# How is the smooth dice loss differentiable?

## Issue

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?

## Solution

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 the element-wise product to approximate the non-differenetiable intersection operation.

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