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Equation 6 · Part 13 · A History of Small and On-Device AI

Denominator: D_K × D_K × M × N × D_F × D_F

DK⋅DK⋅M⋅DF⋅DF+M⋅N⋅DF⋅DFDK⋅DK⋅M⋅N⋅DF⋅DF=1N+1DK2\frac{D_K \cdot D_K \cdot M \cdot D_F \cdot D_F + M \cdot N \cdot D_F \cdot D_F}{D_K \cdot D_K \cdot M \cdot N \cdot D_F \cdot D_F} = \frac{1}{N} + \frac{1}{D_K^2}
DK⋅DK⋅M⋅N⋅DF⋅DFD_K \cdot D_K \cdot M \cdot N \cdot D_F \cdot D_F

What this part means

The complete quantity below the fraction bar; it must be nonzero for this division.

Its job in the formula

DKD_K × DKD_K × M × N × DFD_F × DFD_F occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

The passage around this formula

…on its own, followed by a 1×1 pointwise convolution that recombines the results [ 1 ] . For a DKD_K ×\times DKD_K kernel, M input channels, N output channels and a DFD_F ×\times DFD_F feature map, a standard convolution costs DKD_K ⋅\cdot DKD_K ⋅\cdot M ⋅\cdot N ⋅\cdot DFD_F ⋅\cdot DFD_F multiply-adds. The paper gives the depthwise separable replacement’s cost, and its ratio to the standard cost, directly: DK⋅DK⋅M⋅DF⋅DF+M⋅N⋅DF⋅DFDK⋅DK⋅M⋅N⋅DF⋅DF=1N+1DK2\frac{D_K \cdot D_K \cdot M \cdot D_F \cdot D_F + M \cdot N \cdot D_F \cdot D_F}{D_K \cdot D_K \cdot M \cdot N \cdot D_F \cdot D_F} = \frac{1}{N} + \frac{1}{D_K^2}. With the near-universal 3×\times3 kernel, that ratio works out to roughly an eighth to a ninth of the…

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Learn the underlying idea

A fraction a/b means a divided by b. The top number is the numerator; the bottom number is the denominator, and it cannot be zero.

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Sources cited in the surrounding passage

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