Please see the course page for information on the timing and logistics for the midterm. This page is about what sort of problems appear on our CS exams and how to study for them.
The midterm will be made of little code-writing questions which look similar to the problems seen in lecture and on the homeworks. It will look like a regular college exam, where you write short phrases of code on paper. This may sound a little strange, but in practice it works fine. We do not grade the questions all-or-nothing like a computer — instead, we look at the code carefully, identifying each part that is right to award partial credit. We also don't mark off for syntax. So if a student wrote this for a list
lst = []
lst.add('i'}
We would still give that full credit, even though the name of the function should be .append() and one parenthesis is a curly bracket. If we can see the right idea in there, that's good enough for full credit. We are interested in the algorithm in the solution code, not the syntax. Likewise, if you miss a quote mark or a colon, that is still full credit.
On the other hand, if the correct answer is "while" and you write "if", that is not full credit. If a line has the wrong indentation, we cannot give it full credit.
Topics on the exam: simple Bit (hw1), images/pixels/nested-loops (hw2), 2-d grids (hw3), strings, loops, simple lists (hw4). This is the material through the first four weeks.
Topics not on the exam: bluescreen algorithm, file reading, writing main(). Int division // is not on our midterm, although it appears on some of the old questions here. We've added explanations for those examples. We mostly grade code on producing correct output, not decomposition and style, which are more natural topics for the homeworks than the exams.
We will provide a basic reference for the Python functions we've used (shown below), but other than that the exam is closed-note. Closed-note exams are actually better for the students, although you may be skeptical about that. With an open note exam, the exam questions have to morph to into weird, upside-down versions of the lecture examples. For an open-note exam, perversely, the questions all need to look different from the example code in the notes.
With a closed-note exam, we can take nice, normal example functions from lecture, e.g. brackets(s), and just make the exam out of those. The lecture and section and homework problems tour through important Python coding patterns, so they set the pattern for good exam problems too. You will notice that the practice problems below resemble the lecture examples and homework problems. Ideally, as you practice for the exam, you should feel that you can work out a pretty good solution to any problem we did in lecture, or section, or on a homework.
The exam grading will focus on getting the correct output, so if you use a for/i/range to get the right output, and our solution uses a for/ch/s .. either solution can get full credit so long as it produces the correct output.
The exam will provide enough time to write solutions, but not much extra time. Part of practicing your code writing is building skill and quickness. You want to be able to write the easy parts confidently, so you can take your time on the hard parts.
What is the goal of CS106A? We want everyone to pick up core coding knowledge and skills. These core skills are what we do in lecture, and section, and then on the homeworks, so the exam problems will look similar. The topics on the exam are predictable. Predictable = studying will pay off.
Three levels of working a code concept...
1. See a code in lecture with some understanding
2. Write code in section or on homework while looking at lecture examples
3. Write code on blank sheet of paper without any references
The bad news: It's a trap to get stuck at step 2. Your brain will look at one of our examples and think "oh yeah, I could have written that." On the day of the exam, you may look at that first empty area waiting for you to write something, and find your mind goes blank. The key is, you need to practice with the blank sheet, writing the code out, not just looking at the solution code. As in sports, you need to practice in game-day conditions. The good news..
(1) We publish many examples of the code patterns we care about. (2) We are providing many practice problems for you to use. (3) We will be forgiving of wrong syntax when grading. We want the right algorithmic ideas, and we give partial credit for partially correct code (unlike when the computer grades something!).
How to study:
We have a new feature on the experimental server. Go to a section, like Bit Puzzles:
Go to the first problem. In the Code menu, turn on Practice Mode. In this mode, your previously entered code is temporarily blocked, so you can try practicing things fresh again and again. It keeps a timer of how long you've been coding and how many problems you have solved. Many of the sections on the experimental server are perfect for studying for the midterm: bit puzzles, image nested, image section, strhw, string1, string2, string3, string4, list1, strlisthw. Practice to build speed, skill, and confidence.
The midterm will include printed on it a few reminders of python functions we have used. Of course the key is knowing how to call them. We will not generally grade off for syntax. The emphasis is on passing the right numbers to range() or in a slice, and example problems below all feature that sort of complexity. If a problem requires some boilerplate, the problem statement will likely just include the boilerplate for you. Functions in square brackets below may have been mentioned briefly, but we haven't really used them yet, so none of the problems need them for a solution.
# CS106A Exam Reference Reminder
# [square brackets] denote functions listed here that we have not used yet
# Exam questions will not depend on functions we have not used yet
Bit:
bit = Bit(filename)
bit.front_clear() bit.left_clear() bit.right_clear()
bit.move() bit.left() bit.right()
bit.paint('red') [bit.erase()]
General functions:
len() int() str() range() [list() sorted()]
String functions:
isalpha() isdigit() isspace() isupper() islower()
find() upper() lower() strip()
List functions:
append() index() [extend() pop() insert()]
SimpleImage:
# read filename
image = SimpleImage(filename)
# create blank image
image = SimpleImage.blank(width, height)
# foreach loop
for pixel in image:
# range/y/x loop
for y in range(image.height):
for x in range(image.width):
pixel = image.get_pixel(x, y)
Grid 2D:
grid = Grid.build([['row', '0'], ['row', '1']])
grid.width, grid.height - properties
grid.in_bounds(x, y) - True if in bounds
grid.get(x, y) - returns contents at that x,y, or None if empty
grid.set(x, y, value) - sets new value into grid at x,y
Read lines out of a text fie:
with open(filename) as f:
for line in f:
line = line.strip()
We have a selection of practice problems below, and published previous midterms are a source of realistic practice problems. Homework problems and section problems are also good for practice.
a. For each ??, write what value comes from the Python expression on each line.
>>> 3 + 4 - 1 + 2 ?? >>> >>> 1 + 2 * 3 ?? >>> >>> 47 % 5 ?? >>> >>> s = 'Summer' >>> s[1] ?? >>> >>> s[:2] ?? >>> >>> s[2:] ??
b. Function Call - what 3 numbers print when the caller() function runs. This code runs without error.
def one_more(a):
b = a + 1
return b
def caller():
a = 1
b = 2
c = one_more(b)
print(a, b, c)
Bit is in a world that contains a single rectangle of filled squares. Bit is to the right of the rectangle, facing a block on its right side as shown in this representative example:
Move bit forward to the edge of the rectangle. Then move bit down past the lower right corner of the rectangle. Finally, move bit to the left side of the world, painting every square occupied blue. Your code should work for any position and size rectangle, and bit will always start facing a block on the right side of the rectangle as in the example above.
def do_midterm(filename):
bit = Bit(filename)
pass
image1.
The provided code loads the image for the given filename. The int parameter n will be between 0 and the width of the image. Consider two vertical stripes, n pixels wide, running down the left and right sides of the image. Write nested loop code to set all the red values within the stripes to 0. Return the changed image.
def stripes(filename, n):
image = SimpleImage(filename)
# your code here
return image
image2.
Given an image, create a new 'out' image which is three times as wide as the passed in image. Copy the original image to the rightmost third of the out image, but with left and right reversed. Return the out image. The starter code includes the standard for y/x loops over the original image as a starting point.
def mirrorish(filename):
image = SimpleImage(filename)
# create out image:
# loop x,y over original image
for y in range(image.height):
for x in range(image.width):
# pass
return out
image3.
Given an image and int n which is zero or more. Create a new 'out' image which is n pixels wider than the original. In a vertical stripe, n pixels wide, running down the left side of the output, set all the green values to 0, making it purple. Copy the original image to the area to the right of the stripe. Return the output image. The standard for y/x loop is provided as a start for copying over the original image.
def wider(filename, n):
image = SimpleImage(filename)
# create out image:
# purple stripe at left:
# copy image to space n pixels over
# (loops provided)
for y in range(image.height):
for x in range(image.width):
pass
For the Grid problems you should understand the Peep and Movie lecture examples and your solution to the Sand homework: code that returns a Boolean test about the grid, code that potentially changes the grid, and the ability to write a Doctest about the grid.
Suppose we have a program similar to Sand, but where 's' is smoke. Every square in the grid is either 's' smoke, 'r' rock, or None empty.
The "upright" move, is a diagonal move, moving an 's' up one square and towards the right one square.
Before upright move at 0,1:
--- r r s ---
After upright move:
--- rsr ---
a. Write the code for do_upright():
def do_upright(grid, x, y):
"""
Given an x,y which will be in bounds.
If there is a smoke at that x,y,
and the destination "up-right" square is empty and in-bounds.
Move the smoke from x,y to the upright square.
Otherwise do nothing.
"""
pass
b. Write 1 Doctest for the 3 by 2 case shown above. Some code is included, fill in the ?? of each line.
>>> grid = Grid.build(?? >>> do_upright(grid, ?? >>> ??
There are two functions for this problem, and you can get full credit for each function independent of what you do on the other function.
We have a Grid representing Yellowstone park,
and every square is either a person 'p',
a yelling person 'y', a bear 'b', or empty None.
a. is_scared(grid, x, y): Given a grid and an in-bounds x,y. Return True if there is a person or yelling person at that x,y, and there is a bear to the right of x,y or above x,y. Return False otherwise. (One could check more squares, but we are just checking two.)
b. run_away(grid): This function implements the idea that when a person is scared of a bear they they begin yelling and try to move one square left. Loop over the squares in the grid in the regular top to bottom, left to right order (this loop is provided in the starter code).
For each square, if is_scared() is True, (1) First set the person stored in the grid to 'y' (i.e. they are yelling). (2) Second, if there is an empty square to their left, then move the person to that square leaving their original square empty. You may call is_scared() even if you did not implement it.
def is_scared(grid, x, y):
pass
def run_away(grid):
for y in range(grid.height):
for x in range(grid.width):
pass
There are two functions for this problem, and you can get full credit for each function independent of what you do on the other function.
Bears love honey.
We have a Grid representing Yosemite national park,
and every square is either a bear 'b',
honey 'h', a tree 't', or empty None.
a. honey_dx(grid, x, y): Given a grid and an in bounds x,y. If the square contains a bear, do the following: if the square immediately to the left contains honey, return -1. If the left does not contain honey, but the square to the right contains honey, return 1. In all other cases - no bear or no honey - return 0. So in effect, a 1 or -1 return value indicates that there's a bear, and the x direction of some honey.
b. feed_column(grid, x):
Given a grid and an in-bounds x value. The x
value defines a column within the grid.
This function feeds all
the bears in the column.
Do the following for every square in the column:
Check each square by calling the the honey_dx() helper function.
If the square contains a bear and there is honey to its left or right,
set the honey square to None - in effect the bear eats it. If there is honey on both sides, eat the left side. Return the changed grid.
def honey_dx(grid, x, y):
pass
def feed_column(grid, x):
pass
return grid
str1. Given a string s. If s has a digit at index 1 and also a digit at index 3, then return a string made of the 2 digits. In all other cases, return the empty string. Does not need any loops.
def digits(s):
str2. Given a string s. Return a string made of the upper case chars from s, so 'Hi TherE' returns 'HTE'.
def uppers(s):
str3. Given a string s. Find find the first '<' and the first '>' in the string. If these chars are not present or the '<' is not before the '>', return the empty string. Otherwise, return the chars that are between '<' and '>'. So 'xx<yyy>xxx' returns 'yyy'.
def foo(s):
# your code here
str4. This nutty function returns a string made up of chars from s. For every char in s, if the char is a digit, compute the int value of that lone digit. If that int is a valid index number into s, append the char at that index to the result. So for example 'xyz029', returns 'xz' since 'x' is at index 0 and 'z' is at index 2. (Old exam problem that plays with char/int fluidity).
def nutty(s):
str5. Given a string s of even length, if the string length is 2 or less, return it unchanged. Otherwise take off the first and last chars. Consider the remaining middle piece. Split the middle into front and back halves. Swap the order of these two halves, and return the whole thing with the first and last chars restored. So 'x1234y' yields 'x3412y'.
Note: int division is not required for our midterm.
For this problem, the back half of the string begins at index len(s) // 2,
e.g. 3 for the example string.
def foo(s):
# your code here
str6. Given a string s. Consider all the digit chars in s. Construct and return a list of all the digit chars in s. So for example 'A34and6' returns ['3', '4', '6']. Return the empty list if there are no digits.
def digits(s):
This problem is similar to the Crypto homework with the simplification that the encryption is done entirely with lowercase chars.
This "double" encryption uses two source lists, src1 and src2, and two slug lists, slug1 and slug2. All the lists contain lowercase chars and are the same length. Here is the double_encrypt() algorithm: use the lowercase version of the input char, which may be uppercase. If the char is in src1, return the char at the corresponding index in slug1.
If the char is in src2, return the char at the corresponding "reverse index" in slug2, e.g. like this for lists length 4:
index reverse-index 0 3 1 2 2 1 3 0
If the char is not in src1 or src2, return the char unchanged. No char appears in both src1 and src2.
Double Encrypt Example (len 4) src1 = ['a', 'b', 'c', 'd'] slug1 = ['s', 't', 'u', 'v'] src2 = ['e', 'f', 'g', 'h'] slug2 = ['w', 'x', 'y', 'z'] encrypt 'A' -> 's' encrypt 'e' -> 'z' encrypt 'f' -> 'y' encrypt '$' -> '$'
def double_encrypt(src1, slug1, src2, slug2, char):
pass
First we will describe the double-slug encryption system used in this problem, then you will write the decrypt function below. As a simplification, we will assume that uppercase chars do not exist for this problem.
Double-slug encryption uses one source list and two slug lists, all containing lowercase chars. The source list is even length, and the two slugs are each half its length. The slug1 list holds the encrypted form of chars in the first half of the source list. The slug2 list holds the encrypted form of chars in the second half of the source list. No char is in both slug1 and slug2. Here is an example with a length-4 source list:
Encrypt examples with source/slugs above:
The encrypt of 'a' is 'd' The encrypt of 'c' is 'b'
Write the code for the decrypt function:
Given the lists of lowercase chars: source, slug1, slug2.
And given lowercase ch which is encrypted if possible.
Return the decrypted form of ch, or return ch unchanged if it
is not encrypted (e.g. '$'). The index in the source list where
the slug2-encryption begins is at midway = len(source) // 2,
which is included in the starter code.
2022 note: int division operator // is not required for our midterm, so for
this code it is provided. Int division is simply division that rounds down to the next
round integer, so 57 // 10 is 5.
Decrypt examples with source/slugs above:
The decrypt of 'd' is 'a' The decrypt of 'b' is 'c'
def decrypt(source, slug1, slug2, ch):
midway = len(source) // 2
pass
a. Python expressions
>>>>>> 3 + 4 - 1 + 2 # by default proceeds left-right 8 >>> >>> 1 + 2 * 3 # */ go before +- 7 >>> >>> 47 % 5 # % is remainder 2 >>> >>> s = 'Summer' >>> s[1] 'u' >>> >>> s[:2] 'Su' >>> >>> s[2:] 'mmer' >>>
b. Function-call
The output is 1 2 3 Key idea: the variable names in each function are independent
def do_midterm(filename):
bit = Bit(filename)
# up to block
while bit.front_clear():
bit.move()
# down the side
bit.left()
while not bit.right_clear():
bit.move()
bit.right()
bit.paint('blue')
while bit.front_clear():
bit.move()
bit.paint('blue')
image1.
def stripes(filename):
image = SimpleImage(filename)
for y in range(image.height):
for x in range(n):
# setting both left and right sides
# with one loop. Could be solved with
# multiple loops.
pixel_left = image.get_pixel(x, y)
pixel_left.red = 0
pixel_right = image.get_pixel(image.width - 1 - x, y)
pixel_right.red = 0
return image
image2.
def mirrorish(filename):
image = SimpleImage(filename)
# your code here
out = SimpleImage.blank(image.width * 3, image.height)
# loop x,y over original image
for y in range(image.height):
for x in range(image.width):
pixel = image.get_pixel(x, y)
# key: locate pixel in out image
pixel_out = out.get_pixel(out.width - 1 - x, y)
pixel_out.red = pixel.red
pixel_out.green = pixel.green
pixel_out.blue = pixel.blue
return out
image3.
def wider(filename, n):
image = SimpleImage(filename)
# your code here
out = SimpleImage.blank(image.width + n, image.height)
# purple stripe at left
for y in range(image.height):
for x in range(n):
pixel_stripe = out.get_pixel(x, y)
pixel_stripe.green = 0
# copy image to space n pixels over
for y in range(image.height):
for x in range(image.width):
pixel = image.get_pixel(x, y)
pixel_out = out.get_pixel(n + x, y)
pixel_out.red = pixel.red
pixel_out.green = pixel.green
pixel_out.blue = pixel.blue
return out
a. Write the code for do_upright():
def do_upright(grid, x, y):
"""
Given an x,y which will be in bounds.
If there is a smoke at that x,y,
and the destination "up-right" square is empty and in-bounds.
Move the smoke from x,y to the upright square.
Otherwise do nothing.
"""
if grid.get(x, y) != 's':
return grid
to_x = x + 1 # using vars, though not required
to_y = y - 1
if grid.in_bounds(to_x, to_y):
if grid.get(to_x, to_y) == None:
grid.set(to_x, to_y, 's')
grid.set(x, y, None)
return grid
b. Grid doctest
>>> grid = Grid.build([['r', None, 'r'], ['s', None, None]])
>>> do_upright(grid, 0, 1)
[['r', 's', 'r'], [None, None, None]]
def is_scared(grid, x, y):
# pass
# A person there
if grid.get(x, y) == 'p' or grid.get(x, y) == 'y':
# bear to the right!
if grid.in_bounds(x + 1, y) and grid.get(x + 1, y) == 'b':
return True
# bear above!
if grid.in_bounds(x, y - 1) and grid.get(x, y - 1) == 'b':
return True
return False
def run_away(grid):
for y in range(grid.height):
for x in range(grid.width):
# pass
if is_scared(grid, x, y):
grid.set(x, y, 'y') # begin yelling
if grid.in_bounds(x - 1, y) and grid.get(x - 1, y) == None:
# Move the 'y' one to the left
grid.set(x - 1, y, 'y')
grid.set(x, y, None)
def honey_dx(grid, x, y):
if grid.get(x, y) != 'b':
return 0
if grid.in_bounds(x - 1, y) and grid.get(x - 1, y) == 'h':
return -1
if grid.in_bounds(x + 1, y) and grid.get(x + 1, y) == 'h':
return 1
return 0
def feed_column(grid, x):
pass
for y in range(grid.height):
dx = honey_dx(grid, x, y) # get the dx for this square
# this -1/+1 way is fine
if dx == -1:
grid.set(x - 1, y, None)
if dx == 1:
grid.set(x + 1, y, None)
# the cool way
# if dx != 0:
# grid.set(x + dx, y, None) # nom nom nom nom
return grid
# Subtle issue: our solution calls honey_dx() once
# and stores the returned dx in a var. If a solution calls the function
# twice, it can eat honey twice. This is very subtle, so the deduction
# was very minimal.
str1.
def digits(s):
# Check if index is valid before [ ]
if 3 < len(s):
if s[1].isdigit() and s[3].isdigit():
return s[1] + s[3]
# Could be written as one big "if"
return ''
str2.
def uppers(s):
result = ''
for ch in s:
if ch.isupper():
result += ch
return result
str3.
def foo(s):
start = s.find('<')
end = s.find('>')
if start == -1 or end == -1 or end < start:
return ''
return s[start + 1:end]
str4.
def nutty(s):
result = ''
for i in range(len(s)):
# using for/i/range, but
# for/ch/s works fine too
if s[i].isdigit():
val = int(s[i])
if val < len(s): # valid index
result += s[val]
return result
str5.
def foo(s):
if len(s) <= 2:
return s
first = s[0]
last = s[len(s) - 1]
mid = s[1:len(s) - 1]
halfway = len(mid) // 2 # int div needed
return first + mid[halfway:] + mid[:halfway] + last
str6.
def digits(s):
result = []
for ch in s:
if ch.isdigit():
result.append(ch)
return result
def double_encrypt(src1, slug1, src2, slug2, char):
low = char.lower()
# Try src1, then try src2
if low in src1:
idx = src1.index(low)
return slug1[idx]
if low in src2:
idx = src2.index(low)
rev = len(src2) - idx - 1
return slug2[rev]
return char
def decrypt(source, slug1, slug2, ch):
midway = len(source) // 2
if ch in slug1:
idx = slug1.index(ch)
return source[idx]
if ch in slug2:
idx = slug2.index(ch)
return source[midway + idx]
return ch