If someone could ELI5 this, that'd be great!
The naive approach (which is what I did) is to assign each second to a binary number and work out how high the number goes - this works out at 2^12.
The clever way other people did - and which I've been thinking about all morning, is:
The complete answer to the test is a set of true/false answers that we map to 23-bit vectors.
We know that there 2^23 vectors, which is way more than the 90*60=5400 ways we have to identify each of them.
Pretty sure we're all clear on that.
The hard part comes when we start saying that we can let some of the bits be wrong. This lets us say that some of the 2^23 23-bit vectors are pretty much the same as some others. I think of it like grouping vectors together into groups that are all related to each other by flipping up to 3 bits.
So let's take an example with 4-bit vectors instead because I find it more intuitive. This is a test with 4 true/false answers. Our naive method would make us think we need 2^4=16 seconds to correctly identify true/false answers. But if we look at what happens when we say one of those answers can be wrong, it gets interesting. Let's look at the first 4-bit vector:
[0 0 0 0]
[0 0 0 0]
[0 0 0 1]
[0 0 1 0]
[0 1 0 0]
[1 0 0 0]
If you let 2 answers be wrong you have larger, fewer groups.
If you let 3 answers be wrong you only need one bit of information! Your cheating friend gets up on second 1 or second 2.
This is my intuitive way of looking at it anyway. For the actual maths and such look at other answers...
Just by looking at the first vector we've reduced our space from 2^16 to 2^16-5.
Should be:
Just by looking at the first vector we've reduced our space from 2^4 to 2^4-5.
Can't edit it now - sorry if that confused anyone.
You can use the remaining 1304 seconds to encode some more vectors to increase the possibility of getting more answers correct, but you cannot guarantee 21 or more correct answers in the allotted time. The lower & upper bound (not sure what this part means) for n=23,R=2 is 30686-32768 about 9 hours!
Very simple example with n=3. I have listed all the vectors with Hamming distance of 1 or less (R=1). Looking at the table, we see that for n=3, R=1, we can cover all the vectors with 2 different codes. Heads and tails for example. It's easy to see which combinations we can pick to guarantee coverage, a&h, b&g, c&f, d&e. Using any of those combinations and some way to signal to the other person 'heads' or 'tails' will guarantee they get at least 2 answers correct.
a TTT <= TTT TTF TFT FTT
b TTF <= TTT TTF TFF FTF
c TFT <= TTT TFT TFF FFT
d FTT <= TTT FTT FTF FFT
e TFF <= TTF TFT TFF FFF
f FTF <= TTF FTT FTF FFF
g FFT <= TFT FTT FFT FFF
h FFF <= TFF FTF FFT FFFELI5-s should use cars cookies or candies to explain :)
The test is a list of 23 true/false questions, so you can think of each question as a bit.
You've prearranged a list of 5400 unique 23-bit vectors, so he determines the bit vector with the most correct answers (ie. smallest Hamming distance) and leaves at whatever time offset corresponds to that bit pattern.
It's not a perfect scheme since you only have 5400 possible bit vectors, and 2^23 > 5400.
You ask your buddy what the correct word is, and a dictionary holding all 23 letter combinations would be incredibly long, however in English there's not really all that many words, so your buddy can simply tell you the right word is on page 2352 of the dictionary and the word you pick might not be it, but it'll be close enough.
I tried to avoid using 5 yr old words to explain EE telecom coding, which isn't going to help much, so this is more "in the spirit of the problem" than being an exact analogy.
So you have 23 questions, and on each you can either answer yes or no. You goal is that your friend "transmits" you some information so that you get the most of the answers right. As the problem is set up, what he can transmit to you is just an exact second of a 90 minutes span, that is one number between 0 and 5399.
Having 23 questions, every with 0 or 1, gives 23 bits of information to pass. 4096 is 2 to 12 th power, so the friend can pass you only 12 complete bits.
Now the major point: you're lucky that you don't care which answers exactly you get right. You need 23 bits passed to get the exact 100% right solution. But you don't care. It turns out, if you cleverly write on the paper 4096 different "solutions" before the test and your friend transmits you which one of these 4096 you should take to fill out the form, the mentioned cleverness can guarantee you that you get N answers right, you just don't know which specifically in advance and you don't care as you prepare these 4096 solutions. The friend must also know the same table, and has to select one of the prepared 4096 solutions only once he discovers what the "100%" solution is.
The remaining question was how many answers you can be sure you can get right (N), and it turns out, N is 20.
Why? If you carefully construct each of these 4096 solutions to be "as different" from all the other 4095 solutions as possible, it can be proven that then 20 is the result.
So you have to take to the test a list with 4096 longer binary numbers, carefully prepared, your fried can transmit you which one to write through the time of his leaving of the test, and you'll have 20 answers right. Or you can train yourself to perform a special algorithm to produce a pattern of 23 answers from the time and the friend to do the opposite.
The math details are, thanks to allenz, here (not on "five years old" level): https://en.wikipedia.org/wiki/Binary_Golay_code There are also some algorithms. The 20 in the solution can be figured out from the sentence "G23 is a perfect code. That is, the spheres of radius three around code words form a partition of the vector space." Which for our task means that not more than 3 answers will be "wrong" from 23 answers known to your friend, if the whole trick is performed correctly, therefore 23-3=20.
And of course, this kind of cleverness (but used for error correction, that is, to allow sending more data than minimally necessary, but where some of the data is lost before it reaches the other side and everything still works) is actually built-in in the modern communication and data storage equipment, but we, typically, just reap the benefits and don't think of it.
On another side, human languages also evolved to contain redundancies, so the evolution is very capable to find the acceptable solutions for the message encoding too.
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1) This way you haven't used all 5400 possible values, but now your can split the remaining thousand-something seconds between you and your friend, for him to prepare the transmission, and for you to fill the form after you received the code but before the test is over.
It's not quite ELI5, but I'd be happy to answer any questions.
You'll only have the time at which your friend leaves (in seconds since the start of the exam). Knowing only this, whatever you two scheme, you'll have to write down your answers (from a scheme you both agreed on beforehand like "leaving after 324 seconds means answer true for the first 7 questions and false for the rest", "leaving after 325 seconds means answer false everywhere", ...) once your friend leaves. So really, you're just a transcriber. During the exam, its your friend who decide what you're going to write down by picking the right second to leave at.
Of course, they'll pick the second which makes you write as many correct answers as possible. (And of course, you'll want to pick the best scheme to help with that. But let's see what you can't do, no matter the scheme, first.)
Why can't you just always get all answers right? Because there are too many possible set of answers (2^23 for 23 questions) and your friend can only choose from the 90 * 60 = 5400 you two agreed on before the start of the exam.
In fact, you can't even be sure of getting 22 questions right. For each second (from 0 to 5399), there are only 23 set of true answers to the exam (among 2^23) for which leaving at that second gets you exactly one answer wrong (namely, one of getting each of the questions wrong). So 5400 + 5400 * 23 is an upper on the number of sets of true answers for which you'll get at most one answer wrong (using whatever scheme). Since 5400 + 5400 * 23 is still (much) less than 2^23, there will be many set of true answers to the exam where your friend doesn't have a choice but to make you write down many wrong answers.
The same goes for getting 21 answers right. For each second, there are 23 * 22 / 2 set of true answers for which leaving the exam at that time gets you exactly two questions wrong. (Pick the first question to get wrong, pick the second question to get wrong, divide by 2 because there are two ways to pick these two questions.)
5400 + 5400 * 23 + (5400 * 23 * 22 / 2)
It turns out that getting 20 answers right is possible. Now you have to actually pick a scheme. We can't just make the same calculation and add
5400 * 23 * 22 * 21 / (2 * 3)
1 + 23 + 23 * 22 / 2 + 23 * 22 * 21 / (2 * 3)
In your scheme, the only thing that matters is which 5400 set of answers your friend can make you write (swapping which answer goes with which second doesn't change anything).
As it happens, people have made tables of the minimum and maximum number of seconds needed to cover all 2^23 possibilities for getting at most 3 answers wrong. And its 4096 seconds which is less than 5400. In fact, the table contain ranges for different total number of questions and different number of questions that you can get wrong.
(I think there's something more interesting about this particular scheme that gets you 4096 but I haven't read all the other comments yet and haven't thought about this questions more since.)