CS61A Homework 3
easterling
  2022-09-22 21:55:14 2022-09-22 21:55   2024-03-04 21:09:23  

Homework 3: Recursion, Tree Recursion hw03.zip

Q1: Num eights

Write a recursive function num_eights that takes a positive integer pos and returns the number of times the digit 8 appears in pos.

Important: Use recursion; the tests will fail if you use any assignment statements or loops. (You can however use function definitions if you so wish.)

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def num_eights(pos):
    """Returns the number of times 8 appears as a digit of pos.

    >>> num_eights(3)
    0
    >>> num_eights(8)
    1
    >>> num_eights(88888888)
    8
    >>> num_eights(2638)
    1
    >>> num_eights(86380)
    2
    >>> num_eights(12345)
    0
    >>> from construct_check import check
    >>> # ban all assignment statements
    >>> check(HW_SOURCE_FILE, 'num_eights',
    ...       ['Assign', 'AnnAssign', 'AugAssign', 'NamedExpr', 'For', 'While'])
    True
    """
    "*** YOUR CODE HERE ***"
    if pos == 0:
        return 0
    if pos % 10 == 8:
        return 1 + num_eights(pos//10)
    return 0 + num_eights(pos//10)

Q2: Ping-pong

The ping-pong sequence counts up starting from 1 and is always either counting up or counting down. At element k, the direction switches if k is a multiple of 8 or contains the digit 8. The first 30 elements of the ping-pong sequence are listed below, with direction swaps marked using brackets at the 8th, 16th, 18th, 24th, and 28th elements:

Index 1 2 3 4 5 6 7 [8] 9 10 11 12 13 14 15 [16] 17 [18] 19 20 21 22 23
PingPong Value 1 2 3 4 5 6 7 [8] 7 6 5 4 3 2 1 [0] 1 [2] 1 0 -1 -2 -3
Index (cont.) [24] 25 26 27 [28] 29 30
PingPong Value [-4] -3 -2 -1 [0] -1 -2

Implement a function pingpong that returns the nth element of the ping-pong sequence without using any assignment statements. (You are allowed to use function definitions.)

You may use the function num_eights, which you defined in the previous question.

Important: Use recursion; the tests will fail if you use any assignment statements. (You can however use function definitions if you so wish.)

Hint: If you’re stuck, first try implementing pingpong using assignment statements and a while statement. Then, to convert this into a recursive solution, write a helper function that has a parameter for each variable that changes values in the body of the while loop.

Hint: There are a few pieces of information that we need to keep track of. One of these details is the direction that we’re going (either increasing or decreasing). Building off of the hint above, think about how we can keep track of the direction throughout the calls to the helper function.

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def pingpong(n):
    """Return the nth element of the ping-pong sequence.

    >>> pingpong(8)
    8
    >>> pingpong(10)
    6
    >>> pingpong(15)
    1
    >>> pingpong(21)
    -1
    >>> pingpong(22)
    -2
    >>> pingpong(30)
    -2
    >>> pingpong(68)
    0
    >>> pingpong(69)
    -1
    >>> pingpong(80)
    0
    >>> pingpong(81)
    1
    >>> pingpong(82)
    0
    >>> pingpong(100)
    -6
    >>> from construct_check import check
    >>> # ban assignment statements
    >>> check(HW_SOURCE_FILE, 'pingpong',
    ...       ['Assign', 'AnnAssign', 'AugAssign', 'NamedExpr'])
    True
    """
    "*** YOUR CODE HERE ***"
    if n == 0:
        return 0

    def forward(x):
        if x == 0:
            return 1
        if x % 8 == 0 or num_eights(x) > 0:
            return -1*forward(x-1)
        return forward(x-1)
    return forward(n-1)+pingpong(n-1)

Q3: Count coins

Given a positive integer change, a set of coins makes change for change if the sum of the values of the coins is change. Here we will use standard US Coin values: 1, 5, 10, 25. For example, the following sets make change for 15:

  • 15 1-cent coins
  • 10 1-cent, 1 5-cent coins
  • 5 1-cent, 2 5-cent coins
  • 5 1-cent, 1 10-cent coins
  • 3 5-cent coins
  • 1 5-cent, 1 10-cent coin

Thus, there are 6 ways to make change for 15. Write a recursive function count_coins that takes a positive integer change and returns the number of ways to make change for change using coins.

You can use either of the functions given to you:

  • next_larger_coin will return the next larger coin denomination from the input, i.e. next_larger_coin(5) is 10.
  • next_smaller_coin will return the next smaller coin denomination from the input, i.e. next_smaller_coin(5) is 1.
  • Either function will return None if the next coin value does not exist

There are two main ways in which you can approach this problem. One way uses next_larger_coin, and another uses next_smaller_coin.

Important: Use recursion; the tests will fail if you use loops.

Hint: Refer the implementation of count_partitions for an example of how to count the ways to sum up to a final value with smaller parts. If you need to keep track of more than one value across recursive calls, consider writing a helper function.

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def next_larger_coin(coin):
    """Returns the next larger coin in order.
    >>> next_larger_coin(1)
    5
    >>> next_larger_coin(5)
    10
    >>> next_larger_coin(10)
    25
    >>> next_larger_coin(2) # Other values return None
    """
    if coin == 1:
        return 5
    elif coin == 5:
        return 10
    elif coin == 10:
        return 25

def next_smaller_coin(coin):
    """Returns the next smaller coin in order.
    >>> next_smaller_coin(25)
    10
    >>> next_smaller_coin(10)
    5
    >>> next_smaller_coin(5)
    1
    >>> next_smaller_coin(2) # Other values return None
    """
    if coin == 25:
        return 10
    elif coin == 10:
        return 5
    elif coin == 5:
        return 1

def count_coins(change):
    """Return the number of ways to make change using coins of value of 1, 5, 10, 25.
    >>> count_coins(15)
    6
    >>> count_coins(10)
    4
    >>> count_coins(20)
    9
    >>> count_coins(100) # How many ways to make change for a dollar?
    242
    >>> count_coins(200)
    1463
    >>> from construct_check import check
    >>> # ban iteration
    >>> check(HW_SOURCE_FILE, 'count_coins', ['While', 'For'])
    True
    """
    "*** YOUR CODE HERE ***"
    def get_biggest_smaller_coin(n):
            if n>25:
                return 25
            if n<5:
                return 1
            if next_smaller_coin(n):
                return next_smaller_coin(n)
            else:
                return get_biggest_smaller_coin(n+1)
    def count_partitions(n,m):
        if n<5:
            return 1
        elif n==5:
            return 2
        elif n < 0:
            return 0
        elif m == 0:
            return 0
        elif m == 1:
            return 1
        return count_partitions(n-m,m) + count_partitions(n,next_smaller_coin(m))
    return count_partitions(change,get_biggest_smaller_coin(change+1))
CS61A Homework 3
2022/09/22/CS61A Homework 3/
Author
easterling
Published
2022-09-22 21:55
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  1. 1. Homework 3: Recursion, Tree Recursion hw03.zip
    1. 1.0.1. Q1: Num eights
    2. 1.0.2. Q2: Ping-pong
    3. 1.0.3. Q3: Count coins