/solutions/extra_credit_solutions/Lab04_4.py

  1"""
  2Lab_Python_04
  3Extra Credit
  4Solution to Extra Credit problem 4
  5"""
  6
  7
  8## Part A
  9# The right data structure to store the different market prices
 10# would be a dictionary whose keys are food items, and values
 11# either a list of prices, or another dictionary that maps
 12# the names of different stores to the price that they offer
 13# for example
 14
 15multi_price_with_store_names  = {
 16					'bread' : {
 17							'shoprite' : 4.5,
 18							'maxi-mart' : 5.0,
 19							'r-link' : 0.5
 20						  },
 21					'water' : {
 22							'shoprite' : 0.9,
 23							'maxi-mart' : 0.8,
 24							'r-link' : 0.5
 25						  }
 26				}
 27
 28# or...
 29multi_price_without_store_names = {
 30					'bread' : [
 31							4.5,
 32							5.0,
 33							0.5
 34						  ],
 35					'water' : [
 36							0.9,
 37							0.8,
 38							0.5
 39						  ]
 40				  }
 41
 42# ultimately, what you need to do with the data will determine which data structure
 43# is better to use.
 44
 45
 46## Part B
 47# Binary Search
 48# Binary search insertion is very difficult to get right
 49# and it was awesome that some of you did it!
 50def binary_insert(new_float, some_list_of_floats):
 51	"""
 52	Binary Search works when inserting or searching through
 53	a sorted list. It works by repeatedly cutting the search
 54	space in half, until you find the place for which you look.
 55
 56	It is similar to looking up a word in a dictionary.
 57	First, you open up half way, then check if the word you seek
 58	is before or after a word on that page. If it is after, you
 59	go to the middle of the second half of the dictionary,
 60	and if it is before, you go to the middle of the first
 61	half of the dictionary. This process repeats until you have
 62	found the definition.
 63
 64	If you have questions about this algorithm or algorithms in
 65	general, I love to talk about them, and would definitely
 66	like to chat about algorithms, searching, sorts, runtime,
 67	et cetera!
 68	
 69	Also, this problem is notoriously hard to do correctly,
 70	so if you notice a bug in the code please let me know!!
 71	"""
 72
 73	# upper and lower are variables that bound the range
 74	# of the list in which we are searching
 75
 76	# initially, we are searching the whole list,
 77	# so upper is the last element, and lower is 0, the first
 78	upper = len(some_list_of_floats) - 1
 79	lower = 0
 80	
 81	# mid is the middle of the range 
 82	mid = (upper + lower) / 2
 83
 84	# checking for corner solutions
 85	# if the input list is empty
 86	if upper == -1:
 87		some_list_of_floats.append(new_float)
 88		return
 89
 90	# (when the new element is bigger or smaller than every other one)
 91	if new_float <= some_list_of_floats[lower]:
 92		# then we want to insert it before that element
 93		some_list_of_floats.insert(lower, new_float)
 94		return
 95	elif new_float >= some_list_of_floats[upper]:
 96		#then we want to insert it after that element
 97		some_list_of_floats.insert(upper + 1, new_float)
 98		return
 99
100
101	#while our range contains elements...
102	while upper - lower > 0:
103		
104		if new_float < some_list_of_floats[mid]:
105			# then we want to look at the left half
106			upper = mid
107		
108		elif new_float > some_list_of_floats[mid]:
109			# then we want to look at the right half
110			# the plus one is important, because when
111			# we find out mid, we round down
112			lower = mid + 1
113		
114		else:
115			# then new_float is equal to the middle one,
116			# so we can insert it on either side
117			some_list_of_floats.insert(mid,new_float)
118			return
119		
120		mid = (upper + lower) / 2
121
122	#ok, when we are here, we have found the spot!
123	some_list_of_floats.insert(lower,new_float)
124	return
125
126
127## Part C
128# finding the min cost
129# this problem is not so tricky once you see
130# the the minimum cost is the sum of the minimum costs
131# for each item. The hard thing is understanding
132# and working with the given data structures
133def min_cost(grocery_list, item_to_price_list_dict):
134	
135	total_cost = 0
136	
137	for item in grocery_list:
138		# for each item, we want to get the minimum
139		# price out of all of them
140		
141		# first, get the list of prices for this item
142		list_of_prices = item_to_price_list_dict[item]
143		
144		# then, find the minimum
145		# python has a convenient min() function which
146		# takes a list and returns the smallest element
147		lowest_price_for_item = min(list_of_prices)
148		
149		# now we add it to our running total of cost
150		total_cost += lowest_price_for_item
151		
152	return total_cost
153
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