Non-Max Suppression

Run a real detector over a photo of one dog and look at its raw output before any cleanup. You will not see one box. You will typically see somewhere between five and twenty, clustered tightly around the same animal, each one a slightly different size and position, each one confident that it has found a dog.

This is not a bug in the detector. Every location near the real dog does contain most of a dog, so the network is not wrong to flag it — it is only wrong to flag it a dozen times over.

Non-max suppression’s first move is unconditional: sort every proposed box for the whole image by its confidence score and keep the single highest-scoring one outright, no comparison needed yet.

That box becomes the first entry in the final output. Everything that happens next only decides what else, if anything, gets added to it.

Walk through every remaining box in order of confidence, highest to lowest. For each one, compute its IoU — from the last lesson — against every box you have already kept. If that overlap clears a chosen threshold, discard the new box. If it does not, keep it.

The logic is that much overlap is not two dogs standing precisely on top of each other. It is one dog, described twice. Suppressing the extra box loses nothing; keeping it would double-count the same animal in the final result.

In practice this runs separately for each class. A box labelled “dog” never suppresses a box labelled “person”, however much they overlap — someone holding a puppy is two objects, not one described twice.

The threshold above fails in two directions, and the interactive shows both. Set it too low and boxes get suppressed on the slightest overlap — including a second, separate object sitting near the first one, which gets wrongly merged into a single detection.

Set it too high and almost nothing gets suppressed — a dozen boxes around one dog survive, because none of them overlap each other by quite enough to trip the threshold, and your final output still reports several dogs where there is one. There is no threshold that is safe for every photo; it is tuned against a validation set that looks like the scenes you expect. Crowded scenes, where real objects genuinely overlap, usually want a higher threshold than sparse ones.

The six boxes from the interactive, run through the algorithm at three thresholds. All of them are labelled dog, so every box is compared with every kept box.

def iou(a, b):    ix = max(0, min(a[2], b[2]) - max(a[0], b[0]))    iy = max(0, min(a[3], b[3]) - max(a[1], b[1]))    inter = ix * iy    union = (a[2] - a[0]) * (a[3] - a[1]) + (b[2] - b[0]) * (b[3] - b[1]) - inter    return inter / union # (left, top, right, bottom, confidence), all labelled "dog".# Five boxes crowd one dog; box f sits on a second dog nearby.boxes = {    "a": (60, 60, 200, 160, 0.95), "b": (68, 58, 208, 158, 0.91),    "c": (55, 66, 195, 166, 0.88), "d": (72, 70, 212, 170, 0.82),    "e": (50, 50, 190, 150, 0.77), "f": (130, 100, 270, 200, 0.85),} def nms(boxes, threshold):    kept = []    for name in sorted(boxes, key=lambda n: boxes[n][4], reverse=True):        if all(iou(boxes[name], boxes[k]) <= threshold for k in kept):            kept.append(name)    return kept for threshold in [0.1, 0.5, 0.9]:   # try your own    print(f"threshold {threshold}: kept {nms(boxes, threshold)}")

At 0.5 it keeps a and f, one box per dog. At 0.1 the second dog’s box f is swallowed by a, and at 0.9 all six survive.