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CS 1110 Final Exam, May 2018 with slight edits in Spring 2021
This 150-minute exam has 7 questions worth a total of 79 points. You may tear the pages apart; we have staplers.
You may use any Python introduced so far in this course, but not Python constructs that we have not introduced.
When a question asks for a particular technique to be used, do not expect to receive credit for questions based on a different technique. For example, if we ask you to make effective use of recursion, your solution must be fundamentally based on recursion, even if a recursion-less for-loop would also solve the problem.
The second page of this exam gives you the specifications for some useful functions and methods.
You will be expected to write Python code on this exam. We recommend that you draw vertical lines to make your indentation clear, as follows:
def foo():
if something: do something do more things do something last
It is a violation of the Academic Integrity Code to look at any exam other than your own, to look at any other reference material, or to otherwise give or receive unauthorized help. We also ask that you not discuss this exam with students who are scheduled to take a later makeup.
Academic Integrity is expected of all students of Cornell University at all times, whether in the presence or absence of members of the faculty. Understanding this, I declare I shall not give, use or receive unauthorized aid in this examination.
Signature: Date
s.find(substr) Returns: index of first occurrence of string substr in string s (-1 if not found) s.strip() Returns: copy of string s where all whitespace has been removed from the beginning and the end of s. Whitespace not at the ends is pre- served. s.split(sep) Returns: a list of the “words” in string s, using sep as the word de- limiter (whitespace if sep not given) s.join(slist) Returns: a string that is the concatenation of the strings in list slist separated by string s s[i:j] Returns: if i and j are non-negative indices and i ≤ j-1, a new string containing the characters in s from index i to index j-1, or the substring of s starting at i if j ≥ len(s)
lt.insert(i,item) Insert item into list lt at position i lt.append(item) Adds item to the end of list lt lt.count(item) Returns: count of how many times item occurs in list list(range(n)) Returns: the list [0 .. n-1] lt.remove(item) Removes the first occurrence of item from list lt; raises an error if item not found. lt.index(item) Returns: index of first occurrence of item in list lt; raises an error if item is not found. (There’s no “find” for lists.) lt[i:j] Returns: A new list[lt[i], lt[i+1],.. ., lt[j-1]] under ordinary circumstances. Returns [] if i and j are not both sensible indices. lt.pop(i) Returns: element of list lt at index i and also removes that ele- ment from the list lt. Raises an error if i is an invalid index.
list(map(func, lt)) Returns: A list obtained by applying function func to each element in list lt and concatenating the results of each application. isinstance(o, c) Returns: True if o is an instance of class c, False otherwise.
Question Points Score
Total: 79
This code is copied from the previous page with two additional lines of code. It runs error-free. 1 class A(): 2 x = 1 3 4 def init(self, n): 5 self.y = n 6 A.x += 1 7 8 def p(self): 9 print(self.y) 10 self.y += 3 11 self.r() 12 13 def r(self): 14 self.y += 2 15 print(self.y) 16 17 class B(A): 18 x = 10 19 20 def init(self, n): 21 super().init(n) 22 sum = self.y + B.x 23 self.m = sum 24 25 def r(self): 26 self.y += self.x 27 print(self.m) 28 29 a = A(1) 30 b = B(2) 31 a.p() 32 b.p()
(b) [4 points] What will be printed when Python executes line 31 (assuming lines 29-30 were just executed)?
(c) [4 points] What will be printed when Python executes line 32 (assuming lines 29-31 were just executed)?
- Object Creation and Foor Loops. Consider the following code:
class MenuItem(): """An instance represents an item on a menu.""" def init(self, name, is_veggie, price): """A new menu item called name with 3 attributes: name: a non-empty str, e.g. 'Chicken Noodle Soup' is_veggie: a Bool indicating vegetarian or not price: an int > 0 """ self.name = name self.is_veggie = is_veggie assert price > 0 self.price = price
class LunchItem(MenuItem): """An instance represents an item that can also be served at lunch""" def init(self, name, is_veggie, price, lunch_price): """A menu item with one additional attribute: lunch_price: an int > 0 and <= 10""" super().init(name, is_veggie, price) assert lunch_price > 0 assert lunch_price <= 10 self.lunch_price = lunch_price (a) [2 points] Write a python assignment statement that stores in variable item1 the ID of a new MenuItem object whose name is “Tofu Curry”, a vegetarian dish costing 24 dollars.
(b) [2 points] Write a python assignment statement that stores in variable item2 the ID of a new LunchItem object whose name is “Hamburger”, a non-vegetarian dish that costs 12 dollars, but only 8 dollars at lunch.
(c) [2 points] Class Invariants. Lunch should never cost more than 10 dollars. The init method prevents this. Write a line of python that shows how this invariant can still be broken. You may use any of the global variables you created from the previous parts.
- [13 points] String processing. Complete the function below so that it obeys its specification. Sample inputs, the value of an important local variable, and desired outputs are given at the bottom of the page.
def after_at(s): """Returns a list of every non-empty sequence of non-space, non-@ characters that directly follows an @ in s.
The elements should be ordered by occurrence in s, and there should be no repeats.
Pre: s is a string, possible empty. """ temp = s.split('@') if len(temp) == 1: # There were no @s in s. return [] afters_list = temp[1:] # Drop stuff before 1st @. See table at bottom of page.
DON'T use split. (It sometimes calls strip(), which you don't want here.)
Hint: for each item in afters_list, use string method find() to find the
location of the first space (if any)
s afters_list (local variable) return value '@bill @ted meet @ the Circle K' ['bill ', 'ted meet ', ' the Circle K'] ['bill', 'ted'] 'meet @ the Circle K @Bill@Ted' [' the Circle K ', 'Bill', 'Ted'] ['Bill', 'Ted'] 'hi' [] [] '@martha @Martha @martha' ['martha ', 'Martha ', 'martha'] ['martha', 'Martha'] 'The symbol du jour is an @' [''] [] 'The symbol du jour is an @!' ['!'] ['!']
- [5 points] While Loops. Make effective use of while loops to implement countdown_by_n. Your solution must use a while loop to receive points.
def countdown_by_n(count_from, count_by): """Prints a count down from count_from by count_by. Stops printing before the result goes negative. Note: this function does not return anything.
count_from: the number you're counting down from [int, possibly negative] count_by: the amount you're counting down by [int > 0]
Examples:
countdown_by_n(16, 5) should print: 16 11 6 1
countdown_by_n(21, 7) should print: 21 14 7 0
"""
- Invariants. Let b be a non-empty list of ints, and splitter be an int. We want a for-loop that swaps elements of b and sets the variable i so that the following postcondition holds: i b <= splitter > splitter
Examples: Before After splitter b i b 0 [16, -4, 22] 1 [-4, 22, 16] or [-4, 16, 22] 0 [-10, -20, 15] 2 [-10, -20, 15] or [-20, -10, 15] 0 [-30, -50, -60] 3 any ordering of b works -4 [10, 20, 30] 0 any ordering of b works
The code must maintain the following invariant.
i k b <= splitter > splitter ???
In words, b[0..i-1] are all less than or equal to splitter; b[i..k-1] are all greater than splitter; b[k..len(b)-1] have not yet been processed. (a) [2 points] According to the invariant, should the initialization be i=1?
- If yes, explain why — no credit without correct explanation.
- If no, give the correct initialization; omit explanation in this case.
(b) [4 points] Here is the for-loop header: for k in list(range(len(b))): According to the invariant, is the following the correct and complete for-loop body? (As- sume the helper does what the comment says.) if b[k] <= splitter: swap(b, i, k) # Helper that swaps items at position i and k in b i += 1
- If yes, explain why — no credit without correct explanation.
- If no, give the correct and complete for-loop body; omit explanation in this case.
- [11 points] Recursion. Assume that objects of class Course have two attributes:
- label [non-empty str]: unique identifying string, e.g., 'CS1110'
- prereqs [list of Course, maybe empty]: Courses that one must complete before this one.
Consider the following header and specification of a non -method function.
def requires(c, other_label): """Returns True if Course with label other_label must be taken before c, False otherwise.
Pre: c is a Course. other_label is a non-empty string."""
Example intended operation: suppose c1 is a Course with label 'CS1110' and empty prereqs list; c2 is a Course with label 'CS2110' and prereqs list [c1]; c3 is a Course with label 'CS2800' and prereqs list [c1]; c4 is a Course with label 'CS3110' and prereqs list [c2, c3]
Then, all of the following should evaluate to True: And all of the following should evaluate to False: requires(c4, 'CS2800') requires(c4, 'CS2110') requires(c4, 'CS1110')
requires(c1, 'CS2800') requires(c3, 'CS2800') requires(c4, 'randomstring')
While a majority of the lines are correct, there is at least one error in the proposed implemen- tation below. For each error, circle it, write down/explain the correct version, and draw a line between the circle and the corresponding correction. Responses where the correction is wrong may not receive any credit.
def requires(c, other_label): # STUDENTS: do NOT alter this header. if len(c.prereqs) <= 1: return False else: for p in prereqs: if p.label == other_label: return True elif requires(other_label, c): return True return False
Did you re-read all specs, and check your code works against test cases? Then, have a great summer!
