequent. First, the predicate is evaluated and, if
it is true, the consequent is evaluated and returned. If not, the next clause's
predicate is tried.
Note: While actual
condclauses can contain any number of consequent expressions, we'll assume for the purpose of this quiz that each clause contains exactly one consequent.
Implement the four helper procedures below.
take in a single
cond clause and return the predicate and consequent
first-clause takes in a complete
cond form and returns the
first clause it contains.
rest-clauses also takes in a
cond form, but
instead returns a list of all clauses but the first.
; Returns the predicate expression of a cond clause (define (predicate clause) 'YOUR-CODE-HERE ) ; Returns the consequent expression of a cond clause (define (consequent clause) 'YOUR-CODE-HERE ) ; Returns the first clause of a cond form (define (first-clause expr) 'YOUR-CODE-HERE ) ; Returns all clauses but the first of a cond form (define (rest-clauses expr) 'YOUR-CODE-HERE )
Use OK to unlock and test your code:
python3 ok -q helpers -u python3 ok -q helpers
We can now use our helper procedures as part of our solution for
which takes in a single
cond expression and converts it to a nested structure
Note that the nested structure will display entirely on one line, since there is no way to automatically display it nicely. This is important for unlocking the tests.
At each level of recursion, you should construct an
if form based on the
You should first handle the base case where the current clause's predicate
is the symbol
else, in which case you can just return the consequent,
cond such as
(cond (else 4)) is just equivalent to
If the current predicate is not
else, you should construct an
if of the
(if <predicate> <consequent> <alternative>) where the alternative should
be constructed through a recursive call.
cond-to-if, which does this conversion. You may assume that
consists of a single valid
cond form with no other
cond forms nested inside
of it and that it will always end with an
Hint: You can check if two symbols are equivalent with the
scm> (equal? 'hi 'bye) False scm> (equal? 'bye 'bye) True
; Takes in a single cond form and returns an equivalent nested structure of if ; forms. You may assume that there are no nested cond forms inside of the main ; one, that each clause has a single consequent, and that the last clause is an ; else clause. (define (cond-to-if expr) 'YOUR-CODE-HERE )
Use OK to unlock and test your code:
python3 ok -q cond-to-if -u python3 ok -q cond-to-if