General Code

Consider an army with 10 generals. One wants a security system such that any three of them can determine the code to launch nuclear missiles, but no two of them can. It is possible to devise such a system by using a quadratic polynomial, such as a x^2 + bx + c; to launch the missiles, one must input (a,b,c). One cannot just tell each general one of a, b, or c (as then it is possible that some subset of three generals won’t know a, b and c); however, if a general knows two of (a,b,c), then a set of two generals can launch the missiles! What information should be given to the generals so that any three can find (a,b,c) but no two can? What about the general situation with N generals and any M can launch (but no set of M-1) can?

12 Comments

  1. Steven Miller on September 8, 2012 at 1:11 am

    Jake: well done! not posting as it’s the soln; email me at sjm1 AT williams.edu



  2. Steven Miller on August 24, 2012 at 3:40 pm

    Lukas: looks good, email me (sjm1 AT williams.edu for another soln)



  3. Steven Miller on May 9, 2012 at 1:23 pm

    Monique: correct, well done — not posting as it’s the soln .s



  4. Steven Miller on January 26, 2012 at 4:23 pm

    VERY clever, but can you make it work if you had 1000 generals and same
    constraints?

    One problem is you can’t uniquely determine the values

    say you have ab, a/b and abc

    you can get c from abc / ab

    now you have ab and a/b and need a and b

    you can do ab a/b and get a^2

    you can do ab / a/b and get b^2

    but this gives you sign indeterminacies….

    We’ll assume you only have one chance.

    Again, VERY clever attack!



  5. Steven Miller on January 26, 2012 at 4:23 pm

    ok



  6. Geoff Moss on January 26, 2012 at 4:10 pm

    Ignore this past post, I think I just proved it wrong. Will update after more work.



  7. Geoff Moss on January 26, 2012 at 4:07 pm

    This one is challenging. I’m not 100% certain on the solution, but I’m pretty sure so far. Each general will b given a unique combination. In no particular order, they will be given: a*b, b*c, a*c, a/b, a/c, b/c, a*b/c, a*c/b, b*c/a, and a*b*c. As far as I can tell, no two generals can figure out the code (in some cases, they can solve for a single variable, for example a*b can divide his number by a*b/c, which will give them c, but not a or b), and I believe any combination of 3 should give them all three variables with some basic manipulations.

    Not sure how to approach the general case yet, however.



  8. Steven Miller on January 17, 2012 at 2:17 am

    Sure. This is a fun one, and shows that quadratic equations CAN be fun (ok, I may have a non-standard definition of fun….)



  9. Sara Lublas on January 16, 2012 at 7:24 pm

    Hiya, could you please send me the answer to [email protected]
    Really struggling on this one, and really wanna know how to do this! 😀
    Thanks a lot



  10. Steven Miller on December 17, 2011 at 2:23 pm

    email to you bounced — can you try emailing me at [email protected]



  11. subhas on December 17, 2011 at 11:07 am

    can u send me the solution at [email protected]



  12. Steven Miller on November 29, 2011 at 1:46 am

    NOAH: Email me at [email protected] to discuss the riddles more — you’re definitely on the right track here.



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