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How to transform the vanilla recursive fib function into the iterative DP version through a series of mechanical steps.
# Transforming the vanilla recursive fib into the iterative DP version
# through a series of mechanical steps.
#
# For more on converting recursive algorithms into iterative ones, see:
# http://blog.moertel.com/posts/2013-05-11-recursive-to-iterative.html
# original function
def fib(n):
if n < 2:
return n
return fib(n - 1) + fib(n - 2)
# partition the function into base-case logic and incremental logic
def fib(n):
if n < 2:
return n
return step(n)
def step(n):
return fib(n - 1) + fib(n - 2)
# expand step logic into its fundamental operations
def step(n):
fibn1 = fib(n - 1)
fibn2 = fib(n - 2)
result = fibn1 + fibn2
return result
# extend step logic with optional arg that eliminates recursion if provided
def step(n, fibn1n2=None):
if fibn1n2 is not None:
fibn1, fibn2 = fibn1n2
else:
fibn1 = fib(n - 1)
fibn2 = fib(n - 2)
result = fibn1 + fibn2
return result
# make step logic also return its arguments
def fib(n):
if n < 2:
return n
return step(n)[0] # <-- note [0]
def step(n, fibn1n2=None):
if fibn1n2 is not None:
fibn1, fibn2 = fibn1n2
else:
fibn1 = fib(n - 1)
fibn2 = fib(n - 2)
result = fibn1 + fibn2
return result, n, fibn1n2 # <-- and, correspondingly, here
# now apply to those returned arguments the *opposite* of the
# transformation that was applied to them in the recursive calls
def fib(n):
if n < 2:
return n
return step(n)[0]
def step(n, fibn1n2=None):
if fibn1n2 is not None:
fibn1, fibn2 = fibn1n2
else:
fibn1 = fib(n - 1)
fibn2 = fib(n - 2)
result = fibn1 + fibn2
return result, n + 1, (result, fibn1) # <-- look here
# now the step logic can be run in the *opposite* direction,
# building from the initial conditions upward via iteration;
# we modify the base logic to do this instead of using the
# step logic recursively
def fib(n):
if n < 2:
return n
# initial conditions
N = n
n = 2
fibn1n2 = (1, 0)
# iterate upward from initial conditions
while n <= N:
result, n, fibn1n2 = step(n, fibn1n2)
return result
# since the step logic is always called with the fibn1n2 argument now,
# make that argument required to simplify the step logic
def step(n, fibn1n2):
fibn1, fibn2 = fibn1n2
result = fibn1 + fibn2
return result, n + 1, (result, fibn1)
# inline the step logic back into the original function
def fib(n):
if n < 2:
return n
# initial conditions
N = n
n = 2
fibn1n2 = (1, 0)
# iterate upward from initial conditions
while n <= N:
fibn1, fibn2 = fibn1n2
result = fibn1 + fibn2
result, n, fibn1n2 = result, n + 1, (result, fibn1)
return result
# simplify
def fib(n):
if n < 2:
return n
fibn1, fibn2 = (1, 0)
for _ in xrange(2, n + 1):
result = fibn1 + fibn2
fibn1, fibn2 = result, fibn1
return result
# simplify more
def fib(n):
fibn1, fibn2 = (1, 0)
for _ in xrange(n):
fibn1, fibn2 = fibn1 + fibn2, fibn1
return fibn2
# and we're done!
# tests
def test():
fns = dict(globals())
for fname, f in sorted(fns.iteritems()):
if fname.startswith('fib'):
print('testing {}'.format(fname))
assert map(f, range(10)) == [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
if __name__ == '__main__':
test()
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