CMU 15-112: Fundamentals of Programming and Computer Science
Midterm1 version A

---

Midterm1 Version A

---

See the midterm1 frontmatter. 80 minutes total..

---

PART 1

---

CT1 (8 points):
Indicate what this code prints.

def ct1(s):
    r, n = '', 0
    for t in s.upper().split(' '):
        n += 1
        c = chr(ord('A') + n)
        r += f'{c}{t.count(c)}'
    return r
print(ct1('about baccarat wins!'))

---

CT2 (8 points):
Indicate what this code prints.

def ct2(L):
    M = [ ]
    for i in range(len(L)):
        for j in range(i):
            if (L[i] == L[j]):
                M.append(10*i+j)
    return M[1:] + M[:1]
print(ct2([2,3,3,2,5,3]))

---

CT3 (8 points):
Indicate what this code prints.

def ct3(L):
    M = sorted(L)
    N = L
    M.append(L.pop(0))
    N.append(M.pop(0))
    N = N[:2]
    return M, N
L = [3,1,2]

M, N = ct3(L)
print(L, M, N)

---

CT4 (8 points):
Indicate what this code prints.

from dataclasses import make_dataclass

def ct4():
    A = make_dataclass('A', ['x', 's'])
    M = [ A(x=3,s='a'),
          A(x=1,s='bc'),
          A(x=4,s='de'),
          A(x=2,s='f') ]
    b = A(x=0, s='')
    for a in M:
        if a.x > b.x:
            b.x += a.x
        else:
            b.s += a.s[0]
    return [b.x, b.s]
print(ct4())

---

Note: The ordering of SA-answers was randomized for each exam, and may not match perfectly.

SA1 (3 points):
Which of these does not set M to a copy of L?

• M = copy.copy(L)
• M = L[:]
• M = L + [ ]
• M = L
• M = list(L)

SA2 (3 points):
2. What is the difference between L.append(v) and L.extend(v)?

• There is no difference
• L.append adds one value and L.extend adds a list of values
• L.append adds to the end of L and L.extends adds to the start of L
• There is no such thing as L.extend

SA3 (3 points):
Fill in the blank with ONE line so that M is equal to L, only with the value 5 nondestructively inserted at index 3 (so L is unchanged).
You may assume len(L)>3:

   M = ____________

SA4 (3 points):
Fill in the blank so that T refers to a singleton (length 1) tuple containing just the value 42:

   T = _____________

SA5 (3 points):
Fill in the blank with a list comprehension so that M contains the list of the squares of the values in L, which you may assume is a list of integers:

    M = ____________

SA6 (3 points):
Each of the following are from our Snake case study. For each, indicate if they are part of the Model, the View, or a Controller:

• timerFired(app)
• takeStep(app)
• drawGameOver(app)
• app.foodPosition
• placeFood(app)

---

PART 2

---

Free Response: nthSortOfSquarish(n) [25] points]

Important note: you cannot use strings for this problem but lists are ok.

A number n is sort-of-squarish if all of these are true:

• n is positive
• n contains no zeros
• n is not a perfect square
• the digits of n, when sorted, form a perfect square.

For example, consider 522. The number formed by sorting those digits is 225, and 225 is a perfect square (15**2). Thus, 522 is sort-of-squarish. Since the order of the digits does not matter, 252 is also sort-of-squarish. Note that 225 is a perfect square so is not sort-of-squarish.

Here are the first 10 sort-of-squarish numbers:
[52, 61, 63, 94, 252, 265, 298, 414, 522, 526]

With this in mind, first write isPerfectSquare(n) that takes a non-negative int n and returns True if it is a perfect square (0, 1, 4, 9, 16, etc) and False otherwise.

Next, write the function isSortOfSquarish(n) that takes a possibly-negative int n and returns if n is a sort-of-squarish number.

Then, also write the function nthSortOfSquarish(n) that takes a non-negative int n an returns the nth sort-of-squarish number, so nthSortOfSquarish(0) returns 52.

Hints:

1. Remember, do not use strings here, but lists are ok.
2. You must write isPerfectSquare(n), isSortOfSquarish(n), and nthSortOfSquarish(n) along with any other helper functions you use.
3. In isSortOfSquarish(n) you might want to make a list containing the individual digits in n.

def isPerfectSquare(n):
    return 42

def isSortOfSquarish(n):
    return 42

def nthSortOfSquarish(n):
    return 42



# These test cases are for your use only.  You may comment them out if you
# get stuck and need to move on to the next part.
def testIsPerfectSquare():
    print('Testing isPerfectSquare(n))...', end='')
    assert(isPerfectSquare(4) == True)
    assert(isPerfectSquare(9) == True)
    assert(isPerfectSquare(10) == False)
    assert(isPerfectSquare(225) == True)
    assert(isPerfectSquare(1225) == True)
    assert(isPerfectSquare(1226) == False)
    print('Passed')


def testIsSortOfSquarish():
    print('Testing isSortOfSquarish(n))...', end='')
    assert(isSortOfSquarish(52) == True)
    assert(isSortOfSquarish(16) == False)
    assert(isSortOfSquarish(502) == False)
    assert(isSortOfSquarish(414) == True)
    assert(isSortOfSquarish(5221) == True)
    assert(isSortOfSquarish(6221) == False)
    assert(isSortOfSquarish(-52) == False)
    print('Passed')


def testNthSortOfSquarish():
    print('Testing nthSortOfSquarish()...', end='')
    assert(nthSortOfSquarish(0) == 52)
    assert(nthSortOfSquarish(1) == 61)
    assert(nthSortOfSquarish(2) == 63)
    assert(nthSortOfSquarish(3) == 94)
    assert(nthSortOfSquarish(4) == 252)
    assert(nthSortOfSquarish(8) == 522)
    print('Passed')

testIsPerfectSquare()
testIsSortOfSquarish()
testNthSortOfSquarish()

---

Free Response: animation [25] points]

Write an animation where:

1. The canvas starts empty.
2. Each time the user presses the mouse:
   1. A new green dot of radius 20 is added, centered on the mouse press, and
   2. A line is drawn connecting the center of the green dot to the center of the nearest green dot already drawn, if there is such a dot (breaking ties however you wish).
3. Each time the user presses the 'r' key: the app resets, so that
all the dots and all the lines are deleted.
4. Each time the app has not reset after 5 seconds, it automatically resets
(as though an 'r' was pressed).

Hint: We recommend that you write a helper function that takes app, cx, and cy, and returns the (x, y) location of the nearest existing dot.

---

PART 3

---

BonusCT1 [+2 points]

This is an optional bonus problem. Indicate what this code prints.

def bonusCt1(n):
    return eval(str(list(range(n)))[1:-1]
                .replace(', ','-(') +
                (n-1)*')')
print(bonusCt1(50))

BonusCT2 [+2 points]

This is an optional bonus problem. Indicate what this code prints.

# Note: z() is a helper function for bonusCt2()
# Hint: ord('0') is 48
def z(n):
    m = -n/2
    while n > 0: m, n = m+n, n-1
    return round(2*m)    
def bonusCt2(n):
    while n < 100: n = round(z(z(n))/n)
    s = str(n)
    for i in range(1, 12345):
        d = chr(int(s[:i]))
        if d.isdigit():
            return d * int(s[i:])
    return 42
print(bonusCt2(2))