Composite functions in python - dual compose

Question

I came across the following homework problem:

My code for this problem was marked wrong and when I viewed the suggested solution, I couldn't understand where I went wrong. I ran the codes of both functions in Python IDLE compiler only to see that both functions return the same output as seen below:

>>> def dual_function(f,g,n): #Suggested solution
    def helper(x):
        f1,g1 = f,g
        if n%2==0:
            f1,g1=g1,f1
        for i in range(n):
            x=f1(x)
            f1,g1=g1,f1
        return x
    return helper

>>> def dual_function_two(f,g,n): #My solution
    def helper(x):
        if n%2==0:
            for i in range (n):
                if i%2==0:
                    x = g(x)
                else:
                    x = f(x)
        else:
            for i in range(n):
                if i%2==0:
                    x = f(x)
                else:
                    x = g(x)
        return x
    return helper

>>> add1 = lambda x: x+1
>>> add2 = lambda x: x+2
>>> dual_function(add1,add2,4)(3)
9
>>> dual_function_two(add1,add2,4)(3)
9
>>>

I would appreciate it if someone could identify the mistake in my solution. Thank you.

Do you know the test cases used to grade your problem? I can't find any case where the two functions differ. — Sebastian Mendez
– Sebastian Mendez, Commented Mar 27, 2019 at 4:26
Those aren't good test functions, since addition is commutative -- if you called them in the wrong order you wouldn't notice the difference. But I tried mixing addition and multiplication, and I couldn't find a difference, either. — Barmar
– Barmar, Commented Mar 27, 2019 at 4:57
It's probably much slower than the suggested solution due to the constant modulo operations. Swapping is going to be much faster. — Rick
– Rick, Commented Mar 27, 2019 at 5:02
@Barmar Thank you for letting me know about using commutative operators in such functions, will remember them when working on similar problems in the future. — Prav
– Prav, Commented Mar 27, 2019 at 6:20

Mulan · Accepted Answer · 2019-03-27 06:11:39Z

2

The suggested solution is needlessly complex. Countless reassignments of variables and a loop are a recipe for a headache. Here's a simplified alternative -

def dual (f, g, n):
  if n == 0:
    return lambda x: x
  else:
    return lambda x: f(dual(g, f, n - 1)(x))

add1 = lambda x: 1 + x
add2 = lambda x: 2 + x

print(dual(add1,add2,4)(3))
# 9
# (1 + 2 + 1 + 2 + 3)

print(dual(add1,add2,9)(3))
# 16
# (1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 3)

print(dual(add1,add2,0)(3))
# 3

The reason this works is because in the recursive branch, we call dual with swapped arguments, dual(g,f,n-1). So f and g change places each time as n decrements down to 0, the base case, which returns the identity (no-op) function.

A slightly less readable version, but works identically -

def dual (f, g, n):
  return lambda x: \
    x if n == 0 else f(dual(g, f, n - 1)(x))

answered Mar 27, 2019 at 6:11

Mulan

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1 Comment

Prav Over a year ago

Great, never knew I could be this simple. Thanks for the explanation! :)

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