We study the problem of colouring the vertices of a polygon, such that every viewer can see a unique colour. The goal is to minimize the number of colours used. This is also known as the conflict-free chromatic guarding problem with vertex guards (which is quite different from point guards considered in  other papers). We study the problem in two scenarios of a set of viewers. In the first scenario , we assume that the viewers are all points of the polygon. We  solve the related problem of minimizing the number of guards and approximate (up to only an additive error) the number of colours in the special case of funnels. We also give an upper bound of O(log  n) colours on weak-visibility polygons which generalizes to all simple polygons. In the second scenario, we assume that the viewers are only the vertices of the polygon. We show a lower bound of 3 colours in the general case of simple polygons and conjecture that this is tight. We also prove that already deciding whether 1 or 2 colours are enough is NP-complete.
We study the problem of colouring the vertices of a polygon, such that every viewer can see a unique colour. The goal is to minimize the number of colours used. This is also known as the conflict-free chromatic guarding problem with vertex guards (which is quite different from point guards considered in some other papers). We study the problem in two scenarios of a set of viewers. In the first one , we assume that the viewers are all points of the polygon. We efficiently solve the related problem of minimizing the number of guards and approximate (up to only an additive error) the number of colours in the special case of funnels. We also give an upper bound of O(log ^2 n) colours on weak-visibility polygons which generalizes to all simple polygons. In the second scenario, we assume that the viewers are only the vertices of the polygon. We show a lower bound of 3 colours in the general case of simple polygons and conjecture that this is tight. We also prove that already deciding whether 1 or 2 colours are enough is NP-complete.
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some
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scenario
one
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efficiently
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