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Python - When are the "non-base case" keys/values stored in the cache dictionary?
def getNthFib(n, cache = {1: 0, 2: 1}): if n in cache: return(cache[n]) else: cache[n] = getNthFib(n-1, cache) + getNthFib(n-2, cache) return cache[n] print(getNthFib(10)) #outputs 34 When exactly do n=3,4,5,6,7,8,9 and their corresponding values get stored in the cache dictionary? Does the line "cache[n] = getNthFib(n-1, cache) + getNthFib(n-2, cache)" tell the computer to calculate getNthFib(3), getNthFib(4), getNthFib(5), etc? If so, how come? I'm struggling to understand the function of the cache ARGUMENT.
3 Respuestas
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Solus That's a good question!
I couldn't see how the values of the cache dictionary were being provided either.
I hope you get an answer
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When n becomes equal to 3, the base case will be called because 2 (n-1) and 1 (n-2) are keys of the dictionary, and :
cache[3] = cache[2] + cache[1] = 1 + 0 = 2
cache[4] = cache[3] + cache[2] = 1 + 1 = 3 and so on.
So basically the program will begin by decreasing the value of n from 10 to 3 and after that the dict will be updated and n will increase back to 10 and returns the last value. So from n = 10 to n= 3 cache = {1:0, 2:1} and after that it will be updated.
I think there is a reason why it is called cache since the dictionary is a hash map, so fetching a value from the dict has a time complexity = O(1).
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It's called memoization (google it!).
The values of previously calculated numbers are stored in the cache to avoid recomputation.
I think this program does not memoize correctly as the cache parameter will be reinstantiated everytime. I am not sure though. I am not so talented in python.