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Figure 2.3: (Half padding, unit strides) Convolving a 3 x 3 kernel over a 5 x 5 input using half padding and unit strides (i.e., i= 5, k = 3,5 =1 and p=1).

Figure 2 3: (Half padding, unit strides) Convolving a 3 x 3 kernel over a 5 x 5 input using half padding and unit strides (i.e., i= 5, k = 3,5 =1 and p=1).

Related Figures (21)

Figure 1.1: Computing the output values of a discrete convolution.
Figure 1.1: Computing the output values of a discrete convolution.
Figure 1.2: Computing the output values of a discrete convolution for N = 2, 14 19 5, ky ko 3, S1 S92 2. and Pi = p2 = 1.
Figure 1.2: Computing the output values of a discrete convolution for N = 2, 14 19 5, ky ko 3, S1 S92 2. and Pi = p2 = 1.
Figure 1.3: A convolution mapping from two input feature maps to three output feature maps using a 3 x 2 x 3 x 3 collection of kernels w. In the left pathway, input feature map 1 is convolved with kernel w,,; and input feature map 2 is convolved with kernel wj,2, and the results are summed together elementwise to form the first output feature map. The same is repeated for the middle and right pathways to form the second and third feature maps, and all three output feature maps are grouped together to form the output.
Figure 1.3: A convolution mapping from two input feature maps to three output feature maps using a 3 x 2 x 3 x 3 collection of kernels w. In the left pathway, input feature map 1 is convolved with kernel w,,; and input feature map 2 is convolved with kernel wj,2, and the results are summed together elementwise to form the first output feature map. The same is repeated for the middle and right pathways to form the second and third feature maps, and all three output feature maps are grouped together to form the output.
Figure 1.4: An alternative way of viewing strides. Instead of translating the 3 x 3 kernel by increments of s = 2 (left), the kernel is translated by increments of 1 and only one in s = 2 output elements is retained (right).
Figure 1.4: An alternative way of viewing strides. Instead of translating the 3 x 3 kernel by increments of s = 2 (left), the kernel is translated by increments of 1 and only one in s = 2 output elements is retained (right).
Figure 1.5: Computing the output values of a 3 x 3 average pooling operation ona 5 xX 5 input using 1 x 1 strides.
Figure 1.5: Computing the output values of a 3 x 3 average pooling operation ona 5 xX 5 input using 1 x 1 strides.
Figure 1.6: Computing the output values of a 3 x 3 max pooling operation on a5 x 5 input using 1 x 1 strides.
Figure 1.6: Computing the output values of a 3 x 3 max pooling operation on a5 x 5 input using 1 x 1 strides.
Figure 2.1: (No padding, unit strides) Convolving a 3 x 3 kernel over a 4 x 4 input using unit strides (ie., 2 = 4, k = 3, s=1 and p=0).
Figure 2.1: (No padding, unit strides) Convolving a 3 x 3 kernel over a 4 x 4 input using unit strides (ie., 2 = 4, k = 3, s=1 and p=0).
Figure 2.2: (Arbitrary padding, unit strides) Convolving a 4 x 4 kernel over a 5 x 5 input padded with a 2 x 2 border of zeros using unit strides (i.e., 7 = 5, k=4,s=1 and p=2).
Figure 2.2: (Arbitrary padding, unit strides) Convolving a 4 x 4 kernel over a 5 x 5 input padded with a 2 x 2 border of zeros using unit strides (i.e., 7 = 5, k=4,s=1 and p=2).
Figure 2.4: (Full padding, unit strides) Convolving a 3 x 3 kernel over a 5 x 5 input using full padding and unit strides (i.e., i = 5, k = 3, s = 1 and p = 2).
Figure 2.4: (Full padding, unit strides) Convolving a 3 x 3 kernel over a 5 x 5 input using full padding and unit strides (i.e., i = 5, k = 3, s = 1 and p = 2).
Figure 2.5: (No zero padding, arbitrary strides) Convolving a 3 x 3 kernel over a5 x 5 input using 2 x 2 strides (i.e.,i=5,k =3,s=2 and p=0).
Figure 2.5: (No zero padding, arbitrary strides) Convolving a 3 x 3 kernel over a5 x 5 input using 2 x 2 strides (i.e.,i=5,k =3,s=2 and p=0).
Figure 2.6: (Arbitrary padding and strides) Convolving a 3 x 3 kernel over a 5 x 5 input padded with a 1 x 1 border of zeros using 2 x 2 strides (i.e., i = 5, k =3,s=2 and p=1).
Figure 2.6: (Arbitrary padding and strides) Convolving a 3 x 3 kernel over a 5 x 5 input padded with a 1 x 1 border of zeros using 2 x 2 strides (i.e., i = 5, k =3,s=2 and p=1).
Figure 2.7: (Arbitrary padding and strides) Convolving a 3 x 3 kernel over a 6 x 6 input padded with a 1 x 1 border of zeros using 2 x 2 strides (i.e., i = 6, k = 3, s =2 and p=1). In this case, the bottom row and right column of the zero padded input are not covered by the kernel.
Figure 2.7: (Arbitrary padding and strides) Convolving a 3 x 3 kernel over a 6 x 6 input padded with a 1 x 1 border of zeros using 2 x 2 strides (i.e., i = 6, k = 3, s =2 and p=1). In this case, the bottom row and right column of the zero padded input are not covered by the kernel.
Figure 4.1: The transpose of convolving a 3 x 3 kernel over a 4 x 4 input using unit strides (i.e., i = 4, k = 3, s = 1 and p= 0). It is equivalent to convolving a3 x3 kernel over a 2 x 2 input padded with a 2 x 2 border of zeros using unit strides (i.e., 7’ = 2, k’ =k, s’ =1 and p’ = 2).
Figure 4.1: The transpose of convolving a 3 x 3 kernel over a 4 x 4 input using unit strides (i.e., i = 4, k = 3, s = 1 and p= 0). It is equivalent to convolving a3 x3 kernel over a 2 x 2 input padded with a 2 x 2 border of zeros using unit strides (i.e., 7’ = 2, k’ =k, s’ =1 and p’ = 2).
Figure 4.2: The transpose of convolving a 4 x 4 kernel over a 5 x 5 input padded with a 2 x 2 border of zeros using unit strides (i.e., i = 5, k = 4, s = 1 and p = 2). It is equivalent to convolving a 4 x 4 kernel over a 6 x 6 input padded with a 1 x 1 border of zeros using unit strides (ie., i’ = 6, k’ = k, s’ = 1 and p' =1).
Figure 4.2: The transpose of convolving a 4 x 4 kernel over a 5 x 5 input padded with a 2 x 2 border of zeros using unit strides (i.e., i = 5, k = 4, s = 1 and p = 2). It is equivalent to convolving a 4 x 4 kernel over a 6 x 6 input padded with a 1 x 1 border of zeros using unit strides (ie., i’ = 6, k’ = k, s’ = 1 and p' =1).
Figure 4.3: The transpose of convolving a 3 x 3 kernel over a 5 x 5 input using half padding and unit strides (i.e., 1 = 5, k = 3, s 1 and p = 1). It is equivalent to convolving a 3 x 3 kernel over a 5 x 5 input using half padding and unit strides (ie., 7’ = 5, k’ =k, s’ = 1 and p’ = 1).
Figure 4.3: The transpose of convolving a 3 x 3 kernel over a 5 x 5 input using half padding and unit strides (i.e., 1 = 5, k = 3, s 1 and p = 1). It is equivalent to convolving a 3 x 3 kernel over a 5 x 5 input using half padding and unit strides (ie., 7’ = 5, k’ =k, s’ = 1 and p’ = 1).
Figure 4.4: The transpose of convolving a 3 x 3 kernel over a 5 x 5 input using full padding and unit strides (i.e., 1 = 5, k = 3, s = 1 and p = 2). It is equivalent to convolving a 3 x 3 kernel over a 7 x 7 input using unit strides (i-e., 2’ = 7, k’ =k, s’ =1 and p’ =0).
Figure 4.4: The transpose of convolving a 3 x 3 kernel over a 5 x 5 input using full padding and unit strides (i.e., 1 = 5, k = 3, s = 1 and p = 2). It is equivalent to convolving a 3 x 3 kernel over a 7 x 7 input using unit strides (i-e., 2’ = 7, k’ =k, s’ =1 and p’ =0).
Figure 4.5: The transpose of convolving a 3 x 3 kernel over a 5 x 5 input using 2 x 2 strides (i.e., i= 5, k = 3, s = 2 and p= 0). It is equivalent to convolving a 3 x 3 kernel over a 2 x 2 input (with 1 zero inserted between inputs) padded with a 2 x 2 border of zeros using unit strides (ie., i’ = 2, i = 3, k’! =k, s’ =1 and p’ = 2).
Figure 4.5: The transpose of convolving a 3 x 3 kernel over a 5 x 5 input using 2 x 2 strides (i.e., i= 5, k = 3, s = 2 and p= 0). It is equivalent to convolving a 3 x 3 kernel over a 2 x 2 input (with 1 zero inserted between inputs) padded with a 2 x 2 border of zeros using unit strides (ie., i’ = 2, i = 3, k’! =k, s’ =1 and p’ = 2).
Figure 4.6: The transpose of convolving a 3 x 3 kernel over a 5 x 5 input padded with a 1 x 1 border of zeros using 2 x 2 strides (i.e., i = 5, k = 3, s = 2 and p = 1). It is equivalent to convolving a 3 x 3 kernel over a 3 x 3 input (with 1 zero inserted between inputs) padded with a 1 x 1 border of zeros using unit strides (i.e., i’ = 3, 7’ = 5, k’ =k, s’ =1 and p' = 1).
Figure 4.6: The transpose of convolving a 3 x 3 kernel over a 5 x 5 input padded with a 1 x 1 border of zeros using 2 x 2 strides (i.e., i = 5, k = 3, s = 2 and p = 1). It is equivalent to convolving a 3 x 3 kernel over a 3 x 3 input (with 1 zero inserted between inputs) padded with a 1 x 1 border of zeros using unit strides (i.e., i’ = 3, 7’ = 5, k’ =k, s’ =1 and p' = 1).
Figure 5.1: (Dilated convolution) Convolving a 3 x 3 kernel over a 7 x 7 input with a dilation factor of 2 (ie.,i=7,k =3,d=2,5=1 and p=0).
Figure 5.1: (Dilated convolution) Convolving a 3 x 3 kernel over a 7 x 7 input with a dilation factor of 2 (ie.,i=7,k =3,d=2,5=1 and p=0).
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