Md. Asif Uddin
    I.11.X04

    Update the receptive field before the jump

    numeric▲▲△

    Track receptive-field span and sampling jump through three width-three layers with strides (1,2,1)(1,2,1) and dilation one. All use symmetric padding one. What are the first-output centre and final jump if original pixel centres start at 1/21/2?

    Hint

    Use the preceding layer’s jump when growing the span.

    Solution

    Starting from (R0,J0)=(1,1)(R_0,J_0)=(1,1), (I.11.6) gives

    (R1,J1)=(3,1),(R2,J2)=(5,2),(R3,J3)=(9,2).\begin{aligned} (R_1,J_1)&=(3,1),\\ (R_2,J_2)&=(5,2),\\ (R_3,J_3)&=(9,2). \end{aligned}

    Every layer’s centre increment is ((3−1)/2−1)Jprevious=0((3-1)/2-1)J_{\mathrm{previous}}=0. The first centre remains 1/21/2; neighbouring final outputs are two original pixels apart. At the boundary, part of the nominal nine-position span is padding.

    Updating the second jump before computing its span would give seven instead of five at layer two. The stride changes spacing between outputs, not the taps inside that layer’s first window.

    Draws on