Solutions: You can find the file with solutions for all questions here.
Question 1: Reduce
Write the recursive version of the function reduce
which takes
- reducer - a two-argument function that reduces elements to a single value
- s - a sequence of values
- base - the starting value in the reduction. This is usually the identity of the reducer
If you're feeling stuck, think about the parameters of reduce
.
from operator import add, mul
def reduce(reducer, seq, base):
"""Reduce a sequence under a two-argument function starting from a base value.
>>> def add(x, y):
... return x + y
>>> def mul(x, y):
... return x*y
>>> reduce(add, [1,2,3,4], 0)
10
>>> reduce(mul, [1,2,3,4], 0)
0
>>> reduce(mul, [1,2,3,4], 1)
24
"""
if seq == []:
return base
next_base = reducer(base, seq[0])
return reduce(reducer, seq[1:], next_base)
Use OK to test your code:
python3 ok -q reduce
Question 2: Remove Last from Sequence
Complete the recursive function remove_last
which creates a new list identical to the input list s
but with the last element in the sequence that is equal to x
removed.
def remove_last(x, s):
"""Create a new list that is identical to s but with the last
element from the list that is equal to x removed.
>>> remove_last(1,[])
[]
>>> remove_last(1,[1])
[]
>>> remove_last(1,[1,1])
[1]
>>> remove_last(1,[2,1])
[2]
>>> remove_last(1,[3,1,2])
[3, 2]
>>> remove_last(1,[3,1,2,1])
[3, 1, 2]
>>> remove_last(5, [3, 5, 2, 5, 11])
[3, 5, 2, 11]
"""
if not s:
return []
elif s[-1] == x:
return s[0:-1]
else:
return remove_last(x, s[0:-1]) + [s[-1]]
Illustrated here is a more complete doctest that shows good testing methodology. It is a little cumbersome as documentation, but you'll want to think about it for your projects. Test every condition that might come up. Then you won't be surprised when it does.
Use OK to test your code:
python3 ok -q remove_last
Question 3: Create Number from Lists
Write a recursive function create_num_from_lsts
that creates a number with the elements from lst1
and lst2
as digits in that order.
def create_num_from_lsts(lst1, lst2):
""" Create a number with the elements from lst1 and lst2 as digits in that order.
>>> create_num_from_lsts([1, 2, 3], [4, 5, 6])
123456
>>> create_num_from_lsts([5, 4, 2, 4], [1, 7])
542417
>>> create_num_from_lsts([3], [9, 8])
398
"""
if not lst1 and not lst2:
return 0
elif not lst2:
return create_num_from_lsts(lst2, lst1)
else:
return lst2[-1] + 10 * create_num_from_lsts(lst1, lst2[:-1])
Use OK to test your code:
python3 ok -q create_num_from_lsts
Question 4: Count Change
A set of coins makes change for n
if the sum of the values of the
coins is n
. For example, if you have 1-cent, 2-cent and 4-cent
coins, 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 a list of
the coin denominations and returns the number of ways to make change
for n
using these coins (Hint: You will need to use tree recursion):
def count_change(amount, denominations):
"""Returns the number of ways to make change for amount.
>>> denominations = [50, 25, 10, 5, 1]
>>> count_change(7, denominations)
2
>>> count_change(100, denominations)
292
>>> denominations = [16, 8, 4, 2, 1]
>>> count_change(7, denominations)
6
>>> count_change(10, denominations)
14
>>> count_change(20, denominations)
60
"""
if amount < 0 or denominations == []:
return 0
elif amount == 0:
return 1
using_coin = count_change(amount - denominations[0], denominations)
not_using_coin = count_change(amount, denominations[1:])
return using_coin + not_using_coin
Use OK to test your code:
python3 ok -q count_change
Submit
Make sure to submit this assignment by running:
python3 ok --submit