问这一叠CNN层有效吗？

EN

sεÐgiĀ«]vyн³Dp-Nè©*-yθ/®+ÐïÊyнÈ®_*y`‹«à~i0q]1

s                   # Swap to get the first two (implicit) inputs onto the stack,
                    # with the second input at the top
 ε                  # Map over each layer:
  Ð                 #  Triplicate the layer
   gi               #  If it's length is 1 (thus a convolutional layer):
     Ā              #   Truthify both integers, so we have a pair of 1s: [1,1]
      «             #   Merge it to the layer
 ]                  # Close the if-statement and map
  v                 # Loop over each layer `y`, consisting of two pairs [kernel,stride]:
   yн               #  Get the first pair (the kernel)
     ³              #  Push the third input-list of modes
      Dp-           #  Transform the 2s into 1s (by checking for prime, and subtracting)
         Nè         #  Get the mode at the current loop-index
           ©        #  Store it in variable `®` (without popping)
            *       #  Multiply this mode to the kernel-pair
             -      #  Subtract each from the dimensions-pair
              yθ    #  Get the last pair (the stride)
                /   #  Divide the dimension-pair by the stride-pair
                 ®+ #  And add the modified mode `®` to each
   Ð                #  Triplicate the modified dimensions-pair
    ï               #  Cast the values in the top copy to integers
     Ê              #  Check if the top two pairs are NOT equal
                    #  (1 if the dimension-pair contains decimal values; 0 if integers)
    yн              #  Push the kernel again
      È             #  Check for both values if they're even (1 if even; 0 if odd)
       ®_           #  Check if `®` is 0 (1 if 0; 0 if not)
         *          #  Multiply the checks
    y`              #  Push the kernel-pair and stride-pair separated to the stack
      ‹             #  Check if [kernel-row < stride-row, kernel-column < stride-column]
    «               #  Merge the pairs of checks together
     à              #  Check of any are truthy of this quartet by taking the maximum
    ~               #  Check if either is truthy by taking the bitwise-OR
     i              #  If this is truthy:
      0             #   Push a 0
       q            #   And stop the program
                    #   (after which this 0 is output implicitly as result)
 ]                  # Close the if-statement and loop
  1                 # And push a 1
                    # (which will be output implicitly if we didn't encountered the `q`#qcStackCode#)

_ e.&>(1(+_*[><.)@+(-{.)%(]*>:)/@])`((+_*1>])@+}.(]-~*+_*(2|[)+:|@]){.)@.(]3=#)~&.>/

(…)`(…)@.(]3=#)~&.>/

((+_*1>])@+}.(]-~*+_*(2|[)+:|@]){.)
           }.(                 ){.  split into 3 3 and _1 as arguments
                            |@]     mode != 0?
                      2|[           3 3 even?
                          +:        not-or, so 1 iff mode = 0 and dimension even
                   _*               if this^ returns 1, convert it to infinity
                 *+                 add to this dim * mode (_3 _3)
              ]-~                   subtract the mode (_2 _2)
           +                        add to the image dimension (23 23)
  (+_*1>])                          if the dimensions are less than 1, add infinity

(1(+_*[><.)@+(-{.)%(]*>:)/@])
                   (]*>:)/@]  multiple stride with (window greater/equal stride?)
             (-{.)%           (image - window)% mstride, is infinity iff mstride is 0
 1          +                 add one
  (+_*[><.)                   add infinity if flooring a dimensions changes it

_ e.&>       unbox and check if at least one dimension is infinity