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MaxPool2d backward pass how to? #1
Description
I've implemented the forward pass of the max pooling, but I wasn't able to use the index to perform the backward pass, your version, uses the img_to_row tricks, but I want to implement this without too much indexing and looping. Any idea how you can do this using this implementation?
import numpy as np
as_strided = as_strided
def pool2d(Z, K:tuple=(2,2)):
""" performs the windowing, and padding if needed"""
KH, KW = K # Kernel Height & Width
N, C, ZH, ZW = Z.shape # Input: NCHW Batch, Channels, Height, Width
Ns, Cs, Hs, Ws = Z.strides
EdgeH, EdgeW = ZH%KH, ZW%KW # How many pixels left on the edge
if (EdgeH!=0 or EdgeW!=0): # If there are pixels left we need to pad
PadH, PadW = KH-EdgeH, KW-EdgeW
PadTop, PadBottom = ceil(PadH/2), floor(PadH/2)
PadLeft, PadRight = ceil(PadW/2), floor(PadW/2)
Z = np.pad(Z, ((0,0),(0,0), (PadTop, PadBottom), (PadLeft, PadRight)))
N, C, ZH, ZW = Z.shape #NCHW
Ns, Cs, Hs, Ws = Z.strides
Zstrided = np.lib.stride_tricks.as_strided(Z, shape=(N,C,ZH//KH, ZW//KW, KH, KW), strides=(Ns, Cs, Hs*KH, Ws*KW,Hs, Ws))
return Zstrided.reshape(N,C,ZH//KH, ZW//KW, KH*KW) # reshape to flatten pooling windows to be 1D-vector
def maxpool2d(Z, K:tuple=(2,2)):
ZP = pool2d(Z, K)
MxP = np.max(ZP, axis=(-1))
Inx = np.argmax(ZP, axis=-1)
return MxP, Inx
def unmaxpool2d(ZP, Indx, K:tuple=(2,2)):
ZN,ZC,ZH,ZW = ZP.shape
KH, KW = K
Z = np.zeros((ZN,ZC,ZH*KH,ZW*KW))
_ZP = pool2d(Z, K)
# ... Where to go next
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