Styjun

So i have a homework question, but I'm not sure why I got it wrong / how it works.

once = lambda f: lambda x: f(x)
twice = lambda f: lambda x: f(f(x))
thrice = lambda f: lambda x: f(f(f(x))) 
print(thrice(twice)(once)(lambda x: x + 2)(9))

My ans: 25 -> 8*2 +9

Actual ans: 11 -> 2 + 9

What I was thinking:

thrice -> f(f(f(x))),
let new_x = twice(x)

thrice -> f(f(new_x)),
let new_x2 = twice(new_x)

thrice -> f(new_x2),
let new_thrice = twice(new_x2)

so afterwards I add in the (once) and did
new_thrice(once)(lambda x: x+2)(9)

But answer seems to be that (once) nullifies the earlier thrice(twice) and am lost about. Would be great if someone has an explanation.. Thanks!

once(lambda x: x+2) evaluates to a function that applies lambda x: x+2 to its argument. In other words, it's equivalent to lambda x: x+2.

once(once(lambda x: x+2)) evaluates to a function that applies once(lambda x: x+2) to its argument. In other words, it's also equivalent to lambda x: x+2.

once(once(once(lambda x: x+2))) evaluates to a function that applies once(once(lambda x: x+2)) to its argument. In other words, this is also equivalent to lambda x: x+2. This doesn't change no matter how many times you apply once.

thrice(twice)(once) evaluates to a function that applies once to its argument some number of times. (8 times, not that it matters for the analysis.) once doesn't change a function's behavior. No matter how many times you apply once, the final function only applies the underlying function once.

thrice(twice)(once)(lambda x: x + 2) thus evaluates to a function that does the same thing as lambda x: x + 2.

Now, if it had been thrice(twice)(once(lambda x: x + 2)) (note the moved parentheses), then that would have applied thrice(twice) to once(lambda x: x + 2), and the result would be a function that applies lambda x: x + 2 8 times.

I hope this will help you to figure out what is going on!

once = lambda f: lambda x: f(x)
twice = lambda f: lambda x: f(f(x))
thrice = lambda f: lambda x: f(f(f(x)))

# Created this one to help readability.
custom_func = lambda x: x + 2

print("once:", once(custom_func)(9)) # returns value
print("twice:", twice(custom_func)(9)) # returns value
print("thrice:", thrice(custom_func)(9)) # returns value

print("applying all those:", thrice(custom_func)(twice(custom_func)(once(custom_func)(9))))
# This represents the following: (((9 + 2) + 2 + 2) + 2 + 2 + 2)
# each pair of parenthesis mean one function being applied, first once, then twice, then thrice.

# If I've understood correctly you need to achieve 25
# to achieve 25 we need to apply +4 in this result so, which means +2 +2, twice function...
print("Achieving 25:", twice(custom_func)(thrice(custom_func)(twice(custom_func)(once(custom_func)(9)))))

# That is it! Hope it helps.