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Number of ways to climb a staircase using 1 or 2 steps

If the person takes n steps of ‘2’ and m steps of ‘1’ then to climb ‘N’ steps, the following equation can be written:

2n + m = N

All solutions to the above equation give us a possible way to climb the staircase.

However, solution to the above do not capture the order of climbing and treats all orders of climbing as same if the number of 2 steps and 1 steps are same in them. For example if n=1 and m=2, then the above says that there is only 1 way but if we consider the order too, then there are 3 ways as: 2+1+1, 1+2+1 and 1+1+2

To capture the order we need to consider the possibility of taking 1 or 2 steps at each stage and see which of them leads to the solution.

If we take 1 steps, then N-1 steps are left to be taken.

If we take 2 steps, then N-2 steps are left to be taken.

So, at any stage, no of possible ways to take N steps is:

F(N) = F(N-1) + F(N-2)

This is a fibonacci series, we only need to put some initial values.

Clearly, F(0) = 0 and F(1) = 1

Using the above, F(2) = 1+0 = 1

But we can see that F(2) can be done in 2 ways as: 1+1 and 2

What went wrong?

Turns out that we need to hardcode F(2) also to make this fiboncaai series work.

Let F(2) = 2

Then F(3) = 1+2 = 3 which is correct (1+1+1, 1+2 and 2+1)

F(4) = 2+3 = 5 (1+1+1+1, 2+1+1, 1+2+1, 1+1+2, 2+2)

F(5) = 3+5 = 8 (1+1+1+1+1, 2+1+1+1, 1+2+1+1, 1+1+2+1, 1+1+1+2, 2+2+1, 2+1+2, 1+2+2)

So the above is indeed a fibonacci series with one exception - we need to provide initial values for 3 cases rather than 2. This is so because we have 2 independent variables (1 and 2) which do not depend on each other and have the ability to affect the computation independently. If there were 3 independent steps, then we would have needed to provide 3 initial configurations.

Consider we can take 1,2 or 3 steps at each stage.

Then recursive solution sets in as follows:

F(0) = 0;

F(1) = 1;

F(2) = F(1) + 1 (This one is for the independent step of 2)

F(3) = F(1) + F(2) + 1 (This one is for the independent step of 3)

F(4) = F(1) + F(2) + F(3) (This is where pure recursion sets in)

An interesting case is when steps are says 1, 5, 8 (basically when steps are not just increasing by 1)

Then we have to consider the initial configurations at various different stages.

So F(9) = F(8) + F(4) + F(1)

But F(8) = F(7) + F(3) + 1 (This one is the effect of independent step of 8 steps)

Similarly, F(5) = F(4) + 1

Looking more closely, the above becomes equivalent to the coins problem discussed next.

Also, one more observation is that if we define F(0) = 1 then we do not need to consider the cases separately.

However, if N was really zero, then we should not return 1. This can be solved by creating a wrapper function which return 0 if N is 0 the first time itself. Rest all the times, it simply uses normal recursion of fibonacci series.

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