## The silence of the sheep

Cam has requested this on from the old ToM site. I’ve corrected the worst of the grammatical errors and rephrased it so that dad can have more than 10 sheep; but have left everything else alone.

A guy is owner of a certain number of sheeps for god’s sakes and also the father of three sons who for some reason are expert logicians. So here comes the question. Clever as you are will think to yourself, now all this guy needs is to believe he’s about to die so that he can make a will to divide the sheep among the sons, right? Right!, except he calls them together (the sons, not the sheep) and tells them how many sheep (not sons) he owns and adds that:

1. The eldest will inherit the most sheep

2. The youngest the least

3. No son gets more than 10 sheep, which as we all know would be a crime.

4. He then whispers in each son’s ear how many sheep he personally will inherit.

After that he proceeds from the eldest to the youngest, asking each aloud if he can calculate how many sheep each of his brothers will inherit and each replies, “no”. He does it again and again each replies, “no”, but then the eldest son on being asked the question once more says, “yes, each of the last two ‘noes’ (that’s the plural of ‘no’) gave me some information, and I now know (no plural of ‘knows’) how many sheep each of us will inherit.” What’s the bet you’re already wondering how may sheep each son will get?

So, how many sheep does each brother get?

Explain your answer.

July 10th, 2010 at 8:59 am

This one is very good! I already fell asleep twice while trying to twist my mind in the correct direction and still am clueless… Just to be sure that I am not wasting my time – are the brothers allowed to have the same number of sheep or all the three numbers are different?

July 10th, 2010 at 11:35 am

Hi slavy and everyone. Statements 1 and 2 imply that each son gets a different number of sheep. So assume that is the case. Also assume that the youngest gets at least one sheep.

July 10th, 2010 at 12:38 pm

Well, all I’ve got so far is the maximum no of sheep possible.

No son has 10 or more sheep, therefore it could be 3 x 9 sheep =27, but we know that the eldest gets the most, and the youngest gets the least, so there are 3 different numbers of sheep.

The eldest getting 9 sheep, the other 2 would get 8 and 7 = 24 sheep maximum.

The minimum numbers of sheep would be 6 – 1 for the youngest, 3 for the eldest, and an indeterminate number of sheep between 1 and 3.

So, the minimum no of sheep is 6 and the maximum 24. That’s the easy stuff out of the way.

It is Saturday night…. I’m off for a pint or two, but no more than 10, and it will be more than I had last night, but less than I had on my birthday…. Cheers!

July 10th, 2010 at 12:50 pm

Whilst I await the taxi….

The farmer has already told them the total number of sheep that he owns (his son’s cannot count?) and he tells them that none will receive 10 or more sheep. I would assume (a bad thing to do here on ToM) that the inference is that the person getting the most sheep will receive 9 sheep……? Still no idea about how many the others are going to get….

If the Farmer is fair and reasonable (ish) then I would go with 7,8,9 sheep, but I cannot back this up in any way.

FYI Chis, in Dorset I believe it would be correct, or required to state…..”and I now knows how many…”

July 10th, 2010 at 12:56 pm

Sorry everyone. I’ve reinstated the original statement 3. I’m worried that by changing that statement, that I may have significantly changed the rest of the problem.

July 10th, 2010 at 1:14 pm

I knowded that. Also true in Zomerzet. But, “knows” isn’t a plural, just an (cute) abomination by the bumpkins.

July 10th, 2010 at 2:33 pm

Hm, this is strange. So far I have found only one solution and it is (10,6,4), but this is suspicious due to the two versions of the third condition (my answer doesn’t satisfy the second version and if the two problems are equivalent and the solution is unique then I have a mistake somewhere that I don’t see). Maybe I misinterpreted the text, I don’t know…

July 10th, 2010 at 2:46 pm

OK, I see where my mistake may be – Chris, what do you mean by “each of the last two ‘noes’ (that’s the plural of ‘no’) gave me some information”? Is this about the two rounds or about the last two replies in the second round? If it is the second thing, then my solution doesn’t work since our guy knows the answer after the second “no” of the middle brother and the reply of the youngest doesn’t help him at all…

July 10th, 2010 at 4:40 pm

There is only one case, when both the second answers of his younger brothers are useful for the oldest one and that after that he knows how the sheep have been split: he gets 9 sheep, the middle one gets 7 sheep, while the youngest is left with only 3 sheep… I am too tired to state my arguments now and honestly I don’t want to see this puzzle for a while…

July 10th, 2010 at 7:19 pm

Hmmm, here goes:

The eldest son will receive 10 Sheep since no one can exceed 10 and the eldest gets the most.

The second son will recieve 9 (this one is a gamble, I will think he can receive any number sheep between 1-10.

The youngest and third son will receive 1 Sheep since he has to recieve the least.

Total number of sheep the owner/father has =20

July 10th, 2010 at 7:40 pm

10 >/= a > b > c > 0

a – b = c ?

a – 2 = b or c ?

noes = knows ?

a + b + c = ??

Why can’t b figure out a or c?

If their master logicians, why can’t they just devise a way to take the most sheep?

Who is looking after the sheep whilst this is happening?

Where and when is this happening?

What are the ages of the three sons?

Why isn’t it cattle?

What is with the grammer? and the unneded specifications?

July 10th, 2010 at 8:33 pm

@All. I only reposted the problem. The most significant change that I made was replacing the word “sheeps” with “sheep”. I haven’t spent any time trying to solve it (It’s 4 am in England, and I’ve been asleep for the last six hours or so). I will be going back to bed soon.

I have seen the official answer and Cam’s answer – they both agree. Some of the stuff in the problem is only intended to be humorous (noes, knows) – that humour is not intended to provide you with information. There were two rounds where all 3 sons were unable to deduce all the facts. At the beginning of the third round, the eldest son was able to deduce what each son was going to inherit. But all he knew that you don’t know, is how many sheep he was going to get; so your job (of deduction) is harder than his.

Sadly, there are two solutions. I temporarily tried a fix to force only one solution, but realised that that fix might have side effects and so I undid it. I may as well say that in one solution the eldest does get 10 sheep and in another he doesn’t – but you can’t use that information to solve the problem; if you do, your solution will disqualified. If you guess a value, then your entire solution will be disqualified.

@Jonell, that no son can have more than 10 sheep, does not mean that the eldest will get 10 sheep. It doesn’t mean that he won’t get 10 sheep either. It simply means that he won’t get more than 10 sheep. Your answer is hopeless (sorry )

@Useless or usefull, I take it that most of your response was intended as some form of humour. “Usefull” should be “useful”, “grammer” should be “grammar”, “unneded” should be “unneeded”. What specification do you think is unneeded (other than the anecdotal style)?

July 11th, 2010 at 3:51 am

What a pain html interpreters can be. To reliably get the inequalities >= and =< you can either replace the > and the < with "& g t ;" and "& l t ;" (without the spaces or quotes), or better still use charmap to get, or copy paste the following ≥ ≤

Reserve => and <= for "implies". You can type those directly.

Alternatively put a space between the symbols: > = and = <

>/= and =/< are ambiguous. "/" could mean "not", "and" or "or".

In this problem (whole numbers), 11 > a > b > c > 0 is easier though.

July 11th, 2010 at 4:43 am

I can see why Cam requested this one.

After some thought, I realise that although there are two solutions, the problem is fair as each son knows how many sheep he will get. You don’t have that information, so you have to deal with more than one possibility.

Sorry about the spoilers. But you still have plenty of work to do to crack this one.

July 11th, 2010 at 5:45 am

But I can’t see the point of this problem Unless there is something really clever and/or a very tricky solution that I was not able to find in the last 24 hours… I have tried many different approaches which lead to partial results such as 8>b>3, etc. However it is impossible to connect two partial results into one without losing the important information who and when should have been the brother to first guess the exact number of sheep. Therefore, I gave up and used brute force. Since the triple, consisting of the number of sheep the eldest, the middle, and the youngest brother respectively inherit, is ordered, we have 10 choose 3 possible outcomes, or exactly 120 admissible cases. We write them down in a table in descending order and (always from up to bottom) go through it 6 times (for each time “no” was said) and each time we cross out all the triples which can be guessed at that moment by the corresponding brother. For the last two rounds we use different colors (for example pencil and a red pen) to cross the triples that can be guessed, since these are exactly the possibilities the eldest brother rejects during the second round. One can check, that the triples (10,6,4), (9,8,3), (9,8,2), (8,5,1), (8,4,2), (7,5,1), and (7,4,2) are those which can be guessed by the middle brother after 4 “no”s and cannot be guessed by the eldest brother after only 3 “no”s. Analogously, the triples (10,7,3), (10,7,2), (9,7,4), and (9,6,4) are those who can be determined by the youngest brother after 5″no”s, but cannot be determined by the middle brother after only 4 “no”s. Now, if we interpret the words of the oldest brother “each of the last two ‘noes’ gave me some information” to refer to the last answers of his younger brothers, we are looking for a still-not-crossed-out triple (a,b,c), where such that there is at least one triple (a,b’,c’) and another one (a,b”,c”) in each of the two groups stated above (those after 4 and 5 noes) such that b+c=b’+c’=b”+c”.

The last two conditions give us three possibilities (10,6,4), (10,7,3) (i.e., a=10, b’=6, c’=4, b”=7, c”=3) which leads to (a,b,c)=(10,8,2) or (10,9,1), or (9,8,3), (9,7,4) which leads to (a,b,c)=(9,6,5), and finally, (9,8,2), (9,6,4) which leads to (a,b,c)=(9,7,3). Among (10,8,2), (10,9,1), (9,6,5), and (9,7,3) only the last one is still not crossed out which gives the final answer (9,7,3), as I pointed out in post 9.

If we don’t care whether the oldest had used both his brothers last answers to determine the sheep distribution, but we just care that he should be able to find it on the thirds trial, then we can just go one more time through our table (and disregard the different colors and all the computations above) and see that there are two cases when this is possible: (9,7,3) (no surprise here) and (10,6,3). I assume those are the two answers, Chris is referring to, but we should point out that in the second case ((10,6,3)) the second “no” of the middle brother does not add any info and is useless for the oldest one.

July 11th, 2010 at 6:22 am

Now, let me at least show why (9,7,3) is a solution.

So, all the brothers hear that their father owns 19 sheep, and each of them knows the number of sheep he should get. At the beginning everyone has the following possibilities:

OLdest – (9,8,2), (9,7,3), (9,6,4)

Middle – (10,7,2), (9,7,3), (8,7,4)

Youngest – (10,6,3), (9,7,3)

The oldest has more than one options, so he need more info and at the beginning he answers that he cannot determine the sheep distribution. The middle thinks the following way: if it was (10,7,2), then my oldest brother cannot decide from (10,7,2), (10,8,1), (10,6,3), and (10,5,4). If it was (9,7,3), then he cannot decide from (9,8,2), (9,7,3), and (9,6,4). If it was (8,7,4), then he cannot decide between (8,7,4) and (8,6,5). In all the cases I know he cannot guess the exact number of my sheep, so his “no” is expected and it doesn’t give me any additional information. Hence, I still have two possibilities and don’t know the answer. The same about the youngest – you can check that he was sure both his brothers don’t know the distribution. Now we come back to the oldest and he is again clueless about the exact distribution. Now the middle one – he already knows that it cannot be (8,7,4), since in this case there are only two options for the oldest – (8,7,4) and (8,6,5). If it was (8,6,5) the youngest would have immediately guessed the answer, since he knows we has the least number of sheep, and that the oldest one has the most number of sheep (here a+b=14, a > b > 5 it is a unique solution a=8, b=6). Since he didn’t, the oldest one already knows the second time that (8,6,5) is impossible, so he should have determined the (8,7,4) distribution. Hence, (8,7,4) is impossible, but still (9,7,3) and (10,7,2) are, so even though his info is now improved, the middle brother still doesn’t know the answer. Again, one can check that for the youngest one both the possibilities remain valid and he cannot add more information at that moment, so he answers “no” again.

Now we are back to the oldest for the third decisive time! Now he knows it cannot be (9,8,2), because if the second brother had exactly 8 sheep, the middle one knows that the oldest should have either 10 or 9. If they were 10, the youngest brother knows that in total his two brothers have 18 sheep so from the very beginning he should know the distribution to be (10,8,1). Since he didn’t say anything at the first round, the middle brother knows that the oldest cannot have 10 sheep, and thus he should have 9 sheep and at the second round he should have guessed the answer. Therefore (9,8,2) cannot be the case. Now if it was (9,6,4), then from the beginning the youngest brother should have the trilemma (9,6,4), (10,5,4) or (8,7,4). After the first “no” of the middle brother the option (10,5,4) is not valid (if this was the case, the middle brother knows that a < 11, c < 5, and a+c = 14, which uniquely determines the pair (10,4) from the begining). (8,7,4) cannot survive the forth round, as already discussed in the previous paragraph. Since none of the previous two outcomes happen, the youngest should have been sure that (9,6,4) is the correct answer at his second trial, and since this didn’t happen, as well, the oldest knows that (9,6,4) cannot be the distribution. Therefore, he is left only with (9,7,3) and thus he claims he knows the answer…

Last comment – to give you some insight how the problem was created, what one can check from the table is that if at this seventh round the oldest brother says that he doesn’t know the distribution, nobody ever will know it afterwords. Therefore this is the longest logical path possible (by possible, I mean that it has a positive ending) in this problem and thus, is very appealing for puzzles. After the seventh round there are 28 left sheep distributions which cannot be determined…

Yeah, this is really twisted and crazy, but I hope somebody will be able to follow it anyway…

July 11th, 2010 at 6:22 am

Hi slavy. I just updated my post #14.

Your 9,7,3 and 10,6,3 are the official solutions.

That you got there by disregarding some of the responses (which I will have to take on trust as having been necessary) means that the posed problem must have been faulty in that the eldest said the last two “noes” had helped him.

I’ve only spent 10 minutes trying to solve this puzzle and so am not able to validate or dispute any step of your reasoning. I merely copied and pasted the problem from the old ToM site at Cam’s request.

I am now going to check through Cam’s solution and see if he also made the same observation about the redundancy of some of the noes.

July 11th, 2010 at 6:25 am

Hi again slavy. LOL, we are posting in parallel. I’ll look at your solution before looking at Cam’s.

LOL again; having just taken a look at it, don’t expect a response too soon – I don’t know that I can hold all of that in my brain at the same time.

July 12th, 2010 at 6:21 pm

@slavy. Have I fixed up post 16 correctly this time?

July 13th, 2010 at 2:19 am

Everything is perfect now Thanks, a lot, Chris!

July 13th, 2010 at 3:46 am

@slavy. My pleasure And thanks for your persistence.