英伟达笔试1

1.使用基础逻辑单元2输入与非门(NAND2)，设计1位奇偶发生器。有4个二进制输入A,B,C,D；当输入里面“1”的总个数是奇数时，输出为1；否则输出为0。也就是实现4输入奇校验输出：P=A⊕B⊕C⊕DP = A \oplus B \oplus C \oplus DP=A⊕B⊕C⊕D，全部只用2‑输入NAND门搭建。基础知识XOR（异或）：X⊕Y=X⋅Y‾+X‾⋅Y‾X\oplus Y = \overline{X\cdot \overline Y+\overline X\cdot Y}X⊕Y=X⋅Y+X⋅Y​，可以只用NAND2实现：X⊕Y=NAND(NAND(X,NAND(X,Y)),NAND(Y,NAND(X,Y))) X\oplus Y = NAND\big(NAND(X,NAND(X,Y)),\ NAND(Y,NAND(X,Y))\big)X⊕Y=NAND(NAND(X,NAND(X,Y)),NAND(Y,NAND(X,Y)))一个2输入异或需要5个NAND2门。异或满足结合律：A⊕B⊕C⊕D=(A⊕B)⊕(C⊕D) A\oplus B \oplus C \oplus D = (A\oplus B)\oplus (C\oplus D)A⊕B⊕C⊕D=(A⊕B)⊕(C⊕D)分步构造模块1：X1=A⊕BX_1 = A\oplus BX1​=A⊕B，5个NAND2模块2：X2=C⊕DX_2 = C\oplus DX2​=C⊕D，5个NAND2模块3：P=X1⊕X2P = X_1 \oplus X_2P=X1​⊕X2​，5个NAND2总共有5+5+5=155+5+5 = \boldsymbol{15}5+5+5=15个NAND2门。逻辑表达式（全部用NAND表示）记N(x,y)=x⋅y‾N(x,y)=\overline{x\cdot y}N(x,y)=x⋅y​代表NAND2门输出。G1=N(A,B)X1=A⊕B=N(N(A,G1),N(B,G1))G2=N(C,D)X2=C⊕D=N(N(C,G2),N(D,G2))G3=N(X1,X2)P=X1⊕X2=N(N(X1,G3),N(X2,G3)) \begin{align*} G_1 = N(A,B) \\ X_1=A\oplus B = N(N(A,G_1),\ N(B,G_1)) \\ G_2 = N(C,D) \\ X_2=C\oplus D = N(N(C,G_2),\ N(D,G_2)) \\ G_3 = N(X_1,X_2) \\ P = X_1\oplus X_2 = N(N(X_1,G_3),\ N(X_2,G_3)) \end{align*}G1​X1​=A⊕BG2​X2​=C⊕DG3​P=X1​⊕X2​​=N(A,B)=N(N(A,G