PROBABILITIES

[maiden cobalt]

Funny, I know this with a flashlight. Anyways, if you show your work I'm willing to walk you through.

What did you try? Where did you get stuck? Or do you just need to verify your result?

[proven umbra]

Maryta...

Maryta...

[maiden cobalt]

... So why post the question in help forum when you have the answer anyways? Please go to an appropriate channel like #recreational-math for this. Or try #discussion if you need to.

@near cedar

Seems to me like you used chat gpt or similar to get this. Quite obviously, it is bullshit. Your text says "the radio will work if we try the other 2 uncharged batteries together"

[quaint pewter]

Your arguments is wrong, for n=4 you need 7 attempts

I am pretty sure that this can’t be generalized easily

[maiden cobalt]

!nosols

[cloud heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[maiden cobalt]

If you don't wanna talk, don't post. Look up your solutions online. There's a good video from mind your decisions on youtube.

[quaint pewter]

Do you actually know of an generalization to n?

[maiden cobalt]

Actually no. But it shouldn't be too difficult to figure out I believe.

[quaint pewter]

Hmm I believe that the function f probably doesn’t look very nice, when f(n) is the minimal number of attempts needed

Like at least my sol for n=4 does introduce an useful graph, and case bash pretty brutally after that

[maiden cobalt]

For n=4 we have a first test row of tuples like this (speaking of worst case):
(C, U) (C, U),...
A total of n/2 tuples.

So one can generalize test row number 1 quite easily.

The second test is just cross testing them

Adding the last test then gives an odd result

[quaint pewter]

Yeah but to prove that this is optimal and that there isn’t a better algorithm is the hard part

[maiden cobalt]

It's not that hard. Since we just need to consider the worst case scenario for each test we do (which is quite easy to show) you can argue using logic.

[quaint pewter]

What if I tell you that there exist another completely different algorithm also getting a good pair after 7 attempts

[maiden cobalt]

Then I would be really interested in that one, though finding an algorithm with equal results doesn't disprove me

[quaint pewter]

It kinda does because then there isn’t something like a worst case scenario

Because then you would argue that doing my algorithm leads into a worse case but then I would need more moves

Anyway here is the algorithm

We number the batteries WLOG 1,2,…,8 For example you can also test two triangles so (1,2),(2,3),(3,1) and (4,5),(5,6),(4,6), you know that among 1,2,3 and 4,5,6 there is in both triples there is at most 1 good battery and therefor 7,8 must both be good batteries and testing this one gives the desired result.

[quaint pewter]

This gives also something better in general I think then your algorithm when generalized

[maiden cobalt]

Seems like someone watches mind your decisions, no?

[quaint pewter]

No I don’t, this problem was on some problem sheet which I did like a month ago

[maiden cobalt]

Funny.

So what we'd need to do is to find how those two algorithms correlate.

I'm pretty sure they do

[quaint pewter]

What do you mean by that?

[maiden cobalt]

Oh, forget it, I was not thinking about yours correctly

Crap, OP left and took their original formulation with them. I'd have loved to think up a generalization. Remember what it was?

Did n need to be even or was anything fine?

[quaint pewter]

There were n charged and n uncharged batteries

And the rest is the same I think

[maiden cobalt]

Gotcha. I'll sleep over it. Might find some time to think about it tomorrow.