TheFibonacci Sequence is a series of numbers where each number$F(n)$ is the sum of the two preceding ones, often starting with 0 and 1. That is:

$$F(n) = F(n-1) + F(n-2)$$

with initial conditions

$$F(0) = 0, \quad F(1) = 1$$

The ratio of consecutive Fibonacci numbers approximates the Golden Ratio ($\phi \approx 1.6180339887$):

$$\lim_{{n \to \infty}} \frac{{F(n+1)}}{{F(n)}} = \phi$$

The sequence frequently manifests in nature, such as in flower petals, seedhead spirals, and seashell growth patterns.

Here is the Python code:

# Using recursiondeffibonacci_recursive(n):ifn<=0:return0elifn==1:return1returnfibonacci_recursive(n-1)+fibonacci_recursive(n-2)# Using dynamic programming for efficiencydeffibonacci_dynamic(n):fib= [0,1]foriinrange(2,n+1):fib.append(fib[-1]+fib[-2])returnfib[n]

The task is to write a function thatreturns the $n$th Fibonacci number using arecursive approach.

Thenaive recursive solution for the Fibonacci series, while easy to understand, is inefficient due to its exponential time complexity of$O(2^n)$.

Dynamic Programming methods like memoization and tabulation result in optimized time complexity.

1. Check for the base cases, i.e., if$n$ is 0 or 1.
2. If not a base case,recursively compute$F(n-1)$ and$F(n-2)$.
3. Return the sum of the two recursive calls.

Here is the Python code:

deffib_recursive(n):ifn<=1:returnnreturnfib_recursive(n-1)+fib_recursive(n-2)

• Time Complexity:$O(2^n)$. This is due to the two recursive calls made at each level of the recursion tree, resulting in an exponential number of function calls.

• Space Complexity:$O(n)$. Despite the inefficient time complexity, the space complexity is$O(n)$ as it represents the depth of the recursion stack.

The task is togenerate the Fibonacci sequence to the $n$th number using anon-recursive approach.

While recursion offers a straightforward solution for the Fibonacci sequence, it has performance and stack overflow issues. Anon-recursive approach, often based on aloop oriterative method, overcomes these limitations.

Here, I'll present bothbinet's formula and aniterative method as the non-recursive solutions.

$$F(n) = \frac{{\phi^n - \psi^n}}{{\sqrt 5}}$$

where:

• $\phi = \frac{{1 + \sqrt 5}}{2}$ (the golden ratio)
• $\psi = \frac{{1 - \sqrt 5}}{2}$

1. Compute$\phi$ and$\psi$.
2. Plug values into the formula for$F(n)$.

• Time Complexity:$O(1)$
• Space Complexity:$O(1)$

Note: While Binet's formula offers an elegant non-recursive solution, it's sensitive tofloating-point errors which can impact accuracy, especially for large$n$.

1. Initialize$\text{prev} = 0$ and$\text{curr} = 1$. These are the first two Fibonacci numbers.
2. For$i = 2$ to$n$, update$\text{prev}$ and$\text{curr}$ to be$\text{prev} + \text{curr}$ and$\text{prev}$ respectively. These become the next numbers in the sequence.

• Time Complexity:$O(n)$
• Space Complexity:$O(1)$

Here's the Python code for both methods:

importmathdeffib_binet(n):phi= (1+math.sqrt(5))/2psi= (1-math.sqrt(5))/2returnint((phi**n-psi**n)/math.sqrt(5))# Outputprint(fib_binet(5))# Output: 5

deffib_iterative(n):ifn<=1:returnnprev,curr=0,1for_inrange(2,n+1):prev,curr=curr,prev+currreturncurr# Outputprint(fib_iterative(5))# Output: 5

Both functions will return the 5th Fibonacci number.

Thenaive recursive implementation of the Fibonacci sequence has a time complexity of$O(2^n)$, which can be optimized using techniques such asmemoization or employing aniterative approach.

This is the straightforward, but inefficient, method.

1. Base Case: Return 0 if$n = 0$ and 1 if$n = 1$.
2. Function Call: Recur on the sum of the$n-1$ and$n-2$ elements.

Here is the Python code:

deffibonacci_recursive(n):ifn<=1:returnnreturnfibonacci_recursive(n-1)+fibonacci_recursive(n-2)

• Time Complexity:$O(2^n)$ - As each call branches into two more calls (with the exception of the base case), the number of function calls grows exponentially with$n$, resulting in this time complexity.
• Space Complexity:$O(n)$ - The depth of the recursion stack can go up to$n$ due to the$n-1$ and$n-2$ calls.

Usingmemoization allows for a noticeable performance boost.

1. Initialize a cache,fib_cache, with default values of -1.
2. Base Case: If the $n$th value is already calculated (i.e.,fib_cache[n] != -1), return that value. Otherwise, calculate the $n$th value using recursion.
3. Cache the Result: Once the $n$th value is determined, store it infib_cache before returning.

Here is the Python code:

deffibonacci_memo(n,fib_cache={0:0,1:1}):ifnnotinfib_cache:fib_cache[n]=fibonacci_memo(n-1,fib_cache)+fibonacci_memo(n-2,fib_cache)returnfib_cache[n]

• Time Complexity:$O(n)$ - Each$n$ is computed once and then stored in the cache, so subsequent computations are$O(1)$.
• Space Complexity:$O(n)$ - The space used by the cache.

Theiterative approach shines in terms of time and space efficiency.

1. Initializea andb as 0 and 1, respectively.
2. Loop: Updatea andb to the next two Fibonacci numbers, replacing them as necessary to ensure they represent the desired numbers.
3. Return:a after the loop exits, since it stores the $n$th Fibonacci number.

Here is the Python code:

deffibonacci_iterative(n):a,b=0,1for_inrange(n):a,b=b,a+breturna

• Time Complexity:$O(n)$ - The function iterates a constant number of times, depending on$n$.
• Space Complexity:$O(1)$ - The variablesa andb are updated in place without utilizing any dynamic data structures or recursion stacks.

Memoization is a technique that makes dynamic programming faster by storing the results of expensive function calls and reusing them.

To apply memoization to theFibonacci sequence, a list or dictionary (array orhashmap in terms of computer science) is used to store the intermediate results, essentially turning the calculation process into a more optimized dynamic programming algorithm.

Here is the Python code:

deffibonacci_memo(n,memo={0:0,1:1}):ifnnotinmemo:memo[n]=fibonacci_memo(n-1,memo)+fibonacci_memo(n-2,memo)returnmemo[n]# Testprint(fibonacci_memo(10))# Outputs: 55

The task is togenerate the Fibonacci sequence using aniterative approach, and then analyzing itstime and space complexity.

1. Initializea = 0 andb = 1.
2. Use a loop to updatea andb. On each iteration:
   • Updatea to the value ofb.
   • Updateb to the sum of its old value and the old value ofa.
3. Repeat the loop 'n-1' times, where 'n' is the desired sequence length.

Here's how the first few Fibonacci numbers are computed:

• Iteration 1:$a = 0, , b = 1, , b = 0 + 1 = 1$
• Iteration 2:$a = 1, , b = 1, , b = 1 + 1 = 2$
• Iteration 3:$a = 1, , b = 2, , b = 2 + 1 = 3$
• Iteration 4:$a = 2, , b = 3, , b = 3 + 2 = 5$
• And so on...

• Time Complexity:$O(n)$ — This is more efficient than the recursive approach which is$O(2^n)$.
• Space Complexity:$O(1)$ — The space used is constant, regardless of the input 'n'.

Here is the Python code:

deffibonacci_iterative(n):ifn<=0:return"Invalid input. n must be a positive integer."fib_sequence= [0]ifn>=1else []a,b=0,1for_inrange(2,n+1):fib_sequence.append(b)a,b=b,a+breturnfib_sequence# Example usageprint(fibonacci_iterative(10))# Output: [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

Dynamic Programming is a powerful algorithmic technique that is widely used to optimizerecursive problems, like computing the Fibonacci sequence, by avoiding redundant computations.

The naive method of Fibonacci computation is highly inefficient, often taking exponential time. Dynamic Programming offers better time complexity, often linear or even constant, without sacrificing accuracy.

The most common way of optimizing Fibonacci computations is throughmemoization, where function calls are stored with their results for future reference. In Python, you can employ decorators or dictionaries to achieve this.

Here is the Python code:

defmemoize(fib):cache= {}defwrapper(n):ifnnotincache:cache[n]=fib(n)returncache[n]returnwrapper@memoizedeffib(n):ifn<2:returnnreturnfib(n-1)+fib(n-2)# Testprint(fib(10))# 55

The task is to compute the $n$th Fibonacci number usingmatrix exponentiation and analyze its efficiency.

Matrix exponentiation offers an optimal$O(\log n)$ solution for the Fibonacci sequence, in contrast to the traditional recursive method that has a time complexity of$O(2^n)$.

1. Represent theFibonacci transformation as $F = \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix}$.
2. Utilizeexponentiation by squaring to efficiently compute$F^n$ for large$n$.
3. Extract the$n$-th Fibonacci number as the top right element of the resultant matrix.

• Time Complexity:$O(\log n)$ due to the efficiency of exponentiation by squaring.
• Space Complexity:$O(\log n)$

Here is the Python code:

importnumpyasnpdefmatrix_power(A,n):ifn==1:returnAifn%2==0:half=matrix_power(A,n//2)returnnp.dot(half,half)else:half=matrix_power(A, (n-1)//2)returnnp.dot(np.dot(half,half),A)deffibonacci(n):ifn<=0:return0F=np.array([[1,1], [1,0]])result=matrix_power(F,n-1)returnresult[0,0]# Example usageprint(fibonacci(6))# Output: 8

The$n$-th term of theFibonacci sequence can be approximated using theGolden Ratio$(\phi \approx 1.61803)$ through Binet's formula:

$$ F(n) \approx \frac{{\phi^n}}{{\sqrt{5}}} $$

Here is the Python code:

importmathdefapproximate_fibonacci(n):golden_ratio= (1+math.sqrt(5))/2returnround(golden_ratio**n/math.sqrt(5))# Testprint(approximate_fibonacci(10))# Output: 55

The objective is to compute$\text{Fib}(n)$, the $n$th Fibonacci number, efficiently formultiple queries.

Aprecomputation strategy is suitable for scenarios whereFibonacci numbers are repeatedly requested over afixed range.

1. Generate and store Fibonacci numbers from 0 to the highest$n$ using anarray ordictionary. This operation has a time complexity of$O(n)$.
2. Subsequent queries are answered directly from the precomputed table with a time complexity of$O(1)$.

• Precomputation:$O(\text{max_n})$
• Query time:$O(1)$

Let's consider Python as our programming language.

Here is a Python function that precomputes Fibonacci numbers up to a certain limit and then returns the $n$th Fibonacci number based on the precomputed table.

defprecompute_fibonacci(max_n):fib= [0,1]a,b=0,1whileb<max_n:a,b=b,a+bfib.append(b)returnfibdeffibonacci(n,fib_table):returnfib_table[n]ifn<len(fib_table)else-1# Usagemax_n=100fib_table=precompute_fibonacci(max_n)print(fibonacci(10,fib_table))# 55

Modifying theFibonacci sequence to start with arbitrary initial values still leads to a unique sequence.

The iterative approach can handle custom starting values andcompute any term in the sequence through the specified algorithm.

1. Input: Start valuesa andb (with a not equal to b) and target termn.
2. Check ifn is 1 or 2. If it is, return the corresponding start value.
3. Otherwise, execute a loopn-2 times and update the start values.
4. Compute then-th term once the loop concludes.

Here is the Python code:

defcustom_fibonacci(a,b,n):ifn==1:returnaelifn==2:returnbfor_inrange(n-2):a,b=b,a+breturnb# Example with start values 3 and 4 for the 6th termresult=custom_fibonacci(3,4,6)# Output: 10

TheFibonacci sequence is a classic mathematical series defined by the following:

$$F(n) =\begin{cases}0 & \text{if } n = 0 \\1 & \text{if } n = 1 \\F(n-1) + F(n-2) & \text{if } n > 1\end{cases}$$

1. Sum of First n Numbers: The sum of the first$n$ Fibonacci numbers is equal to$F(n+2) - 1$.

This can be computed using a simple,iterative function:

$$ \text{Sum}(n) = F(n+2) - 1 $$

2. n-th Fibonacci Number: It can be calculated usingtwo seed values$F(0)$ and$F(1)$.

The general form is:

$$ F(n) = F(0)A + F(1)B $$

\begin{bmatrix}1 & 1 \1 & 0 \\end{bmatrix}^{(n-1)}\begin{bmatrix}F(1) \F(0) \\end{bmatrix}$$

Where:

• $A$ and$B$ vary based on the initial seed values.
• The matrix is multiplied by itself$(n-1)$ times.

Here is the Python code:

deffibonacci_sum(n):fib_curr,fib_next,fib_sum=0,1,0for_inrange(n):fib_sum+=fib_currfib_curr,fib_next=fib_next,fib_curr+fib_nextreturnfib_sum

The time complexity is$O(n)$, which is significantly better than$O(n\log n)$ when computing the$n$-th Fibonacci number using matrix exponentiation.

TheLucas Sequence, a variant of the Fibonacci sequence, starts with 2 and 1, rather than 0 and 1, and follows the same recursive structure as the classic sequence:

$$\text{Lucas}(n) = \begin{cases}2, & \text{if } n = 0, \\1, & \text{if } n = 1, \\\text{Lucas}(n-1) + \text{Lucas}(n-2), & \text{otherwise}.\end{cases}$$

Compared to the Fibonacci sequence, the Lucas sequence offers:

• Simpler Recurrence Relation: The Lucas sequence uses only addition for its recursive relation, which can be computationally more efficient than Fibonacci's addition and subtraction.

• Alternate Closed-Form Expression: While the closed-form formula for the $n$th term of the Fibonacci sequence involves radicals, the Lucas sequence provides an alternate expression that can be easier to work with.

TheFibonacci sequence is closely related to the problem of covering a 2xN rectangle with 2x1 tiles, often referred to as the "tiling problem". It in fact offers a direct solution to this problem.

Covering a 2xN rectangle$R_{1}$ with 2x1 tiles can be understood in terms of the number of ways to cover the last column, whether with a single vertical tile or two horizontal tiles:

$$W_{1}(2xN) = W(2x(N-1)) + W(2x(N-2))$$

This is a recursive relationship similar to the one used to define the Fibonacci numbers, making it clear that there is a connection between the two.

Here is the Python code:

deftiling_ways(n):a,b=1,1for_inrange(n):a,b=b,a+breturnan=5ways_to_tile=tiling_ways(n)print(f'Number of ways to tile a 2x{n} rectangle:{ways_to_tile}')

In this example,a, b = b, a + b is a compact way toupdatea andb. It relies on the fact that the right-hand side of the assignment isevaluated first before being assigned to the left-hand side.

This approach has atime complexity of$O(N)$ and aspace complexity of$O(1)$ since it only requires two variables.

Fibonacci numbers grow at a rapid rate, which can lead tointeger overflow. Numerous strategies exist to circumvent this issue.

1. Using a Data Type with Larger Capacity:Thelong long int data type in C/C++ provides up to 19 digits of precision, thus accommodating Fibonacci numbers up to$F(92)$.

2. Using Built-in Arbitrary Precision Libraries:Certain programming languages such as Python and Ruby come witharbitrary-precision arithmetic support, making them suited for such computations.

3. Implementing Custom Arithmetic:Libraries like GMP (GNU Multiple Precision Arithmetic Library) andbigInt in Java enable the handling of arbitrary-precision operations. Additionally, rolling out a custom arithmetic procedure througharrays, linked lists, or strings is viable.

Here is the C++ code:

  #include<iostream>  #include<vector>usingnamespacestd;longlongintfibonacci(int n) {if (n <=1)return n;longlongint a =0, b =1, c;for (int i =2; i <= n; ++i) {          c = a + b;          a = b;          b = c;      }return b;  }intmain() {int n =100;      cout <<"Fibonacci number at position" << n <<" is:"            <<fibonacci(n) << endl;return0;  }