IOU Calculations

 1. check https://www.programcreek.com/python/?CodeExample=get+iou

Example 51 : properly explains the concept, only include  a small number in denominator 1e-8 to avoid divide by zero 

iou = intersection / (area1 + area2 - intersection + 1e-8)

Example 54 : provides a good implementation using tf

2. lu and rd refer to left up and right down points. 

    lu = tf.maximum(boxes1_corners[:, None, :2], boxes2_corners[:, :2])
    rd = tf.minimum(boxes1_corners[:, None, 2:], boxes2_corners[:, 2:])
    intersection = tf.maximum(0.0, rd - lu)

3. The following is an implementation from https://github.com/srihari-humbarwadi/YOLOv1-TensorFlow2.0/blob/master/yolo_v1.py

def compute_iou(boxes1, boxes2):
    boxes1_t = tf.stack([boxes1[..., 0] - boxes1[..., 2] / 2.0,
                         boxes1[..., 1] - boxes1[..., 3] / 2.0,
                         boxes1[..., 0] + boxes1[..., 2] / 2.0,
                         boxes1[..., 1] + boxes1[..., 3] / 2.0],
                        axis=-1)

    boxes2_t = tf.stack([boxes2[..., 0] - boxes2[..., 2] / 2.0,
                         boxes2[..., 1] - boxes2[..., 3] / 2.0,
                         boxes2[..., 0] + boxes2[..., 2] / 2.0,
                         boxes2[..., 1] + boxes2[..., 3] / 2.0],
                        axis=-1)
    lu = tf.maximum(boxes1_t[..., :2], boxes2_t[..., :2])
    rd = tf.minimum(boxes1_t[..., 2:], boxes2_t[..., 2:])

    intersection = tf.maximum(0.0, rd - lu)
    inter_square = intersection[..., 0] * intersection[..., 1]

    square1 = boxes1[..., 2] * boxes1[..., 3]
    square2 = boxes2[..., 2] * boxes2[..., 3]

    union_square = tf.maximum(square1 + square2 - inter_square, 1e-10)
    return tf.clip_by_value(inter_square / union_square, 0.0, 1.0)