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Easier solution #14

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35eb0de
Easier solution
madamak Dec 13, 2019
e25054e
Update ch09-lists-tuples-and-dictionaries/9a-cats-with-hats
madamak Dec 14, 2019
e8687aa
Update and renaming of the 9a-challenge-cats-with-hats
madamak Dec 23, 2019
48a5e72
Forgot to correct end loop
madamak Dec 23, 2019
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16 changes: 16 additions & 0 deletions ch09-lists-tuples-and-dictionaries/9a-cats-with-hats
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# At the start all cats are hatless.
cats = [False]*100

# We do a loop over each step size from 1 to 100
for step in range(1,101):

# We only look at cat indices that are a multiple of our step size
for i in range(1, 100//step + 1):

# We change the hat status for each cat we loop over
cats[i*step-1] = not cats[i*step-1]
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@somacdivad somacdivad Dec 13, 2019 •
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I like the trick of using not to reverse the value at the index, but I'm not convinced that the way you calculate indices here is easier to read than using the % operator.

Personally, I think the following is easier to understand:

for i in range(1, 101):
    # If the index is a multiple of the step, reverse the
    # hat status of the cat at that index
    if i % step == 0:
        cats[i-1] = not cats[i-1]

Granted, that solution requires an extra line of code, but IMO it's a lot easier to digest than first computing the number of multiples of step there are between 1 and 100, then computing those multiples in the indices.

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@madamak madamak Dec 14, 2019 •
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I agree with you, the way I wrote this part is clearly not easy to read.

I have a new suggestion as I don't like the fact that we loop over each cat and test them when we can know from the start exactly which ones we need to visit.

# For each step size we calculate how many cats we are going to visit
cats_to_visit = 100 // step_size

for i in range (1, cats_to_visit + 1):
   # We reverse the hat status for each cat we visit
    cats[i*step] = not cats[i*step]

What do you think?

If you agree, I am a beginner with Github, should I directly edit my file with this?


# Print the list of cats that have a hat
for i in range(100):
if cats[i]:
print(f"Cat {i+1} has a hat!")
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+1 on this part of the solution. The statement of the problem says:

Write a program that simply outputs which cats have hats at the end.

I think the other solutions actually don't do a good job of this. They print a list of indices of cats that have hats.

Printing something like what you have here seems more in line with what the challenge asks for. 👍