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There are 12 parrots in a zoo. 11 parrots out of them have an amazing ability to eat guavas at exactly the same speed. Just one of them is special and has a different speed of eating guavas. There are no resources available (like knife, watch etc) except the standard identical sized guavas. How many minimum guavas do you need to find out the special parrot if,

  • a. It is known that the special parrot eats the guava faster.

  • b. It is known that the special parrot eats the guava slower.

Hint: You may remove the parrots anytime away from the guavas and use the left-over guavas later.

Have no source link for the same

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    $\begingroup$ Welcome to Puzzling! Did you have permission to share the last four questions to outside sources (which includes this site)? $\endgroup$
    – oAlt
    Sep 5 at 11:33
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    $\begingroup$ Even if you don't have a link to the source, you could tell us in words where it's from and how you accessed it $\endgroup$
    – bobble
    Sep 5 at 13:52

3 Answers 3

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Taking idea from similar well-known puzzle about burning rope, we can use

5 guavas

without any measuring tool, and without eyeballing how much the guava has been finished.

First, divide the parrots into 4 groups of 3 parrots each. Then take 3 guavas, and give one each for the first three groups. Then let each group of 3 parrots eat the guava at the same time. In case (a) the group who finished first must contain the special parrot. In case (b) it's the one finished last. If all finish at the same time, it must be at the fourth group of 3 parrots.

Now we have a group of 3 parrots which we know must contain the special parrot. Similar to previous step, take 2 guavas, let one parrot eat one each. If they finish at the same time, the special parrot is the third one. Otherwise, we can know which parrot depends on case (a) or case (b) based on which one finishes first.

(This assumes one guava can be eaten at the same time by multiple parrots, which, at the time of this writing, hasn't been clarified in the question, but also not forbidden).

If eating at the same time is forbidden, then in the first step we can simply rotate the guava in the group of 3 after it has eaten about a third (doesn't have to be exact, the rotations don't even have to be done at the same time for all groups). The point is to let each parrot in the group have some of the guava. If there is any special parrot in the group, the total time to finish the guava will be different. For example, we can rotate the guava to the next parrot in the group every 5 seconds. As long as it takes more than 10s for any parrot to finish a guava, this will work.

Additional note: upperbound for the answer would be 12 guavas, since we can just give one each, and see which one finishes in different time. So any answer should be less than 12.

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5 (or 2, depending on interpretation)

I one can see if from one partial finished guava more is eaten than from another one, use two and have pairs of parrots eat a short time, and check if the same amount is eaten after that time. The two guavas can be reused and thus 2 are enough.

If not: One can use 2 guavas for 3 outcomes, 3 guavas for 4 outcomes (note: This would be 7 outcomes if slower/faster was not known, but then you also would have twice the possibilities, i.e. 24 for 12 parrots), 4 for 5 etc.
So the parrot can be identified from 12 (at most) with 2+3 guavas. example: Have 4 parrots eat from guava 1, 4 from guava 2, and keep 4 in reserve. The group of 4 the special parrot belongs to is now known. (the group depending on guava 1 being finished earlier, later or at the same time)
Then have 3 eat their own guava, and keep 1 in reserve. The special parrot is now known => 5 guavas are needed

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  • $\begingroup$ "3 guavas for 4 outcomes (this would be 7 if slower/faster was not known)", if slower/faster is not known, you can do better than 7. If we only have 4 parrots, instead of 4 groups of parrots, just give one guava each, total 4. If we have more, put 3 guavas on 3 groups, if one is different, we know that's the special group, and now we know whether it's faster/slower. If all the same, then we know the parrot on the last group, but we still don't know faster/slower, but we can delay that knowledge until later, by repeating this step. So at most you add one more guavas compared to if you know. $\endgroup$
    – justhalf
    Sep 5 at 8:55
  • $\begingroup$ Ah, so by outcome you mean not the number of groups, but the number of distinct results. Understood. $\endgroup$
    – justhalf
    Sep 5 at 9:04
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    $\begingroup$ I hope my edit clarifies things. $\endgroup$
    – Retudin
    Sep 5 at 9:04
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I think the minimum is

six. Give a guava to half of the parrots at the same time, and let them eat until the guavas are half-finished. If one of the parrots has eaten more or less than the others, then that's the special parrot. If not, then remove the parrots and let the other six eat the rest of the guavas. One of them will finish before or after the other parrots, and that's the special parrot.

You could, hypothetically, extrapolate this approach:

four guavas, with each parrot getting 1/3; three guavas, with each parrot getting 1/4; or two guavas, with each parrot getting 1/6.

However,

we have no measuring tools, and anything smaller than 1/2 would be impractical to measure just by eyeballing it.

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  • $\begingroup$ What's stopping you from rot13(gnxvat guerr thninf naq yrggvat guerr cneebgf rnpu rng n dhnegre)? $\endgroup$
    – xyldke
    Sep 5 at 8:19
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    $\begingroup$ @xyldke We have no measuring tools, so we're going entirely off of sight, and rot13(bar unys vf zhpu rnfvre gb rlronyy guna bar dhnegre). I'll edit to address that. $\endgroup$
    – F1Krazy
    Sep 5 at 8:21

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