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Comments on Reduce over the range [1..n]

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Reduce over the range [1..n]

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Task

I often need to find the factorial of a number or the sum of all numbers up to a number when cheating on math tests. To help me with this, your task is to write $F$, a generalized version of those functions:

$$F(n) = 1 * 2 * \space ... \space * (n-1) * n$$

Please note that the operator $ * $ does not necessarily represent multiplication here, but stands for a commutative, associative operator that will be an input to your program/function. This means that $a * b$ is the same as $b * a$, and $a * (b * c)$ is the same as $(a * b) * c$. Its inputs are positive integers, and its outputs are integers.

Rules

  • $n$ will be a positive integer.
  • $*$ is a binary function/operator that can be taken in any convenient format, including but not limited to:
    • A function object
    • A function pointer
    • An object with a method with a specific name (e.g. Java's BiFunction)
    • A string that can be evaluated to get a function
  • $*$ is a blackbox function. That means that you will not be able to examine it to see how it works; all you can do is feed it two positive integers and get an integer back.
  • The output of your function will be an integer (not necessarily positive).
  • This is code golf, so shortest code in bytes wins!

Testcases

f         | n  | F(f, n)
Add       | 1  | 1
Add       | 5  | 15
Multiply  | 1  | 1
Multiply  | 5  | 120
XOR       | 1  | 1
XOR       | 2  | 3
XOR       | 5  | 1
XOR       | 10 | 11
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1 comment thread

Function constraint (2 comments)
Function constraint
Hakerh400‭ wrote over 3 years ago

"... you will not be able to examine it to see how it works" - that's an uncomputable requirement. Determining whether a program examines the given function (assuming it's able to do so) is equivalent to solving the halting problem, especially in languages that need to parse a function from a string.

user‭ wrote over 3 years ago

¯\(ツ)/¯ Blackbox functions have been used a lot Somewhere Else without any problems (afaik). I'll remove that part if you like, though.