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I am soon to start my compulsory military service. I applied to the Cyber Warfare Unit of Finnish army. There was a test for applicants. Since the test is done the questions have now been published here: http://erityistehtavat.puolustusvoimat.fi/cyberchallenge.html

Here is question 4:

Two completely isolated programs get one random bit each from different hardware random number generators. After getting the bit each program guesses what the other program's random bit was. Programs can be different and use different strategies for guessing. After running them once, if at least one program guessed correctly, the author of the programs receives a prize.

Is it possible to devise a strategy that provides a 100% chance of winning the prize? If yes, explain the strategy.

I answered no because I couldn't figure out winning strategy. Was that the right answer or did I miss something?

Community
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JV JV
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    Might want to ask this on the Cryptography SE. – ztk Jul 17 '15 at 15:07
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    Probability of success is 75% is both programs answer randomly (since the prize is obtained when either or both guesses correctly). – Tom Leek Jul 17 '15 at 15:14
  • No way, zero chance –  Jul 17 '15 at 15:18
  • I doubt this is a basic probability question based on the objectives of the test. – Eric G Jul 17 '15 at 15:26
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    Considering the answer I gave, the puzzle can been reformulated to be completely unrelated to computer science. For this reason I think that https://puzzling.stackexchange.com/ probably would have been the fastest SE site to give the correct answer. – helm Jul 17 '15 at 16:10
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    That questionnaire is pretty cool, and chance there is an answer sheet somewhere? – David says Reinstate Monica Jul 17 '15 at 17:58
  • This is a math question, actually. It's off-topic here, off-topic on [so]. It could be asked on [math.se] (surprisingly, I can't find it; this question shows the solution for the N-program case and asks a follow-up question) or, since it doesn't require advanced mathematics, on [puzzling.se] (where this question asks for a non-math-heavy explanation of the N-program case). – Gilles 'SO- stop being evil' Jul 17 '15 at 21:16
  • The title of the question shows the misunderstanding of the test. The test problem isn't about a way to guess a random bit with 100% accuracy: it's a math/logic puzzle. The computer and programming aspects of the problem could be removed without changing the core of the problem. – AlannaRose Jul 17 '15 at 21:49

5 Answers5

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There is actually a solution that will always succeed: Program A will guess the opposite of the value it receives, program B will guess the same value as the one it receives.

You can also think of that as such: A guesses that they will receive different numbers; B guesses they receive the same. One of them is bound to be correct. If you look at the following table (r for receive, g for guess), you will see that either A or B is always right (* denotes correct response):

rA | rB | gA | gB
0     0    1    0*
0     1    1*   1
1     0    0*   0
1     1    0    1*
blahdiblah
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helm
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    And this is 100% the right answer :) – pineappleman Jul 17 '15 at 15:45
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    @pineappleman and this Q&A is a textbook example why the green checkbox of accepted answers mean less and less in SE sites. – Mindwin Remember Monica Jul 17 '15 at 17:51
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    @Mindwin OP was rather quick with selecting the accepted answer and had done so before I posted mine. – helm Jul 17 '15 at 17:58
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    Never has an answer felt so wrong, but been so right. – Jeff Meden Jul 17 '15 at 20:14
  • This is a very simple form of the 100 prisoners and a lightswitch problem, or also the "one person always lies, one tells the truth" riddle. – Jason Coyne Jul 17 '15 at 21:17
  • Follow-up: what if there are N programs and the RNG returns values between 1 and N? (Solution) – Gilles 'SO- stop being evil' Jul 17 '15 at 21:19
  • How does this differ from program A always guessing 1 and program B always guessing 0? – Dancrumb Jul 17 '15 at 21:51
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    @Dancrumb: See Lynxy's answer. This will fail if A receives 1 and B receives 0. – helm Jul 17 '15 at 21:54
  • I thought the question required that A and B couldn't communicate? "Two completely isolated programs" If you have the ability to send values between the programs then that is no different then looking at system artifacts, outputs, etc. Is this not a trick question then? – Eric G Jul 17 '15 at 22:21
  • @EricG : ​ ​ ​ You thought correctly. ​ For this answer, you don't "have the ability to send values between the programs". ​ ​ ​ ​ ​ ​ ​ –  Jul 17 '15 at 22:48
  • @RickyDemer I see now, he doesn't mean receive in his explanation as in sending between, he just means predictions, etc. – Eric G Jul 18 '15 at 04:39
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Since this looks like a trick question, here is a trick answer (i.e. it is "cheating").

Let program A do the following:

  • If it receives a 0, then output "1" and exit.
  • Otherwise, loop forever, until a "Ctrl-C" signal is received, in which case the program outputs "0" and exits.

Let program B do the following:

  • If it receives a 0, then output "0" and exit.
  • Otherwise, loop forever, until a "Ctrl-C" signal is received, in which case the program outputs "1" and exits.

Analysis:

  • If both programs get a 0, program B outputs "0", which is correct -> win.
  • If program A gets a 0 and program B gets a 1, then program A outputs "1", which is correct -> win.
  • If program A gets a 1 and program B gets a 0, then program A loops forever while program B outputs "0" (which is wrong); at some point, the operator who runs the experiment loses patience, types Ctrl-C in the terminal running program A, and sees "0" which is a correct answer -> win.
  • If both programs get a 1, then both programs loop forever. The operator tries to kill both with Ctrl-C, at which point program A outputs 0 and program B outputs 1; the latter is correct -> win.

And voilà! a 100% winning strategy.

Edit: and if you forget the whole "wait until Ctrl-C" thing, then you get essentially @Helm's answer, which is correct and a lot less tricky. The only point of that waiting process is to get a side-channel by which a program may not only guess the value, but also "know" that it guessed the value (which is not part of the challenge anyway).

Tom Leek
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    "at some point, the operator who runs the experiment loses patience" +1 for using the operator :) – mikeazo Jul 17 '15 at 18:03
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    Yeah, I was so pleased with myself for thinking about this (I mean, even more pleased that I usually am) that I failed to notice that without this side-channel I actually had the correct answer. – Tom Leek Jul 17 '15 at 18:12
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You are correct. The closest you could get would be a 75% chance of at least one of them being correct, with both computers always guessing the negation of the other.

Travis Gunderson
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Since this is information warfare, I wonder if scenarios with "cheating" are allowed. There may be a system level artifact which the programs can check after the random number is generated or one of the programs could hook into that process, monitor it.

The requirements say the programs are isolated from each other, but are not necessarily on separate systems. You may be able to detect or influence the seed value used by the other application as well if these are both on the same system. If you are writing both applications, you could have each application even print out the value and let the other system read that value in. Even if these are on separate boxes, you could theoretically propose a solution that used some type of heat or acoustic monitoring.

I think the word "guessing" is a throw off, you cannot have any certainty if you are just guessing. This question sounds more like a way of thinking about different ways of breaking of influencing the system and also about understanding how to influence or predict random values.

Eric G
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NO it is not possible to guess a randomly generated bit, with 100% accuracy, you will be right on average half of the time.

Edit: I agree I'm wrong, I read the question as only owning one of the programs. If you own both, each will be correct half of the time, thus on average always.

Pieter
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    The correct answer has already been given, and explained how to be right 100% of the time. – Xander Jul 17 '15 at 21:56
  • If you still want to assert your 50% even after a clear answer has been demonstrated, then you need to explain how you arrived at your conclusion. – schroeder Jul 17 '15 at 22:57