darkshadows's blog

Gordian Knot(Math-Programming contest)
By darkshadows, 10 years ago, In English

Although two hours late,
I invite you to the Project Euler-styled mathematical contest "Gordian Knot" by IIIT-Hyderabad.
See timings here: http://bit.ly/1x9GA9P
Total duration of contest: 48hrs.
For further details, visit: http://felicity.iiit.ac.in/threads/gordian-knot/
Register yourself at: http://felicity.iiit.ac.in/register/ Prizes worth INR 12k/-

UPD:
Here is the final leaderboard.

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10 years ago, # |
  Vote: I like it +16 Vote: I do not like it

How to solve problem about x^4 — 2*(k+2)*x^2 + (k-2)^2 polynomial?

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    10 years ago, # ^ |
    ← Rev. 2   Vote: I like it +16 Vote: I do not like it

    There are two forms:

    (x^3+ax^2+bx+c)(x+d) or (x^2+ax^2+bx)(x^2+cx+d)

    For the first equation, you get a+d=0, b+ad=-2(k+2), c+bd=0, cd=(k-2)^2 This simplifies to (k-2)^2-2d^2(k+2)+d^4 = 0, a = -d, b = d^2-2(k+2), c = -d(d^2-2(k+2)), which are all integer if d is an integer. You can check if d has an integer solution by solving this quadratic equation.

    For the second equation, you get a+c=0, b+ca+d=-2(k+2), cb+ad=0, bd = (k-2)^2

    Using first/third equation, we know c=-a, so a(b-d)=0, so either a=0 or b=d If a=0, then b+d=-2(k+2) and bd = (k-2)^2, which you can check for efficiently (using binary search for instance).

    Otherwise, b=d and c=-a, so we have ac=2(k+2+b) and b=(k-2) so a^2=4k. Thus, if k is a perfect square there exists a solution.

    Now, it turns out you don't really need the first equation at all and can get away with just the second one, so you just need to check if k is a perfect square or there exists integers with m+n=2(k+2) and mn = (k-2)^2.

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      10 years ago, # ^ |
      ← Rev. 3   Vote: I like it +21 Vote: I do not like it

      Just to clarify, the part "...or there exists integers with m+n=2(k+2) and mn = (k-2)^2" is equal to stating that equation x^2 — 2(k+2)x + (k-2)^2 has integer solutions for given k. By finding discriminant, you could see that it happens iff 8 * k is a square of integer.
      So, overall, if 8 * k or 4 * k is perfect square, then this k is "good", otherwise not.

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      10 years ago, # ^ |
      ← Rev. 2   Vote: I like it +5 Vote: I do not like it

      Is there any k for which the original polynomial will be a required product but won't be a product of polynomials like (x2 - A)·(x2 - B) for some A and B? I can't see any counterexample but my answer based on this idea wasn't accepted.

      UPD: Nevermind, got my mistake.

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