How is the smooth dice loss differentiable?

问题

I am training a U-Net in keras by minimizing the dice_loss function that is popularly used for this problem: adapted from here and here

def dsc(y_true, y_pred):
     smooth = 1.
     y_true_f = K.flatten(y_true)
     y_pred_f = K.flatten(y_pred)
     intersection = K.sum(y_true_f * y_pred_f)
     score = (2. * intersection + smooth) / (K.sum(y_true_f) + K.sum(y_pred_f) + smooth)
     return score

def dice_loss(y_true, y_pred):
    return (1 - dsc(y_true, y_pred))

This implementation is different from the traditional dice loss because it has a smoothing term to make it "differentiable". I just don't understand how adding the smooth term instead of something like 1e-7 in the denominator makes it better because it actually changes the loss values. I have checked this by using a trained unet model on a test set with a regular dice implementation as follows:

def dice(im1,im2):
     im1 = np.asarray(im1).astype(np.bool)
     im2 = np.asarray(im2).astype(np.bool)
     intersection = np.logical_and(im1, im2)
     return np.float(2. * intersection.sum()) / (im1.sum() + im2.sum() + 1e-7))

Can someone explain why the smooth dice loss is conventionally used?

回答1:

Adding smooth to the loss does not make it differentiable. What makes it differentiable is
1. Relaxing the threshold on the prediction: You do not cast y_pred to np.bool, but leave it as a continuous value between 0 and 1
2. You do not use set operations as np.logical_and, but rather use element-wise product to approximate the non-differenetiable intersection operation.

You only add smooth to avoid devision by zero when both y_pred and y_true do not contain any foreground pixels.

来源：https://stackoverflow.com/questions/51973856/how-is-the-smooth-dice-loss-differentiable