Hacker News Reader

Comment by bofaGuy | Hacker News Reader

Let m = n. For m^n = n^m. There are infinite solutions. Am I missing something?

Edit: “Distinct”

m = n would be trivial and pointless. Also, I'm curious why he didn't specify precisely clarify "distinct" to mean "pairs of unequal values" (e.g., exclude {n,m} where n=m) and "pairs not equal to others" (e.g., {x,y} excludes {y,x}).

The condition is as precise as possible "pairs of distinct integers" is not equal to "distinct pairs of integers". Don't blame the writer when you fail to read what is written.

Relax and be civilized.

It's common for the obvious cases and obvious constraints to be assumed.

Adding "distinct" would have made it clearer, but it was obvious that the author meant distinct from reading the title alone.

> Adding "distinct" would have made it clearer, but it was obvious that the author meant distinct from reading the title alone.

It's math. No assumption is ever obvious. Only parts of some proofs are ever allowed to be called obvious.

Would you really call that a "pair of integers"? Especially when it makes the solution so trivial?

Yes. (1,1) is a pair of integers, and so is (17,17). Why wouldn't they be?

But if you make it a sentence, "my pair of integers is 1 and 1", it sounds a little bit off. Because "pair" for the same concept twice is often an error.

Picking a mathematical formulation like you did can avoid that kind of implication, but there wasn't a template for which formulation to use and the way you wrote that makes it look like order matters which isn't right either.

In binary, you will probably get 1 next to another 1 all the time.

Those are digits, not integers?

And that also sounds like two ordered numbers which is not what the question calls for.

int

Reply on news.ycombinator.com