You just have to implement one line of code in the bottom and rest is just infor

Question

You just have to implement one line of code in the bottom and rest is just information. ### First read the following function that caculates the Fib numbers def fib(k) if k in (0, 1): returnk return fib(k-1) fib(k-2) for i in range(20) print(i, fib(i)) ### The above function correctly calculates the Fib numbers as shown ### by the print statements above ### we are interested in how many function calls are made to calculate ### each Fib number ### we modify the above function to accomplish that ### Try to undertand the following code and see how the above goal ### is accomplished def fib1(k) if k == 0: return 0,1 if k == 1: return 1,1 u = fibl(k-1) v = fibl (k-2) return u[0] v[0], u[1]v[1]1 for i in range (20) print(i, fib1(i)) ### From the printout above, you should see that the number of function calls

Explanation / Answer

#this is compiled in python 3 the one line of code you asked for

return u[2],u[1]+1,u[2]+u[0]

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