Exam Prep 2: Higher-Order Functions, Self Reference
Students from past semesters wanted more content and structured time to prepare for exams. Exam Prep sections are a way to solidify your understanding of the week's materials. The problems are typically designed to be a bridge between discussion/lab/homework difficulty and exam difficulty.
Reminder: There is nothing to turn in and there is no credit given for attending Exam Prep Sections.
We try to make these problems exam level , so you are not expected to be able to solve them coming straight from lecture without additional practice. To get the most out of Exam Prep, we recommend you try these problems first on your own before coming to the Exam Prep section, where we will explain how to solve these problems while giving tips and advice for the exam. Do not worry if you struggle with these problems, it is okay to struggle while learning.
You can work with anyone you want, including sharing solutions. We just ask you don't spoil the problems for anyone else in the class. Thanks!
You may only put code where there are underscores for the codewriting questions.
You can test your functions on their doctests by clicking the red test tube in the top right corner after clicking Run in 61A Code. Passing the doctests is not necessarily enough to get the problem fully correct. You must fully solve the stated problem.
We recommending reading sections 1.1-1.6 from the textbook for these problems. We also recommend reviewing Hog project announce_lead_changes and announce_highest for question 2 and 3.
Test your work! For example, for match_k, you can type test(match_k) in the python interpreter you get once you click Run in 61A Code to verify if you pass the doctests or not.
Q1: Match Maker
Difficulty: ⭐
Implement match_k, which takes in an integer k and returns a function that takes in a variable x and returns True if all the digits in x that are k apart are the same.
For example, match_k(2) returns a one argument function that takes in x and checks if digits that are 2 away in x are the same.
match_k(2)(1010) has the value of x = 1010 and digits 1, 0, 1, 0 going from left to right. 1 == 1 and 0 == 0, so the match_k(2)(1010) results in True.
match_k(2)(2010) has the value of x = 2010 and digits 2, 0, 1, 0 going from left to right. 2 != 1 and 0 == 0, so the match_k(2)(2010) results in False.
RESTRICTION: You may not use strings or indexing for this problem.
IMPORTANT: You do not have to use all the lines, one staff solution does not use the line directly above the while loop.
HINT: Floor dividing by powers of 10 gets rid of the rightmost digits.
Run in 61A CodeQ2: Natural Chainz
Difficulty: ⭐⭐
For this problem, a chain_function is a higher order function that repeatedly accepts natural numbers (positive integers).
The first number that is passed into the function that chain_function returns initializes a natural chain,
which we define as a consecutive sequence of increasing natural numbers (i.e., 1, 2, 3). A natural chain
breaks when the next input differs from the expected value of the sequence. For example, the sequence
(1, 2, 3, 5) is broken because it is missing a 4.
Implement the chain_function so that it prints out the
value of the expected number at each chain break as well as the number of chain breaks seen so far,
including the current chain break. Each time the chain breaks, the chain restarts at the most recently input number.
For example, the sequence (1, 2, 3, 5, 6) would only print 4 and 1. We print 4 because there is a missing 4, and we print 1 because the 4 is the first number to break the chain. The 5 broke the chain and restarted the chain, so from here on out we expect to see numbers increasingly linearly from 5. See the doctests for more examples. You may assume that the higher-order function is never given numbers ≤ 0.
IMPORTANT: For this problem, the starter code template is just a suggestion. You are welcome to add/delete/modify the starter code template, or even write your own solution that doesn’t use the starter code at all.
Run in 61A CodeQ3: CS61 - NAY
Difficulty: ⭐⭐⭐
Part A: Implement cs61nay, which takes a two argument function combiner and positive integer n and returns a function.
The returned function then takes n arguments, one at a time, and computes
combiner(...(combiner(combiner(arg1, arg2), arg3)...), arg_n). Notice combiner takes in two integers and returns one integer.
For example, the first doctest has the returned function f = cs61nay(lambda x, y: x * y, 3). Now when f is applied to three arguments, like f(2)(3)(4), it multiplies them together, 2*3*4 to get 24.
IMPORTANT: For this problem, the starter code template is just a suggestion. You are welcome to add/delete/modify the starter code template, or even write your own solution that doesn’t use the starter code at all.
HINT: For the n = 1 case, the returned function doesn't use combiner.
Difficulty: ⭐⭐⭐
Part B: Somebody who writes very complicated code has given you a challenge! You would hopefully never see something so hard to comprehend in the real world.
Complete the expression below by writing one integer in each blank so that the whole expression evaluates to 2021. Assume cs61nay is implemented correctly.
HINTS:
- You can fill the blanks and test your code with a python interpreter
- Try to understand what all the subparts and smaller calls do
- You can split this up into multiple lines to make it more readable
- You can add spaces/indentation to make it easier to read
Difficulty: ⭐⭐
Part C: All those lines of code are unnecessary! Solve cs61NAY but only using one line.
RESTRICTION: You may not use the python ternary operator (the one line if/else statment).
HINT: Use short circuiting and boolean operators to your advantage
Run in 61A CodeJust for Fun
This is a challenge problem and not reflective of exam difficulty. We will not be going over this problem in examprep section, but we will be releasing solutions.Q4: Abusing the Call Stack
(a) Implement the following higher order functions so that we can simulate append and get behavior. As the
name suggests, the get function should get the ith element that was appended (the first element that
was appended is element 0). For example, if I append 2, append 30, and then append 4, then get(0) is
2 (the first element appended), get(1) is 30 (the second element appended), and get(2) is 4 (the third
element appended). If I append more items, get(0) through get(2) should not be affected. Assume all get
calls ask for non-negative indices (i.e., you’d never do get(-1)). If the argument to get would go out of
bounds otherwise, the call should return the string "Error: out of bounds!".
(b) Build on your solution to the previous question to implement insert functionality! As the name suggests,
the insert function inserts a value into an existing sequence of numbers. The function takes an insertion
index, the value to insert into that index, as well as two other arguments whose purpose is left for you
to determine. When the value is inserted into the provided index, all numbers from that index and to
the right are shifted one element right. For example, if my current sequence is 5, 9, 14, 3 and I specify
an insertion index of 1 with value 100, then my updated sequence should be 5, 100, 9, 14, 3. The 100 is
inserted at index 1, and all numbers from the original index 1 to the end are shifted to the right by one
position. You can always assume that the provided insertion index will be within bounds.
You don’t actually have to represent the sequence as a contiguous block of numbers that need to shift
around though. As long as the get(i) call returns the correct value, that’ll do.