2
$\begingroup$

I have promised 250 rep bounties to five questions here. Recently I got 2 answers on this question from @DrXorile and @Hackiisan. However, I could not understand one of them, and am unsure of the other. Anyways, I don't want to rely entirely on my own math for such questions.

I put a 50 bounty, hoping that others would up/down vote the answers, so I would get to know which answers are, in fact, correct (and the award the remaining 200). However, this has not happened and there are only 1-2 upvotes on each answer.

What should I do know?

(a) I could just retract on the bounty offer of 250, and award the current 50 to the other previously verified answer. But this seems somewhat unfair.

(b) I could trust that the answers are right and award the bounty. But this has its own risks. Not only does it give my rep to a user for a potentially false answer, it also misdirects future users who will be more likely to believe in the possibly dubious answer and not bother posting or even searching for a true answer with a higher upper bound.

(c) I could ignore the current 50, and put a condition that I will award 250 rep when either answer receives $\geq5$ upvotes. But this also seems unfair, as it could possibly deny them from the bounty forever.

(d) Or you could come up with another better option. Please hurry, I have only 3 days left.

$\endgroup$
2
  • $\begingroup$ If it helps, looking at Dr Xorile's answer, I believe that it is an effective solution. $\endgroup$
    – Gordon K
    Nov 2, 2015 at 20:14
  • $\begingroup$ Oops, I did not notice this discussion until now. Let me see if I can clarify my answer over the weekend. (Dr Xorile's answer is correct in my opinion as well). $\endgroup$
    – Hackiisan
    Nov 5, 2015 at 18:41

2 Answers 2

1
$\begingroup$

I would suggest to the answerer(s) that they edit their answers to make them more understandable.

The answers look like they would likely be sufficient on Physics.SE or Electrical Engineering.SE, but the average Puzzling.SE user probably doesn't have the knowledge of electrical circuits and components required to understand these answers.

To be fair, the question implied a fair amount of electronics knowledge on your own part, which means the answerers might have assumed you would understand their answers without a detailed explanation.

All that being said, I think that an answer that cannot be clearly understood is not a good answer, regardless of its correctness, and therefore should not be awarded a bounty. Give them a chance to improve their answers, and see where that gets you.

$\endgroup$
4
  • $\begingroup$ It's not that I lack electronics knowledge; it's simply that I couldn't understand the maths. $\endgroup$ Nov 5, 2015 at 15:17
  • $\begingroup$ @ghosts_in_the_code, I'd appreciate if you could let me know where I should give a more detailed explanation (the 501-fuse solution? Or the 32-fuse one?) $\endgroup$
    – Hackiisan
    Nov 5, 2015 at 18:43
  • 1
    $\begingroup$ @Hackiisan I didn't understand either of them, since some of the maths is too complicated for me. I don't know how you even got $I_j = \frac{V_0}{R(mk+A)+1}a_j$. I'm only a school student, so I don't know if you should even bother trying to simplify it. Maybe what you could do is add a diagram with the actual values for the $32$ fuses and resistors. I'll try some random values for $m$ and see if it works by manually calculating. If they do, I'll give you the bounty. $\endgroup$ Nov 6, 2015 at 6:06
  • $\begingroup$ @Hackiisan DrXorile is also unable to understand your answer. However, if you want provide a circuit with the actual resistor values that works,you have only 5 days left. $\endgroup$ Nov 7, 2015 at 3:43
0
$\begingroup$

I've decided to trust @DrXorile 's answer for 44 fuses (which I somewhat understand) and award the 250 rep. After this, I'm removing the bounty offer on the Q

I don't mean to justify my decision. And I'm not suggesting that anybody else do the same.

P.S. If you beat his answer within the week and get sufficient up-votes, you can get the bounty instead.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .