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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 1+1+1+1 ? 1 way 2 1+2 1+1+2 2+1 1+2+1 Too much work! 2 ways n = 3: n = 4: 2 +1+1 3 ways 4 ways? 2+2 5 ways! n = 5: 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 0 1 0 1 0 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. Problem restated: Count all sequences of 0s and 1s of length n with no two 1s in a row 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 00101000100 -----------------00000000000 & 00101100100 01011001000 -----------------00001000000 Final program: 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)) 11 22 33 45 58 6 13 ... 100 573147844013817084101 Fibonacci numbers! The answer is 101th Fibonacci number! There is an easier way to compute it, of course, for example: 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] Back to math: Show mathematically that the number of paths for n steps is the (n+1)th Fibonacci number. [email protected]