What's the input shape in AC loss? about active-contour-loss HOT 6 CLOSED

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Hi @xuuuuuuchen ,

Thanks for your quick reply.

In eq. (11), length term requires to compute the derivative of u. However, the derivative has been computed by forward difference here.

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The purpose for line 17 & 18 line is to compute within (a-1, a-1) shape, because after line 14 & 15, the shape of x & y are like (a-1, a) and (a, a-1),

Hope it answers your question.

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Hi @xuuuuuuchen ,
Thanks for your reply.
I understand, but I think the shape of delta_x/y is (a-2, a-2) in your code. For ACDC, it is (254, 254).

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@JunMa11 Yes, it is (a-2, a-2) not (a-1, a-1) in this code. I'm trying to make sure the shape of delta_x/delta_y is always square as well as contain meaningful info., but you could try on (a-1, a-1). I think there's no big difference if the image size is large enough.

Hope it answers your question.

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Got it. Yes, a-1 or a-2 is not critical.
Thanks for your reply very much.

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