Question:

How many different ways are there to climb a staircase with n steps (for example, 100 steps) if you are allowed to skip steps, but not more than one at a time?

Explore by hand, look for a pattern:

n = 1:

1

n = 2:

1+1

2

n = 3:

1+1+1

1+2

2+1

1 way

2 ways

3 ways

n = 4:

1+1+1+1

1+1+2

1+2+1

2 +1+1

4 ways?

2+2

5 ways!

n = 5:

?

Too much work!

Use a computer:

Generate all sequences of 1s and 2s of length from 1 to n, and count the sequences for which the sum of the elements is equal to n.

Generate... – how?!

A better approach:

Model the situation in a different way (isomorphism):

0

1

0

0

0

1

0

0 marks a step we step on; 1 marks a step we skip.

A valid path cannot have two 1s in a row, ends with a 0.

Binary number system

for x in range (2**n): # Binary digits of x are used as a # sequence of 0s and 1s of length n

Bitwise logical operators

if x & (x << 1) == 0: # If the binary representation of x# has no two 1s in a row...

00010100010 00101100100 00101000100 01011001000 ------------------ ------------------ 00000000000 00001000000

Problem restated: Count all sequences of 0s and 1s of length n with no two 1s in a row

& &

def countPaths(n): """ Returns the number of sequences of 0s and 1s of length n with no two 1s in a row """ count = 0 for x in range(2**n): if x & (x << 1) == 0: count += 1 return count

for n in range(101): print(n+1, countPaths(n))

1 12 23 34 55 86 13...100 573147844013817084101

Fibonacci numbers!

Finalprogram:

def fibonacciList (n): "Returns the list of the first n Fibonacci numbers" fibs = [1, 1] while len(fibs) < n: fibs.append (fibs[-1] + fibs[-2]) return fibs

print (fibonacciList (101))

[1, 1, 2, 3, 5, 8, 13, ...,...... 573147844013817084101]

The answer is 101th Fibonacci number!

There is an easier way to compute it, of course, for example:

Back to math:

Show mathematically that the number of paths for n steps is the (n+1)th Fibonacci number.

mlitvin@andover.edu