20150728, 18:01  #1 
"Robert Gerbicz"
Oct 2005
Hungary
1531_{10} 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 20150728 at 18:03 Reason: grammar 
20150729, 02:46  #2  
Jun 2003
12233_{8} Posts 
Quote:
Also, maybe some restriction on the values of the coordinates? Otherwise it would be trivially solvable (straight line). 

20150729, 04:18  #3  
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 
Quote:
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 loopfree. (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 20150729 at 04:19 

20150729, 04:54  #4 
Jun 2003
5^{2}×211 Posts 
Missed the "loop of string" bit. So basically, starting point = end point.

20150729, 14:03  #5 
Jul 2015
3^{2} 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? 
20150801, 03:13  #6 
Jul 2015
3^{2} Posts 
Never mind.
It was tricky and I solved. 
20150904, 12:02  #7 
Aug 2002
2^{2}×3×701 Posts 

20150904, 13:18  #8 
"Robert Gerbicz"
Oct 2005
Hungary
1531_{10} 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 loopfree, 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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