Non-Maximum Suppression

Remove duplicate detections of the same object.

A detector fires many times around each object. NMS keeps the most confident box and discards anything that overlaps it too much:

1. Sort the boxes by score, highest first. Break ties towards the lower index.
2. Take the highest scoring box that is still available and keep it.
3. Discard every remaining box whose IoU with it is greater than iou_threshold.
4. Repeat until nothing is left.

Return the kept indices as a list, in the order they were selected, which is descending by score.

Boxes are (n, 4) as [x1, y1, x2, y2] and scores are (n,).

Note the strict comparison: a box is dropped when IoU is strictly greater than the threshold, so at exactly the threshold it survives. That is the convention torchvision uses, and it is the kind of off-by-one that silently changes a detector's precision.

Suppression is not transitive. A box only ever competes with a box that was actually kept, never with one that was already suppressed. Comparing against suppressed boxes would remove far too much.

Function signature

def non_max_suppression(boxes, scores, iou_threshold):

Example

Input

boxes  = np.array([[0, 0, 10, 10], [1, 1, 11, 11], [20, 20, 30, 30]])
scores = np.array([0.9, 0.8, 0.7])
non_max_suppression(boxes, scores, 0.5)

Output

[0, 2]

Reasoning

Box 0 has the top score and is kept. Box 1 overlaps it with IoU 81 / 119 = 0.68, which is above 0.5, so it is dropped. Box 2 does not overlap box 0, so it is kept.

Constraints and notes

• Drop a box only when IoU is strictly greater than the threshold.
• Compare only against boxes that were kept.
• Return the indices in the order they were kept.