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 2015-07-28, 18:01 #1 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 153110 Posts August 2015 The problem is out... https://www.research.ibm.com/haifa/p...ugust2015.html "Challenge: 02/08/2015 @ 12:00 PM EST" This time is clearly wrong. (in the past it was also wrong, in general came out earlier, there should be a much better policy on this) Note that the July problem is still living! Last fiddled with by R. Gerbicz on 2015-07-28 at 18:03 Reason: grammar
2015-07-29, 02:46   #2
axn

Jun 2003

122338 Posts

Quote:
 Originally Posted by R. Gerbicz The problem is out... https://www.research.ibm.com/haifa/p...ugust2015.html
Shouldn't there be a length for the string? Otherwise, how many points are you going to give?

Also, maybe some restriction on the values of the coordinates? Otherwise it would be trivially solvable (straight line).

2015-07-29, 04:18   #3
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

1,531 Posts

Quote:
 Originally Posted by axn Shouldn't there be a length for the string? Otherwise, how many points are you going to give? Also, maybe some restriction on the values of the coordinates? Otherwise it would be trivially solvable (straight line).
OK, you are tricky. So to restrict the problem:
find a simple, closed polygonal chain (ref. https://en.wikipedia.org/wiki/Polygonal_chain), so p[0],p[1],...,p[n], where
p[i]==p[j] if and only if i=0,j=n or i=n,j=0. And n>1, the rest is the same p[i] is an integer point in 3D, in p[0],p[1],..,p[n] each (consecutive!) pair differs in exactly one coordinate and all three projections should be loop-free.

(You can see that you can get these simple closed polygonal chains by folding a loop.)

There is no given length for the string, so you can use any (finite) value of n. And no restriction for the coordinate values.

Last fiddled with by R. Gerbicz on 2015-07-29 at 04:19

 2015-07-29, 04:54 #4 axn     Jun 2003 52×211 Posts Missed the "loop of string" bit. So basically, starting point = end point.
 2015-07-29, 14:03 #5 KangJ   Jul 2015 32 Posts I saw August 2015 problem yesterday, and I tried to solve the problem with many different ways. However, I could not figure out the solution. If the loop is divided by two parts arbitrarily, then the two parts should be connected by at least two line segments. In that case, I think that the two line segments always make a loop on at least one projection plane. Therefore, I think that I misunderstood something... How can I get close to the solution?
 2015-08-01, 03:13 #6 KangJ   Jul 2015 32 Posts Never mind. It was tricky and I solved.
 2015-09-04, 12:02 #7 Xyzzy     Aug 2002 22×3×701 Posts
 2015-09-04, 13:18 #8 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 153110 Posts My sent solution was: "One solution: [0,1,2],[0,0,2],[1,0,2],[1,1,2],[1,2,2],[0,2,2],[0,2,1],[1,2,1],[2,2,1],[2,2,2],[2,1,2],[2,1,1],[2,1,0],[2,2,0],[1,2,0],[1,1,0],[1,0,0],[2,0,0],[2,0,1],[1,0,1],[0,0,1],[0,0,0],[0,1,0],[0,1,1],[0,1,2] To see the solution in Mathematica use: Graphics3D[{Thick, Line[{{0,1,2},{0,0,2},{1,0,2},{1,1,2},{1,2,2},{0,2,2}, {0,2,1},{1,2,1},{2,2,1},{2,2,2}, {2,1,2},{2,1,1},{2,1,0},{2,2,0}, {1,2,0},{1,1,0},{1,0,0},{2,0,0}, {2,0,1},{1,0,1},{0,0,1},{0,0,0}, {0,1,0},{0,1,1},{0,1,2} }]}] This was also a recent problem on Komal, see http://www.komal.hu/verseny/feladat....at&f=A638&l=hu The book's picture shows the projections, one can easily verify that these projections are loop-free, so contains no cycle." ps. for the English version see: http://www.komal.hu/verseny/feladat....at&f=A638&l=en (but it does not show the picture), there was only one solver on the competetion (it was on the math section), pretty hard problem.

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