Problem 17
Question
Write an algorithm to compute the \(n\) th Lucas number \(L_{n}\) using recursion.
Step-by-Step Solution
Verified Answer
To compute the $n$th Lucas number $L_n$ using recursion, we can write a function `lucas_number(n)` with the following implementation:
```python
def lucas_number(n):
if n == 0:
return 2
elif n == 1:
return 1
return lucas_number(n - 1) + lucas_number(n - 2)
```
This function uses the base cases $L_0 = 2$ and $L_1 = 1$ to return the corresponding values and computes $L_n = L_{n-1} + L_{n-2}$ for other values of $n$ through recursion.
1Step 1: Define the base cases
First, we need to define the base cases for the recursive function. In the case of Lucas numbers, the base cases are L_0 = 2 and L_1 = 1. We will return these values when n = 0 or n = 1.
2Step 2: Write the recursive function
Now, we will write the recursive function to find the nth Lucas number. The function takes n as input and returns L_n. Inside the function, we will use the formula (L_n = L_{n-1} + L_{n-2}) by calling the function with n - 1 and n - 2 as input.
3Step 3: Test the algorithm with sample input
Finally, we will test the algorithm with a few sample input values to verify the correctness of the implementation.
Here's the implementation:
```python
def lucas_number(n):
# Base cases
if n == 0:
return 2
elif n == 1:
return 1
# Recursive formula
return lucas_number(n - 1) + lucas_number(n - 2)
# Test the algorithm
print("L_4 =", lucas_number(4))
print("L_6 =", lucas_number(6))
print("L_10 =", lucas_number(10))
```
The output of the above code will be:
```
L_4 = 7
L_6 = 18
L_10 = 123
```
The algorithm correctly computes the nth Lucas number using recursion.
Key Concepts
Recursive FunctionBase CaseLucas Number
Recursive Function
A recursive function is a powerful tool in programming that solves problems by calling itself. It breaks down a complex problem into smaller, more manageable parts. Here's how it works:
- A recursive function always has one or more base cases to stop recursion, preventing the function from calling itself indefinitely.
- Each recursive call reduces the problem into a smaller version of the original problem.
- The function calls itself with these smaller versions until it reaches a base case, at which point it stops and returns a result.
Base Case
In recursion, the base case is crucial as it provides the condition to stop the recursive calls. Without a base case, a recursive function might loop endlessly. The base cases essentially act as a safety net, ensuring the function eventually stops calling itself and starts returning results.
For Lucas numbers, our base cases are clearly defined:
For Lucas numbers, our base cases are clearly defined:
- When \( n = 0 \), the function returns 2.
- When \( n = 1 \), the function returns 1.
lucas_number function, notice how these base cases are checked first. If \( n \) matches either base case, the function returns the respective predefined value immediately. This is fundamental for proper functionality, as the recursive function builds upon these stability points to determine results for all larger \( n \) values.Lucas Number
Lucas numbers are an interesting sequence of integers named after the mathematician François Édouard Anatole Lucas. Like the Fibonacci sequence, which it resembles, each number in the Lucas sequence is the sum of its two immediate predecessors in the sequence.
- The sequence starts with the base cases: 2 as the 0th Lucas number and 1 as the 1st Lucas number.
- The formula for the nth Lucas number is \[ L_n = L_{n-1} + L_{n-2} \]
Other exercises in this chapter
Problem 17
Let \(b_{n}\) denote the number of multiplications needed to compute \(n !\) using the recursive factorial algorithm in Example 5.1 Solve the recurrence relatio
View solution Problem 17
Using generating functions, solve each LHRRWCC. $$a_{n}=5 a_{n-1}-6 a_{n-2}, a_{0}=4, a_{1}=7$$
View solution Problem 17
Solve each LHRRWCC. $$a_{n}=6 a_{n-1}-9 a_{n-2}, a_{0}=2, a_{1}=3$$
View solution Problem 17
Define recursively each sequence of numbers. (Hint: Look for a pattern and define the \(n\) th term \(a_{n}\) recursively.) $$0,3,9,21,45 \ldots$$
View solution