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    <p>There will not be an unique solution for any particular setup, but you can easily find one of the solutions with this algorithm: </p> <ol> <li>Find a rectangle within A that is above rectangle B. If the top of A is higher than B (i.e. has a lower value in px), there is such an rectangle. This rectangle is defined by: (left edge of A, top edge of A) to (right edge of A, top edge of B). </li> <li>If the left edge of B is to the right of the left edge of A, the next rectangle is: (left edge of A, min(top edge of A, top edge of B)) to (left edge of B, max (bottom edge of A, bottom edge of B)) </li> <li>If the right edge of B is to the left of the right edge of B, similar to above</li> <li>...and the possible rectangle below B</li> </ol> <p>In total, you will get from 0 to 4 rectangles. </p> <p>Pseudocode with a somewhat unusual, but for this purpose clear, definition of rectangle: </p> <pre><code>function getClipped(A, B) { var rectangles = []; if (A.top &lt; B.top) { rectangles.push({ left: A.left, top: A.top, right: A.right, bottom: B.top }); } if (A.left &lt; B.left) { rectangles.push({ left: A.left, top: max(A.top, B.top), right: B.left, bottom: min(A.bottom, B.bottom) }); } if (A.right &gt; B.right) { rectangles.push({ left: B.right, top: max(A.top, B.top), right: A.right, bottom: min(A.bottom, B.bottom) }); } if (A.bottom &gt; B.bottom) { rectangles.push({ left: A.left, top: B.bottom, right: A.right, bottom. A.bottom }); } return rectangles; } var rectA = { left: nn, top: nn, right: nn, bottom: nn}; var rectB = { left: nn, top: nn, right: nn, bottom: nn}; var clipped = getClipped(rectA, rectB) ; </code></pre>
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    1. COThere are some cases not covered - e.g. if B is wider than A and cuts A in two...
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    2. CO@boisvert, it is covered. All your nine regions are fully covered. Not mathematically elegant like your explanation, but with a very simple algorithm.
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    3. COOops... I didn't pay attention to your use of min and max. You're right, it works very well. Don't see anything inelegant about it either.
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