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Notes and code for Week 2 of Informatics at PLC Sydney
# -----------------------------------------------------------------------------
# Informations 2019 Week 2
# -----------------------------------------------------------------------------
#
# After Week 1 we had the following functions:
# is_divisible(number, factor) --> True/False
# factors(n) --> list
# is_prime(n) --> True/False
#
# This week we calculate the HIGHEST COMMON FACTOR of two numbers.
# E.g.
# hcf(28,70) --> 14
# hcf(60,120) --> 60
# hcf(8,11) --> 1
#
# For that, we need (at least to begin with) our factors(n) function.
# I am using a new function is_factor instead of the old is_divisible.
# I am also rewriting factors(n) to be more efficient.
def is_factor(f, n):
return (n % f == 0)
def factors(n):
result = []
for f in range(1,n//2):
if is_factor(f, n):
result.append(f)
result.append(n)
return result
# Now consider the highest common factor of 16 and 40.
# factors(16) --> [1,2,4,8,16]
# factors(40) --> [1,2,4,5,8,10,20,40]
#
# The factors in common are [1,2,4,8] and the highest of these is 8.
# The way we will find the HCF is to start with the 16 and ask "is 16 in the
# other list?". The answer is False. Then go to the 8 and ask "is 8 in the other
# list?", and the answer is True. Therefore, the HCF is 8.
# We need a function that tells us whether a given number is present in a list.
def contains(target, xs):
for x in xs:
if x == target:
return True
return False
def test_contains():
assert(contains(5, [1,3,5,7,9]) == True)
assert(contains(5, [2,4,6,8]) == False)
assert(contains(3, []) == False)
# We introduce a function hce (highest common element), which takes two sorted
# lists and returns the first common thing it finds when searching backwards
# through the first list. Here is an algorithm description.
#
# hce(xs, ys) xs and ys are sorted lists
# For each x in xs, going backwards
# If x is in the list ys
# Return x
# If we get this far, there is nothing in common, so...
# ...Return None
#
# To go backwards through a list, we use index notation. The biggest index of
# the list [1,3,7,21] is 3 -- counting starts at zero -- and this is found by
# taking the length of the list (4) and subtracting one.
def hce(xs, ys):
# Return the highest common element between the two lists, searching backwards in xs.
index = len(xs) - 1 # Our starting index
while index >= 0: # Loop while we have a valid index
x = xs[index] # x is the element in xs
if contains(x, ys): # See if x is in ys
return x # Return x if it is
index = index - 1 # Decrease the index and repeat the loop
# If we get this far...
return None
def test_hce():
list1 = [1,3,7,21]
list2 = [1,2,5,10]
list3 = [1,2,3,5,6,10,15,30]
list4 = [99, 100]
assert(hce(list1, list2) == 1)
assert(hce(list1, list3) == 3)
assert(hce(list2, list3) == 10)
assert(hce(list1, list4) == None)
# Now we have enough pieces to implement hcf(a,b).
# We get the factors of a and the factors of b, and get the highest common
# element of the two lists.
def hcf(a,b):
return hce(factors(a), factors(b))
def test_hcf():
assert(hcf(21,30) == 3)
assert(hcf(10,30) == 10)
assert(hcf(10,21) == 1)
assert(hcf(21,10) == 1)
assert(hcf(4200,3528) == 168)
# This function will run all our tests.
# If an assertion is wrong, we get an error.
# If nothing happens, all the tests passed.
def test():
test_contains()
test_hce()
test_hcf()
# So we have written a function that will calculate the HCF of two numbers. Is
# this an efficient way to calculate the HCF? Not really. Next week we will look
# at two more approaches:
# * the intersection of two sets
# * Euclid's algorithm
#
# Feel free to do some research on those things before Fri 22 March.
# * https://www.thoughtco.com/intersection-in-set-theory-3126587
# * https://brilliant.org/wiki/greatest-common-divisor/
# * https://www.programiz.com/python-programming/set
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