====== 3 Combinatorial Logic ======

<callout  title="introductional example">

<WRAP><well>
<imgcaption pic1|Simulation of a 7-segment encoder and display></imgcaption> \\
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The combinatorial logic shown in <imgref pic1> enables to output of distinct logic values for each logic input. When you change the ''input nibble'', you can see that the correct number appears on the 7-segment display. By clicking on the bits of the input, you can change the number. 

Tasks:
  - Which output $Y_0$ ... $Y_6$ is generated from the input nibble ''1000''? Which from ''1001''?
  - Is the output only depending on the input? Is there a dependence on history?
</callout>


===== 3.1 Combinatorial Circuits =====

Up to now, we looked at simple logic circuits. These are relatively easy to analyze and synthesize (=develop). The main question in this chapter is: how can we set up and optimize logic circuits?

In the following, we have a look at combinatorial circuits. These are generally logic circuits with 
    * $n$ inputs $X_0$, $X_1$, ... $X_{n-1}$ 
    * $m$ outputs $Y_0$, $Y_1$, ... $Y_{m-1}$ 
  * no "memory", that is: a given set of input bits results in a distinct output
They can be described by
    * truth table
    * boolean formula
    * hardware description language 
The ladder one is not the focus of this course. 

The applications range:
  * (simple) half/full adder
  * [[http://www.falstad.com/circuit/e-digcompare.html|digital comparator]]s (logic circuit to compare 2 values)
  * Multiplexer/demultiplexer
  * Arithmetic logic units in microcontrollers and processors
  * much more

==== 3.1.1 Example ====

To understand the synthesis of combinatoric logic we will follow a step-by-step example for this chapter.

Imagine you are working for a company called "mechatronics and Robotics". One customer wants to have an intelligent switch as an input device connected to a microcontroller for controlling an oven. For this project "Therm-o-Safety" he needs a combinatoric logic:
  * The intelligent switch has 4 user selectable positions: $1$, $2$, $3$, $4$
  * Additionally there are 2 non-selectable positions in the case of failure. 
  * The output $Y=1$ will activate temperature monitoring.
  * The temperature monitoring has to be active for $3$ and $4$ in case of a major failure. A major failure is for example when the switch position is unclear. In this case, the input of the combinatorial circuit is "ON".
  * There are no other cases of inputs.

These requirements are put into a truth table:
<WRAP center>
<imgcaption pic01 | Therm-o-Safety truth table>
</imgcaption>
{{drawio>ToStruthtable1.svg}}
</WRAP>

<imgref pic01> shows one implementation of these requirements. The inputs ''001'' ... ''011'' represent the inputs $1$...$4$. The cases of failure are coded with ''110'' and ''111''. \\
The output $Y$ is activated as requested. For the two combinations ''000'' and ''101'' there is no output value defined. Depending on the requirements for a project these shall either better be ''0'' or ''1'' or the output of these does not matter. We had this "does not matter" before: the technical term is "I don't care", and it is written as a ''-'' or a ''x''. 

By this, we have done the first step to synthesize the requested logic.

==== 3.1.2 Sum-of-Products ====

Now, we want to investigate some of the input combinations (= lines in the truth table). At first, we have a look at the input combination ''011'', where the output has to be $Y=1$. 

If this input combination would be the only one for the output of $Y=1$, the following could be stated: \\ 
"$Y=1$ (only) when the $X_0$ is $1$ AND $X_1$ is $1$ AND $X_2$ is $0$ ". It can also be re-arranged to: \\ 
"$Y=1$ (only) when the $X_0$ is $1$ AND $X_1$ is $1$ AND $X_2$ is not $1$ ". 

This statement is similar to $X_0 \cdot X_1 \cdot \overline{X_2}$. The used conjunction results only in $1$ when all inputs are $1$. 
The negation of $X_2$ takes account of the fact, that $X_2$ has to be $0$. 

<WRAP center>
<imgcaption pic02 | Therm-o-Safety truth table - first analysis>
</imgcaption>
{{drawio>ToStruthtable2.svg}}
</WRAP>

<imgref pic02> shows the boolean expression for this combination. In <imgref pic03>, this boolean expression is converted into a structure with logic gates. 

<WRAP><well>
<imgcaption pic03|logic circuit for the combination '011'></imgcaption> \\
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With the same idea in mind, we can have a look at the other combinations resulting in $Y=1$. These are the combinations ''100'' and ''111'': 
  * For ''100'' The statement would be: "$Y=1$ (only) when the $X_0$ is $0$ AND $X_1$ is $0$ AND $X_2$ is $1$". Similarly to the combination above this leads to: $\overline{X_0} \cdot \overline{X_1} \cdot {X_2}$.
  * For ''111'', the boolean expression is ${X_0} \cdot {X_1} \cdot {X_2}$.

<WRAP column 100%> 
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  * Each row in a truth table (=one distinct combination) can be represented by a **minterm** or **maxterm**
  * A **minterm** is the conjunction (AND'ing) of all inputs, where under certain instances a negation has to be used
  * In a minterm an input variable with ''0'' has to be negated, to use it as an input for the AND. \\ e.g. $X_0 = 0$ AND $X_1 = 1$$ \quad \rightarrow \quad \overline{X_0} \cdot X_1$
  * A minterm results in an output of ''1''

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<WRAP center>
<imgcaption pic04 | Therm-o-Safety truth table - the sum-of-products>
</imgcaption>
{{drawio>ToStruthtable3.svg}}
</WRAP>

In <imgref pic04> all minterms for $Y=1$ are shown. The <imgref pic05> depicts all the logic circuits for the three minterms. These lead to the outputs $Y'$, $Y''$, and $Y'''$.

<WRAP><well>
<imgcaption pic05|logic circuit for the combinations '100', '110', '111'></imgcaption> \\
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For the final step, we have to combine the single results for the minterms. The output has to be $1$ when at least one of the minterms is $1$. Therefore, the minterms have to be connected disjunctive: 

\begin{align*}
Y &=  & Y'  & \quad + & Y'' & \quad + & Y'''  \\
Y &=  & (X_0 \cdot X_1 \cdot \overline{X_2}) & \quad + & (\overline{X_0} \cdot \overline{X_1} \cdot {X_2}) & \quad  + & ({X_0} \cdot {X_1} \cdot {X_2})  \\
\end{align*}

This leads to the logic circuit shown in <imgref pic06>. Here, you can input the different combinations by clicking on the bits of the input nibble. 

<WRAP><well>
<imgcaption pic06|logic circuit for therm-o-safety></imgcaption> \\
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<panel type="danger" title="Note!"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

  * The disjunction of the minterms is called **sum-of-products**, **SoP**, **disjunctive normal form** or **DNF**.
  * When all inputs are used in each of the minterms the normal form is called **full disjunctive normal form**
  * When synthesizing a logic circuit by the sum-of-products, all 'don't care' terms output $0$. 

</WRAP></WRAP></panel>
 </WRAP>

We have seen, that the sum-of-products is one tool to derive a logic circuit based on a truth table. Alternatively, it is also possible to insert an intermediate step, where the logic formula is simplified.

In the following one possible optimization is shown:

<WRAP>
\begin{align*}
Y &=  & (X_0 \cdot X_1 \cdot \overline{X_2}) & \quad + & (\overline{X_0} \cdot \overline{X_1} \cdot {X_2}) & \quad  + & ({X_0} \cdot {X_1} \cdot {X_2}) & \quad |  \text{associative law} \\
Y &=  & (\overline{X_0} \cdot \overline{X_1} \cdot {X_2}) & \quad  + & (X_0 \cdot X_1 \cdot \overline{X_2}) & \quad + & ({X_0} \cdot {X_1} \cdot {X_2}) & \quad | \text{associative law	}  \\
Y &=  & (\overline{X_0} \cdot \overline{X_1} \cdot {X_2}) & \quad  + & ((X_0 \cdot X_1) \cdot \overline{X_2}) & \quad + & (({X_0} \cdot {X_1}) \cdot {X_2}) & \quad | \text{distributive law	}  \\
Y &=  & (\overline{X_0} \cdot \overline{X_1} \cdot {X_2}) & \quad  + & ((X_0 \cdot X_1) \cdot (\overline{X_2} + {X_2})) & & & \quad | \text{complementary element}  \\
Y &=  & (\overline{X_0} \cdot \overline{X_1} \cdot {X_2}) & \quad  + & (X_0 \cdot X_1)    \\
\end{align*}
</WRAP>

==== 3.1.3 Product-of-Sums ====

In the sub-chapter before we had a look at the combinations which generate an output of $Y=1$ using the AND operator. Now we are investigating the combinations with $Y=0$. Therefore, we need an operator, which results in $0$ for only one distinct combination.

The first combination to look for is ''001''.
If this input combination would be the only one for the output of $Y=0$, the following could be stated: \\ 
"$Y=0$ (only) when the $X_0$ is $1$ AND $X_1$ is $0$ AND $X_2$ is $0$ ". \\ With having the duality in mind (see chapter [[boolean_algebra#The Set of Rules]]) the opposite is also true: \\ 
"$Y=1$ when $X_0$ is $0$ OR $X_1$ is $1$ OR $X_2$ is $1$ "

This is the same as: $\overline{X_0} + X_1 + X_2$ \\ The boolean operator we need here is the OR-operator.

Similarly, the combinations ''010'' and ''110'' can be transformed. Keep in mind, that this time we are looking for a formula with results in $0$ only for the given one distinct combination.

<WRAP column 100%> <panel type="danger" title="Note!"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

  * A **maxterm** is the disjunction (OR'ing) of all inputs, where under certain instances a negation has to be used.
  * In a maxterm an input variable with ''1'' has to be negated, to use it as an input for the OR.
  * A maxterm results in an output of ''0''

</WRAP></WRAP></panel> </WRAP>

The <imgref pic07> shows all the maxterms for the Therm-o-Safety example.

<WRAP center>
<imgcaption pic07 | Therm-o-Safety truth table>
</imgcaption>
{{drawio>SoTtruthtable1.svg}}
</WRAP>

The formulas of <imgref pic07> can again be transformed into gate circuits (<imgref pic08>). Here, only for the inputs ''001'', ''010'', ''110'' one of the outputs $Y'$, $Y''$ or $Y'''$ is $0$. 

<WRAP><well>
<imgcaption pic08|logic circuit for the combinations '001', '010', '110'></imgcaption> \\
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When these intermediate outputs $Y'$, $Y''$, and $Y'''$ are used as an input for an AND-gate the resulting output will get $0$ when at least one of the intermediate outputs is $0$. This results in another way to synthesize the Therm-o-Safety (see <imgref pic09>)

<WRAP><well>
<imgcaption pic09|logic circuit for therm-o-safety></imgcaption> \\
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Also, the product-of-sum can be simplified:

\begin{align*}
Y &=  & (\overline{X_0} + X_1 + X_2) & \quad \cdot & ({X_0} + \overline{X_1} + X_2) & \quad  \cdot & (\overline{X_0} + \overline{X_1} + X_2)  \\
  &=  & ...  \\
Y &=  & (\overline{X_0} + X_1 + X_2) & \quad \cdot & (\overline{X_0} + \overline{X_1})   \\
\end{align*}

This result from $Y$ by the sum-of-products is different compared to the result in a product-of-sums:
  * product-of-sums: $Y = (\overline{X_0} \cdot \overline{X_1} \cdot {X_2})  +  (X_0 \cdot X_1)$ 
  * sum-of-products: $Y = (\overline{X_0} + X_1 + X_2) \cdot (\overline{X_0} + \overline{X_1})$

In this case, these results cannot be transformed into each other with the means of boolean rules. 

<WRAP column 100%> <panel type="danger" title="Note!"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

  * The disjunction of the maxterms is called **products-of-sum**, **PoS**, **conjunctive normal form** or **CNF**.
  * When all inputs are used in each of the minterms the normal form is called **full conjunctive normal form**
  * When synthesizing a logic circuit by the sum-of-products, all "don't care" terms outputting $1$
  * The products-of-sum is the DeMorgan dual of the sum-of-products **__if__** there are no don't care terms. Otherwise, the results cannot be transformed into each other with the means of boolean rules. 

</WRAP></WRAP></panel> </WRAP>

===== 3.2 Karnaugh Map =====

==== 3.2.1 Introduction with the two dimensional Map ====

For a simple introduction, we take one step back and look at a simple example. The formula $Y(X_1, X_0) = X_0 \cdot X_1$ combines two variables. Therefore, it has two dimensions.
In <imgref pic13> (a) the truth table of this is shown. The leftmost column shows the decimal interpretation of the binary numeral given by $X_1$, $X_0$ (e.g. $(X_1=1, X_0=0) \rightarrow 10_2=2_{10}$).

The given logic expression can also be interpreted in a coordinate system, with the following conditions:
  * There are only two coordinates on each axis possible: ''0'' and ''1''. 
  * There are as many axes as variables given in the logic: For the example, we have two variables $X_0$ and $X_0$. These are spanning a two-dimensional system.
  * On the possible positions, the results $Y$ have to be shown.

In the following pictures of this representation, the values are shown:
  * green dot, when the result is ''1''
  * red dot, when the result is ''0''
  * grey dot, when the result is ''don't care'' 

For the given example the coordinate system shows four possible positions: These are the edges of a square (<imgref pic13> (b)). 

We will in the future write this as in <imgref pic13> (c). This diagram is also called **{{wp>karnaugh map}}** (often called k-map or KV map).
In the shown Karnaugh map the coordinate $X_0$ is shown vertically and $X_1$ horizontally. Similar to the coordinate system the upper left cell is for $X_1=0$ and $X_0=0$. The upper right cell is for  $X_1=1$ and $X_0=0$, and the lower right one is for $X_1=1$ and $X_0=1$. In each cell the result $Y(X_1, X_0)$ is shown as a large number - similar to the color code in the coordinate system. The small number is the decimal representation of the number given by $X_1$ and $X_0$. This index is often __not__ explicitly shown in the Karnaugh map, but simplifies the fill-in of the map and helps for the start.

<WRAP center>
<imgcaption pic13 | two dimensional Karnaugh map>
</imgcaption>
{{drawio>multicube2.svg}}
</WRAP>

The Karnaugh map will help us in the following to find simplifications of more complex logic expressions.

==== 3.2.2 The three-dimensional Karnaugh Map ====

In this subchapter, we will have a look at our example of thermo--safety. For this example, we found two possible gate logic which can produce the required output. We have also seen, that optimizing the terms (i.e. the min- or maxterms) is often possible. However we do not know how we can find the optimum implementation.

For this, we try to interpret the inputs of our example again as dimensions in a multidimensional space. The three input variables $X_0$, $X_1$, $X_2$ span a 3-dimensional space. The point ''000'' is the origin of this space. The three combinations ''001'', ''010'', ''100'' are onto the $X_0$-, $X_1$-, and $X_2$-axis, respectively (see <imgref pic10> (a)). The other combinations can be reached by adding these axis values together (see <imgref pic10> (b)+(c)). This is similar to the situation of a two-dimensional or three-dimensional vector. Three inputs result in this representation in the edges of a cube.

In the <imgref pic10> (d) the situation $X_0=1$, $X_1=1$, $X_2=0$ is shown. 

<WRAP center>
<imgcaption pic10 | 3 dimensional cube represetation>
</imgcaption>
{{drawio>multicube0.svg}}
</WRAP>

There is also an alternative way to look at this representation:
  * The formula $Y=1$ (independent from inputs $X_0...X_{n-1}$) lead to all positions are ''1''
  * A single input equal ''1'' ( $Y:\, X_0=1$, independent from all other inputs, i.e. all others are ''don't care'') leads in the three-dimensional example to the edges of a side surface of the cube. \\ In out example $\color{violet}{X_1=1}$ lead to the situation shown in <imgref pic11> (a). When investigating the shown green dots ''0**__1__**0'', ''0**__1__**1'', ''1**__1__**0'', ''1**__1__**1'' it is visible that the middle value (= the value for $X_1$) is the same.  \\ Generally: A single input equal ''1'' (independent from all other inputs) leads to a structure one dimension smaller than the number of inputs (In our example: 3 inputs $\rightarrow$ two-dimensional structure = surface).
  * Multiple given inputs equal ''1'' lead to smaller structures correspondingly. In our example: $\color{blue}{X_0=1}$ and $\color{violet}{X_1=1}$ ($=\color{blue}{X_0}\cdot \color{violet}{X_1}$) result in the two edges on a corner of the cube (<imgref pic11> (b)). For these coordinates (''0**__11__**'', ''1**__11__**'') the last two values are the same.
  * A minterm (=''1'' as an output) in our example is given by the intersection of all surfaces for the individual dimensions. In our example:$\color{blue}{X_0=1}$ and $\color{violet}{X_1=1}$ and $\color{brown}{X_2=0}$ ($=\color{blue}{X_0}\cdot \color{violet}{X_1}\cdot \color{brown}{\overline{X_2} }$) result in the two edges on a corner of the cube (<imgref pic11> (c))


On the right side of <imgref pic11> also the truth table is shown. There, the combinations for each side surface of the cube are marked with the corresponding color. 

<WRAP center>
<imgcaption pic11 | examples in 3-dimensional cube represetation>
</imgcaption>
{{drawio>multicube01.svg}}
</WRAP>

With this representation in mind, we can simplify other representations much more simply. \\
One example for this would be to represent the formula: $Y= X_0 \cdot X_1 \cdot \overline{X_2} + X_2 \cdot X_1 + X_0 \cdot X_1 $. By drawing this into the cube one will see that it represents only a side surface of the cube. It can be simplified into $Y=X_1$.

We can also try to interpret our Therm-o-Safety truth table. The <imgref pic12> shows the corresponding cube. The problem here is, that it is a bit unhandy to reduce a three-dimensional cube onto a flat monitor or paper. It will also get more stressful for higher dimensions.

<WRAP center>
<imgcaption pic12 | Therm-o-Safety in multi-dimensional space>
</imgcaption>
{{drawio>multicube1.svg}}
</WRAP>

Therefore, we try to find a better way to sketch the coordinates, before we simplify our Therm-o-Safety. 
For the three-dimensional Karnaugh map, it is a good idea to "unwrap" the cube. This can be done as shown in <imgref pic14>. 

<WRAP center>
<imgcaption pic14 | flattening the 3-dimensional cube represetation>
</imgcaption>
{{drawio>kmap3D.svg}}
</WRAP>


Out of the flattened cube, we can derive the three-dimensional Karnaugh map (see <imgref pic15>). Generally, there are different ways to show the Karnaugh map.
  - One way is to show the variable names of the dimensions in the corner. Be aware, that the order of the numbers in the horizontal direction is ''0,0'', ''0,1'', ''1,1'', ''1,0'' and not in ascending order! 
  - In other ways, the columns/rows related to the dimension are marked with lines. In this representation, only the ''TRUE'' (=''1'') position of the coordinate is highlighted.
In <imgref pic15> the dimensions are additionally marked with colors. 

In the following chapters mainly the visualization shown in <imgref pic15> (b) is used. 

<WRAP center>
<imgcaption pic15 | different representation of the three-dimensional Karnaugh map I>
</imgcaption>
{{drawio>kmap3D2.svg}}
</WRAP>

With this representation, we can now try to read out the logic terms for the Therm-o-Safety from its Karnaugh map. <imgref pic18> shows the neighboring combinations:
  * In light brown the group (''011'' + ''111'') resp. position 3 and position 7 are shown. This can be simplified into $X_0 \cdot X_1$
  * In light blue the group  (''101'' + ''111'') resp. position 5 and position 7 are shown. This can be simplified into $X_0 \cdot X_2$. This group seems not to be needed, since ''111'' is already in the light brown group.
  * In light violet the group  (''000'' + ''100'') resp. position 0 and position 4 are shown. This can be simplified into $\overline{X_0} \cdot \overline{X_1}$ 

The first two groups are also neighboring cells in the Karnaugh map. For the last one, we have to keep in mind, that we had to cut the surface of the cube to flatten it. 
Therefore, these cells are also neighboring. This leads to the conclusion, that the borders of the Karnaugh map are connected to opposite borders!

For our example, the result would be: (light brown the group) + (light violet the group) $= X_0 \cdot X_1 + \overline{X_0} \cdot \overline{X_1}$

<WRAP center>
<imgcaption pic18 | simplification of the three-dimensional Karnaugh map>
</imgcaption>
{{drawio>kmap3D4.svg}}
</WRAP>

We also have to remember, that there are multiple permutations to show exactly the same logic assignment. This can be interpreted as other ways to unwrap the cube. The <imgref pic17> shows the variant from before at (a). In the image (b) the coordinates are mixed ($X_0 \rightarrow X_1$, $X_1 \rightarrow X_2$, $X_2 \rightarrow X_0$). In image (c) the position of the origin is not in the upper left corner anymore. \\ 
Independent of the permutation, the grouped cells are always neighboring each other. 

<WRAP center>
<imgcaption pic17 | different representation of the three-dimensional Karnaugh map II>
</imgcaption>
{{drawio>kmap3D3.svg}}
</WRAP>


<WRAP column 100%> <panel type="danger" title="Note!"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

  * The Karnaugh map is an alternative way to represent a logical relation.
  * There are possible groups to simplify the logic.
  * We need to take care of whether a group is needed or not. (will be done in chapter 3.3)

</WRAP></WRAP></panel> </WRAP>

==== 3.2.3 Four-Dimensional Karnaugh Map ====

For the four-dimensional Karnaugh map, the situation becomes more complicated in the classical coordinate system. 
The respective object would be a four-dimensional hypercube (see <imgref pic16>). This is hard to print in a two-dimensional layout like on a website or a page.
Here, a four-dimensional Karnaugh map might be a good representation.

<WRAP center>
<imgcaption pic16 | four-dimensional hypercube>
</imgcaption>
{{drawio>kmap4D1.svg}}
</WRAP>

To create a four-dimensional Karnaugh map, we look at first how the two and three-dimensional Karnaugh maps can be derived. The <imgref pic19> (a) shows, that the two-dimensional Karnaugh map can be created from a one-dimensional one by folding the table on the x-axis and adding $10_2$ to all values. The additional line is marked with the new dimension $X_1$. 

The three-dimensional one is created by folding the table on the __y-axis__ and adding $100_2$ to all values. The additional line is marked with the new dimension $X_2$ (<imgref pic19> (b) ). Be aware, that the $X_0$ marking now has to be extended - it is also folded to the right.

The four-dimensional one is created by folding the table again on the __x-axis__ and adding $100_2$ to all values. The additional line is marked with the new dimension $X_2$ (<imgref pic19> (c) ). 

By this, we can visually derive the four-dimensional Karnaough map. 

<WRAP center>
<imgcaption pic19 | four dimensional Karnaugh map>
</imgcaption>
{{drawio>kmap4D3.svg}}
</WRAP>

Again, there are alternative ways to show the Karnaugh map. To get the index of each cell one can easily add up the values of the dimension $X_0 ... X_3$. This is shown in an example in <imgref pic20>. 


<WRAP center>
<imgcaption pic20 | different representations of a four-dimensional Karnaugh map>
</imgcaption>
{{drawio>kmap4D4.svg}}
</WRAP>

To get used to the four-dimensional Karnaugh map, we expand our Therm-o-Safety: Instead of four user-selectable levels for the oven, version 2.0 will have seven. The temperature monitoring ($Y=1$) has to be active starting with level $4$. Additionally, the Therm-o-Safety 2.0 has three non-selectable positions for the case of failure, where the last one needs active temperature monitoring. Some of the combinations (''0000'', ''1000'', ''1001'', ''1010'', ''1011'', ''1100'') are not needed. 

The truth table of the new Therm-o-Safety 2.0 is shown in <imgref pic16>

<WRAP center>
<imgcaption pic21 | truth table of Therm-o-Safety 2.0>
</imgcaption>
{{drawio>kmap4D2.svg}}
</WRAP>

Now we can fill the Karnaugh map:

<WRAP center>
<imgcaption pic22 | Karnaugh map of Therm-o-Safety 2.0>
</imgcaption>
{{drawio>kmap4D5.svg}}
</WRAP>

=== Disjunctive Form ===

The Karnaugh map can now be used to either get the disjunctive form (= sum-of-products) when looking at the groups of $1$s, or the conjunctive form (= product-of-sums) out of the groups of $0$s. 
We will first look at the disjunctive form. There are multiple ways to group the minterms. One is shown here:

<WRAP center>
<imgcaption pic23 | Disjunctive solution of Therm-o-Safety 2.0>
</imgcaption>
{{drawio>kmap4D6.svg}}
</WRAP>

The group $I$ in <imgref pic23> can be expanded since there are don't care states nearby:  

<WRAP center>
<imgcaption pic24 | optimized disjunctive solution of Therm-o-Safety 2.0>
</imgcaption>
{{drawio>kmap4D7.svg}}
</WRAP>

This can be transformed back into a formula. We can derive the boolean terms from the Karnaugh map with the following steps (see <imgref pic26>):
  - Investigate every single group individually. 
  - For the sum-of-products, each group has to be the product (AND-combination) of the inputs/coordinates.
  - Then the formula can be derived as the sum-of-products...
 
<WRAP center>
<imgcaption pic26 | Interpretation of the formula>
</imgcaption>
{{drawio>kmap4D9.svg}}
</WRAP>

The given groups are created as a product as follows:
  * group $I$: all of this minterms are in the columns of $\color{green}{X_3}$. They are also all in the columns of $\color{blue}{X_1}$ and in the rows of $X_0$. This leads to  $X_0 \cdot \color{green}{X_3} \cdot \color{blue}{X_1}$
  * group $II$: all of this minterms are in the columns of $\color{green}{\overline{X_3}}$. They are also all in the rows of $\color{magenta}{X_2}$. This leads to  $\color{green}{\overline{X_3}} \cdot \color{magenta}{X_2}$

Out of these groups, we can get the full formula by disjunctive combination:
\begin{align*}
Y &= X_0 \cdot \color{green}{X_3} \cdot \color{blue}{X_1} + \color{green}{\overline{X_3}} \cdot \color{magenta}{X_2}
\end{align*}

=== Conjunctive Form ===

Similarly, we look at the conjunctive form. Here we have to find groups of $0$s. One way of grouping the maxterms is shown here:

<WRAP center>
<imgcaption pic25 | Conjunctive solution of Therm-o-Safety 2.0>
</imgcaption>
{{drawio>kmap4D8}}
</WRAP>

The optimization would be:

<WRAP center>
<imgcaption pic28 |  optimized conjunctive solution of Therm-o-Safety 2.0>
</imgcaption>
{{drawio>kmap4D11}}
</WRAP>

Also here, we can derive the boolean terms from the Karnaugh map with the following steps (see <imgref pic27>):
  - Investigate every single group individually. 
  - For the product-of-sum, each group has to be the sum (OR-combination) of the inputs/coordinates. \\ Be aware, that for the OR-combination we get the groups of $0$s as the "rest" of not-marked cells! 
  - In the end, the formula can again be derived, now as the product-of-sums.

<WRAP center>
<imgcaption pic27 | Interpretation of the formula>
</imgcaption>
{{drawio>kmap4D10.svg}}
</WRAP>

\begin{align*}
Y &= ( \color{magenta}{X_2}) \cdot( \color{green}{\overline{X_3}} + \color{blue}{X_1})  \cdot( \color{green}{\overline{X_3}} + {X_0})
\end{align*}

Beyond the maxterms this formula can be optimized to 

\begin{align*}
Y &=  \color{magenta}{X_2} \cdot( \color{green}{\overline{X_3}} + (\color{blue}{X_1} \cdot {X_0}) )
\end{align*}

When comparing the disjunctive solution ($X_3 \cdot X_1 \cdot X_0 + \cdot \overline{X_3} \cdot X_2 $) with the conjunctive one ($X_2 \cdot X_1 \cdot X_0 + \cdot \overline{X_3} \cdot X_2 $) we see, that these are definitely different. The given truth table had don't care states. This can be taken as  $0$s or $1$s - and therefore combined in groups of $0$s or groups of $1$s. 


==== 3.2.4 Rules for the Karnaugh map ====

We saw, that with the Karnaugh map, we can analyze logic combinations much better. But to use this tool right we have first to look at some definitions.

=== Allowed Groups ===

As we have seen, the groups in the Karnaugh map are created by the combination of the inputs. By this, only distinct groups are allowed. These can only have $2^{n-1}$ cells for $n$ inputs (= dimensions).

For a four-dimensional Karnaugh map only the following groups are possible:
  * groups of 1: Derived from 4 inputs, e.g. $X_3 \cdot \overline{X_2} \cdot \overline{X_1} \cdot \overline{X_0}$  
  * groups of 2: Derived from 3 inputs, e.g. $ \overline{X_3} \cdot {X_1} \cdot \overline{X_0}$  
  * groups of 4: Derived from 2 inputs, e.g. ${X_2} \cdot {X_0}$  
  * groups of 8: Derived from 1 input , e.g. $\overline{X_1} $  

<WRAP center>
<imgcaption pic29 | allowed groups in a four-dimensional Karnaugh map>
</imgcaption>
{{drawio>kmapallowedgroups.svg}}
</WRAP>

Keep in mind, that not all groups of 2, 4, or 8 cells are allowed. In <imgref pic30> some not allowed groups are shown

<WRAP center>
<imgcaption pic30 | examples for not allowed groups>
</imgcaption>
{{drawio>kmapnotallowedgroups.svg}}
</WRAP>

=== Necessary and Important Groups ===

The sub-chapter before the creation of the groups was done rather intuitively. This shall be explained more structured here. This needs some more definitions.

<callout icon="fa fa-exclamation" color="red" title="Note!">

  * **implicant** (I) is what we called "group" up to now. An implicant can be any allowed combination of minterms ($1$s) or maxterms ($0$s).
  * **prime implicant** (PI) is an implicant, which is not completely included in any other possible implicant.
  * **core prime implicant** (CPI) is a prime implicant, containing at least one term, which is not covered with any other prime implicant.

The (non-core) prime implicant additionally are separated into:
  * **Redundant prime implicants**: These are prime implicants, which are fully contained in a disjunction of multiple core prime implicants.
  * **Selective prime implicants**: These are prime implicants, which are neither core prime implicants nor redundant prime implicants.

</callout>

<imgref pic31> depicts the different implicants in one example.

<WRAP center>
<imgcaption pic31 | different implicants>
</imgcaption>
{{drawio>kmapimplicants}}
</WRAP>

To get all the necessary implicants the following has to be considered:
  * **All core prime implicant** are needed. They contain terms, which are not covered anywhere else.
  * **No redundant prime implicant** is needed. They are covered by core prime implicants
  * **Selective prime implicant** need to be investigated. Some of them are necessary, depending on the solution (see <imgref pic32>)

<WRAP center>
<imgcaption pic32 | selective prime implicants>
</imgcaption>
{{drawio>kmapspi.svg}}
</WRAP>

=== higher dimensional Karnaugh Maps ===

For higher dimensional Karnaugh maps the implicants look more am more unintuitive (see <imgref pic33>).
The Karnaugh map is in our course used to understand the different types of implicants (there will be only three and four-dimensional maps in the exam).

<WRAP center>
<imgcaption pic33 | five dimensional Karnaugh map>
</imgcaption>
{{drawio>kmap5D.svg}}
</WRAP>

When solving higher dimensional boolean problems the [[https://www.mathematik.uni-marburg.de/~thormae/lectures/ti1/code/qmc/index.html|Quine–McCluskey algorithm]] or a [[https://en.wikipedia.org/wiki/Espresso_heuristic_logic_minimizer|heuristic approach, like the ESPRESSO algorithm]] can be used. Languages like the Hardware Description Language (HDL) often come with implemented optimizers in the development environment.

----
==== Exercises ====

<callout>
Tipp: You can check your answer yourself.
  * Online: [[https://www.mathematik.uni-marburg.de/~thormae/lectures/ti1/code/normalform/index.html|Normalforms]] and [[https://www.mathematik.uni-marburg.de/~thormae/lectures/ti1/code/karnaughmap/index.html|K-map]]
  * in the [[https://wiki.mexle.org/introduction_to_digital_systems/tools|Tool digital]]: via 
    * ''Analysis >> Synthesis'' 
    * filling the truth table 
    * change the variable names by right click on the name
    * ''Kmap >> Kmap''

</callout>

<panel type="info" title="Exercise 3.1.1 CNF, DNF, Optimization"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

The following truth table is given.

<WRAP center>
<imgcaption pic50 | Exercise 3.1.1 CNF, DNF, Optimization>
</imgcaption>
{{drawio>Exer311.svg}}
</WRAP>

1. Write down the DNF for the function table. \\ <button size="xs" type="link" collapse="Solution_3_1_1_1_Final_result">{{icon>eye}} Result </button><collapse id="Solution_3_1_1_1_Final_result" collapsed="true">
For the DNF, one has to only use the following lines to get the minterms:
{{drawio>Exer311_res1.svg}}
</collapse> 

2. Minimize the DNF step by step, starting with the used boolean rules \\ <button size="xs" type="link" collapse="Solution_3_1_1_2_Solution_Path">{{icon>eye}} Solution steps </button><collapse id="Solution_3_1_1_2_Solution_Path" collapsed="true">
\begin{align*}
Y &= \overline{X2}\cdot \overline{X1} \cdot X0 + \color{blue}{{X2}\cdot \overline{X1} \cdot X0} + {X2}\cdot {X1} \cdot X0                                          && | \color{blue}{\text{ Idempotence: Double the second term}} \\
  &= \overline{X2}\cdot \overline{X1} \cdot X0 + {X2}\cdot \overline{X1} \cdot X0 + {X2}\cdot \overline{X1} \cdot X0 + {X2}\cdot {X1} \cdot X0 \quad &&  \\
  &= \overline{X2}\cdot \color{blue}{\overline{X1} \cdot X0} + {X2}\cdot \color{blue}{\overline{X1} \cdot X0} + {X2}\cdot \color{violet}{\overline{X1} \cdot X0} + {X2}\cdot  \color{violet}{{X1} \cdot X0} \quad && |  \color{blue}{\text{Distributive Law }} \color{violet}{\text{(twice)}} \\
  &= (\overline{X2}+ X2) \cdot \overline{X1} \cdot X0 + (\overline{X1}+ X1) \cdot {X2} \cdot X0                                                      &&  \\
  &= \color{blue}{(\overline{X2}+ X2)} \cdot \overline{X1} \cdot X0 + \color{violet}{(\overline{X1}+ X1)} \cdot {X2} \cdot X0                        && | \color{blue}{\text{ Complementary Element + Neutral Element }} \color{violet}{\text{(twice)}}\\
  &= \overline{X1} \cdot X0 +  {X2} \cdot X0                                                                                                         && | \color{blue}{\text{ Distributive Law}} \\
  &= (\overline{X1} +  {X2}) \cdot X0                                                                                                                && \\
\end{align*}
</collapse>

<button size="xs" type="link" collapse="Solution_3_1_1_2_Final_result">{{icon>eye}} Result </button><collapse id="Solution_3_1_1_2_Final_result" collapsed="true">
The final result is 
\begin{align*}
Y = (\overline{X1} +  {X2}) \cdot X0
\end{align*}
</collapse> 

3. Write down the CNF for the function table \\ <button size="xs" type="link" collapse="Solution_3_1_1_3_Final_result">{{icon>eye}} Result </button><collapse id="Solution_3_1_1_3_Final_result" collapsed="true">
For the CNF, one has to only use the following lines to get the maxterms:
{{drawio>Exer311_res2.svg}}
</collapse> 

4. Minimize the CNF step by step, starting with the used boolean rules \\
<button size="xs" type="link" collapse="Solution_3_1_1_4_Solution_Path">{{icon>eye}} Solution steps</button><collapse id="Solution_3_1_1_4_Solution_Path" collapsed="true">

\begin{align*}
Y &= (X2 + X1 + X0)_{\color{blue}{I}} \cdot \color{blue}{(X2 + \overline{X1} + X0)}_{\color{blue}{II}} &&\cdot (X2 + \overline{X1} + \overline{X0})_{\color{blue}{III}} \cdot (\overline{X2} + X1 + X0)_{\color{blue}{IV}} &&\cdot (\overline{X2} + \overline{X1} + X0)_{\color{blue}{V}}                                          && | \color{blue}{\text{ Idempotence: Double the second term (II)}} \\
  &= (X2 + X1 + X0)_{\color{blue}{I}} \cdot (X2 + \overline{X1} + X0)_{\color{blue}{II}}               &&\cdot (X2 + \overline{X1} + \overline{X0})_{\color{blue}{III}} \cdot (\overline{X2} + X1 + X0)_{\color{blue}{IV}} &&\cdot (\overline{X2} + \overline{X1} + X0)_{\color{blue}{V}} \cdot (X2 + \overline{X1} + X0)_{\color{blue}{II}} &&| \text{Commutative Law: rearrange the terms} \\
  &= (X2 + X1 + X0)_{\color{blue}{I}} \cdot (\overline{X2} + X1 + X0)_{\color{blue}{IV}}               &&\cdot (X2 + \overline{X1} + X0)_{\color{blue}{II}} \cdot (X2 + \overline{X1} + \overline{X0})_{\color{blue}{III}} &&\cdot (\overline{X2} + \overline{X1} + X0)_{\color{blue}{V}} \cdot (X2 + \overline{X1} + X0)_{\color{blue}{II}}\\
  &= (X2 + \color{blue}{X1 + X0})_{\hphantom{I}} \cdot (\overline{X2} + \color{blue}{X1 + X0})_{\hphantom{IV}} &&\cdot (\color{violet}{X2 + \overline{X1}} + X0)_{\hphantom{II}} \cdot (\color{violet}{X2 + \overline{X1}} + \overline{X0})_{\hphantom{III}} &&\cdot (\overline{X2} + \color{green}{\overline{X1} + X0})_{\hphantom{V}} \cdot (X2 + \color{green}{\overline{X1} + X0})_{\hphantom{II}} &&| \color{blue}{\text{Distributive Law (}} \color{violet}{\text{three }} \color{green}{\text{times}} \color{black}{\text{)}} \\
  &= (X1 + X0) \cdot (X2 + \overline{X2}) &&\cdot (X2 + \overline{X1}) \cdot (X0 + \overline{X0}) &&\cdot (\overline{X2} + X2) \cdot (\overline{X1} + X0) \\
  &= (X1 + X0) \cdot \color{blue}{(X2 + \overline{X2})} &&\cdot (X2 + \overline{X1}) \cdot \color{violet}{(X0 + \overline{X0})} &&\cdot \color{green}{(\overline{X2} + X2)} \cdot (\overline{X1} + X0) &&| \color{blue}{\text{ Complementary Element + Neutral Element (}} \color{violet}{\text{three }} \color{green}{\text{times}} \color{black}{\text{)}} \\
  &=  (X1 + X0) \cdot  (X2 + \overline{X1}) \cdot  (\overline{X1} + X0) \\
  &=  (\color{blue}{X1} + X0) \cdot  (X2 + \overline{X1}) \cdot  (\color{blue}{\overline{X1}} + X0) && && &&| \color{blue}{\text{Distributive Law}}\\
  &=  (X1 + \overline{X1}) \cdot X0 \cdot  (X2 + \overline{X1}) \\
  &=  \color{blue}{(X1 + \overline{X1})} \cdot X0 \cdot  (X2 + \overline{X1}) && && &&| \color{blue}{\text{Complementary Element + Neutral Element}}\\
  &=  X0 \cdot  (X2 + \overline{X1}) \\
  \end{align*}
</collapse>

<button size="xs" type="link" collapse="Solution_3_1_1_4_Final_result">{{icon>eye}} Result </button><collapse id="Solution_3_1_1_4_Final_result" collapsed="true">
The final result is 
\begin{align*}
Y = (\overline{X1} +  {X2}) \cdot X0
\end{align*}
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5. Show, that the un-minimized DNF can be converted into the un-minimized CNF (duality principle) \\ <button size="xs" type="link" collapse="Solution_3_1_1_5_path">{{icon>eye}} Solution Steps</button><collapse id="Solution_3_1_1_5_path" collapsed="true">

The distributive law ($a\cdot(b+c)=a\cdot b + a \cdot c$) can be generalized - similar to the conventional algebra:
\begin{align*}
(a+b)\cdot(c+d) = a\cdot c + a \cdot d + b\cdot c + b\cdot d
\end{align*}

This also works for the dual situation:
\begin{align*}
(a\cdot b)+(c \cdot d) = (a + c) \cdot  (a + d) \cdot  (b + c) \cdot  (b + d)
\end{align*}

This can be applied to the DNF:
\begin{align*}
Y &= \overline{X2}\cdot \overline{X1} \cdot X0 + \color{blue}{{X2}\cdot \overline{X1} \cdot X0} + {X2}\cdot {X1} \cdot X0 
\end{align*}

</collapse> 

6. Create a Karnaugh map and mark the smallest number of the largest prime implicants. Derive the minimized DNF and CNF from the map. \\ <button size="xs" type="link" collapse="Solution_3_1_1_6_Final_result">{{icon>eye}} Result </button><collapse id="Solution_3_1_1_6_Final_result" collapsed="true">
{{drawio>Exer311_res6.svg}}
</collapse> 

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<panel type="info" title="Exercise 3.1.2 Implementation of a BCD-to-7-Segment Decoder"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A full BCD-to-7-Segment Decoder with a positive logic shell be developed
As a base, the following truth table shall be used.
  - Write down the DNF and CNF for the function table and the outputs a...g. Use the don't care states wisely.
  - Optimize the functions by the use of 7 Karnaugh maps. 
  - Show, that the minimized DNF can be converted into the CNF (duality principle)
  - Create a Karnaugh map and mark the smallest number of the largest prime implicants. Derive the minimized DNF and CNF from the map. Mark the smallest number of the largest prime implicants. \\ Write down the derived, smallest formula based on CNF and DNF.
  - Check the expressions with the program "Digital". 
    - Use at first only the DNF and CNF. 
    - After this, use the minimized DNF and the minimized DNF
    - Check whether a 7-Segment-Display works fine with the created logic


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<imgcaption pic51 | Exercise 3.1.2 Implementation of a BCD-to-7-Segment Decoder>
</imgcaption>
{{drawio>Exer312.svg}}
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<panel type="info" title="Exercise 3.1.3 Further Questions"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

  * Write a random allocation of $0$s and $1$s to the variations of inputs. Write the min- and maxterms
  * Compare the results with the output given [[https://www.mathematik.uni-marburg.de/~thormae/lectures/ti1/code/normalform/index.html|here]] (the output $y$ can be changed by clicking onto the grey cells)
  * Use the [[https://www.mathematik.uni-marburg.de/~thormae/lectures/ti1/code/karnaughmap/index.html|interactive example]] and activate "Hide result" plus the push of the "Random example" button to get an individual test for a minterm optimization via a Karnaugh map.
  
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