3-input XOR using NAND gates only

A 3-input XOR gives $Y = A \oplus B \oplus C = (A \oplus B) \oplus C$. Each 2-input XOR needs 4 NAND gates, so cascade two 4-NAND XOR blocks: 8 NAND gates.

Block 1: $P = A \oplus B$ $G_1 = \overline{AB}$, $G_2 = \overline{A G_1}$, $G_3 = \overline{B G_1}$, $P = \overline{G_2 G_3}$

Block 2: $Y = P \oplus C$ $G_5 = \overline{PC}$, $G_6 = \overline{P G_5}$, $G_7 = \overline{C G_5}$, $Y = \overline{G_6 G_7}$

 A --+-------[G2]--+
     +--[G1]--+    +--[G4]--P--+-------[G6]--+
 B --+-------[G3]--+           +--[G5]--+    +--[G8]-- Y
                           C --+-------[G7]--+
 (all gates are 2-input NAND)

Y = 1 when an odd number of inputs are 1.