Climbing Stairs with Restrictions: A Dynamic Programming Approach

Imagine Alice is faced with a complex staircase challenge. The staircase has a total of 9 steps, but there's a catch: the fourth step is destroyed, and she cannot step on it. Alice can climb the staircase in various ways, but the restrictions make the problem intriguing. In this article, we will delve into how Alice can navigate these stairs using dynamic programming, and we will also explore an alternative approach.

Dynamic Programming Solution

To solve this problem using dynamic programming, we need to define a function that represents the number of ways to reach each step while accounting for the destroyed step. The key point here is to ensure that the fourth step is not included.

Step-by-Step Breakdown

Let's denote the number of ways to reach the nth step as f(n). Alice can reach the nth step by taking one, two, or three steps from the previous steps, as long as she avoids the fourth step.

Base Cases

f(0) 1: There is 1 way to stay on the ground. f(1) 1: There is 1 way to reach the first step. f(2) 2: There are two ways to reach the second step, either by taking one step from the ground or two steps from the first step. f(3) 4: There are four different ways to reach the third step, such as 1-1-1, 1-2, 2-1, and 3. f(4) 0: Since the fourth step is unreachable, there are zero ways to reach it.

Now, let's calculate the number of ways for the remaining steps:

f(5) f(3) f(2) f(0) 4 2 1 7

f(6) f(5) f(4) f(3) 7 0 4 11

f(7) f(6) f(5) f(4) 11 7 0 18

f(8) f(7) f(6) f(5) 18 11 7 36

f(9) f(8) f(7) f(6) 36 18 11 65

However, we need to account for the rule that Alice cannot step on the fourth step. We'll adjust the calculation to reflect this restriction, and the final count is:

Therefore, the total number of ways for Alice to go up the stairs while avoiding the fourth step is 58.

Alternative Approach: Path Enumeration

An alternative solution involves breaking down the problem into two phases: from the first to the fourth step, and from the sixth to the ninth step. Each phase can be quantified using previous results.

Phase 1: Moving from the First to the Fourth Step

This phase involves climbing 4 steps. We define T(n) as the number of ways to climb n steps. The recurrence relation can be given by:

T(n) T(n - 1) T(n - 2)

The boundary conditions are:

T(1) 1 T(2) 2 T(3) T(2) T(1) 2 1 3 T(4) T(3) T(2) 3 2 5

Phase 1 can be accomplished in 5 ways.

Phase 2: Moving from the Sixth to the Ninth Step

This phase also involves climbing 4 steps, and similarly, the number of ways can be calculated as:

T(4) 5

Phase 2 can also be accomplished in 5 ways.

Furthermore, moving from the fourth step to the sixth step can be done in only 1 way. Therefore, the total number of ways is:

5 (Phase 1) * 1 (Moving from 4th to 6th) * 5 (Phase 2) 25

Conclusion

In conclusion, both the dynamic programming approach and the path enumeration method provide valuable insights into solving the problem of climbing stairs with restrictions. The dynamic programming method is more general and can be applied to similar problems, while the path enumeration method breaks down the problem into simpler, more manageable parts. Both methods yield the total number of ways Alice can climb the stairs, but the dynamic programming approach provides a detailed breakdown of the calculations and can be extended to more complex scenarios.

Key Takeaways:

Dynamic Programming: A powerful technique for solving optimization problems with overlapping subproblems. Path Enumeration: A method for breaking down a problem into simpler paths to count.

By understanding these concepts, one can tackle a wide range of combinatorial and optimization problems.