Implementation 1 def PF1x n lenx m n2 top listxm bot listxm for k in rangem x2k

# Implementation 1 def pf1x n lenx m n2 top listxm bot

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Implementation 1 def PF1(x): n = len(x) m = n/2 top = list(x[:m]) bot = list(x[m:]) for k in range(m): x[2*k] = top[k] x[2*k+1] = bot[k] This is a Void function. It returns None. However, it permutes the values in the list referenced by x according to the perfect shuffle. Make a copy of the top and bottom halves They become the even-indexed and odd-indexed entries
Implementation 2 def PF2(x): n = len(x) m = n/2 y = [] for k in range(m): y.append(x[k]) y.append(x[k+m]) return y This is a fruitful function. It returns a reference to a list that is the perfect shuffle of the list referenced by x Build y up through repeated appending x[k] comes from the top half of the list, x[k+m] comes from the bottom half.
Perfect Shuffle Cycles Question: Given a length-n list x where n is even, how many perfect shuffle updates are required before we cycle back to the original x?
Perfect Shuffle Cycles # Assume x0 is a given list x = list(x0) PF1(x) numPFs = 1 while x!=x0: PF1(x) numPFs+=1 print numPFs Solution Using the Void function PF1:
Perfect Shuffle Cycles # Assume x0 is a given list x = PF2(x0) numPFs = 1 while x!=x0: x = PF2(x) numPFs+=1 print numPFs Solution Using the Fruitful function PF2:
Sample Outputs n numPFs -------------------- 8 3 52 8 444 442 1000 36 10000 300 100000 540

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• Fall '19