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fibonacci sequence speed with lru cache shorts

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the fibonacci sequence is a classic example in computer science that illustrates the concept of recursion and dynamic programming. the fibonacci numbers are defined by the recurrence relation:
- f(0) = 0
- f(1) = 1
- f(n) = f(n-1) + f(n-2) for n 1
problem with naive recursion
a naive recursive approach to calculate fibonacci numbers has exponential time complexity o(2^n) because it recalculates the same values multiple times. this is highly inefficient for larger values of `n`.
optimizing with memoization
to optimize the calculation, we can use **memoization**, which is a technique where we store the results of expensive function calls and return the cached result when the same inputs occur again.
in python, the `functools` module provides a convenient decorator called `lru_cache` (least recently used cache) that can be used to cache the results of function calls automatically. this eliminates the need for manual caching and significantly speeds up the fibonacci calculation.
implementation
here's how you can implement the fibonacci sequence using `lru_cache`:
```python
import time
from functools import lru_cache
@lru_cache(maxsize=none) no limit on the size of the cache
def fibonacci(n):
if n 0:
raise valueerror("fibonacci number is not defined for negative numbers.")
if n == 0:
return 0
elif n == 1:
return 1
else:
return fibonacci(n - 1) + fibonacci(n - 2)
example usage
if __name__ == "__main__":
n = 35 you can change this to test with larger numbers
result = fibonacci(n)
print(f"fibonacci({n}) = {result}")
print(f"time taken: {end_time - start_time:.6f} seconds")
```
explanation of the code
1. **importing libraries**: we import `time` to measure execution time and `lru_cache` from `functools`.
#Fibonacci #LRUCache #CodingShorts
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- f(0) = 0
- f(1) = 1
- f(n) = f(n-1) + f(n-2) for n 1
problem with naive recursion
a naive recursive approach to calculate fibonacci numbers has exponential time complexity o(2^n) because it recalculates the same values multiple times. this is highly inefficient for larger values of `n`.
optimizing with memoization
to optimize the calculation, we can use **memoization**, which is a technique where we store the results of expensive function calls and return the cached result when the same inputs occur again.
in python, the `functools` module provides a convenient decorator called `lru_cache` (least recently used cache) that can be used to cache the results of function calls automatically. this eliminates the need for manual caching and significantly speeds up the fibonacci calculation.
implementation
here's how you can implement the fibonacci sequence using `lru_cache`:
```python
import time
from functools import lru_cache
@lru_cache(maxsize=none) no limit on the size of the cache
def fibonacci(n):
if n 0:
raise valueerror("fibonacci number is not defined for negative numbers.")
if n == 0:
return 0
elif n == 1:
return 1
else:
return fibonacci(n - 1) + fibonacci(n - 2)
example usage
if __name__ == "__main__":
n = 35 you can change this to test with larger numbers
result = fibonacci(n)
print(f"fibonacci({n}) = {result}")
print(f"time taken: {end_time - start_time:.6f} seconds")
```
explanation of the code
1. **importing libraries**: we import `time` to measure execution time and `lru_cache` from `functools`.
#Fibonacci #LRUCache #CodingShorts
in cache zoeken
in cache meaning
in cache wordpress plugin
in cache database
in cache definition
cache in a sentence
in cache
in cache google
cache in the castle
in cache memory
fibonacci in nature
fibonacci in forex
fibonacci in java
fibonacci in art
fibonacci in trading
in fibonacci sequence
fibonacci in javascript
fibonacci in c