Imagine deciphering an ancestral code where each number is the key to a mathematical mystery 🕵️♂️. Ready to unravel the secret of a peculiar series?
🔮 Problem Statement
Alfredo, an enthusiast of numeric sequences, came across an intriguing challenge. He was assigned the task of discerning the underlying pattern of the following series:
3, 4, 6, 9, 14, 22, ...
The mission is to implement a function that, given an index n, determines the corresponding value in this enigmatic series.
Parameters:
int a: The first number of the series.int b: The second number of the series.int n: The index of the desired number in the series.
Returns:
int: The n-th number of the series.
Example:
>>> pattern2(1, 2, 2)
2
>>> pattern2(3, 5, 1)
3
>>> pattern2(5, 6, 3)
10
>>> pattern2(2, 6, 20)
23490
>>> pattern2(1, 1, 1000000000)
1This task requires a meticulous analysis of the pattern and an efficient implementation to handle even large values of n.
🧩 Step-by-Step Solution
The key to solving this problem lies in understanding how each term of the series is generated from the previous ones. Observing the sequence, we see that each new number is calculated as the sum of the two previous numbers, minus one. To do this, we use a list that stores the two last numbers calculated, allowing us to iterate and update the series efficiently.
def pattern2(a, b, n):
"level: medium; points: 4"
serie = [a, b]
if n <= 2:
return serie[n - 1]
for _ in range(n - 2):
serie[0], serie[1] = serie[1], sum(serie) - 1
return serie[1]Initialization of the Series: We begin by initializing a list called serie with the two first numbers given a and b. This list will act as a sliding window over the series, maintaining only the two last values relevant to the calculation.
serie = [a, b]Base Case: If n is less than or equal to 2, the result is directly the first or second number of the initial series. This avoids unnecessary calculations and provides the correct answer for the base cases.
if n <= 2:
return serie[n - 1]Iteration for Calculation: The core of the solution resides in a for loop that runs n - 2 times. Inside this loop, we use Python’s simultaneous assignment to update the values of the serie list. The first element takes the value of the second, and the second element becomes the sum of the two original elements minus 1.
for _ in range(n - 2):
serie[0], serie[1] = serie[1], sum(serie) - 1Return of the Result: Finally, after the loop has completed its execution, the second element of the serie list will contain the n-th number of the sequence. This value is returned as the result of the function.
return serie[1]Complete Solution:
def pattern2(a, b, n):
"level: medium; points: 4"
serie = [a, b]
if n <= 2:
return serie[n - 1]
for _ in range(n - 2):
serie[0], serie[1] = serie[1], sum(serie) - 1
return serie[1]🧠 Key Concepts
Simultaneous assignment is a fundamental concept in Python that allows exchanging or updating variable values concisely and efficiently. In this case, we use it to update the elements of the serie list in a single step, avoiding the need for temporary variables. This technique not only improves the readability of the code but can also optimize performance by performing the assignment more efficiently internally.
Lists are versatile data structures that allow storing ordered collections of elements. In this solution, the serie list is used as a sliding window that maintains the two last numbers of the sequence, facilitating the calculation of the next number. The ability of lists to store and access elements efficiently is crucial for the performance of the function, especially when working with large values of n.
The pattern of the sequence can be interpreted as a form of implicit recursion. Although the solution does not use explicit recursion, the way each new term is calculated from the previous ones reflects an underlying recursive relationship. Each term depends on the two previous ones, which creates a chain dependency similar to that observed in recursive functions. Did you know that this “implicit recursion” can be analyzed mathematically to find a closed-form formula that avoids the need to iterate for very large values of n? Although the given solution is more direct and understandable, the existence of a closed-form formula is an example of how mathematics can optimize algorithms.
💫 Final Thoughts
This solution, although functional, has limitations in terms of efficiency for extremely large values of n. A possible improvement would be to explore the derivation of a closed-form formula to calculate the n-th term directly, avoiding the need to iterate. Another optimization could be to use generators instead of lists to reduce memory consumption, especially if you expect to handle very long sequences. In addition, the function could benefit from the addition of error handling to validate the input parameters and avoid unexpected behaviors.
I hope this journey through the numeric labyrinth has been as intriguing for you as it was for Alfredo. If you are passionate about algorithmic challenges and code optimization, I invite you to explore more articles on our blog. The next puzzle awaits you! 🚀