10 ms·

Cheating on a string theory exam

210 points by cperciva 10y ago

pizza 10y ago
My attempt: - need 23 bit string to fill 23 yes/no answers - the test is 90min * 60sec/min = 5400 sec - log2 5400 time-indexes = ~13 bit time-address capability 23 bits required - 13 bits given = 10 bits that are unavailable. You can answer 13 questions.
- cperciva 10y ago
  I rephrased the question slightly for clarity after you posted this: You're looking for the largest value N such that you can guarantee that you get at least N questions correct. Since log(5400)/log(2) is only ~12.3987, your approach only counts as 12 correct answers. (You can do better.)
  - yakult 10y ago
    We can start with the prior that there will probably be a similar number of trues and falses. We are mapping 23 bits of state into 12 bits, we can leave out those states where T>>F or vice versa. If the test turns out to be one of those unexpected states, pick the answer that gives the best score. Edit: Rereading, it looks like we're optimizing the worst case, rather than average. So we're looking for a 12:23 map where at most n bits are inaccurate, minimizing n. I'm sure there's a signals alg that does this...
szemet 10y ago
Reed Solomon for example? Or anything to increase Hamming distance between legal codes.
steventhedev 10y ago
A simple approach that gives 13 correct answers minimum is to have him walk out after X seconds, where X is the number of true answers. Answer true for everything if X is greater than 12. Otherwise answer false for everything. This uses only 6 bits of information, and you should be able to pack some extra info rather easily into the remaining 7. Thinking about alternative approaches leads me to think that the only cases that you should focus on to solve this are those where the number of true/false is almost equal. Figure out how to pack them together, layer a few solutions together, and you should be able to answer the exam to within some N guaranteed. EDIT: 12 correct answers. not 13. And you can easily pack this into 1 bit. Leaving my wrong answer for posterity.
- SomeStupidPoint 10y ago
  Doesn't your method only guarantee 12 answers -- half of 23 (rounded up)? That can be reduced to a single bit: If you should mark 'true' on all the answers, he leaves during the first half of the exam; if you should mark 'false' on all the answers, he leaves during the second part of the exam. This gets you at least 12 correct answers in 1 bit. This leaves you with 11 bits to modify the pattern with to improve your score.
  - szemet 10y ago
    Using this on only 3 questions, you get 2/3 success rate: 000,001,010,100,000 -> buddy signals 0, I write 000 111,110,101,011,111 -> buddy signals 1, I write 111 with 7 bit you definitely win 14 questions from 21
    - SomeStupidPoint 10y ago
      You can get at least 17 by using the 1 bit to express preference for just the first 12 questions, and the remaining 11 bits the answers to the last 11 questions exactly. I've been trying to think if there's a clever scheme to spread the knowledge around to beat 17 with statistical tricks. (As a fun aside: 17/23 is about 73%, which typically gets you a C -- so you'd pass the exam.)
      - szemet 10y ago
        6 bit to win 12 question from the first 18 (group by 3) 5 bit for the remaining 5 17 answer and one bit remained
        SomeStupidPoint 10y ago
        Which answer do you use the extra bit on that's guaranteed to give you one more correct? For any you choose, there are answer patterns where you already counted that answer being correct in your guaranteed number as part of the 2/3rds correct. So I don't think you can (trivially) use that bit to guarantee another correct answer.
unabridged 10y ago
I'm thinking create set of 90 * 60 = 5400 different True False patterns. Pick the one that gets the most answers correct. I'm assuming there is some algorithm with a name I don't know that can select the optimal set of patterns. Then the minimum they get correct in any situation probably has something to do with shannon entropy ...
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rinovan 10y ago
I would say that the lower bound is 18. Because we want to represent a 23 bits string by 12 bits codewords, the max distance between 2 codewords is 11 bits. So the protocol would be to agree on a dictionary of 12 bits codewords and send the closest codeword to the actual answer. The max hamming distance between any answers and a codeword would be floor(11/2) = 5. Which corresponds to 18 correctly answered questions.
- blauditore 10y ago
  I'm not sure I understand the max hamming distance part. From what I understand, the hamming distance would only be relevant for expectations with a certain probability, but worst case will still be 12.
wnoise 10y ago
log_2 of 60*90 is about 12.4. This shows that a naive scheme would expect to get about 12.4 + (23 - 12.4)/2 = 17.7 questions right. I would be very skeptical about any scheme that purports to guarantee better than this. A reasonable approach would be trade the possibility of doing better than this with the possibility of doing worse. In fact, I can guarantee 17 correct questions, with only 11 bits which has the same floor for expected value (11 + (12/2) = 17): Divide the first 18 questions into 6 groups of three. Use the first 6 bits of time to indicate the majority in each group of three, giving you 12 guaranteed answers. The last 5 questions can be encoded directly. It seems like there should be room for improvement, as this does very well on each group in the 1/4 time that each is uniform. I also worry that the average here is 12 + 1.5 + 5 > 17.7. Perhaps overlapping the groups could help, which starts to look like an LDPC code, but with max, rather than parity. The difficulty is that overlapping bits can no longer guarantee more, because the previous ones could have already done so.
asrp 10y ago
Pick 5400 23-bit vectors (the answers you will write) and your friend picks the one that minimizes its Hamming distance to the true answer. 5400*23 + 5400*23*22/2 < 2**23 so answering 21 questions with certainty is impossible. (Pick the true answers so as to maximize the Hamming distance between it and whatever answer you decide to write, regardless of the scheme used.) And there is a size 4096 < 5400 covering code [1] so answering 20 question is possible (by picking the 23-bit vectors in that code). [1] see n = 23, R = 3 here http://www.sztaki.hu/~keri/codes/2_tables.pdf http://www.sztaki.hu/~keri/codes/2_tables.pdf Edit: formatting
- cperciva 10y ago
  Correct! You're the second person to find the answer (the first was not on HN). Now for the extra credit part: What's the connection between this solution and string theory?
  - dvt 10y ago
    This was a neat brain-teaser (I took an information theory class years ago so it was a nice refresh), but the answer ended up being a lot less interesting than I had hoped :P
    - cperciva 10y ago
      I said it was cute, not that it was interesting. :-) (And while asrp's answer is correct, it would have been nice if he had stated what the code is rather than merely citing the fact that it exists.)
  - asrp 10y ago
    Sadly, I can only give the Google answer to this one: > Working on a branch of physics called supersymmetry, Dr. James Gates Jr., discovered what he describes as the presence of what appear to resemble a form of computer code, called error correcting codes, embedded within, or resulting from, the equations of supersymmetry that describe fundamental particles.
deepnotderp 10y ago
Could something similar to "illegal primes" be used?
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SandB0x 10y ago
I like the double entendre in the title.
xarope 10y ago
TIL about Golay codes; interesting read, thanks!
frivoal 10y ago
Unless I am misunderstanding something, all the answers here seem to assume that the friend is either able finish instantly, or to leave before he is done. That isn't stated in the question, so I don't think it's valid to assume you can use all of 90 minutes * 60 seconds to transmit information. "a fraction of the time" isn't a very clear problem statement, as 9999999/1000000 is a fraction, but that's certainly not what is meant by the common sense understanding of this phrase. I am wondering what we can do without clarifying the statement further. We can still encode 90*60 23-bit vectors and have the friend pick the one, in the remaining time after he finishes, that minimizes the distance to the true answer, but without knowing how early he finishes. That is probably still the way you can get the best score on average, but I don't think it does better for the guaranteed number of correct answers than just transmitting 1 bit of information (which guarantees at least 12 correct answers). Would it help to use odd vs even seconds to indicate whether the fiend finished early enough to use the clock to indicate an arbitrary time vs the time at which he should have left was already gone when he finished his exam? I can't think of how to use this (or a similar) signaling to improve the worse case odds. On the other hand, while I cannot think of a way to improve your worse case situation, you should be able to improve average score by sorting your 5400 23-bit vectors so that the distances between the last ones are as large as possible.
bicepjai 10y ago
Could someone elaborate the solution ? Thnx
allenz 10y ago
Here's my writeup, spoilers: We're looking for a lossy compression function that maps strings of length 23 (answers) to strings of length 12 (bits of information in 5400 min). We seek to minimize the number of errors, aka the Hamming distance. Under the Hamming distance metric, the space of strings forms a hypercube where two strings are adjacent if they are related by a bit flip (one error).[1] This gives an elegant visual interpretation: we're searching for a way to partition the hypercube of 23-bit strings into 2^12 balls[2] containing 2^11 elements each. What is the minimum radius of the balls? There is 1 element in the center of the ball, 23C1=23 elements at distance 1, 23C2=253 elements at distance 2, and 23C3=1771 at distance 3, which sums to precisely 2^11. This combinatorial coincidence makes it possible to build a beautifully symmetric 23->12 compression function! In order to avoid overlap, the centers of any two balls must be 7 bits apart. One idea to evenly space the centers: write all possible first 12 bits (1 bit distance), then distribute the remaining bits 6 bits apart. We can also see that inverting the compression function gives an error-correcting code going 12->23, mapping 12-bit inputs to optimally-spaced 23-bit strings. There is a duality between lossy compression (aka rate-distortion theory[3]) and error correction. As it turns out, the optimal 12->23 error-correcting code is the perfect binary Golay code discovered in 1949.[4] The inverse of the Golay code is the compression function we're looking for. A bit of historical trivia: Golay's 1949 paper was reviewed by Berlekamp, who in 1974 called it the "best single published page" in coding theory. At the time, Berlekamp was working as a code breaker with Jim Simons at the Institute for Defense Analyses. Later, Berlekamp would help Simons found Renaissance Technologies, which remains today the most successful quant hedge fund in history.[6] Renaissance was famous, of course, for hiring many of the best minds in string theory. [1] https://en.wikipedia.org/wiki/Hamming_distance https://en.wikipedia.org/wiki/Hamming_distance [2] https://en.wikipedia.org/wiki/Ball_(mathematics) https://en.wikipedia.org/wiki/Ball_(mathematics) [3] https://en.wikipedia.org/wiki/Rate%E2%80%93distortion_theory https://en.wikipedia.org/wiki/Rate%E2%80%93distortion_theory [4] https://en.wikipedia.org/wiki/Binary_Golay_code https://en.wikipedia.org/wiki/Binary_Golay_code [5] https://en.wikipedia.org/wiki/Hamming_bound https://en.wikipedia.org/wiki/Hamming_bound [6] https://en.wikipedia.org/wiki/More_Money_Than_God https://en.wikipedia.org/wiki/More_Money_Than_God
- kowdermeister 10y ago
  Thanks for the clearest (and still mysterious) explanation so far in this thread.
owenversteeg 10y ago
This is really cool, since I'm currently working on a way to transmit extremely small amounts of data (~30 bits) in a situation where you cannot use any RF waves. Of course, it's banned here but the simplest thing (for the person taking the test) would be to take a short amount of time - or a long amount of time - to fill in each bubble. A short scribble = false, a long scribble = true. To thwart the proctors you could swap the keys every other question, such that for question 1 short=false, long=true; question 2 short=true, long=false, etc etc. Of course, for any of these answers the proctors could trivially catch you cheating. The best way to prevent this would be applying a simple one-time pad to them - exchange a small number beforehand and use that.
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botexpert 10y ago
Similar to the lottery problem and other covering problems. Let's say lottery has N numbers. Tickets contain K numbers. What is the least amount of tickets M that you need to buy so you guarantee when the winning ticket is pulled that you matched at least R numbers on that winning ticket with your pool of tickets? LP(N, K, R) = M. LP(N, K, K) = (N choose K), is winning the lottery and for that to be sure we need to buy all tickets. LP(N, K, 2) is solved, I believe, for many values (theoretically). LP(N, K, 3) is already a problem.
xfs 10y ago
Solvable by basic information analysis > answer at least N out of the 23 questions correctly This represents S(N) = \sum_{i=N}^23 \binom{23}{i} acceptable states of correct answers out of 2^23 all possible states, which requires -log_2(S(N)/2^23) bits of self-information transmittable with a code of alphabet size of 5400. Therefore the largest N that satisfies this is 20. 23 - log_2(S(20)) = 12, log_2(5400) = 12.399
- mrkgnao 10y ago
  But you can use those 12 bits as a key into some pre-arranged set of information, each of which carries more information.
  - xfs 10y ago
    Actually you can't. There is no prior information about the distribution of correct answers given in the question so everything is uniform distribution and there is nothing to set up "pre-arranged" entropy encoding with.
    - Robin_Message 10y ago
      The trick (to get around the information theoretic limit of 12) is to be uncertain about which of the answers are correct. For example, with 1 bit, you can transmit if the majority are 1 or 0, already getting you at least 11 right answers but no information about which ones are right.
seesomesense 10y ago
Binary Golay code...
blobman 10y ago
The friend could easily submit far more answers than 23 if need be. He just needs to go to the toilet. Step 1: 10 bits information - First 10 answers as seconds from start of exam. After x seconds, friend goes to toilet. Step 2: Restart counter Step 3: On return from the toilet, take time taken in toilet. You may want to limit this to 6 bits of information / ~ 1 minute in the toilet. Anything longer might be suspicious. Step 4: Restart counter Step 5: Only 7 bits of information to hand over to get 23 questions. Count time till he leaves. Step 6: Repeat toilet-going process for more answers in different situations. There should be no suspicions, as the answer-receiver will not go to the toilet. Do not drink tea when attempting this method.
dbfclark 10y ago
Presumably the extra credit is that the binary Golay code is closely related to the Leech Lattice, and thus to the entire moonshine situation, which gets you to string theory. See: https://en.wikipedia.org/wiki/Leech_lattice https://en.wikipedia.org/wiki/Leech_lattice https://en.wikipedia.org/wiki/Monstrous_moonshine https://en.wikipedia.org/wiki/Monstrous_moonshine http://motls.blogspot.com/2015/03/umbral-moonshine-and-golay-code.html http://motls.blogspot.com/2015/03/umbral-moonshine-and-golay...
- cperciva 10y ago
  Yes. In particular, the Leech Lattice gives rise to Bosonic String Theory.
  - dbfclark 10y ago
    Hey, I'm a mathematician, not a physicist. We tend only to get as far as "oh yeah, here's all this pretty math we care about. I hear it has some applications to physics..."
    - cperciva 10y ago
      Yeah, that's pretty much where I'm at too. :-) I originally wrote the puzzle without reference to string theory, then added the "extra credit" part mainly as a hint.
      - dbfclark 10y ago
        Sure. Moonshine I know a lot about, relative to physics at least. Given that I now work somewhere that specializes in term embeddings, it kind of feels like I've spent half my life injecting stuff into vector spaces...
skizm 10y ago
More practically (and specifically ruled out in the question): you can see when your friend "starts" the exam (picks up his pencil) and when he ends (gets up from his seat). Also it probably takes him at least 10 minutes to read the whole test and figure out the answers. This makes it easy mode: friend takes 10 minutes to read test and get answers in his head, start the timer (0 seconds) at the 10 minute mark. Friend "starts" sometime in the next 34 minutes. 34 minutes is 2048 seconds or 11 bits of info. Then "finishes" between 44 minutes and 78 minutes: another 34 minute window and another 11 bits of info. 22 of 23 questions answered. Friend coughs on way out if last answer is true. Boom, all questions answered. Yes, OP said we can't do this, but it is more practical than memorizing 5400 bit vectors, doesn't assume the friend can instantly finish the test, gives the friend some lead time to actually do some binary conversions, and ensure the correct times down to the second.
SFJulie 10y ago
Encoding time as a one hot https://en.wikipedia.org/wiki/One-hot https://en.wikipedia.org/wiki/One-hot Getting 12 out of 23 should get him the average on the lucky assumption the teacher was not a joker that put all answers higher than bit 6 to false ...
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anitasvasu 10y ago
ELI5 Version: In order to understand the solution, we need to first understand basics of coding theory. Unlike envelopes when we send messages over a wire, or through air (wireless) it is possible that the messages we send get corrupted. Specifically if we are sending “Hello” it can end up becoming “Hgllo” where e became g because of some electric/electromagnetic distortion. If we typed incorrectly using our mobile keyboards it would have auto corrected “Hgllo” to “Hello”. This is possible, because our natural languages have a lot of redundant information. Otherwise if we lost one word in a conversation whole sentence would not make sense. Coding theory is about how do we systematically put redundant information so that computers can correct for errors. We are only going to talk about one type of error, where a letter gets replaced by another letter. If ABCDEF becomes GBCDEH ; here two letters are wrong, and we say they have a distance of 2. What will happen if we simply repeat each letter? Message ABC becomes AABBCC. Now if a single error happens GABBCC we know original message got corrupted. We still can’t figure out what the original message is. It could be either ABC or GBC. So a better technique would be to repeat each letter thrice. Message ABC becomes AAABBBCCC. Now if a single error happens GAABBBCCC, then we can say if there is only one error then the correct message is ABC by taking majority. Note that this cannot correct for two errors because if you see GGABBBCCC then you will think real message is GBC. What if the message has 4 letters, ABCD becomes AAABBBCCCDDD? This scheme requires you to send 12 letters just to correct 1 error in 4 letters. We denote this by saying [12, 4, 1] — where 12 is the coding distance, 4 is the original message length, and 1 is the number of letters we can correct for. For now let us temporarily restrict our focus to binary letters (0, 1). This is because in computers we communicate purely using 0s and 1s. For example letter A would be represented by a number 65, which would be represented as 0100 0001 an 8-bit message, or simply called a byte. It turns out if we want to send 4 bits, then using 12 bits for it is actually not very efficient. You could do better. In fact just 7 bits is enough to represent 4 bits where you can correct for a single error. This is called the hamming code. Let us understand why our repeating scheme works in more detail. Consider any two valid code sequence, for example AAABBBCCCDDD and AAAAEEECCCDDD, even though only one character changed in the input, 3 letters changed in the output. So if we make sure distance between any two coded sentences is 3 then we can correct for one error. There are 16 possible words using 4 bits (0000, 0001, 0010, … 1111). If we can assign code words for each of them such that distance between any two is 3, then we can say that it can correct for 1 error. Hamming code essentially achieves it. If are encoding ABCD, then first 4 bits is the same as the the input viz ABCD, 5th bit is A + B + D; 6th bit is A + C + D, and 7th bit is B + C + D. Of course sum like 3 will be replaced by 1 again by taking remainder to 2. Such codes are also called linear codes, where each bit can be written as linear combination of input bits. You can read chapter 5 of Jiri Matousek’s 33 Miniatures to see a quick proof on why hamming code is indeed a valid code using a bit of linear algebra. Or you can simply write down 16 of those codes, and check every two combination and make sure they differ in at least three places. (You will need to compare 120 pairs though). http://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf http://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf What if we want to correct for more than 1 error? Golay code is one such code. G23 and G24 are two codes that encodes 12 bits in 23 and 24 bits respectively. In the first case hamming distance is 7 and in second case it is 8. There is no reason to have a distance 8, distance 7 is enough because Imagine two codes valid codes (x0) and (x7) which are 7 distances apart. Then using 7 changes you can reach from x0 to x7. (x0) — x1 — x2 — x3* — x4* — x5 — x6 — (x7) Even if (x0) makes 3 errors, and (x7) makes 3 errors you will only end up with (x3) or (x4). And never get confused. So if we want to correct for k errors then hamming distance between any two code words must be 2k + 1 or higher. It turns out that G23 is not just a good code, it is a “perfect” code. Meaning if we want to send 12 bits, with 3 bits of correction, then you must need 23 bits. Now back to our problem, you have 23 answers in front of you, assume it is G23 code and decode back to 12 bits. You get up at a time to represent these 12 bits. We know that when your friend encodes this 12 bits back to G23 code then at most 3 bits will be out of place. In other words you friend will get at most 3 answers wrong. Hence getting at least 20 of them correct. Now relationship with String theory requires a bigger write up. But essentially it is because the internal structure of field theory on some kind of donut shaped universe, the type of transformations you can do there will look something like the type of transformations you can do on the Golay codes.
BrandoElFollito 10y ago
Shouldn't the friend also communicate the time he actually finished the exam (which can be different depending on the difficulty of the questions)? In other words, it took him 23 minutes to answer, left after 53, which means that the information about the answers is encoded in "30" (53-23). If it would have taken him 17 min to solve the questions, he would have left after 17+30=47 min. This also means that the more difficult the questions, the less information he can provide (because the "time" allocated to coding the answers is smaller (as in smaller keyspace)

210 points by cperciva 10y ago

pizza 10y ago

My attempt: - need 23 bit string to fill 23 yes/no answers - the test is 90min * 60sec/min = 5400 sec - log2 5400 time-indexes = ~13 bit time-address capability 23 bits required - 13 bits given = 10 bits that are unavailable. You can answer 13 questions.

cperciva 10y ago
I rephrased the question slightly for clarity after you posted this: You're looking for the largest value N such that you can guarantee that you get at least N questions correct. Since log(5400)/log(2) is only ~12.3987, your approach only counts as 12 correct answers. (You can do better.)
- yakult 10y ago
  We can start with the prior that there will probably be a similar number of trues and falses. We are mapping 23 bits of state into 12 bits, we can leave out those states where T>>F or vice versa. If the test turns out to be one of those unexpected states, pick the answer that gives the best score. Edit: Rereading, it looks like we're optimizing the worst case, rather than average. So we're looking for a 12:23 map where at most n bits are inaccurate, minimizing n. I'm sure there's a signals alg that does this...

szemet 10y ago

Reed Solomon for example? Or anything to increase Hamming distance between legal codes.

steventhedev 10y ago

A simple approach that gives 13 correct answers minimum is to have him walk out after X seconds, where X is the number of true answers. Answer true for everything if X is greater than 12. Otherwise answer false for everything. This uses only 6 bits of information, and you should be able to pack some extra info rather easily into the remaining 7. Thinking about alternative approaches leads me to think that the only cases that you should focus on to solve this are those where the number of true/false is almost equal. Figure out how to pack them together, layer a few solutions together, and you should be able to answer the exam to within some N guaranteed. EDIT: 12 correct answers. not 13. And you can easily pack this into 1 bit. Leaving my wrong answer for posterity.

SomeStupidPoint 10y ago
Doesn't your method only guarantee 12 answers -- half of 23 (rounded up)? That can be reduced to a single bit: If you should mark 'true' on all the answers, he leaves during the first half of the exam; if you should mark 'false' on all the answers, he leaves during the second part of the exam. This gets you at least 12 correct answers in 1 bit. This leaves you with 11 bits to modify the pattern with to improve your score.
- szemet 10y ago
  Using this on only 3 questions, you get 2/3 success rate: 000,001,010,100,000 -> buddy signals 0, I write 000 111,110,101,011,111 -> buddy signals 1, I write 111 with 7 bit you definitely win 14 questions from 21
  - SomeStupidPoint 10y ago
    You can get at least 17 by using the 1 bit to express preference for just the first 12 questions, and the remaining 11 bits the answers to the last 11 questions exactly. I've been trying to think if there's a clever scheme to spread the knowledge around to beat 17 with statistical tricks. (As a fun aside: 17/23 is about 73%, which typically gets you a C -- so you'd pass the exam.)
    - szemet 10y ago
      6 bit to win 12 question from the first 18 (group by 3) 5 bit for the remaining 5 17 answer and one bit remained
      - SomeStupidPoint 10y ago
        Which answer do you use the extra bit on that's guaranteed to give you one more correct? For any you choose, there are answer patterns where you already counted that answer being correct in your guaranteed number as part of the 2/3rds correct. So I don't think you can (trivially) use that bit to guarantee another correct answer.

unabridged 10y ago

I'm thinking create set of 90 * 60 = 5400 different True False patterns. Pick the one that gets the most answers correct. I'm assuming there is some algorithm with a name I don't know that can select the optimal set of patterns. Then the minimum they get correct in any situation probably has something to do with shannon entropy ...

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rinovan 10y ago

I would say that the lower bound is 18. Because we want to represent a 23 bits string by 12 bits codewords, the max distance between 2 codewords is 11 bits. So the protocol would be to agree on a dictionary of 12 bits codewords and send the closest codeword to the actual answer. The max hamming distance between any answers and a codeword would be floor(11/2) = 5. Which corresponds to 18 correctly answered questions.

blauditore 10y ago
I'm not sure I understand the max hamming distance part. From what I understand, the hamming distance would only be relevant for expectations with a certain probability, but worst case will still be 12.

wnoise 10y ago

log_2 of 60*90 is about 12.4. This shows that a naive scheme would expect to get about 12.4 + (23 - 12.4)/2 = 17.7 questions right. I would be very skeptical about any scheme that purports to guarantee better than this. A reasonable approach would be trade the possibility of doing better than this with the possibility of doing worse. In fact, I can guarantee 17 correct questions, with only 11 bits which has the same floor for expected value (11 + (12/2) = 17): Divide the first 18 questions into 6 groups of three. Use the first 6 bits of time to indicate the majority in each group of three, giving you 12 guaranteed answers. The last 5 questions can be encoded directly. It seems like there should be room for improvement, as this does very well on each group in the 1/4 time that each is uniform. I also worry that the average here is 12 + 1.5 + 5 > 17.7. Perhaps overlapping the groups could help, which starts to look like an LDPC code, but with max, rather than parity. The difficulty is that overlapping bits can no longer guarantee more, because the previous ones could have already done so.

asrp 10y ago

Pick 5400 23-bit vectors (the answers you will write) and your friend picks the one that minimizes its Hamming distance to the true answer. 5400*23 + 5400*23*22/2 < 2**23 so answering 21 questions with certainty is impossible. (Pick the true answers so as to maximize the Hamming distance between it and whatever answer you decide to write, regardless of the scheme used.) And there is a size 4096 < 5400 covering code [1] so answering 20 question is possible (by picking the 23-bit vectors in that code). [1] see n = 23, R = 3 here http://www.sztaki.hu/~keri/codes/2_tables.pdf http://www.sztaki.hu/~keri/codes/2_tables.pdf Edit: formatting

cperciva 10y ago
Correct! You're the second person to find the answer (the first was not on HN). Now for the extra credit part: What's the connection between this solution and string theory?
- dvt 10y ago
  This was a neat brain-teaser (I took an information theory class years ago so it was a nice refresh), but the answer ended up being a lot less interesting than I had hoped :P
  - cperciva 10y ago
    I said it was cute, not that it was interesting. :-) (And while asrp's answer is correct, it would have been nice if he had stated what the code is rather than merely citing the fact that it exists.)
- asrp 10y ago
  Sadly, I can only give the Google answer to this one: > Working on a branch of physics called supersymmetry, Dr. James Gates Jr., discovered what he describes as the presence of what appear to resemble a form of computer code, called error correcting codes, embedded within, or resulting from, the equations of supersymmetry that describe fundamental particles.

deepnotderp 10y ago

Could something similar to "illegal primes" be used?

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SandB0x 10y ago

I like the double entendre in the title.

xarope 10y ago

TIL about Golay codes; interesting read, thanks!

frivoal 10y ago

Unless I am misunderstanding something, all the answers here seem to assume that the friend is either able finish instantly, or to leave before he is done. That isn't stated in the question, so I don't think it's valid to assume you can use all of 90 minutes * 60 seconds to transmit information. "a fraction of the time" isn't a very clear problem statement, as 9999999/1000000 is a fraction, but that's certainly not what is meant by the common sense understanding of this phrase. I am wondering what we can do without clarifying the statement further. We can still encode 90*60 23-bit vectors and have the friend pick the one, in the remaining time after he finishes, that minimizes the distance to the true answer, but without knowing how early he finishes. That is probably still the way you can get the best score on average, but I don't think it does better for the guaranteed number of correct answers than just transmitting 1 bit of information (which guarantees at least 12 correct answers). Would it help to use odd vs even seconds to indicate whether the fiend finished early enough to use the clock to indicate an arbitrary time vs the time at which he should have left was already gone when he finished his exam? I can't think of how to use this (or a similar) signaling to improve the worse case odds. On the other hand, while I cannot think of a way to improve your worse case situation, you should be able to improve average score by sorting your 5400 23-bit vectors so that the distances between the last ones are as large as possible.

bicepjai 10y ago

Could someone elaborate the solution ? Thnx

allenz 10y ago

Here's my writeup, spoilers: We're looking for a lossy compression function that maps strings of length 23 (answers) to strings of length 12 (bits of information in 5400 min). We seek to minimize the number of errors, aka the Hamming distance. Under the Hamming distance metric, the space of strings forms a hypercube where two strings are adjacent if they are related by a bit flip (one error).[1] This gives an elegant visual interpretation: we're searching for a way to partition the hypercube of 23-bit strings into 2^12 balls[2] containing 2^11 elements each. What is the minimum radius of the balls? There is 1 element in the center of the ball, 23C1=23 elements at distance 1, 23C2=253 elements at distance 2, and 23C3=1771 at distance 3, which sums to precisely 2^11. This combinatorial coincidence makes it possible to build a beautifully symmetric 23->12 compression function! In order to avoid overlap, the centers of any two balls must be 7 bits apart. One idea to evenly space the centers: write all possible first 12 bits (1 bit distance), then distribute the remaining bits 6 bits apart. We can also see that inverting the compression function gives an error-correcting code going 12->23, mapping 12-bit inputs to optimally-spaced 23-bit strings. There is a duality between lossy compression (aka rate-distortion theory[3]) and error correction. As it turns out, the optimal 12->23 error-correcting code is the perfect binary Golay code discovered in 1949.[4] The inverse of the Golay code is the compression function we're looking for. A bit of historical trivia: Golay's 1949 paper was reviewed by Berlekamp, who in 1974 called it the "best single published page" in coding theory. At the time, Berlekamp was working as a code breaker with Jim Simons at the Institute for Defense Analyses. Later, Berlekamp would help Simons found Renaissance Technologies, which remains today the most successful quant hedge fund in history.[6] Renaissance was famous, of course, for hiring many of the best minds in string theory. [1] https://en.wikipedia.org/wiki/Hamming_distance https://en.wikipedia.org/wiki/Hamming_distance [2] https://en.wikipedia.org/wiki/Ball_(mathematics) https://en.wikipedia.org/wiki/Ball_(mathematics) [3] https://en.wikipedia.org/wiki/Rate%E2%80%93distortion_theory https://en.wikipedia.org/wiki/Rate%E2%80%93distortion_theory [4] https://en.wikipedia.org/wiki/Binary_Golay_code https://en.wikipedia.org/wiki/Binary_Golay_code [5] https://en.wikipedia.org/wiki/Hamming_bound https://en.wikipedia.org/wiki/Hamming_bound [6] https://en.wikipedia.org/wiki/More_Money_Than_God https://en.wikipedia.org/wiki/More_Money_Than_God

kowdermeister 10y ago
Thanks for the clearest (and still mysterious) explanation so far in this thread.

owenversteeg 10y ago

This is really cool, since I'm currently working on a way to transmit extremely small amounts of data (~30 bits) in a situation where you cannot use any RF waves. Of course, it's banned here but the simplest thing (for the person taking the test) would be to take a short amount of time - or a long amount of time - to fill in each bubble. A short scribble = false, a long scribble = true. To thwart the proctors you could swap the keys every other question, such that for question 1 short=false, long=true; question 2 short=true, long=false, etc etc. Of course, for any of these answers the proctors could trivially catch you cheating. The best way to prevent this would be applying a simple one-time pad to them - exchange a small number beforehand and use that.

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botexpert 10y ago

Similar to the lottery problem and other covering problems. Let's say lottery has N numbers. Tickets contain K numbers. What is the least amount of tickets M that you need to buy so you guarantee when the winning ticket is pulled that you matched at least R numbers on that winning ticket with your pool of tickets? LP(N, K, R) = M. LP(N, K, K) = (N choose K), is winning the lottery and for that to be sure we need to buy all tickets. LP(N, K, 2) is solved, I believe, for many values (theoretically). LP(N, K, 3) is already a problem.

xfs 10y ago

Solvable by basic information analysis > answer at least N out of the 23 questions correctly This represents S(N) = \sum_{i=N}^23 \binom{23}{i} acceptable states of correct answers out of 2^23 all possible states, which requires -log_2(S(N)/2^23) bits of self-information transmittable with a code of alphabet size of 5400. Therefore the largest N that satisfies this is 20. 23 - log_2(S(20)) = 12, log_2(5400) = 12.399

mrkgnao 10y ago
But you can use those 12 bits as a key into some pre-arranged set of information, each of which carries more information.
- xfs 10y ago
  Actually you can't. There is no prior information about the distribution of correct answers given in the question so everything is uniform distribution and there is nothing to set up "pre-arranged" entropy encoding with.
  - Robin_Message 10y ago
    The trick (to get around the information theoretic limit of 12) is to be uncertain about which of the answers are correct. For example, with 1 bit, you can transmit if the majority are 1 or 0, already getting you at least 11 right answers but no information about which ones are right.

seesomesense 10y ago

Binary Golay code...

blobman 10y ago

The friend could easily submit far more answers than 23 if need be. He just needs to go to the toilet. Step 1: 10 bits information - First 10 answers as seconds from start of exam. After x seconds, friend goes to toilet. Step 2: Restart counter Step 3: On return from the toilet, take time taken in toilet. You may want to limit this to 6 bits of information / ~ 1 minute in the toilet. Anything longer might be suspicious. Step 4: Restart counter Step 5: Only 7 bits of information to hand over to get 23 questions. Count time till he leaves. Step 6: Repeat toilet-going process for more answers in different situations. There should be no suspicions, as the answer-receiver will not go to the toilet. Do not drink tea when attempting this method.

dbfclark 10y ago

Presumably the extra credit is that the binary Golay code is closely related to the Leech Lattice, and thus to the entire moonshine situation, which gets you to string theory. See: https://en.wikipedia.org/wiki/Leech_lattice https://en.wikipedia.org/wiki/Leech_lattice https://en.wikipedia.org/wiki/Monstrous_moonshine https://en.wikipedia.org/wiki/Monstrous_moonshine http://motls.blogspot.com/2015/03/umbral-moonshine-and-golay-code.html http://motls.blogspot.com/2015/03/umbral-moonshine-and-golay...

cperciva 10y ago
Yes. In particular, the Leech Lattice gives rise to Bosonic String Theory.
- dbfclark 10y ago
  Hey, I'm a mathematician, not a physicist. We tend only to get as far as "oh yeah, here's all this pretty math we care about. I hear it has some applications to physics..."
  - cperciva 10y ago
    Yeah, that's pretty much where I'm at too. :-) I originally wrote the puzzle without reference to string theory, then added the "extra credit" part mainly as a hint.
    - dbfclark 10y ago
      Sure. Moonshine I know a lot about, relative to physics at least. Given that I now work somewhere that specializes in term embeddings, it kind of feels like I've spent half my life injecting stuff into vector spaces...

skizm 10y ago

More practically (and specifically ruled out in the question): you can see when your friend "starts" the exam (picks up his pencil) and when he ends (gets up from his seat). Also it probably takes him at least 10 minutes to read the whole test and figure out the answers. This makes it easy mode: friend takes 10 minutes to read test and get answers in his head, start the timer (0 seconds) at the 10 minute mark. Friend "starts" sometime in the next 34 minutes. 34 minutes is 2048 seconds or 11 bits of info. Then "finishes" between 44 minutes and 78 minutes: another 34 minute window and another 11 bits of info. 22 of 23 questions answered. Friend coughs on way out if last answer is true. Boom, all questions answered. Yes, OP said we can't do this, but it is more practical than memorizing 5400 bit vectors, doesn't assume the friend can instantly finish the test, gives the friend some lead time to actually do some binary conversions, and ensure the correct times down to the second.

SFJulie 10y ago

Encoding time as a one hot https://en.wikipedia.org/wiki/One-hot https://en.wikipedia.org/wiki/One-hot Getting 12 out of 23 should get him the average on the lucky assumption the teacher was not a joker that put all answers higher than bit 6 to false ...

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anitasvasu 10y ago

ELI5 Version: In order to understand the solution, we need to first understand basics of coding theory. Unlike envelopes when we send messages over a wire, or through air (wireless) it is possible that the messages we send get corrupted. Specifically if we are sending “Hello” it can end up becoming “Hgllo” where e became g because of some electric/electromagnetic distortion. If we typed incorrectly using our mobile keyboards it would have auto corrected “Hgllo” to “Hello”. This is possible, because our natural languages have a lot of redundant information. Otherwise if we lost one word in a conversation whole sentence would not make sense. Coding theory is about how do we systematically put redundant information so that computers can correct for errors. We are only going to talk about one type of error, where a letter gets replaced by another letter. If ABCDEF becomes GBCDEH ; here two letters are wrong, and we say they have a distance of 2. What will happen if we simply repeat each letter? Message ABC becomes AABBCC. Now if a single error happens GABBCC we know original message got corrupted. We still can’t figure out what the original message is. It could be either ABC or GBC. So a better technique would be to repeat each letter thrice. Message ABC becomes AAABBBCCC. Now if a single error happens GAABBBCCC, then we can say if there is only one error then the correct message is ABC by taking majority. Note that this cannot correct for two errors because if you see GGABBBCCC then you will think real message is GBC. What if the message has 4 letters, ABCD becomes AAABBBCCCDDD? This scheme requires you to send 12 letters just to correct 1 error in 4 letters. We denote this by saying [12, 4, 1] — where 12 is the coding distance, 4 is the original message length, and 1 is the number of letters we can correct for. For now let us temporarily restrict our focus to binary letters (0, 1). This is because in computers we communicate purely using 0s and 1s. For example letter A would be represented by a number 65, which would be represented as 0100 0001 an 8-bit message, or simply called a byte. It turns out if we want to send 4 bits, then using 12 bits for it is actually not very efficient. You could do better. In fact just 7 bits is enough to represent 4 bits where you can correct for a single error. This is called the hamming code. Let us understand why our repeating scheme works in more detail. Consider any two valid code sequence, for example AAABBBCCCDDD and AAAAEEECCCDDD, even though only one character changed in the input, 3 letters changed in the output. So if we make sure distance between any two coded sentences is 3 then we can correct for one error. There are 16 possible words using 4 bits (0000, 0001, 0010, … 1111). If we can assign code words for each of them such that distance between any two is 3, then we can say that it can correct for 1 error. Hamming code essentially achieves it. If are encoding ABCD, then first 4 bits is the same as the the input viz ABCD, 5th bit is A + B + D; 6th bit is A + C + D, and 7th bit is B + C + D. Of course sum like 3 will be replaced by 1 again by taking remainder to 2. Such codes are also called linear codes, where each bit can be written as linear combination of input bits. You can read chapter 5 of Jiri Matousek’s 33 Miniatures to see a quick proof on why hamming code is indeed a valid code using a bit of linear algebra. Or you can simply write down 16 of those codes, and check every two combination and make sure they differ in at least three places. (You will need to compare 120 pairs though). http://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf http://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf What if we want to correct for more than 1 error? Golay code is one such code. G23 and G24 are two codes that encodes 12 bits in 23 and 24 bits respectively. In the first case hamming distance is 7 and in second case it is 8. There is no reason to have a distance 8, distance 7 is enough because Imagine two codes valid codes (x0) and (x7) which are 7 distances apart. Then using 7 changes you can reach from x0 to x7. (x0) — x1 — x2 — x3* — x4* — x5 — x6 — (x7) Even if (x0) makes 3 errors, and (x7) makes 3 errors you will only end up with (x3) or (x4). And never get confused. So if we want to correct for k errors then hamming distance between any two code words must be 2k + 1 or higher. It turns out that G23 is not just a good code, it is a “perfect” code. Meaning if we want to send 12 bits, with 3 bits of correction, then you must need 23 bits. Now back to our problem, you have 23 answers in front of you, assume it is G23 code and decode back to 12 bits. You get up at a time to represent these 12 bits. We know that when your friend encodes this 12 bits back to G23 code then at most 3 bits will be out of place. In other words you friend will get at most 3 answers wrong. Hence getting at least 20 of them correct. Now relationship with String theory requires a bigger write up. But essentially it is because the internal structure of field theory on some kind of donut shaped universe, the type of transformations you can do there will look something like the type of transformations you can do on the Golay codes.

BrandoElFollito 10y ago

Shouldn't the friend also communicate the time he actually finished the exam (which can be different depending on the difficulty of the questions)? In other words, it took him 23 minutes to answer, left after 53, which means that the information about the answers is encoded in "30" (53-23). If it would have taken him 17 min to solve the questions, he would have left after 17+30=47 min. This also means that the more difficult the questions, the less information he can provide (because the "time" allocated to coding the answers is smaller (as in smaller keyspace)