[Sentience]

This gives me a recursion error:

Code

def f(n: int) -> bool:
return f(n) >= f(n+1)

How can I write this so it runs?

Thanks,
Oliver

[SIM-574702]

Funtion f is defined with int parameter so you can call it with f(0) or any other integer. If you really want to call it as f(n) you need to define n.

Code

n=0
f(n)

I'm not sure what you're trying to do but this will always give you maximum recursion depth exceeded because you don't have an exit condition.

[Athr]

This gives me a recursion error:

Code

def f(n: int) -> bool:
return f(n) >= f(n+1)

This says n is undefined:

Code

f(n)

How can I write this so it runs?

Thanks,
Oliver

You have not defined n before calling the function.
Write the following line to define n.

Code

#Sample n value
n = 5
print(f(n))

EDIT:

SIM-574702 is right about the exit condition.
Its like a base case for an induction problem in math.

[Sentience]

Code

# We represent elements of N_∞ as functions f : int -> bool, where we only care about
# the values of those functions for nonnegative inputs n, and the functions are
# assumed to satisfy f(n) >= f(n+1) for all n >= 0.

This is an excerpt of what I am talking about

It’s not supposed to call itself…

---

How can I write this:

Code

def f(n: int) -> bool:
return f(n) >= f(n+1)

Without recursion?

---

Edit: Solved…

Code

def g(f: typing.Callable[[int], bool], n: int) -> bool:
return f(n) >= f(n + 1)

[Sentience]

Wait wtf how do call g()?

---

Turns out I can call g just by going

g

[SIM-4763645]

```
def f():
def g():
f()
# then call g() inside main thread
```

[Sentience]

Wait wtf how do call g()?

---

Turns out I can call g just by going

g

Edit: This thread is solved