C4 W4 P1 (UNQ_C1) triplet_loss() - Calculation of Loss & Cost

Course Q&A Deep Learning Specialization Convolutional Neural Networks
1

Hello there!

I am working on the triplet_loss() function in C4 W4 P1 UNQ_C1, and I have a conceptual question on the following test case:

y_true = (None, None, None)
y_pred_perfect = (
    [1., 1.], #Anchor
    [1., 1.], #Positive
    [0., 0.,] #Negative
    )
loss = triplet_loss(y_true, y_pred_perfect, 3) # alpha = 3
assert loss == 1., "Wrong value. Check that pos_dist = 0 and neg_dist = 2 in this example"
  • We have a perfect match between y_pred_perfect[0] and y_pred_perfect[1]
  • We have a perfect non-match between y_pred_perfect[0] and y_pred_perfect[2]
  • We have two samples of each (m = 2)

Thus, when indeed pos_dist = 0 and neg_dist = 2, the result of loss is 0 instead of 1:

Question

basic_loss (loss) is -1.0 = 0.0 - 4.0 + 3 #diff of the squares plus alpha
loss (āˆ‘ of m losses) should be 1. an it is 0. #grader wants 1 but max(-1., 0.) = 0.

Why does the grader expects a loss of +1 (assert loss == 1) when the max of the loss in this test case is 0, is it not?

Thanks for any feedback!

2

Hi @Jonathanlugo,

The neg_dist is 2. The pos_dist is 0. The loss is max(pos_dist - neg_dist + alpha,0) = max(0-2+3,0) = max(1,0). Thus the loss is 1.

Hope this helps!

3

Thank you @thearkamitra,

but arenā€™t the squares missing in the formula? :thinking:

Never mind, in order for the formula to work 0 and 2 are already the squares of the differences. Thanks!

4

The distance can be anything you want; obviously some things distances should follow are:

  1. D(x,y)>=0
  2. D(x,y) = D(y,x)
  3. D(x,y)+ D(y,z)>= D(x,z).
    Where D is the dist function.
5

Got you.

Also, I actually fixed the issue for all cases once I understood the idea. I was squaring the difference AFTER the SUM, which was incorrect. Now I do it immediately after the subtraction. That worked.

But I think I got the concept :sunglasses:

2 Likes
6

Right! The point is that the loss is computed ā€œper sampleā€ and then summed. Thatā€™s why the ā€œorder of operationsā€ is that way. And of course the sum of the squares is not the same as the square of the sum.