it feels like overhead but still clever

how dare this be so beautiful

Thank you!

Ok, thanks.

Please, use spoiler flag next time.

Hi @Rockkley.

Lambdas are anonymous callable objects. If you want call it later, you must use a variable, no other choice to do that. And here (in codewars) if there no name, unit test cannot check the solution.

For your information, in python, a def (and so, a fonction) defined like that create a variable:

def foobar(): pass

The variable created is foobar, the proof, you can do that:

def foobar(): pass
foobar = 1234

It's the aim of the mechanism under the concept of decorators.

If you prefer i can use these solutions instead of lambda, the solution will be acepted:

# Common accepted solution
def check(s, e): return e in s

# Another solution
def singleton(c): return c()
@singleton
class check:
  def __call__(self, s, e):
    return e in s

I know that lambda is anonymous function, and there's no need to create variable for lambda, but here we need to meet the interface requirements. Requirement here is that check is callable.

In real world I would not create function for this one liner, that's why I'm showing solution with lambda.

variable "check" holds the function. and ambda describes the oneliner function.

mult = lambda a,b:a*b

is the same as

def mult(a,b):
return a*b

What do you mean? Do you know how lambdas work? Do you know you can copy the code and make trials on your own to understand how it works? On your own computer, or just here, by clicking on Forkjust below any solution.

Why creating a variable for lambda?

clever use of mapping :)

After finding prime p with another prime at p + g, your program checks next g numbers for in-between primes, but if it actaully finds prime in-between, then that check is redundant since the outer for loop will go over it anyway.

So, your program basically does some extra looping.

Nice to see that Golden Ratio and Fibonacci properties are interleaving to produce O(1) and 0(log n) algorithms

How to simplify Bessel+Kepler's formula...

N = Log(Sqrt(X / Phi) * Phi * Sqrt(5)) / Log(Phi)

into your form

N = Log(5 * X / math.Phi) / (2 * Log(Phi))

to get the index of the nearest Fib number?

Or this is to get the shortest term instead of the largest?

Moreover, fib function needs apply a correction factor to avoid loss precision and get the floor directly

F(n) = ((phi^n) + (1 - phi)^n) / sqrt(5)

Please fork and re-publish your translation. Cannot approve because there are some merge conflicts.