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CS 1110 Final Exam Solutions May 2018
- Object Diagramming and Terminology. (a) [10 points] The questions on the right pertain to the code on the left. Some questions may have multiple correct answers. Write down only one answer.
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)
an object folder is created when Python exe- cutes line
a class folder is created when Python executes line
an object attribute is created on line
a class attribute is created on line
a superclass de nition begins on line
a class method de nition begins on line
an attribute de nition that overrides another begins on line
a method de nition that overrides another be- gins on line
a local variable is created on line
a global variable is created on line
Solution: object folder created: 29 or 30 method override: 20 or 25 class folder created: 1 or 17 local variable: 22 object attribute created: 5 or 23 global variable: 1, 17, 29, or 30 class attribute created: 2 or 18 superclass begins: 1 class method begins: 4, 8, 13, 20, 25 attribute override: 18
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? Solution: 1 6 (c) [4 points] What will be printed when Python executes line 32? Solution: 2 12
The same code has been copied to this page for your convenience: 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 (d) [8 points] For Loops. Make effective use of a for-loop to write the body of the function audit menu according to its speci cation. def audit_menu(the_menu): """Performs an audit of each LunchItem on the_menu, making sure that each lunch_price is never more than 10 dollars. A lunch_price of 11 dollars is changed to 9. An item whose lunch_price is more than 11 is too expensive to be offered at lunch; it must be replaced with a new, equivalent MenuItem (that has no lunch price). Items that are not LunchItems are unchanged. Modifies the_menu; does not create or return a new menu/list the_menu: possibly empty list of MenuItem """ Solution:
for i in list(range(len(the_menu))): item = the_menu[i] if isinstance(item, LunchItem): if item.lunch_price == 11: item.lunch_price = 9 elif item.lunch_price > 11: newItem = MenuItem(item.name, item.is_veggie, item.price) the_menu[i] = newItem
- [13 points] String processing. Complete the function below so that it obeys its speci cation. 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 characters that 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, find where its first space is.
Solution: Code for testing your own implementation: http://www.cs.cornell.edu/courses/cs1110/2018sp/exams/final/2018_spring_string_processing.py
BEGIN REMOVE
outlist = [] for item in afters_list: space_pos = item.find(' ') if space_pos == -1: if item != '' and item not in outlist: outlist.append(item) elif space_pos > 0: follower = item[:space_pos] if follower not in outlist: outlist.append(follower)
If space_pos == 0, don't append anything
Alternate solution by Kevin Cook, with some variable-name changes to match the above: outlist = [] for item in afters_list: i = 0 follower = '' while i < len(item) and item[i] != ' ': follower += item[i] i += 1 if follower not in outlist and len(follower) > 0: outlist.append(follower) return outlist
b. If you made the mistake of removing from after_list in a for loop, you used after_list as your accumulator, so you will still get this point.
- (+1.0) Loops through afters_list (index or element OK) and refer to the element in the loop correctly (e.g. would not get point if do "for x in after_list" and "after_list[x]"). a. If you did “for i in range(len(after_list))” and refer to elements of after_list as “afterlist[i]”, you will get this point. b. If you did “for x in after_list” and refer to elements of after_list as “x”, you will get this point.
- (+1.0) Finds location of first space (if use index() without guard do not get point) a. If you used string.find(“ “) you will get this point. b. If you put string.index(“ “) in a try catch block, you will get this point. c. If you put string.index(“ “) in a “if ‘ ‘ in string”, you will get this point. d. If you simply did string.index(“ “) you will lose this point because your code will throw an not found error.
- (+2.0) Correctly processes string to obtain new string up to (and not including) the first space. a. If the position you got from 3 is named “pos” (e.g. pos = string.find(“ “), then you will get 2 points if you did string[:pos]. b. If you did string[:pos+1] or string[:pos-1] you will get 2 points for this rubric item but you will also get 1 point deduction for off by 1.
- (+2.0) Correctly adds processed string to accumulator (in the case that there is a space AND it’s not at position 0) a. This corresponds to correct implementation rule 3. If string = “Hello World”, and your implementation appended “Hello” to the output_list, you will get 2 points. b. If string = “Hello World”, and your implementation removed “Hello World” from the after_list and added “Hello” back to the after_list, you will get these 2 points, but will get a 3 point deduction (which will only be applied one time even though you remove something else for later rubric items) for removing item from after_list while going through it.
- (+2.0) Correctly adds string to accumulator (in the case that there is no space AND it’s not empty a. This corresponds to correct implementation rule 4. If string = “HelloWorld”, and your implementation appended “HelloWorld” to the output_list, you will get 2 points. b. If string = “HelloWorld”, and your implementation used after_list as accumulator and you chose to do nothing in this case, you will get these 2 points.
- (+1.0) Does NOT add processed string if it’s empty. a. This corresponds to correct implementation rule 1 and 2. If you added any empty string to the output_list you will lose this point. b. If you used remove() on empty string from after_list, you will get this point, but will get a 3 point deduction if you haven’t got it from previous rubric item for removing item from after_list while going through it.
- (+1.0) Checks for duplicates at ANY place before adding to accumulator (if string is already in accumulator list) a. If you successfully removed at least 1 duplicate, you will get this point. b. If you tried to remove duplicates from the after_list while going through after_list in a for loop, you will still get this point, but will get a 3 point deduction if you haven’t got it from previous rubric item for removing item from after_list while going through it. c. If you did not check for duplicates when you loop through after_list and appended everything to a temporary_list, and correctly removed duplicates in temporary_list, you will get this point. d. If you did not check for duplicates when you loop through after_list and appended everything to a temporary_list, but attempted to remove items from temporarily_list in a for loop, you are still getting this point. e. If you did not check for duplicate at all, you are not getting this point.
- (+1.0) Checks for duplicates at ALL places before adding to accumulator (if string is already in accumulator list) a. If you successfully removed all duplicates by checking if element is in the output_list, you will get this point. b. If you did not check for duplicates when you loop through after_list and appended everything to a temporary_list, and correctly removed duplicates in temporary_list, you will get this point. c. If you tried to remove duplicates from the after_list while going through after_list in a for loop, you will not get this point. d. If you did not check for duplicates when you loop through after_list and appended everything to a temporary_list, but attempted to remove items from temporarily_list in a for loop, you are not getting this point. e. If you did not check for duplicate at all, you are not getting this point.
- (+1.0) Returns accumulator list. a. If you returned output_list, you will get this point. b. If you used after_list as accumulator by removing elements from it, and you returned after_list, you will get this point.
The other mistake we saw was using strip(). Strip() would get rid of the empty spaces, but also “\n”, “\t”, etc, which we would like to keep in this case. If you are appending string.strip() to the output_list, you are possibly adding an empty string to the output_list, and will lose the point for rubric item 7.
- [12 points] Call Frames. On the right, draw the full call stack as it would look after all of the code on the left has executed. Include crossed-out frames. Do not worry about drawing any variables outside the call frames.
1 def fib(n): 2 sum = 0 3 if n == 0 or n == 1: 4 return 1 5 6 sum += fib(n-1) 7 sum += fib(n-2) 8 return sum 9 10 x = fib(2)
Solution:
- 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. Solution: This invariant was inspired by Section 2.2 of Kernighan and Pike, The Practice of Programming (1999). (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. Solution: No, should be i = 0. Optional explanation: we don't know if b[0..i-1] == b[0] is <= splitter. See 4th example. (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. Solution: The update of i should happen in the if-statement: if b[k] <= splitter: swap(b, i, k) # Helper that swaps items at position i and k in b i += 1
