Due by 11:59pm on Wednesday, 2/20
Submission. See the online submission instructions. We have provided a starter file for the questions below.
Readings. Section 1.7 of the online lecture notes.
Q1. In the lecture notes, we defined a pig_latin function to compute the Pig Latin equivalent of an English word. The code that we used is as follows:
def pig_latin_original(w):
"""Return the Pig Latin equivalent of a lowercase English word w."""
if starts_with_a_vowel(w):
return w + 'ay'
return pig_latin_original(rest(w) + first(w))
def first(s):
"""Returns the first character of a string."""
return s[0]
def rest(s):
"""Returns all but the first character of a string."""
return s[1:]
def starts_with_a_vowel(w):
"""Return whether w begins with a vowel."""
c = first(w)
return c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u'
This code repeatedly moves a letter from the beginning of a word to the end, until the first letter is a vowel, at which point it adds on 'ay' to the end. However, this code fails when the original word has no vowels in the set {a, e, i, o, u}, such as the word "sphynx." Write a new version of the pig_latin function that just adds 'ay' to the original word if it does not contain a vowel in this set. Use only the first, rest, and starts_with_a_vowel functions to access the contents of a word, and use the built-in len function to determine its length. Do not use any loops.
def pig_latin(w):
"""Return the Pig Latin equivalent of a lowercase English word w.
>>> pig_latin('pun')
'unpay'
>>> pig_latin('sphynx')
'sphynxay'
"""
"*** YOUR CODE HERE ***"
Q2. 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 the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.
The objective of the puzzle is to move the entire stack to another rod, obeying the following rules:
Complete the definition of towers_of_hanoi which prints out the steps to solve this puzzle for any number of n disks starting from the start rod and moving them to the end rod:
def towers_of_hanoi(n, start, end):
"""Print the moves required to solve the towers of hanoi game if we start
with n disks on the start pole and want to move them all to the end pole.
The game is to assumed to have 3 poles (which is traditional).
>>> towers_of_hanoi(1, 1, 3)
Move 1 disk from rod 1 to rod 3
>>> towers_of_hanoi(2, 1, 3)
Move 1 disk from rod 1 to rod 2
Move 1 disk from rod 1 to rod 3
Move 1 disk from rod 2 to rod 3
>>> towers_of_hanoi(3, 1, 3)
Move 1 disk from rod 1 to rod 3
Move 1 disk from rod 1 to rod 2
Move 1 disk from rod 3 to rod 2
Move 1 disk from rod 1 to rod 3
Move 1 disk from rod 2 to rod 1
Move 1 disk from rod 2 to rod 3
Move 1 disk from rod 1 to rod 3
"""
"*** YOUR CODE HERE ***"
def move_disk(start, end):
print("Move 1 disk from rod", start, "to rod", end)
Q3. The number of partitions of a positive integer n is the number of ways in which n can be expressed as the sum of positive integers in increasing order. For example, the number 5 has 7 partitions.
5 = 5
5 = 1 + 4
5 = 2 + 3
5 = 1 + 1 + 3
5 = 1 + 2 + 2
5 = 1 + 1 + 1 + 2
5 = 1 + 1 + 1 + 1 + 1
Write a tree-recursive function part(n) that returns the number of partitions of n.
Hint: Introduce a helper function that computes partitions of n using only a subset of the integers less than or equal to n. Once you have done so, you can use very similar logic to the count_change function from the lecture notes:
def part(n):
"""Return the number of partitions of positive integer n.
>>> part(5)
7
>>> part(10)
42
>>> part(15)
176
>>> part(20)
627
"""
"*** YOUR CODE HERE ***"
Q4. (Extra for experts) Recall that the summation function computes the sum of a sequence of terms from 1 to n:
def summation(n, term):
"""Return the sum of the first n terms of a sequence.
>>> summation(5, lambda x: pow(x, 3))
225
"""
total, k = 0, 1
while k <= n:
total, k = total + term(k), k + 1
return total
Write a function interleaved_sum that similarly computes the sum of a sequence of terms from 1 to n, but uses different function to compute the terms for odd and even numbers. Do so without using any loops or testing in any way if a number is odd or even. (You may test if a number is equal to 0, 1, or n.)
def interleaved_sum(n, odd_term, even_term):
"""Compute the sum odd_term(1) + even_term(2) + odd_term(3) + ..., up
to n.
>>> # 1 + 2^2 + 3 + 4^2 + 5
... interleaved_sum(5, lambda x: x, lambda x: x*x)
29
"""
"*** YOUR CODE HERE ***"