Difference between revisions of "Aufgaben:Exercise 1.2: Decimal/Binary Converter"

From LNTwww
Line 72: Line 72:
  
  
⇒  &nbsp;<u>Proposed solutions 2 and 4</u> are correct.
+
⇒  &nbsp;<u>Proposed solutions 2 and 4</u>&nbsp; are correct.
  
  
'''(2)'''&nbsp; The event&nbsp; $V$&nbsp; consists of the two numbers&nbsp; $4$&nbsp; (binary 0100) and&nbsp; $6$&nbsp; (binary 0110)  &nbsp;
+
'''(2)'''&nbsp; The event&nbsp; $V$&nbsp; consists of the two numbers&nbsp; $4$&nbsp; (binary 0100) and&nbsp; $6$&nbsp; (binary 0110)  &nbsp; ⇒  &nbsp;  The correct <u>solutions are 1 and 3</u>.
 
⇒  &nbsp;  The correct <u>solutions are 1 and 3</u>.
 
  
  
[[File:P_ID2848__Sto_A_1_2c.png|right|frame|Auxiliary Venn-diagram]]
+
[[File:P_ID2848__Sto_A_1_2c.png|right|frame|Auxiliary Venn diagram]]
'''(3)'''&nbsp; For the event&nbsp; $W$&nbsp;, de Morgan's theorem holds:
+
'''(3)'''&nbsp; For the event&nbsp; $W$,&nbsp; de Morgan's theorem holds:
  
 
:$$\overline W = \overline A \cup \overline D \cup (\overline B \cap C) \cup (B \cap \overline C)
 
:$$\overline W = \overline A \cup \overline D \cup (\overline B \cap C) \cup (B \cap \overline C)
 
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} W = \overline{\overline W} = A \cap D \cap (\overline{\overline B \cap C}) \cap (\overline{B \cap \overline C}).$$
 
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} W = \overline{\overline W} = A \cap D \cap (\overline{\overline B \cap C}) \cap (\overline{B \cap \overline C}).$$
  
*Using de Morgan's theorems, it further follows:
+
*Using de Morgan's theorems,&nbsp; it further follows:
  
 
:$$ W = A \cap D \cap (B \cup \overline C) \cap (\overline B \cup C).$$
 
:$$ W = A \cap D \cap (B \cup \overline C) \cap (\overline B \cup C).$$
  
*Finally, using the Boolean relation&nbsp; $(B \cup \overline C) \cap (\overline B \cup C) = (B \cap C) \cup (\overline B \cap \overline C)$&nbsp; we obtain (see sketch):
+
*Finally,&nbsp; using the Boolean relation&nbsp; $(B \cup \overline C) \cap (\overline B \cup C) = (B \cap C) \cup (\overline B \cap \overline C)$&nbsp; we obtain&nbsp; (see sketch):
  
 
:$$W = (A \cap B \cap C \cap D) \cup (A \cap \overline B \cap \overline C \cap D).$$
 
:$$W = (A \cap B \cap C \cap D) \cup (A \cap \overline B \cap \overline C \cap D).$$
  
*Thus,&nbsp; $W$&nbsp; contains the numbers&nbsp; $15$&nbsp; and&nbsp; $9$&nbsp;  &nbsp;  
+
*Thus,&nbsp; $W$&nbsp; contains the numbers&nbsp; $15$&nbsp; and&nbsp; $9$&nbsp;  &nbsp; ⇒  &nbsp;  only the&nbsp; <u>proposed solution 1</u>&nbsp; is correct.
  
 
⇒  &nbsp;  only the <u>proposed solution 1</u> is correct.
 
  
  
 
'''(4)'''&nbsp; The union of&nbsp; $U$,&nbsp; $V$&nbsp; and&nbsp; $W$&nbsp; contains the following numbers: &nbsp; $4, 6, 8, 9, 10, 12, 14, 15$.
 
'''(4)'''&nbsp; The union of&nbsp; $U$,&nbsp; $V$&nbsp; and&nbsp; $W$&nbsp; contains the following numbers: &nbsp; $4, 6, 8, 9, 10, 12, 14, 15$.
  
*Accordingly, the set&nbsp; $P$&nbsp; as the complement of this union is: &nbsp; $P \in {\{1, 2, 3, 5, 7, 11, 13\}}$.
+
*Accordingly, the set&nbsp; $P$&nbsp; as the complement of this union is: &nbsp;  
 +
:$$P = {\{1, 2, 3, 5, 7, 11, 13\}}.$$
  
 
*These are exactly the prime numbers which can be represented with four bits  &nbsp; ⇒  &nbsp;  <u>Proposed solution 2</u>.
 
*These are exactly the prime numbers which can be represented with four bits  &nbsp; ⇒  &nbsp;  <u>Proposed solution 2</u>.

Revision as of 14:06, 19 November 2021

Logical circuit for D/B converting

A number generator  $Z$  supplies decimal values in the range  $1$  to  $15$.

  • These are converted into binary numbers  (block outlined in red).
  • The output consists of the four binary values  $A$,  $B$,  $C$  and  $D$  with decreasing significance.
  • For example  $Z = 11$  delivers the binary values
$$ A = 1, \ B = 0, \ C = 1, \ D = 1. $$

Set-theoretically,  this can be represented as follows:

$$ Z = 11\qquad\widehat{=}\qquad A \cap\overline{ B} \cap C \cap D.$$

Three more Boolean expressions are formed from the binary quantities  $A$,  $B$,  $C$  and  $D$  and their union set is denoted  $X$ :

\[ U = A \cap \overline{D} \]
\[ V = \overline{A} \cap B \cap \overline{D} \]
$$W,\; {\rm where} \; \, \overline{W} = \overline{A} \cup \overline{D} \cup (\overline{B} \cap C) \cup (B \cap \overline{C}). $$
  • Note that  $Z = 0 \ ⇒ \ A = B = C = D = 0$  is already excluded by the number generator.
  • Note also that not all input quantities  $A$,  $B$,  $C$  and  $D$  are used to calculate all intermediate quantities  $U$,  $V$  and  $W$,  resp.



Hints:

  • The exercise belongs to the chapter  Set theory basic.
  • The topic of this chapter is illustrated with examples can be found in the (German language) learning video: 
"Mengentheoretische Begriffe und Gesetzmäßigkeiten"   ⇒   "Set-theoretical terms and laws".



Questions

1

Which statements are true regarding the random variable  $U$?

$U$  contains two elements.
$U$  contains four elements.
The smallest element of  $U$  is  $4$.
The largest element of  $U$  is  $14$.

2

Which statements are true regarding the random variable  $V$?

$V$  contains two elements.
$V$  contains four elements.
The smallest element of  $V$  is  $4$.
The largest element of  $V$  is  $14$.

3

Which statements are true regarding the random variable  $W$?

$W$  contains two elements.
$W$  contains four elements.
The smallest element of  $W$  is  $4$.
The largest element of  $W$  is  $14$.

4

Which statements are true regarding the random quantity  $P$ ?

$P$  contains all powers of two.
$P$  contains all prime numbers.
$P$  describes the empty set  $\phi$.
$P$  is identical with the universal set  $G = {1,2, \ \text{...} \ , 15}$.


Solution

(1)  The event  $U$  contains

  • those numbers greater/equal to eight  $(A = 1)$, 
  • which are even  $(D = 0)$:  $8, 10, 12, 14$  


⇒  Proposed solutions 2 and 4  are correct.


(2)  The event  $V$  consists of the two numbers  $4$  (binary 0100) and  $6$  (binary 0110)   ⇒   The correct solutions are 1 and 3.


Auxiliary Venn diagram

(3)  For the event  $W$,  de Morgan's theorem holds:

$$\overline W = \overline A \cup \overline D \cup (\overline B \cap C) \cup (B \cap \overline C) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} W = \overline{\overline W} = A \cap D \cap (\overline{\overline B \cap C}) \cap (\overline{B \cap \overline C}).$$
  • Using de Morgan's theorems,  it further follows:
$$ W = A \cap D \cap (B \cup \overline C) \cap (\overline B \cup C).$$
  • Finally,  using the Boolean relation  $(B \cup \overline C) \cap (\overline B \cup C) = (B \cap C) \cup (\overline B \cap \overline C)$  we obtain  (see sketch):
$$W = (A \cap B \cap C \cap D) \cup (A \cap \overline B \cap \overline C \cap D).$$
  • Thus,  $W$  contains the numbers  $15$  and  $9$    ⇒   only the  proposed solution 1  is correct.


(4)  The union of  $U$,  $V$  and  $W$  contains the following numbers:   $4, 6, 8, 9, 10, 12, 14, 15$.

  • Accordingly, the set  $P$  as the complement of this union is:  
$$P = {\{1, 2, 3, 5, 7, 11, 13\}}.$$
  • These are exactly the prime numbers which can be represented with four bits   ⇒   Proposed solution 2.