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    <p>There are multiple problems here. They aren't specifically related to Qt or to Python, but to general computer science.</p> <p>You have floating point geometrical shapes that you want to display on a raster device, so somehow there has to be a floating point to integer conversion. It's not in your code, so it will happen at a lower level: in the graphics library, the display driver or whatever. Since you're not happy with the result, you have to handle this conversion yourself.</p> <p>There's no right or wrong way to do this. For example, take your case of a hex tile that has a “radius” of 50. The hexagon is oriented so that the W vertex is at (-50,0) and the E vertex is at (50,0). Now the NE vertex of this hexagon is at approximately (25,0,43.3). The hexagon that's adjacent to this one in the N direction has its center at about y=86.6 and its top edge at 129.9. How would you like to pixellate this? If you round 43.3 down to 43, now you no longer have a mathematically exact regular hexagon. If you round 129.9 up to 130, your first hexagon is 86 pixels in total height but the one on top of it is 87. This is an issue that you must resolve based on the project's requirements.</p> <p>And this is just one case (radius=50). If you allow the radius to be variable, can you come up with an algorithm to handle all cases? I couldn't. I think you need to use a fixed screen dimension for your hexagons, or at least reduce the possibilities to a small number.</p> <p>Nowhere in your code do you determine the size of the display window, so I don't understand how you intend to handle scaling issues, or determine how many hexes are needed to show the full map. </p> <p>As to your first question, I am certain that the performance will be poor. The constructor for QRegularPolygon is inside the loop that creates the hexes, so it gets called many times (800 in your example). It does two trig calculations for each vertex, so you perform 9600 trig calculations as you build your list of hexes. You don't need ANY of them. The calculations are the sine and cosine of 0 degrees, 60 degrees, 120 degrees and so on. Those are so easy you don't even need sin and cos. </p> <p>The use of the trig functions exacerbates the floating point/integer problem, too. Look at this:</p> <pre><code>&gt;&gt; int(50.0*math.sin(math.pi/6)) 24 </code></pre> <p>We know it should be 25, but the computer figures it as int(24.999999999996) – I may have left out a few 9's.</p> <p>If you calculate the vertex positions of just one hexagon, you can get all the other ones by a simple translation. See the useful Qt functions QPolygon->translate or QPolygon->translated.</p> <p>It seems that you don't need a constructor that can handle any type of polygon when your design concept absolutely needs hexagons. Did you just copy it from somewhere? I think it's mostly clutter, which always opens the door to errors. </p>
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