I am trying to understand the Yolo v2 loss function but I don't understand this function, if any person can detail the function, please help me.

\begin{align} &\lambda_{coord} \sum_{i=0}^{S^2}\sum_{j=0}^B \mathbb{1}_{ij}^{obj}[(x_i-\hat{x}_i)^2 + (y_i-\hat{y}_i)^2 ] \\&+ \lambda_{coord} \sum_{i=0}^{S^2}\sum_{j=0}^B \mathbb{1}_{ij}^{obj}[(\sqrt{w_i}-\sqrt{\hat{w}_i})^2 +(\sqrt{h_i}-\sqrt{\hat{h}_i})^2 ]\\ &+ \sum_{i=0}^{S^2}\sum_{j=0}^B \mathbb{1}_{ij}^{obj}(C_i - \hat{C}_i)^2 + \lambda_{noobj}\sum_{i=0}^{S^2}\sum_{j=0}^B \mathbb{1}_{ij}^{noobj}(C_i - \hat{C}_i)^2 \\ &+ \sum_{i=0}^{S^2} \mathbb{1}_{i}^{obj}\sum_{c \in classes}(p_i(c) - \hat{p}_i(c))^2 \\ \end{align}

but I don't understand this function, if any person can detail the function, please help me.

  • 4
    nobody can help you without context...at least tell us what paper this is from. – bdeonovic Jun 27 '17 at 2:04
  • 1
    "I don't understand" and "detail the function" are overly broad. Please try to identify particular questions. Note that there are numerous questions relating to Yolo already, some of which may provide you with at least part of what you seek – Glen_b Jun 27 '17 at 5:23
  • 1
    easy: YOLO = you only live once – Aksakal Mar 19 at 1:24

Explanation of the different terms :

  • The 3 $\lambda$ constants are just constants to take into account more one aspect of the loss function. In the article $\lambda_{coord}$ is the highest in order to have the more importance in the first term
  • The prediction of YOLO is a $S*S*(B*5+C)$ vector : $B$ bbox predictions for each grid cells and $C$ class prediction for each grid cell (where $C$ is the number of classes). The 5 bbox outputs of the box j of cell i are coordinates of tte center of the bbox $x_{ij}$ $y_{ij}$ , height $h_{ij}$, width $w_{ij}$ and a confidence index $C_{ij}$
  • I imagine that the values with a hat are the real one read from the label and the one without hat are the predicted ones. So what is the real value from the label for the confidence score for each bbox $\hat{C}_{ij}$ ? It is the intersection over union of the predicted bounding box with the one from the label.
  • $\mathbb{1}_{i}^{obj}$ is $1$ when there is an object in cell $i$ and $0$ elsewhere
  • $\mathbb{1}_{ij}^{obj}$ "denotes that the $j$th bounding box predictor in cell $i$ is responsible for that prediction". In other words, it is equal to $1$ if there is an object in cell $i$ and confidence of the $j$th predictors of this cell is the highest among all the predictors of this cell. $\mathbb{1}_{ij}^{noobj}$ is almost the same except it values 1 when there are NO objects in cell $i$

Note that I used two indexes $i$ and $j$ for each bbox predictions, this is not the case in the article because there is always a factor $\mathbb{1}_{ij}^{obj}$ or $\mathbb{1}_{ij}^{noobj}$ so there is no ambigous interpretation : the $j$ chosen is the one corresponding to the highest confidence score in that cell.

More general explanation of each term of the sum :

  1. this term penalize bad localization of center of cells
  2. this term penalize the bounding box with inacurate height and width. The square root is present so that erors in small bounding boxes are more penalizing than errors in big bounding boxes.
  3. this term tries to make the confidence score equal to the IOU between the object and the prediction when there is one object
  4. Tries to make confidence score close to $0$ when there are no object in the cell
  5. This is a simple classification loss (not explained in the article)
  • Is the second point supposed to be B*(5+C)? Atleast that's the case for YOLO v3. – sachinruk Sep 17 at 6:55

\begin{align} &\lambda_{coord} \sum_{i=0}^{S^2}\sum_{j=0}^B \mathbb{1}_{ij}^{obj}[(x_i-\hat{x}_i)^2 + (y_i-\hat{y}_i)^2 ] \\&+ \lambda_{coord} \sum_{i=0}^{S^2}\sum_{j=0}^B \mathbb{1}_{ij}^{obj}[(\sqrt{w_i}-\sqrt{\hat{w}_i})^2 +(\sqrt{h_i}-\sqrt{\hat{h}_i})^2 ]\\ &+ \sum_{i=0}^{S^2}\sum_{j=0}^B \mathbb{1}_{ij}^{obj}(C_i - \hat{C}_i)^2 + \lambda_{noobj}\sum_{i=0}^{S^2}\sum_{j=0}^B \mathbb{1}_{ij}^{noobj}(C_i - \hat{C}_i)^2 \\ &+ \sum_{i=0}^{S^2} \mathbb{1}_{i}^{obj}\sum_{c \in classes}(p_i(c) - \hat{p}_i(c))^2 \\ \end{align}

Doesn't the YOLOv2 Loss function looks scary? It's not actually! It is one of the boldest, smartest loss function around.

Let's first look at what the network actually predicts.

If we recap, YOLOv2 predicts detections on a 13x13 feature map, so in total, we have 169 maps/cells.

We have 5 anchor boxes. For each anchor box we need Objectness-Confidence Score (whether any object was found?), 4 Coordinates ($t_x, t_y, t_w,$ and $t_h$) for the anchor box, and 20 top classes. This can crudely be seen as 20 coordinates, 5 confidence scores, and 100 class probabilities for all 5 anchor box predictions put together.

We have few things to worry about:

  • $x_i, y_i$, which is the location of the centroid of the anchor box
  • $w_i, h_i$, which is the width and height of the anchor box
  • $C_i$, which is the Objectness, i.e. confidence score of whether there is an object or not, and
  • $p_i(c)$, which is the classification loss.
  • We not only need to train the network to detect an object if there is an object in a cell, we also need to punish the network, it if predicts an object in a cell, when there wasn't any. How do we do this? We use a mask ($𝟙_{i}^{obj}$ and $𝟙_{i}^{noobj}$) for each cell. If originally there was an object $𝟙_{i}^{obj}$ is 1 and other no-object cells are 0. $𝟙_{i}^{noobj}$ is just inverse of $𝟙_{i}^{obj}$, where it is 1 if there was no object in the cell and 0 if there was.
  • We need to do this for all 169 cells, and
  • We need to do this 5 times (for each anchor box).

All losses are mean-squared errors, except classification loss, which uses cross-entropy function.

Now, let's break the code in the image.

  • We need to compute losses for each Anchor Box (5 in total)

    • $\sum_{j=0}^B$ represents this part, where B = 4 (5 - 1, since the index starts from 0)
  • We need to do this for each of the 13x13 cells where S = 12 (since we start index from 0)

    • $\sum_{i=0}^{S^2}$ represents this part.
  • $𝟙_{ij}^{obj}$ is 1 when there is an object in the cell $i$, else 0.

  • $𝟙_{ij}^{noobj}$ is 1 when there is no object in the cell $i$, else 0.
  • $𝟙_{i}^{obj}$ is 1 when there is a particular class is predicted, else 0.
  • λs are constants. λ is highest for coordinates in order to focus more on detection (remember, in YOLOv2, we first train it for recognition and then for detection, penalizing heavily for recognition is waste of time, rather we focus on getting best bounding boxes!)
  • We can also notice that $w_i, h_i$ are under square-root. This is done to penalize the smaller bounding boxes as we need better prediction on smaller objects than on bigger objects (author's call). Check out the table below and observe how the smaller values are punished more if we follow "square-root" method (look at the inflection point when we have 0.3 and 0.2 as the input values) (PS: I have kept the ratio of var1 and var2 same just for explanation):

        var1        | var2        | (var1 - var2)^2 | (sqrtvar1 - sqrtvar2)^2

        0.0300    | 0.020      | 9.99e-05          | 0.001

        0.0330    | 0.022      | 0.00012           | 0.0011

        0.0693    | 0.046      | 0.000533         | 0.00233

        0.2148    | 0.143      | 0.00512           | 0.00723

        0.3030    | 0.202      | 0.01                 | 0.01

        0.8808    | 0.587      | 0.0862             | 0.0296

        4.4920    | 2.994      | 2.2421             | 0.1512

Not that scary, right!

Read HERE for further details.

  • Should i and j in \sum start from 1 instead of 0? – webbertiger Mar 16 at 22:48
  • Yes, that's correct webertiger, have updated the answer accordingly. Thanks! – RShravan Mar 19 at 0:49
  • Isnt $\mathbb{1}_{ij}^{obj}$ 1 when there is an object in cell i of bounding box j? and not for all j? how do we choose which j to set to one and the rest to zero. i.e. what is the correct scale/ anchor where it is turned on. – sachinruk Sep 17 at 12:21
  • I believe S should still be 13 but if the summation starts in 0 it should end in $S^2 -1$ – Julian Sep 28 at 18:15
  • 1
    @RShravan, you say: "All losses are mean-squared errors, except classification loss, which uses cross-entropy function". Could you explain? In this equation, it looks as MSE also. Thanks in advance – Julian Oct 1 at 4:56

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.