Homework 2: Recursion
Due by 11:59pm on Thursday, July 18th
Instructions
Download hw2.zip. Inside the archive, you will find a file called hw2.py.
Submission: When you are done, submit with the hw2.py to Gradescope. You may submit more than once before the deadline; only the final submission will be scored
Readings: You might find the following references useful:
Grading: Homework is graded based on correctness. Each incorrect problem will decrease the total score by five point. There is a homework recovery policy as stated in the syllabus. This homework is out of 20 points.
Required questions
Question 1: G function
A mathematical function G
on positive integers is defined by two
cases:
G(n) = n, if n <= 3
G(n) = G(n - 1) + 2 * G(n - 2) + 3 * G(n - 3), if n > 3
Write a recursive function g
that computes G(n)
. Then, write an
iterative function g_iter
that also computes G(n)
:
def g(n):
"""Return the value of G(n), computed recursively.
>>> g(1)
1
>>> g(2)
2
>>> g(3)
3
>>> g(4)
10
>>> g(5)
22
"""
"*** YOUR CODE HERE ***"
def g_iter(n):
"""Return the value of G(n), computed iteratively.
>>> g_iter(1)
1
>>> g_iter(2)
2
>>> g_iter(3)
3
>>> g_iter(4)
10
>>> g_iter(5)
22
"""
"*** YOUR CODE HERE ***"
Question 2: 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 7 or contains the digit 7. The first 30 elements of the ping-pong
sequence are listed below, with direction swaps marked using brackets at the
7th, 14th, 17th, 21st, 27th, and 28th elements:
1 2 3 4 5 6 [7] 6 5 4 3 2 1 [0] 1 2 [3] 2 1 0 [-1] 0 1 2 3 4 [5] [4] 5 6
Implement a function pingpong
that returns the nth element of the
ping-pong sequence. Do not use any assignment statements; however, you
may use def
statements.
Hint: If you're stuck, try implementing
pingpong
first using assignment and awhile
statement. Any name that changes value will become an argument to a function in the recursive definition.
def pingpong(n):
"""Return the nth element of the ping-pong sequence.
>>> pingpong(7)
7
>>> pingpong(8)
6
>>> pingpong(15)
1
>>> pingpong(21)
-1
>>> pingpong(22)
0
>>> pingpong(30)
6
>>> pingpong(68)
2
>>> pingpong(69)
1
>>> pingpong(70)
0
>>> pingpong(71)
1
>>> pingpong(72)
0
>>> pingpong(100)
2
"""
"*** YOUR CODE HERE ***"
You may use the function has_seven
, which returns True if a number k
contains the digit 7 at least once.
def has_seven(k):
"""Returns True if at least one of the digits of k is a 7, False otherwise.
>>> has_seven(3)
False
>>> has_seven(7)
True
>>> has_seven(2734)
True
>>> has_seven(2634)
False
>>> has_seven(734)
True
>>> has_seven(7777)
True
"""
if k % 10 == 7:
return True
elif k < 10:
return False
else:
return has_seven(k // 10)
Question 3: Count change
Once the machines take over, the denomination of every coin will be a power of two: 1-cent, 2-cent, 4-cent, 8-cent, 16-cent, etc. There will be no limit to how much a coin can be worth.
A set of coins makes change for n
if the sum of the values of the
coins is n
. For example, the following sets make change for 7
:
- 7 1-cent coins
- 5 1-cent, 1 2-cent coins
- 3 1-cent, 2 2-cent coins
- 3 1-cent, 1 4-cent coins
- 1 1-cent, 3 2-cent coins
- 1 1-cent, 1 2-cent, 1 4-cent coins
Thus, there are 6 ways to make change for 7
. Write a function
count_change
that takes a positive integer n
and returns the number
of ways to make change for n
using these coins of the future:
def count_change(amount):
"""Return the number of ways to make change for amount.
>>> count_change(7)
6
>>> count_change(10)
14
>>> count_change(20)
60
>>> count_change(100)
9828
"""
"*** YOUR CODE HERE ***"
Q4: Towers of Hanoi
A classic puzzle called the Towers of Hanoi is a game that consists of three
rods, and a number of disks of different sizes which can slide onto any rod.
The puzzle starts with n
disks in a neat stack in ascending order of size on
a start
rod, the smallest at the top, forming a conical shape.
The objective of the puzzle is to move the entire stack to an end
rod,
obeying the following rules:
- Only one disk may be moved at a time.
- Each move consists of taking the top (smallest) disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod.
- No disk may be placed on top of a smaller disk.
Complete the definition of move_stack
, which prints out the steps required to
move n
disks from the start
rod to the end
rod without violating the
rules. The provided print_move
function will print out the step to move a
single disk from the given origin
to the given destination
.
Hint: Draw out a few games with various
n
on a piece of paper and try to find a pattern of disk movements that applies to anyn
. In your solution, take the recursive leap of faith whenever you need to move any amount of disks less thann
from one rod to another. If you need more help, see the following hints.
print
effectively "collects" all the results in the terminal as long as you make sure that the moves are printed in order.
def print_move(origin, destination):
"""Print instructions to move a disk."""
print("Move the top disk from rod", origin, "to rod", destination)
def move_stack(n, start, end):
"""Print the moves required to move n disks on the start pole to the end
pole without violating the rules of Towers of Hanoi.
n -- number of disks
start -- a pole position, either 1, 2, or 3
end -- a pole position, either 1, 2, or 3
There are exactly three poles, and start and end must be different. Assume
that the start pole has at least n disks of increasing size, and the end
pole is either empty or has a top disk larger than the top n start disks.
>>> move_stack(1, 1, 3)
Move the top disk from rod 1 to rod 3
>>> move_stack(2, 1, 3)
Move the top disk from rod 1 to rod 2
Move the top disk from rod 1 to rod 3
Move the top disk from rod 2 to rod 3
>>> move_stack(3, 1, 3)
Move the top disk from rod 1 to rod 3
Move the top disk from rod 1 to rod 2
Move the top disk from rod 3 to rod 2
Move the top disk from rod 1 to rod 3
Move the top disk from rod 2 to rod 1
Move the top disk from rod 2 to rod 3
Move the top disk from rod 1 to rod 3
"""
assert 1 <= start <= 3 and 1 <= end <= 3 and start != end, "Bad start/end"
"*** YOUR CODE HERE ***"
Submit
Make sure to submit the hw2.py on Gradescope.
Just for Fun Questions
Q5: Anonymous factorial
This question demonstrates that it's possible to write recursive functions without assigning them a name in the global frame.
The recursive factorial function can be written as a single expression by using a conditional expression.
>>> fact = lambda n: 1 if n == 1 else mul(n, fact(sub(n, 1)))
>>> fact(5)
120
However, this implementation relies on the fact (no pun intended) that
fact
has a name, to which we refer in the body of fact
. To write a
recursive function, we have always given it a name using a def
or
assignment statement so that we can refer to the function within its
own body. In this question, your job is to define fact recursively
without giving it a name!
Write an expression that computes n
factorial using only call
expressions, conditional expressions, and lambda expressions (no
assignment or def statements). Note in particular that you are not
allowed to use make_anonymous_factorial
in your return expression.
The sub
and mul
functions from the operator
module are the only
built-in functions required to solve this problem:
from operator import sub, mul
def make_anonymous_factorial():
"""Return the value of an expression that computes factorial.
>>> make_anonymous_factorial()(5)
120
"""
return 'YOUR_EXPRESSION_HERE'