Question How many different ways are there to

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: n = 2: 1 1+1+1+1 ? 1 way 2 1+1+2 2+1 1+2+1 Too much work! 2 ways n = 3: 3 ways n = 4: 2 +1+1 4 ways? 2+2 5 ways! n = 5:

Use a computer: Generate all sequences of 1 s and 2 s 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 0 1 0 0 0 marks a step we step on; 1 marks a step we skip. A valid path cannot have two 1 s in a row, ends with a 0.

Problem restated: Count all sequences of 0 s and 1 s of length n with no two 1 s in a row Binary number system for x in range (2**n): # Binary digits of x are used as a # sequence of 0 s and 1 s of length n Bitwise logical operators if x & (x << 1) == 0: # If the binary representation of x # has no two 1 s in a row. . . & 00010100010100 ---------000000 & 001011001000 ---------00001000000

Final program: def count. Paths(n): """ Returns the number of sequences of 0 s and 1 s of length n with no two 1 s 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, count. Paths(n)) 11 22 33 45 58 6 13 . . . 100 573147844013817084101 Fibonacci numbers!

The answer is 101 th Fibonacci number! There is an easier way to compute it, of course, for example: def fibonacci. List (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 (fibonacci. List (101)) [1, 1, 2, 3, 5, 8, 13, . . . 573147844013817084101]

Back to math: Show mathematically that the number of paths for n steps is the (n+1)th Fibonacci number. mlitvin@andover. edu