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Fibonacci Modified
Fibonacci Modified
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To address the task of computing the nth term of a modified Fibonacci sequence in Ruby, I devised a function named fibonacci_modified. This function takes three integers as input parameters: t1 and t2, representing the initial terms of the sequence, and n, indicating the term to be computed. The function first handles the base cases: if n equals 1, it returns t1, and if n equals 2, it returns t2. For any other value of n, the function constructs the sequence iteratively. It initializes an array, fib_sequence, to store the sequence, with the first two elements set as t1 and t2. It then iterates from the third term up to the nth term, computing each term based on the modified Fibonacci formula: the sum of the two preceding terms, where the second preceding term is squared. This approach ensures an efficient calculation of the desired term in the sequence.
For instance, consider the example where t1 = 0, t2 = 1, and n = 5. Here, the function generates the sequence [0, 1, 1, 2, 5], and returns 5 as the 5th term. Similarly, in the case of t1 = 2, t2 = 1, and n = 6, the function produces the sequence [2, 1, 3, 10, 103, 10519], yielding 10519 as the 6th term. It's essential to note that for larger values of n, the resulting term may exceed the range of typical integers, necessitating the use of appropriate data structures or libraries to handle such large results.
Here's the implementation of the fibonacci_modified function in Ruby:
def fibonacci_modified(t1, t2, n) return t1 if n == 1 return t2 if n == 2
end
Example usage
t1, t2, n = 0, 1, 5 puts fibonacci_modified(t1, t2, n) # Output: 5
t1, t2, n = 2, 1, 6 puts fibonacci_modified(t1, t2, n) # Output: 10519
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Sure, here's a structured paragraph that explains your approach, followed by example explanations, and finally, your code:
To tackle the problem of computing the nth term of a modified Fibonacci sequence, I devised a function named fibonacciModified(t1, t2, n). The function takes three integers as input: t1 and t2 represent the initial terms of the sequence, while n denotes the term to be computed. The function first handles the base cases: if n equals 1, it returns t1, and if n equals 2, it returns t2. For any other value of n, the function constructs the sequence iteratively. It initializes a list, fib_sequence, to store the sequence, with the first two elements set as t1 and t2. It then iterates from the third term up to the nth term, computing each term based on the modified Fibonacci formula: the sum of the two preceding terms, where the second preceding term is squared. This approach ensures an efficient calculation of the desired term in the sequence.
For instance, consider the example where t1 = 0, t2 = 1, and n = 5. Here, the function generates the sequence [0, 1, 1, 2, 5], and returns 5 as the 5th term. Similarly, in the case of t1 = 2, t2 = 1, and n = 6, the function produces the sequence [2, 1, 3, 10, 103, 10519], yielding 10519 as the 6th term. It's worth noting that for larger values of n, the resulting term may exceed the range of typical integers, necessitating the use of appropriate data structures or libraries to handle such large results.
Here's the implementation of the fibonacciModified function in Python:
sys.set_int_max_str_digits(0)
def fibonacciModified(t1, t2, n): if n == 1: return t1 elif n == 2: return t2 else: fib_sequence = [0] * n fib_sequence[0] = t1 fib_sequence[1] = t2 for i in range(2, n): fib_sequence[i] = fib_sequence[i-2] + fib_sequence[i-1] ** 2 return fib_sequence[-1]
Sure, here's a structured paragraph that explains your approach, followed by example explanations, and finally, your code:
To tackle the problem of computing the nth term of a modified Fibonacci sequence, I devised a function named fibonacciModified(t1, t2, n). The function takes three integers as input: t1 and t2 represent the initial terms of the sequence, while n denotes the term to be computed. The function first handles the base cases: if n equals 1, it returns t1, and if n equals 2, it returns t2. For any other value of n, the function constructs the sequence iteratively. It initializes a list, fib_sequence, to store the sequence, with the first two elements set as t1 and t2. It then iterates from the third term up to the nth term, computing each term based on the modified Fibonacci formula: the sum of the two preceding terms, where the second preceding term is squared. This approach ensures an efficient calculation of the desired term in the sequence.
For instance, consider the example where t1 = 0, t2 = 1, and n = 5. Here, the function generates the sequence [0, 1, 1, 2, 5], and returns 5 as the 5th term. Similarly, in the case of t1 = 2, t2 = 1, and n = 6, the function produces the sequence [2, 1, 3, 10, 103, 10519], yielding 10519 as the 6th term. It's worth noting that for larger values of n, the resulting term may exceed the range of typical integers, necessitating the use of appropriate data structures or libraries to handle such large results.
Here's the implementation of the fibonacciModified function in Python:
sys.set_int_max_str_digits(0)
def fibonacciModified(t1, t2, n): if n == 1: return t1 elif n == 2: return t2 else: fib_sequence = [0] * n fib_sequence[0] = t1 fib_sequence[1] = t2 for i in range(2, n): fib_sequence[i] = fib_sequence[i-2] + fib_sequence[i-1] ** 2 return fib_sequence[-1]
This implementation efficiently computes the nth term of the modified Fibonacci sequence based on the given initial terms t1 and t2, providing accurate results even for large values of n.