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: n = 3: way 2 ways 3 ways n = 4: ways? 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): 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 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)) Fibonacci numbers! Final program:

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,..., ] 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.