程序代写代做代考 CS 61A Structure and Interpretation of Computer Programs Fall 2020 Midterm 1 Solutions – cscodehelp代写
CS 61A Structure and Interpretation of Computer Programs Fall 2020 Midterm 1 Solutions
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Preliminaries
You can complete and submit these questions before the exam starts.
(a) What is your full name?
(b) What is your student ID number?
Exam generated for
1. (14.0 points) Down for the Count
Definition. A digit is a non-negative integer less than 10. Integers contain digits. Examples.
• The integer 21 contains the digits 1 and 2.
• The integer 474 contains the digit 4 twice and the digit 7 once. • The integer 400 contains the digit 4 once and the digit 0 twice. • The integer -77 contains the digit 7 twice.
• The integer 0 is a 0-digit number that contains no digits.
Reminders.
• You may call built-in functions that do not require import, such as min, max, abs, and pow.
• You may call functions defined in earlier parts of the question in your implementation for later parts, and
you may assume that the functions you call are implemented correctly. RESTRICTION. You may not call str or repr or use [ or ] in any part of this question.
(a) (4.0 points)
Implement count, which takes a digit element and an integer box. It returns the number of times that element appears in box.
Warning: n % d and n // d may not behave as you expect for negative n. For example, -123 % 10 evaluates to 7. -1 // 10 evaluates to -1. You do not need to know how these operators apply to negative n in order to solve this problem.
def count(element, box):
“””Count how many times digit element appears in integer box.
>>> count(2, 222122)
5
>>> count(0, -2020)
2
>>> count(0, 0) # 0 has no digits
0
“””
assert element >= 0 and element < 10
_________
(a)
total = 0
while box > 0:
if _________:
(b)
total = _________
(c)
box = box // 10
return total
Exam generated for
ii. (1.0 pt) Which of these could fill in blank (b)? box == element
box % 10 == element
box % element == 0
box % element > 0
iii. (1.0 pt) Which of these could fill in blank (c)? total + 1
element
total + element box % 10 total + box % 10
box = abs(box)
Exam generated for
Implement count_nine, which takes a digit element and a non-negative integer box. It returns the number of times that element appears in box and is not adjacent to a 9.
def count_nine(element, box):
“””Count how many times digit element appears in the non-negative integer
box in a place that is not next to a 9.
>>> count_nine(2, 222122)
5
>>> count_nine(1, 1911191) # Only the middle 1 is not next to a 9
1
>>> count_nine(9, 9)
1
>>> count_nine(9, 99)
0
>>> count_nine(3, 314159265359)
2
>>> count_nine(5, 314159265359)
1
>>> count_nine(9, 314159265359)
2
>>> count_nine(0, 0) # No digits are in 0
0
“””
assert element >= 0 and element < 10
assert box >= 0
nine, total = False, 0
while box > 0:
if _________ and not (nine or _________):
(a) (b)
total = _________
(c)
nine = _________ == 9
(d)
box = box // 10
return total
i. (1.0 pt) Which of these could fill in blank (a)? box == element
box % 10 == element
box % element == 0
box % element > 0
Exam generated for
iii. (1.0 pt) Which of these could fill in blank (c)? total + 1
element
total + element box % 10 total + box % 10
iv. (1.0 pt) Fill in blank (d).
(box // 10) % 10 == 9
box % 10
Exam generated for
Implement fit, which takes two non-negative integers pegs and holes. It returns whether every digit in pegs appears at least as many times in holes as it does in pegs.
def fit(pegs, holes):
“””Return whether every digit in pegs appears at least as many times in
holes as it does in pegs.
>>> fit(123, 321) # Each digit appears once in pegs and in holes.
True
>>> fit(1213, 33221) # 1 appears twice in pegs, but only once in holes.
False
>>> fit(12, 22) # 1 appears once in pegs, but not at all in holes.
False
>>> fit(314159, 112233456789)
True
“”” i=0
while i <= _________:
(a)
if _________:
(b)
_________
(c)
i=i+1
return _________
(d)
i. (1.0 pt) Fill in blank (a).
ii. (2.0 pt) Fill in blank (b).
iii. (1.0 pt) Fill in blank (c).
9
count(i, pegs) > count(i, holes)
return False
Exam generated for
iv. (1.0 pt) Which of these could fill in blank (d)? True
False
holes > pegs pegs > holes holes >= pegs pegs >= holes
Exam generated for
Assume mystery is a deterministic pure function that takes one integer argument, returns an integer, and never errors.
def mystery(n):
…
Assume the following functions are also defined:
def add_two(y):
return y + 2
def two(y):
return 2
def constant(k):
def ignore(x):
return k
return ignore
def diff(f, g):
return lambda z: abs(f(z) – g(z))
Definition. Two functions f and g have identical behavior if f(x) and g(x) return equal values or return functions with identical behavior, for every x that does not cause an error.
Complete each statement below so that it is true for all possible deterministic pure mystery functions.
(a) (2.0 pt) The result of evaluating constant(2) has identical behavior to the result of evaluating the
expression. . . … add_two … add_two(0) … add_two(2) … two
… two(0) … two(2)
None of these
(b) (2.0 pt) The result of evaluating diff(constant(1), constant(-1)) has identical behavior to the result of evaluating the expression. . .
… constant
… constant(0)
… constant(2)
… diff(constant, constant) None of these
Exam generated for
evaluating the expression. . .
… constant
… constant(0)
… constant(2)
… diff(constant, constant) … constant(mystery)
… mystery None of these
(d) (2.0 pt) The result of evaluating diff(mystery, diff(mystery, mystery)) has identical behavior to the result of evaluating the expression. . .
… mystery
… abs(mystery)
… lambda y: abs(mystery(y))
… lambda y: mystery(abs(y))
… lambda y: lambda z: mystery(abs(y)) – mystery(abs(z)) … lambda y: lambda z: abs(mystery(y)) – abs(mystery(z)) … lambda y: lambda z: abs(mystery(y) – mystery(z))
None of these
Exam generated for
3. (8.0 points) Please Register to Vote
Fill in each blank in the code example below so that its environment diagram is the following. RESTRICTIONS. You must use all of the blanks. Each blank can only include one statement or expression. Click here to open the diagram in a new window
def vote(vote):
please = _________
(a)
_________ = ty + 3
(b)
return please
ty = 1
register = _________(lambda nov: nov + ty)
(c)
Exam generated for
(d)
register(_________)
(e)
(a) (2.0 pt) Which of these could fill in blank (a)? vote(ty)
vote(30)
vote
lambda nov: vote(nov) + third lambda nov: vote(nov + third) lambda nov: vote(nov) + ty
lambda nov: vote(nov + ty)
(b) (1.0 pt) Which of these could fill in blank (b)? third
ty
please vote
(c) (1.0 pt) Which of these could fill in blank (c)? third
ty
please vote
(d) (2.0 pt) Fill in blank (d).
(e) (2.0 pt) Which of these could fill blank (e)? Check all that apply. ty * 10
ty – 1 + 30
30
third + 26
(lambda x: x + x)(15)
ty = 3
Exam generated for
4. (10.0 points) Amazing Job Growth
Definition. A repeatable function is a function that returns a repeatable function.
Reminder. You may call built-in functions that do not require import, such as min, max, abs, and pow.
(a) (4.0 points)
Implement growth, which takes a number baseline and returns a repeatable function increase. When increase is called on a number observed, it prints the difference between observed and the smallest argument passed to growth or increase so far among the repeated calls.
def growth(baseline):
“””Return a function that can be called repeatedly on numbers and prints
the difference between its argument and the smallest argument used so far
(including baseline).
>>> job = growth(148)(149)(150)(130)(133)(139)(137)
1
2
0
3
9
7
“””
def increase(observed):
under = _________
(a)
print(observed – under)
return _________
(b)
return increase
i. (2.0 pt) Fill in blank (a).
ii. (2.0 pt) Which of these could fill in blank (b)? increase
increase(under)
increase(observed) increase(baseline) growth
growth(under)
growth(observed) growth(baseline)
min(observed, baseline)
Exam generated for
Implement maxer, a higher-order function that takes a function smoke, which takes a number and returns a number. The maxer function returns a repeatable function fire that takes a number y and prints the largest result of calling smoke on any value of y passed to fire so far among the repeated calls.
Assume that smoke is a deterministic pure function. def square(x):
return x * x
def maxer(smoke):
“””Return a repeatable function fire(y) that prints the largest smoke(y) so far.
>>> g = maxer(square)
>>> h = g(2)(1)(3)(2)(-4) # print the largest square(y) so far
4
4
9
9
16
>>> h = maxer(abs)(2)(1)(3)(2)(-4) # print the largest abs(y) so far
2
2
3
3
4
“””
def fire(y):
_________
(a)
def haze(z):
if _________:
(b)
z=y
return _________
(c)
return haze
return fire
i. (2.0 pt) Fill in blank (a). You may not write a return statement for this blank.
print(smoke(y))
Exam generated for
iii. (2.0 pt) Which of these could fill in blank (c)? y
smoke(y)
fire(y)
fire(smoke(y)) haze
haze(y)
haze(smoke(y)) z
smoke(z)
fire(z)
fire(smoke(z)) haze(z)
haze(smoke(z))
smoke(y) > smoke(z)
Exam generated for
16
No more questions.
