# Problem 1: Entropy In this problem, be calculating what’s ca...

## Transcribed Text

Problem 1: Entropy In this problem, be calculating what’s called Entropy. This is related, but not directly, to the entropy you’ve seen in physics. Some time ago, Claude Shannon at IBM was working on making a mathematical model of communication. Imagine, you’d have a piece of text, and then his function would give a number indicating how important or significant it was. What Shannon observed was that importance was really the degree of how surprising the message was. So, in a sense, the message itself wasn’t as important as how unexpected it was. He then realized that he could utilize probability as a measure of surprise: Message Entropy Probability not surprising ↓ high surprising ↑ low For this problem, we are only working with finite probabilities. We can think of probabilities as a list of numbers p0, p1, . . . , pn such that pi ≥ 0, i = 0, 1, . . . , n (1) 1 = p0 + p1 + p2 + · · · + pn (2) = Xn i=0 pi (3) Line (3) is usually how it’s written using the P as a shorthand for addition. We’ll learn about this later, but I thought it might be interesting to see it now. Observe it looks a lot like range! You might think about what’s similar and what’s different. To say in words, they are a finite collection of numbers that are non-negative that sum to exactly one. We can make a list (we’ll assume the items are of the same type and immutable) into a probability. Consider x = ["a","b","a","c","c","a"] 1. gather the items uniquely (hint: dictionary) 2. count each time the item occurs 3. create a new list of the count/len(dictionary) The list of probabilities would be y = [3/6,1/6,2/6]. We still need to show you how to calculate entropy: entropy = −(p0 log2 (p0) + p1 log2 (p1) + · · · pn log2 (pn) (4) entropy(y) = −(3/6 log2 (3/6) + 1/6 log(1/6) + 2/6 log2 (2/6) (5) = −(.5(−1.0) + 0.17(−2.58) + .33(−1.58)) (6) = 1.46 (7) Because of continuity arguments we treat log2 (0) = 0. Python’s math module will correctly state this math error, so you’ll have to simply add 0 if the probability is 0. Listing 1: entropy.py 1 import math 2 def makeProbability(xlst): 3 #TO DO: IMPLEMENT FUNCTION 4 5 def entropy(xlst): 6 #TO DO: IMPLEMENT FUNCTION 7 8 9 s1 = [’a’,’b’,’a’,’b’,’a’,’b’,’b’,’b’] 10 s2 = [(1),(2),(3),(4)] 11 s3 = [1] 12 s4 = [1,2] 13 xlst = [s1,s2,s3,s4] 14 for i in xlst: 15 print(entropy(makeProbability(i))) Output function.py 0.954434002924965 2.0 0.0 1.0 Deliverables Problem 1 • Complete the two TO DO’s. • Put this code in a new module named entropy.py. Problem 2: Magick You’ve encountered a soothsayer named Soothy McSoothface. He will magically determine any whole, positive number you guess by asking you to perform a few operations and tell him the result. It’s quite amazing. Modify the previous code to show he’s not so magic after all. Magical Encantation Pick a number x Add fifteen to x Multiply the sum by three Subtract nine from the product Divide the difference by three Subtract 12 from the quotient Hocus Pocus–that is your number Table 1: Encantation (operations) on a secret number that yields the secret number at the end. To give you a start, here’s the code and where you’ll write yours: 1 # Input value: some number 2 # Output value: the result of the magical encantation 3 def magic(x): 4 #TODO: implement the magical encantation 5 6 if __name__=="__main__": 7 #get input 8 x = input("Pick any positive whole number: ") 9 10 #change from string to integer 11 x = int(x) 12 13 print("Your number was", magic(x)) Ponder this a bit, and try your best. When you’re done, and it works, take a look on the next page. To help you out, we’ll remind you of some of the vocabulary: Operation Result Name Division Quotient Addition Sum Multiplication Product Subtraction Difference Power Exponetiation Table 2: Vocabulary for mathematical operations. Deliverables Problem 2 • Complete the program. • Put your program in a new module named magic.py

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import math

def makeProbability(xlst):
# TO DO: IMPLEMENT FUNCTION
# gather the items uniquely (hint: dictionary)
unique = set(xlst)
# count each time the item occurs
counter = [0.0] * len(unique)
# iterate over the main list
for item in xlst:
# get index and value the unique list
for index, val in enumerate(xlst):
# if this value equals to the current item in the list
# count it up
if val == item:
counter[index] = counter[index] + 1.0
break
pass
pass
pass

# create a new list of the count/len(dictionary)
for index, val in enumerate(counter):
# get probability of each unique item
counter[index] = (counter[index]) / float(len(xlst))
pass
return counter
pass...

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