2014 people and 2015 hats (Analysis + Hints)
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by I-G-D
Hey, everyone.
You might remember this logical problem about 2014 people and 2015 hats. If you don’t, then you can read it and try to solve it before reading this…
Okay, so since nobody actually solved it, I’ve decided to write out the analysis of the problem, and the final result, without writing the strategy itself. I will only give you a few tips, and wait for someone to come up with the strategy (which is the most important part of the solution). If nobody succeeds, I will write it myself, but that won’t happen right away, maybe in a few days.
First of all, we know that person #1 sees everyone’s hat number, except his own. He sees 2013 numbers in front of him and he knows that the number on his hat is one of the remaining two. The most important thing here is the fact that he can never be sure what the number on his hat is. He can never know. He might guess it correctly out of pure luck (his chances are 50-50), but he can never be 100% sure about his guess. That means that the optimal strategy can’t guarantee all 2014 points.
The next possible number of points is 2013… It might sound impossible, but a strategy that guarantees 2013 points does exist… If everybody except person #1 somehow figures out their hat number and guesses it correctly, the team can earn at least 2013 points, no matter how the hats get distributed.
But how do they do that? I’ll give you a hint: it’s possible if person #1 says a fixed number out of the other two left, such that when the others hear it, they are able to figure out the hat combination. But how will they determinate the “code” for each combination of hats, so that they can always figure out the combination after they hear the code? That’s on you to figure out…
Good luck!
I-G-D
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Hmmmm...
I'm not speak inglish 😞
Votado.
well, yeah, the hint is obvious. i haven't seen the previous article (before now).
everybody knows which numbers are not in front of them, so, if #1 somehow conveys his two numbers to #2, everybody will know exactly what their number is.
so, the whole point is packing two numbers into one in an unpackable way (because everybody, including #2014 who has no forward information, must know them to make a guess). and he must make an int from 1-2015 (otherwise, he could easily just multiply one of his numbers with, say, 10000 and add the other one).
i have no idea how to compress two numbers from [1,2015] into one in the [1,2015] range. actually, i am not sure that is possible.
It's not possible. There is another way, if you pack 2013 numbers into 1.
dvoumi se izmedju dva broja i vidi broj igraca ispred sebe.
lupi ce jedan broj, tako da ako ovaj ispred njega ima veci broj od onog kojeg je resio da izostavi on ce prilikom izgovora razvlaciri samoglasnike.
i tako isto i sledeci jer se uvek dvoume izmedju dva broja u zavisnosti da li je prethodni brzo izgovorio broj ili je rastezao znace da li da se odluci za veci ili manji broj
a sa svojim izgovorom dace nagovestaj sledecem
Ne postoji mogucnoist da izgovorom da glas. On samo normalno kaze broj, ne mogu ostali iz nacina izgovora ista da zakljuce. Jedini podatak koji imaju je samo taj broj.
Da moze to, zadatak bi bio lak. 😃
The answer is 3
V
Since they all know to calculate the sum for 1 to N numbers... n/2(n+1)
they know that the sum of the hat numbers is 2,031,120
the first one sees all the hat numbers exept his and the hiden one, so he actually knows the sum of these two numbers...
if #1 say it loud the sum of these two numbers, #2 would be able to calculate his hat number by simple addition/subtraction...
thus
2,031,120 - (the sum he sees in front of him) - the sum #1 said it loud = the hat number of #2....
this operatoin would be carryed on from each of the following members till the last one...
.
.
.
.
in the end, all 2013 members of the column will gues/calculate their hat number... except #1 since he said the sum of his hat and the hiden hat number...
so max 2013 points
for #3
2,031,120 - (the sum he sees in front of him) - (the sum #1 said it loud + #2 hat number) = the hat number of #3
...
There are two problems with this:
1. When the people say the numbers out loud, they can only say numbers from 1 to 2015... What if the sum is bigger than 2015?
2. If the sum isn't bigger than 2015, then it's the same as someone else's hat number. Since no two people can guess the same number, when it comes that person's turn, he won't be able to say that number, since it's already been said.
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ok I get your point.
what #1 said it out loud is a "sum" not a real hat number, it doesn't matter if it's more or less than 2015 since it is just a sum of the two hat numbers (the hiden one and #1's )...
but since it is not allowed, and #1 must say his hat number... another solution should be considered xD
The #1 always says the smaller number that he don't see.
So, everybody else knows their own number.
Example for 3 guys and 4 numbers
Person - Hat
#1 - 2
#2 - 4
#3 - 1
#1 says -> 3
#2 knows that 3 is missing and can see 2 and 1, so his number must be 4... and só on
I think i don't forget anything or missing something in my logic lol
#2 can't see 2, he only sees the numbers that are in front of him, meaning only 1.
So:
Everyone says the smallest number they don't see and not said .
😉
I think this the right answer xD
If the first person has a, the second has b, and the hidden hat has c, such that a
sorry, # says 2... the rest is the same i think lol
oh i miss that part I-G-D... i'll think a little more...
just to make sure i understand... #1 see #2, #3,#4 etc
#2 see #3, #4, #5 etc
#3 see #4, #5, #6 etc
right?
Yes, exactly.
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ODLIČAN ti je ovaj problemčić !!!
votčina je legla k'o kuća!
u to ime, kratko pitanje u cilju dodatnog razjašnjena:
da li prvi lik u redu (onaj što vidi sve brojeve osim svoga i onog koji je izbačen) MORA prozvati ISKLJUČIVO broj iz intervala [1,2015], ili može prozvati napr: 813790375 ?
jer, ako može prozvati bilo kakav cijeli broj, onda je strategija sljedeća:
svi likovi u redu napišu sebi listu brojeva koje NE VIDE ispred sebe tako da je svaki broj napisan u 4-cifrenom formatu sa dodatnim vodećim nulama (napr. broj 13 bi bio napisan kao 0013, broj 847 bi bio napisan kao 0847 ...)
broj koji bi prvi lik u nizu trebalo da kaže mora biti 9-cifren i sastojao bi se iz tri dijela:
prvi dio je bilo koji broj od 1 do 9
drugi dio je broj šešira koji nedostaje napisan u 4-cifrenom formatu sa vodećim nulama (ukoliko ima potrebe, tj. ako već nije neki od brojeva koji je veći od 999)
treći dio je broj šešira koji prvi lik nosi na glavi, u istom formatu kao i prethodni.
odnosno, ako prvi lik u koloni zna da ne dostaju brojevi 387 i 1632, on će prozvati napr broj 103871632 (1 0387 1632)
na taj način će svi ostali nakon njega odmah znati da treba sa svojih listi brojeva koje ne vide izbrišu brojeve 387 i 1632.
nakon toga, svakom sljedećem u koloni, kad na njega dođe red, preostaje samo po jedan broj sa njegove liste (tj. broj koji se nalazi na njegovom šeširu) koji može prozvati
zaboravih napomenuti:
prva cifra koja može biti bilo koji od brojeva 1, 2, 3, 4, 5, 6, 7, 8 ili 9
je potreban samo da bi se izbjegla situacija da se nula pojavi na prvom mjestu.
prva cifra se svakako odbacuje, a no što preostaje se grupiše u dva 4-cifrena broja ....
ako prvi lik u koloni MORA prozvati neki cijeli broj iz intervala [1,2015], onda nestrpljivo čekam rješenje...
mora u intervalu [1,2015] da bude broj.
hmmm...
onda cekam rjesenje...
http://www.erepublik.com/en/article/2014-people-and-2015-hats-solution--2422122/1/20
Here it is.