Construct circuit that implements truth table

[I like Serena]

We can add a third input R to the A we found, can't we?
And afterwards add both carries together?

 

[evinda]

We can add a third input R to the A we found, can't we?
And afterwards add both carries together?

You mean that we pick the outputs of the circuit as inputs at an other circuit? Or do we add the third input R at the existing circuit? :unsure:

 

[evinda]

So is it somehow as below? If so, what do we do with the carries? Why do we add them? And also, if we use the sum, don't they have to be connected somehow at the circuits? :unsure: :unsure:

 

[I like Serena]

So is it somehow as below?

If so, what do we do with the carries? Why do we add them? And also, if we use the sum, don't they have to be connected somehow at the circuits?

Shouldn't the block in the left top be the same as the one at the right bottom? (Worried)
I'll assume that it is, since otherwise it won't work.

Suppose we calculate $P+Q+R=1+1+0=10$.
Then we would get the outputs $K=1$, $K_2=0$, and $A_2=0$, wouldn't we?
How are those supposed to represent the result?
Shouldn't we have just 2 outputs?

 

[evinda]

Shouldn't the block in the left top be the same as the one at the right bottom? (Worried)
I'll assume that it is, since otherwise it won't work.

Oh yes, sorry! That's what I wanted to draw

Suppose we calculate $P+Q+R=1+1+0=10$.
Then we would get the outputs $K=1$, $K_2=0$, and $A_2=0$, wouldn't we?

Yes (Nod)

How are those supposed to represent the result?
Shouldn't we have just 2 outputs?

We do not get the result because the carries are not added, right?

Should we have in general only two outputs? So do we connect the arrows of the first circuit directly with the second circuit? Also do we have to connect in the second circuit the results $K_1$ and $K_2$ in order to get the right carry as output?

In general, the only possible results are $1+0+0=1$, $1+1+0=0$ with carry $1$ and $1+1+1=1$ with carry $1$ and we can get it with any possible order of $0,1$ at the sum, right? :unsure::unsure:

 

[I like Serena]

We do not get the result because the carries are not added, right?

Should we have in general only two outputs? So do we connect the arrows of the first circuit directly with the second circuit? Also do we have to connect in the second circuit the results $K_1$ and $K_2$ in order to get the right carry as output?

How about something like this:
\begin{tikzpicture}
\draw (0,0) rectangle (8,6);
\draw (0,5) -- +(-1,0) node[ left ] {P};
\draw (0,3) -- +(-1,0) node[ left ] {Q};
\draw (0,1) -- +(-1,0) node[ left ] {R};
\draw (8,4) -- +(1,0) node[ right ] {K};
\draw (8,2) -- +(1,0) node[ right ] {A};
\draw (2,2) node {+} +(-0.5,+0.2) -- +(-1,+0.2) +(-0.5,-0.2) -- +(-1,-0.2) +(0.5,+0.2) -- +(1,+0.2) +(0.5,-0.2) -- +(1,-0.2) +(-0.5,-0.5) rectangle +(0.5,0.5);
\draw (3,3.5) node {+} +(-0.5,+0.2) -- +(-1,+0.2) +(-0.5,-0.2) -- +(-1,-0.2) +(0.5,+0.2) -- +(1,+0.2) +(0.5,-0.2) -- +(1,-0.2) +(-0.5,-0.5) rectangle +(0.5,0.5);
\draw (6,1) node {+} +(-0.5,+0.2) -- +(-1,+0.2) +(-0.5,-0.2) -- +(-1,-0.2) +(0.5,+0.2) -- +(1,+0.2) +(0.5,-0.2) -- +(1,-0.2) +(-0.5,-0.5) rectangle +(0.5,0.5);
\draw (5,5) node {+} +(-0.5,+0.2) -- +(-1,+0.2) +(-0.5,-0.2) -- +(-1,-0.2) +(0.5,+0.2) -- +(1,+0.2) +(0.5,-0.2) -- +(1,-0.2) +(-0.5,-0.5) rectangle +(0.5,0.5);
\end{tikzpicture}
That leaves how to connect the lines so that all possible sums are summed correctly.

In general, the only possible results are $1+0+0=1$, $1+1+0=0$ with carry $1$ and $1+1+1=1$ with carry $1$ and we can get it with any possible order of $0,1$ at the sum, right?

What do you mean?

 

[evinda]

How about something like this:
\begin{tikzpicture}
\draw (0,0) rectangle (8,6);
\draw (0,5) -- +(-1,0) node[ left ] {P};
\draw (0,3) -- +(-1,0) node[ left ] {Q};
\draw (0,1) -- +(-1,0) node[ left ] {R};
\draw (8,4) -- +(1,0) node[ right ] {K};
\draw (8,2) -- +(1,0) node[ right ] {A};
\draw (2,2) node {+} +(-0.5,+0.2) -- +(-1,+0.2) +(-0.5,-0.2) -- +(-1,-0.2) +(0.5,+0.2) -- +(1,+0.2) +(0.5,-0.2) -- +(1,-0.2) +(-0.5,-0.5) rectangle +(0.5,0.5);
\draw (3,3.5) node {+} +(-0.5,+0.2) -- +(-1,+0.2) +(-0.5,-0.2) -- +(-1,-0.2) +(0.5,+0.2) -- +(1,+0.2) +(0.5,-0.2) -- +(1,-0.2) +(-0.5,-0.5) rectangle +(0.5,0.5);
\draw (6,1) node {+} +(-0.5,+0.2) -- +(-1,+0.2) +(-0.5,-0.2) -- +(-1,-0.2) +(0.5,+0.2) -- +(1,+0.2) +(0.5,-0.2) -- +(1,-0.2) +(-0.5,-0.5) rectangle +(0.5,0.5);
\draw (5,5) node {+} +(-0.5,+0.2) -- +(-1,+0.2) +(-0.5,-0.2) -- +(-1,-0.2) +(0.5,+0.2) -- +(1,+0.2) +(0.5,-0.2) -- +(1,-0.2) +(-0.5,-0.5) rectangle +(0.5,0.5);
\end{tikzpicture}
That leaves how to connect the lines so that all possible sums are summed correctly.

You mean that we conctruct for example a circuit that contains each logic gate of the above circuit twice? :unsure: :unsure:

What do you mean?

I meant that these are the possible operations that we could have, given that we take into consideration the sum of three one-digit binary numbers... Or am I wrong? (Thinking)(Thinking)

 

[I like Serena]

You mean that we conctruct for example a circuit that contains each logic gate of the above circuit twice?

I meant that we have a 'global' $K$ and $A$ as output, and we have to figure out how to get them.
We might also call them $K_{\text{3 digit sum}}$ and $A_{\text{3 digit sum}}$ to make it more explicit.
Actually, let's make them $K_3$ and $A_3$ for short.

I've simplified the circuit we are supposed to use as a building block by representing it as a square with a $\boxed{+}$ sign in it.
The $\boxed{+}$ represents the combination of the AND gate and the XOR gate.

I think we can also assume that we will start with adding $P$ and $Q$ together, can't we?
If we do, then perhaps we can call its intermediate outputs $K_1$ and $A_1$.

I meant that these are the possible operations that we could have, given that we take into consideration the sum of three one-digit binary numbers... Or am I wrong?

In general, the only possible results are $1+0+0=1$, $1+1+0=0$ with carry $1$ and $1+1+1=1$ with carry $1$ and we can get it with any possible order of $0,1$ at the sum, right?

Did you mean "any possible order of $0,1$ at the first sum"?
That is, we treat (1+0)+R the same as (0+1)+R? :unsure:

If so, isn't $(1+0)+1$ missing then? :unsure:

 

[evinda]

I meant that we have a 'global' $K$ and $A$ as output, and we have to figure out how to get them.
We might also call them $K_{\text{3 digit sum}}$ and $A_{\text{3 digit sum}}$ to make it more explicit.
Actually, let's make them $K_3$ and $A_3$ for short.

I've simplified the circuit we are supposed to use as a building block by representing it as a square with a $\boxed{+}$ sign in it.
The $\boxed{+}$ represents the combination of the AND gate and the XOR gate.

I think we can also assume that we will start with adding $P$ and $Q$ together, can't we?
If we do, then perhaps we can call its intermediate outputs $K_1$ and $A_1$.

So is the desired circuit the following? :unsure: Then the sum P+Q+R is $A_4$.

Did you mean "any possible order of $0,1$ at the first sum"?
That is, we treat (1+0)+R the same as (0+1)+R?

If so, isn't $(1+0)+1$ missing then? :unsure:

So at the first step the possible sums are 0+0, 0+1 and 1+0, 1+1 and then at the second step the possible sums are the results of these + 0 or 1, right? :unsure:

 

[I like Serena]

So is the desired circuit the following? Then the sum P+Q+R is $A_4$.

What about the carry?
Shouldn't we have $K_4$ as output then as well?

Suppose $+Q+R=0+1+1=10$. Then we get $A_4=1$ don't we? (Worried)

So at the first step the possible sums are 0+0, 0+1 and 1+0, 1+1 and then at the second step the possible sums are the results of these + 0 or 1, right?

Yep. (Nod)

 

[evinda]

What about the carry?
Shouldn't we have $K_4$ as output then as well?

Suppose $+Q+R=0+1+1=10$. Then we get $A_4=1$ don't we? (Worried)

Oh yes, right... :unsure::unsure: So do we have to add an other circuit in order to include $K_4$?

 

[I like Serena]

Oh yes, right... So do we have to add an other circuit in order to include $K_4$?

We already have $K_4$ don't we?

But I think the circuit is not correct yet.
If we have $P+Q+R=0+1+1$ we should get $10$ shouldn't we? That is, $K_4=1$ and $A_4=0$.
But don't we get $K_4=0$ and $A_4=1$? (Worried)

 

[evinda]

We already have $K_4$ don't we?

But I think the circuit is not correct yet.
If we have $P+Q+R=0+1+1$ we should get $10$ shouldn't we? That is, $K_4=1$ and $A_4=0$.
But don't we get $K_4=0$ and $A_4=1$? (Worried)

Oh yes, you are right! (Wasntme) I have noticed that the second circuit gives the right result...
Does this happen for any possible case? So is $A_2$ with carry $K_2$the desired sum? :unsure::unsure:

 

[evinda]

I have thought about it again... :unsure:

Is the desired circuit the following?

 

[I like Serena]

I have thought about it again...

Is the desired circuit the following?

Looks good to me! (Nod)

Just for fun, I made a drawing as well. (Blush)

\begin{tikzpicture}
\usetikzlibrary{shapes.gates.logic.US}
\tikzset{
add/.pic = {
\draw[thick] (-0.8,-0.8) rectangle (0.8,0.8);
\node[and gate US,draw] at (0,0.35) (And1) {};
\node[xor gate US,draw] at (0,-0.35) (Xor1) {};
\path
(And1.input 1)+(-1,0) coordinate (-in 1)
(Xor1.input 2)+(-1,0) coordinate (-in 2)
(And1.output)+(1,0) coordinate (-K)
(Xor1.output)+(1,0) coordinate (-A);

\draw (-in 1) -- (And1.input 1) +(-0.35,0) |- (Xor1.input 1);
\draw (-in 2) -- (Xor1.input 2) +(-0.2,0) |- (And1.input 2);
\draw (And1.output) -- (-K);
\draw (Xor1.output) -- (-A);
}
}

\node[draw,minimum width=10cm,minimum height=5cm] (Block) {};

\draw (-3.5,1) pic (Add1) {add};
\draw (-0.5,-1) pic (Add2) {add};
\draw (2.5,0.5) pic (Add3) {add};

\path
(Add1-in 1 -| Block.west)+(-1,0) coordinate[label=left: P] (P)
(Add2-in 2 -| Block.west)+(-1,0) coordinate[label=left:R] (R)
(Add2-A -| Block.east)+(1,0) coordinate[label=right:A] (A)
(Add3-A -| Block.east)+(1,0) coordinate[label=right:K] (K)
(P) -- node[shape=coordinate,label=left:Q] (Q) {} (R)
;

\draw (P) -| (Add1-in 1);
\draw (Q) -| (Add1-in 2);
\draw (R) -| (Add2-in 2);
\draw (Add1-A) -| (Add2-in 1);
\draw (Add1-K) -| (Add3-in 1);
\draw (Add2-A) -| (A);
\draw (Add2-K) -| (Add3-in 2);
\draw (Add3-A) -| (K);

\node[above] at (Add1-A) {$A_1$};
\node[above] at (Add1-K) {$K_1$};
\node[above] at (Add2-A) {$A_2$};
\node[above] at (Add2-K) {$K_2$};
\node[above] at (Add3-A) {$A_3$};
\node[above] at (Add3-K) {$K_3$};

\end{tikzpicture}

 

[evinda]

Looks good to me! (Nod)

Just for fun, I made a drawing as well. (Blush)

\begin{tikzpicture}
\usetikzlibrary{shapes.gates.logic.US}
\tikzset{
add/.pic = {
\draw[thick] (-0.8,-0.8) rectangle (0.8,0.8);
\node[and gate US,draw] at (0,0.35) (And1) {};
\node[xor gate US,draw] at (0,-0.35) (Xor1) {};
\path
(And1.input 1)+(-1,0) coordinate (-in 1)
(Xor1.input 2)+(-1,0) coordinate (-in 2)
(And1.output)+(1,0) coordinate (-K)
(Xor1.output)+(1,0) coordinate (-A);

\draw (-in 1) -- (And1.input 1) +(-0.35,0) |- (Xor1.input 1);
\draw (-in 2) -- (Xor1.input 2) +(-0.2,0) |- (And1.input 2);
\draw (And1.output) -- (-K);
\draw (Xor1.output) -- (-A);
}
}

\node[draw,minimum width=10cm,minimum height=5cm] (Block) {};

\draw (-3.5,1) pic (Add1) {add};
\draw (-0.5,-1) pic (Add2) {add};
\draw (2.5,0.5) pic (Add3) {add};

\path
(Add1-in 1 -| Block.west)+(-1,0) coordinate[label=left: P] (P)
(Add2-in 2 -| Block.west)+(-1,0) coordinate[label=left:R] (R)
(Add2-A -| Block.east)+(1,0) coordinate[label=right:A] (A)
(Add3-A -| Block.east)+(1,0) coordinate[label=right:K] (K)
(P) -- node[shape=coordinate,label=left:Q] (Q) {} (R)
;

\draw (P) -| (Add1-in 1);
\draw (Q) -| (Add1-in 2);
\draw (R) -| (Add2-in 2);
\draw (Add1-A) -| (Add2-in 1);
\draw (Add1-K) -| (Add3-in 1);
\draw (Add2-A) -| (A);
\draw (Add2-K) -| (Add3-in 2);
\draw (Add3-A) -| (K);

\node[above] at (Add1-A) {$A_1$};
\node[above] at (Add1-K) {$K_1$};
\node[above] at (Add2-A) {$A_2$};
\node[above] at (Add2-K) {$K_2$};
\node[above] at (Add3-A) {$A_3$};
\node[above] at (Add3-K) {$K_3$};

\end{tikzpicture}

Nice, thanks a lot! (Party)