如何仅使用4个NAND门构造XOR门？

17

xor门，现在我只需要使用4个nand门来构造此门

的xor = (a and not b) or (not a and b)，这是

\begin{aligned} \bar{A} B + A \bar{B} \end{aligned}

$\begin{split}\overline{A}{B}+{A}\overline{B}\end{split}$

我知道答案，但是如何从公式中获得门图？

异或门

编辑

我的直觉是，对我来说，如果我逐步进行定义，然后再进行定义，则应该理解这一点xor = (a and not b) or (not a and b)。

\begin{aligned} \bar{\bar{\bar{A} B} \cdot \bar{A \bar{B}}} \end{aligned}

$\begin{split}\overline{\overline{\overline{A}{B}}\cdot\overline{{A}\overline{B}}}\end{split}$

并且xor将与5个构成nand栅极（第一＃1下图）

异或门2

我的问题更像是：想象历史上第一个人想出这个公式，他或她（思考过程）如何nand逐步地从这个公式中得到4个答案。

\begin{aligned} \bar{A} 乙 + 一种 \bar{B} \end{aligned}

$\begin{split}\overline{A}{B}+{A}\overline{B}\end{split}$

logic boolean-algebra

— 永恒的
source

我确定您知道如何进行XOR（或任何其他函数）并将其转换为仅使用NAND的等效电路（由于NAND已完成，因此这始终是可能的）。但是如果你问怎么这个公式减少到只用4个NAND，或在一般情况下，不到

非门，以及它是否是可能获得的等效电路与

与非门-我不知道有一个简单的为此。

k

$k$

\leq k

$\le k$

— Ran G.

以下是该问题的两个答案。我的想法很坦率，您可以设计（后验）一种方法，通过预先知道最终结果来找到所需的结构，这是问题中给出的，并且可以在Internet上找到。显然，这是一种更简单的做事方式，看起来很荒谬，没有给出一个通用的程序，而这是没有答案的。因此，我很想知道为什么选民在投票时会选择一个答案而不是另一个答案……如果您愿意抽时间发表简短评论。提前致谢。

— babou 2015年

这个问题尚待解决，不清楚。我想可能是相当清楚什么OP是问，多i8nteresting，如果OP不屑于向谁试图回答他的各种用户的反应，

— babou

electronics.stackexchange.com/questions/84714/…-这个问题较为笼统，答案为解决此问题的一般方法提供了更多信息，此答案electronics.stackexchange.com/a/84803显示了如何导出NAND XOR运算符的表示形式

— Anton Trunov

我遇到了一些类似的问题，只是编写了一个程序来对所有内容进行系统地尝试...最多可以输入四个输入，其中只有65,536个可能的函数。对于稍微复杂一些的电路，这还使我能够优化延迟，并在一个或两个输入比其他输入晚的情况下找到最佳电路。具有5个输入= 2 ^ 32种可能功能的电路可能会使用蛮力来完成。

— gnasher729

13

从那个公式？可以办到。但是从这开始比较容易：（在这里使用不同的表示法）

a ^ b = ~(a & b) & (a | b)

好吧，现在呢？最终，我们应该派生它~(~(~(a & b) & a) & ~(~(a & b) & b))（看起来它有5个NAND，但就像电路图一样，它有一个被两次使用的子表达式）。

因此，制作看起来像~(a & b) & a（和相同的东西，但b结尾是a）的东西，并希望它能坚持下去：（and分布在or）

(~(a & b) & a) | (~(a & b) & b)

现在已经很接近了，只需应用DeMorgan即可将中间部分or变为and：

~(~(~(a & b) & a) & ~(~(a & b) & b))

就是这样。

— 哈罗德
source

9

我认为您正在要求此证明：

A^B = (!A)B + A(!B)
    = !!((!A)B) + !!(A(!B))
    = !(!!A + !B) + !(!A + !!B)
    = !(A + !B) + !(!A + B)
    = !((A + !B)(!A + B))
    = !(A(!A) + AB + (!A)(!B) + B(!B))
    = !(AB + (!A)(!B))
    = !(AB)(!(!A)(!B))
    = !(AB)(!!A + !!B)
    = !(AB)(A+B)
    = !(AB)A + !(AB)B
    = !!(!(AB)A + !(AB)B)
    = !((!(!(AB)A))(!(!(AB)B)))

尽管显然NAND在合成方程中使用了5 s，但是在!(AB)设计其电路时，该副本仅使用一次。

— 蒙塔西尔
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抱歉，A ^ B不是A AND B吗？您似乎打算证明XOR哪个符号应为should或⊻。但是这个证明是我真正想要的，谢谢！

— osiixy

5

由于您已经有了图表答案，可以通过在Google中键入问题标题从Wikipedia轻松唤醒，作为与您相同的.png图表，您应该很容易通过从该图中提取公式来查找公式。给出的定义如NAND $\text{NAND}(A,B)=\overline{AB}\;$ ：

最左边的栅极给出 ; $C=\overline{AB}$
顶栅给出 ; $D_1=\overline{AC}$
顶栅给出，因为NAND是commutatve像AND; $D_2=\overline{BC}$
最右边的栅极给出。 $E=\overline{D_1D_2}$

综上所述，我们首先注意到

$C=\overline{AB}=\overline A+\overline B$

$\begin{align} \overline{D_1}&=AC\\ &=A(\overline A+\overline B)\\ &=A\overline A+A\overline B\\ &=0+A\overline B\\ &=A\overline B\\ \end{align}$

类似地： $\overline{D_2}=B\overline A$

因此
$\begin{align} E&=\overline{D_1D_2}\\ &=\overline{D_1}+\overline{D_2}\\ &=A\overline B + B\overline A \end{align}$

这正是XOR的定义。如果您想从初始数据开始，则可以全部撤消，而不仅仅是检查答案。

在没有先验知识的情况下找到答案

这旨在回答作为对问题的编辑而添加的显式请求，以便从头开始查找解决方案。考虑到问题与思考过程有关，我将提供所有细节。

$A$ $B$

$\text{XOR}(A,B)=A\overline B + B\overline A\;$ 。

因此，我们可以尝试猜测此门的哪种输入将产生所需的输出。

$\text{NAND}(X,Y)=\overline{XY}= \overline X+\overline Y\;$

将最后一个公式与我们必须得到的结果统一起来，我们得到：

$\overline X=A\overline B\;$ $X=\overline{A\overline B}=\overline A+B\;$ 。
$Y=\overline{\overline A B}=A+\overline B\;$ 。

请注意，这只是最简单的可能性。由于NAND具有方程式属性，因此我们没有统一自由代数，因此还有其他输入对将给出期望的结果。但我们首先尝试一下。

$X$ $Y$ $A$ $B$

我们可以尝试重复统一过程（我曾经做过），但这自然会导致我们再使用四个门，从而得出5个门的解决方案。

$X$ $Y$ $Z$ $A$ $B$

$X$ $Y$ $Z$ $A$ $B$ $A$ $B$

$A$ $B$ 作为输入，产生作为输出：

$Z=\text{NAND}(A,B)=\overline{AB}= \overline A+\overline B\;$

Now, we have to check whether combining $Z$ with itself, $A$ , $B$ , 0, or 1 through a NAND gate can produce $X$ , and also $Y$ .

We know that combining a value with itself, 0 or 1 through a NAND gate is either the identity function or the negation. So the only remaining candidates are $A$ and $B$ .

It is easy to check that

$\begin{align} \text{NAND}(Z,A)&=\overline{ZA}\\ &=\overline{\overline{AB}\;A}\\ &=\overline{(\overline A+\overline B)\;A}\\ &=\overline{\overline AA+\overline BA}\\ &=\overline{0+\overline BA}\\ &=\overline{\overline BA}\\ &=\overline{A\overline B}\\ &=X \end{align}$

Similarly $\text{NAND}(Z,B)=Y$

Hence we can compose these four gates to get the desired result, i.e., the XOR function.

— babou
source

Not in a reverse way to prove that they are equal. But image that you don't know the diagram but to construct the gate using minimum nand gate.

— Timeless

1

What do you expect as an answer? A systematic technique for doing that. I do not know that there is any that is tractable enough to be worth using in complex cases. Given that I know the answer I can just lie to you and pretend to have found by reasonning what I discovered by checking the answer. This said, looking at what I get with NAND(A,B) is all that seems useful for a start. Then NANDing the result with one argument A or B, is also one thing to look at, to get a view of where I am. From there, one is pretty close to the final answer.

— babou

1

@Timeless Another way to go about it is backward from the answer, knowing that the answer is fron a NAND gate. If you assume that the solution is symmetrical in A and B, it gives you a likely form of the inputs to the last NAND gate. There are many way to go about it, either to find the answer, or to justify finding it a posteriory. But a proof is a proof, whether found by your ingenuity, or given by some oracle or a good friend. And at some point no one can tell the difference. Actually, the backward proof I give could be the best proof, even if the solution was found some other way.

— babou

Actually, it is quite common in math to have an analysis part to find a solution, then a synthesis part where you prove it is the solution. One usually gives both, but only the second part is really necessary.

— babou

@Timeless Both answers were based on the knowledge of a formula to obtain, deduced from the diagram to be obtained. Your edit asked for a plausible intuitive scenario to find the answer without any prior knowledge of the result. I did add that to my answer, but it would be nice to know whether it fits what you expected.

— babou 2015年

0

I take the input $(0,0)$ as an example.

For $\text{XOR}$ , the desired output is 0. However, $\text{NAND}(0,0) = 1$ .

Because the only way to get a 0 using $\text{NAND}$ is (at the last layer) $\text{NAND}(1,1) = 0$ , you should first produce two 1's.
- According to $\text{NAND}(0,1) = 1$ or $\text{NAND}(1,0) = 1$ , you produce a 1 using one $\text{NAND}(0,0)$ at the first layer and feed it, along with one input 0, into a second layer $\text{NAND}$ .

Only four $\text{NAND}$ s are involved. But it is only correct for the input $(0,0)$ so far. So you need to check other inputs $(0,1), (1,0),$ and $(1,1)$ against the solution and find that it just works. Lucky.

— hengxin
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0

I tried my best to give the answer using formula as asked.Hope you appreciate it.
Z=AB'+A'B
Z=AA'+AB'+BB'+A'B --->BB'=AA'=0
Z=A(A'+B')+B(B'+A')
Z=A(AB)'+B(AB)' --> Hint
so now (AB)' can get through 1st NAND gate,then in 2nd and third NAND gate the output of 1st NAND gate pass through with one of the input as A and B.After this we need one more complement so use fourth NAND gate.
NAND(1st)=(AB)'=A'+B'
NAND(2nd)=(A(AB)')'=(A(A'+B'))'=(AB')'=A'+B
NAND(3rd)=(B(AB)')'=(B(A'+B'))'=(A'B)'=A+B'
NAND(4th)=[(A'+B)(A+B')]' =[A'B'+AB]'=(A+B)(A'+B')=AB'+A'B

Happy!

— MANVENDRA SINGH MANOHAR
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0

The formula: XOR = (a and not b) or (not a and b).

Thats' not what you want, you want a formula that is a NAND. Remember that not (a or b) = not a and not b, and therefore (a or b) = not (not a and not b). Therefore

(a and not b) or (not a and b) =

not (not (a and not b) and not (not a and b)) =

not ((not a or b) and (a or not b)) =

NAND (not a or b, a or not b).

So we used one NAND gate, and have to calculate (not a or b) and (a or not b) using three NANDs. We turn each expression into a NAND:

not a or b = not (a and not b) = NAND (a, not b)

a or not b = not (not a and b) = NAND (not a, b)

Now we observe that (x and y) = x and (not x or y): If x is false then both sides are false. If x is true then (not x or y) = (false or y) = y. This is true for NAND just as it's true for AND. Therefore

NAND (a, not b) = NAND (a, not a or not b) = NAND (a, NAND (a, b))

NAND (b, not a) = NAND (b, not b or not a) = NAND (b, NAND (a, b)).

So we first find mid = NAND (a, b), left = NAND (a, mid) and right = NAND (b, mid), finally XOR = NAND (left, right).

— gnasher729
source

-2

*From left to right--D1,D2,D3,D4 ** D1=(A.B)' OR(A'+B')

suppose

(A.B)'=C

D2=(A.C)'=A'+C'

D3=(B.C)'=B'+C' then

D4=(D2.D3)'

D4=((A.C)'.(B.C)')'

D4=(A.C)''+(B.C)''

D4=(A.C)+(B.C)

D4=A.(A'+B')+B.(A'+B')

D4=AB'+BA' {A.A'=B.B'=0}**

— Bivash
source

2

I find it hard to follow this answer or understand what process you are using. Can you add some text sentences to explain the approach, so this isn't just a sequence of equations?

— D.W.