他们如何计算Caffe中这个convnet示例的输出量? [英] How did they calculate the output volume for this convnet example in Caffe?

问题描述

在此教程中，输出量在输出[25]中说明，接受区域在输出[26]中指定.

In this tutorial, the output volumes are stated in output [25], and the receptive fields are specified in output [26].

好吧，输入体积[3, 227, 227]与大小为[3, 11, 11]的区域卷积.

Okay, the input volume [3, 227, 227] gets convolved with the region of size [3, 11, 11].

使用此公式 (W−F+2P)/S+1 ，其中:
W =输入音量大小
F =接收域大小
P =填充
S =步幅

Using this formula (W−F+2P)/S+1, where:
W = the input volume size
F = the receptive field size
P = padding
S = stride

...结果为(227 - 11)/4 + 1 = 55，即 [55 * 55 * 96] .到目前为止一切都很好:)

...results with (227 - 11)/4 + 1 = 55 i.e. [55*55*96]. So far so good :)

对于'pool1'，他们使用了F=3和S=2我认为吗?计算将检出:55-3/2+1=27.

For 'pool1' they used F=3and S=2 I think? The calculation checks out: 55-3/2+1=27.

从这一点上我有点困惑.第二个convnet层的接收字段为[48, 5, 5]，但是'conv2'的输出等于[256, 27, 27].这里发生了什么计算?

From this point I get a bit confused. The receptive field for the second convnet layer is [48, 5, 5], yet the output for 'conv2' is equal to [256, 27, 27]. What calculation happened here?

然后，'conv3'到'conv4'的输出量的高度和宽度都相同[13, 13]?这是怎么回事?

And then, the height and width of the output volumes of 'conv3' to 'conv4' are all the same [13, 13]? What's going on?

谢谢！

推荐答案

如果您仔细观察

If you look closely at the parameters of conv2 layer you'll notice

   pad: 2

也就是说，输入Blob周围被额外的2个像素填充，因此公式为

That is, the input blob is padded by 2 extra pixels all around, thus the formula now is

27 + 2 + 2 - ( 5 - 1 ) = 27

在两侧都填充5像素的5内核大小会产生相同的输出大小.

Padding a kernel size of 5 with 2 pixels from both sides yields the same output size.

这篇关于他们如何计算Caffe中这个convnet示例的输出量?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！