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POFinding the distance between two hexagons
 

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  1. POFinding the distance between two hexagons
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    <p>I have two hexagons which I am trying to make snap together when the edges hit a certain tolerance.</p> <p>How can I find which edges are the closest?</p> <p><img src="https://i.stack.imgur.com/VObAh.png" alt="Hexagon 1, Side 1 is closest to Hexagon 3, Side 4"></p> <p><img src="https://i.stack.imgur.com/q1Prt.png" alt="Hexagon 1, Side 2 is closest to Hexagon 3, Side 5"></p> <p>Here is the code returning the two closest Hexagons:</p> <pre><code>Canvas.getClosestPiece = function(){ var current = {}; current.x = selection.MidPoint.X; current.y = selection.MidPoint.Y; smallestDistance = null; closestHex = null; hexagons.forEach(function(hexagon){ if(hexagon !== selection){ testPiece = {}; testPiece.x = hexagon.MidPoint.X; testPiece.y = hexagon.MidPoint.Y; if((lineDistance(current, testPiece) &lt; smallestDistance) || smallestDistance === null){ smallestDistance = lineDistance(current, testPiece) closestHex = hexagon hexagons.forEach(function(hexagon){ hexagon.lineColor = 'grey' }) hexagon.lineColor = 'red'; } } }) // console.log(smallestDistance) return [selection, closestHex] } </code></pre> <p>Distance between two hexagon midpoints:</p> <pre><code>function lineDistance( point1, point2 ){ var xs = 0; var ys = 0; xs = point2.x - point1.x; xs = xs * xs; ys = point2.y - point1.y; ys = ys * ys; return Math.sqrt( xs + ys ); } </code></pre> <p>And here is a standard point array for one of the hexagons that getClosestPiece returns:</p> <pre><code>Point {X: 658, Y: 284} Point {X: 704, Y: 304} Point {X: 704, Y: 354} Point {X: 658, Y: 375} Point {X: 613, Y: 354} Point {X: 613, Y: 304} </code></pre>
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    1. COThis sounds more like a math problem than a programming problem. Regardless, what have you done so far? You need to show code to get help.
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    2. COIt is a math problem, I have a bunch of code not related to the question that handles dragging and distance between two points. What would you like to see?
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    3. COIf 100% precise results are not required, you can just compute the distance between the hexagon's centers, thus approximating them as circles. As soon as (x1-x2)^2 + (y1-y2)^2 is under a threshold value, let them snap.
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