Hopping Rabbit

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Hopping Rabbit

Problem Statement

A rabbit sits at the bottom of a staircase with 10 stairs. The rabbit can hop up only one or two stairs at a time. How many different ways are there for the rabbit to ascend to the top of the stairs?

Solution

This problem exemplifies how dynamic programming and recurrence relations can solve combinatorial counting problems. The solution reveals a connection to the well-known Fibonacci sequence through systematic decomposition.

Step 1: Establish a Counting Function

We define f(n) as the total number of distinct hop sequences that allow the rabbit to reach exactly the n-th stair from the bottom.

Step 2: Identify Boundary Conditions

Working from the simplest scenarios upward:

Single stair (n = 1):

  • Only one valid sequence exists: a single hop of size 1.

  • Result: f(1) = 1

Two stairs (n = 2):

  • Two distinct sequences are possible:

  • Sequence A: Two consecutive 1-stair hops

  • Sequence B: One 2-stair hop

  • Result: f(2) = 2

Step 3: Derive the Recurrence Pattern

For any staircase with n \geq 3 stairs, we analyze the possible final moves in any valid sequence.

Every complete sequence ending at stair n must conclude with either:

  • A 1-stair hop from position (n-1), or

  • A 2-stair hop from position (n-2)

These two cases are exhaustive and mutually exclusive.

Scenario A: Final move is a 1-stair hop

  • To execute this, the rabbit must first reach stair (n-1) using some valid sequence.

  • There are f(n-1) such sequences.

  • Each can be extended by one 1-stair hop to complete a path to stair n.

  • Contribution: f(n-1) sequences

Scenario B: Final move is a 2-stair hop

  • The rabbit must first reach stair (n-2) using some valid sequence.

  • There are f(n-2) such sequences.

  • Each can be extended by one 2-stair hop to complete a path to stair n.

  • Contribution: f(n-2) sequences

By the addition principle for mutually exclusive cases:

f(n) = f(n-1) + f(n-2)

Step 4: Recognize the Fibonacci Sequence

This recurrence relation, with base cases f(1) = 1 and f(2) = 2, defines a sequence that is closely related to the Fibonacci sequence.

The standard Fibonacci sequence is:

  • F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, …

Our sequence is:

  • f(1) = 1 = F_2

  • f(2) = 2 = F_3

  • f(3) = f(2) + f(1) = 2 + 1 = 3 = F_4

  • f(4) = f(3) + f(2) = 3 + 2 = 5 = F_5

  • f(5) = f(4) + f(3) = 5 + 3 = 8 = F_6

In general: f(n) = F_{n+1} where F_k is the k-th Fibonacci number.

Step 5: Calculate Specific Values

Using the recurrence relation:

n f(n) Calculation
1 1 Base case
2 2 Base case
3 3 f(2) + f(1) = 2 + 1
4 5 f(3) + f(2) = 3 + 2
5 8 f(4) + f(3) = 5 + 3
6 13 f(5) + f(4) = 8 + 5
7 21 f(6) + f(5) = 13 + 8
8 34 f(7) + f(6) = 21 + 13

Final Answer

For a staircase with 10 stairs, there are 89 different ways for the rabbit to reach the top.

The solution follows the recurrence relation:

f(n) = f(n-1) + f(n-2)

with base cases:

  • f(1) = 1

  • f(2) = 2

This defines the Fibonacci sequence (shifted): f(n) = F_{n+1}.

Computing the sequence:

  • f(1) = 1

  • f(2) = 2

  • f(3) = f(2) + f(1) = 2 + 1 = 3

  • f(4) = f(3) + f(2) = 3 + 2 = 5

  • f(5) = f(4) + f(3) = 5 + 3 = 8

  • f(6) = f(5) + f(4) = 8 + 5 = 13

  • f(7) = f(6) + f(5) = 13 + 8 = 21

  • f(8) = f(7) + f(6) = 21 + 13 = 34

  • f(9) = f(8) + f(7) = 34 + 21 = 55

  • f(10) = f(9) + f(8) = 55 + 34 = 89

Therefore, f(10) = 89.

Closed-Form Formula

The Fibonacci sequence has a closed-form expression using Binet’s formula:

F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}

where:

  • \phi = \frac{1 + \sqrt{5}}{2} \approx 1.618 (golden ratio)

  • \psi = \frac{1 - \sqrt{5}}{2} \approx -0.618

Therefore:

f(n) = F_{n+1} = \frac{\phi^{n+1} - \psi^{n+1}}{\sqrt{5}}

For large n, since |\psi| < 1, we have the approximation:

f(n) \approx \frac{\phi^{n+1}}{\sqrt{5}}

Key Insights

  1. Recursive Thinking: Breaking down the problem by considering the final step reveals the recurrence structure.

  2. Induction Approach: The hint suggests using induction, which naturally leads to the recurrence relation.

  3. Fibonacci Connection: This problem is a classic example of how Fibonacci numbers appear in combinatorial problems.

  4. Dynamic Programming: This problem is often used to teach dynamic programming, as it can be solved efficiently using memoization or bottom-up approaches.

  5. Exponential Growth: The number of ways grows approximately as \phi^n, where \phi is the golden ratio.

Alternative Interpretations

This problem is equivalent to:

  • Coin Change Problem: How many ways to make n cents using 1-cent and 2-cent coins?

  • Binary Strings: How many binary strings of length n have no consecutive zeros? (with some interpretation)

  • Tiling Problem: How many ways to tile a 1 \times n board using 1 \times 1 and 1 \times 2 tiles?

Applications

  1. Dynamic Programming: Fundamental example for teaching DP concepts

  2. Algorithm Design: Illustrates recursive thinking and optimization

  3. Combinatorics: Demonstrates how recurrence relations arise naturally

  4. Computer Science: Used in coding interviews to test problem-solving skills

  5. Mathematical Modeling: Appears in various optimization and counting problems

Computational Complexity

Naive Recursive Approach: O(\phi^n) - exponential time due to redundant calculations

Memoization (Top-Down DP): O(n) time, O(n) space

Bottom-Up DP: O(n) time, O(1) space (only need to store last two values)

Matrix Exponentiation: O(\log n) time - can compute f(n) in logarithmic time

Generalization

Problem: Rabbit can hop 1, 2, or 3 stairs at a time.

Recurrence: f(n) = f(n-1) + f(n-2) + f(n-3) with base cases f(1) = 1, f(2) = 2, f(3) = 4.

Problem: Rabbit can hop up to k stairs at a time.

Recurrence: f(n) = f(n-1) + f(n-2) + \cdots + f(n-k) for n > k.

Common Pitfalls

  1. Off-by-one errors: Be careful with base cases. Some definitions use f(0) = 1, which would make f(n) = F_{n+2}.

  2. Forgetting base cases: The recurrence only works for n > 2; base cases must be defined separately.

  3. Confusing with permutations: This counts sequences of hops, but order matters (1,2) is different from (2,1) in terms of the sequence, though both reach the same destination.

  4. Exponential complexity: Naive recursion recalculates the same values many times; use memoization or dynamic programming.

Summary

The hopping rabbit problem demonstrates how a seemingly complex counting problem can be solved elegantly using recurrence relations. By considering the rabbit’s position before the final hop, we derive f(n) = f(n-1) + f(n-2), which is the Fibonacci recurrence. This problem is a fundamental example in combinatorics, dynamic programming, and algorithmic thinking, showing how recursive decomposition leads to efficient solutions.