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Timeline for Simple cake cutting puzzle

Current License: CC BY-SA 3.0

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Dec 12, 2016 at 20:10 comment added Gerhard Paseman I think I am misunderstanding your intent, or you are not communicating it. I ask you if you can get 5 pieces under your present scheme (with all intersection points of cuts inside the circle). I maintain that you will always get 7 with three such cuts, as you are asking every two cuts to intersect uniquely on the cake. The problem I asked (given K the number of pieces, find the minimal number of cuts) is not much better or harder, but feels more like a complexity problem than this latest version. Gerhard "Perhaps I Am Missing Something" Paseman, 2016.12.12.
Dec 12, 2016 at 17:36 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Dec 12, 2016 at 17:27 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Dec 12, 2016 at 17:20 comment added Mohammad Al-Turkistany @GerhardPaseman Please review the edited post and give me your feedback :)
Dec 12, 2016 at 17:19 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Dec 12, 2016 at 17:08 comment added Mohammad Al-Turkistany @GerhardPaseman I have a fix :). Any instance of my problem can be converted to an instance of the problem that Joseph mentioned. In other words, for any set of non-parallel cuts, enlarge the circle so that all intersection points become inside the circle.
Dec 11, 2016 at 18:06 comment added Gerhard Paseman One problem that I hoped you would mention is to find the minimum number of cuts to produce K pieces. I know of no polytime (in log K) algorithm to answer that. (Actually, one just occurred to me after I hit 'send'.) Gerhard "Don't Ask For Three Pieces" Paseman, 2016.12.11.
Dec 11, 2016 at 16:32 comment added Gerhard Paseman Unfortunately, that restriction does not reduce the triviality. I can take my concurrent solution and tweak the lines so that all points of intersection lie outside the circle, resulting in K pieces. To a casual observer the portion inside the circle will look unchanged. If you insist that all intersection points are strictly inside, you get what Joseph provided in his post. Tweaking a solution to allow more than two lines concurrent inside the circle allows you fewer regions. I don't think you've captured a good problem yet. Gerhard "Is Reading Between The Lines" Paseman, 2016.12.11.
Dec 11, 2016 at 12:52 answer added Joseph O'Rourke timeline score: 4
Dec 11, 2016 at 12:40 comment added Joseph O'Rourke If a parallel pair's "extensions outside the circle on both ends do not cross," doesn't that imply that they are simply parallel lines clipped to the disk?
Dec 11, 2016 at 8:48 comment added Mohammad Al-Turkistany @GerhardPaseman To eliminate all trivial cases, any crossing point (inside or outside the circle) is the result of the intersection of exactly two cuts.
Dec 11, 2016 at 8:47 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Dec 11, 2016 at 7:26 comment added Gerhard Paseman OK. What about K-1 many concurrent lines, with common point far away from the circle, and all of them intersecting the circle non trivially (non tangentially)? Gerhard "What Does Parallel Really Mean?" Paseman, 2016.12.10.
Dec 11, 2016 at 6:02 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Dec 11, 2016 at 5:54 comment added Mohammad Al-Turkistany @GerhardPaseman Crossings are NOT allowed on the circle boundary. So, two cuts either cross inside or outside the circle but they can not cross each other on the circle boundary.
Dec 11, 2016 at 5:50 comment added Gerhard Paseman Assuming crossings are allowed on the circle boundary, K-1 cuts emanating from a single point do the job. Did you want general position, or a limit on the number of cuts through a point? Gerhard "Seems Too Easy To Me" Paseman, 2016.12.10.
Dec 11, 2016 at 5:41 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Dec 11, 2016 at 5:27 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Dec 11, 2016 at 5:18 history asked Mohammad Al-Turkistany CC BY-SA 3.0