Started onWednesday, 6 December 2023, 8:12 PM
StateFinished
Completed onWednesday, 6 December 2023, 8:38 PM
Time taken26 mins 47 secs

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Z uporabo Quinove metode določite minimalno disjunktivno obliko funkcije f(x1,x2,x3,x4)=a{2,3}(x1,x2,¯x3)b{1,2}(¯x2,x3,x4). Svoje rešitve po korakih vnesite v vprašanja kviza.

Using the Quine method determine the minimal disjunctive normal form of f(x1,x2,x3,x4)=a{2,3}(x1,x2,¯x3)b{1,2}(¯x2,x3,x4). Submit your step-by-step solution by answering the questions of this quiz.

Question 1

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Marked out of 2.00

Question text

Dopolnite pravilnostno tabelo za $f(x_1, x_2, x_3, x_4)$.

$x_1$ 
$x_2$
$x_3$
 $x_4$  
$f(x_1, x_2, x_3, x_4)$
0 0 0 0
0
0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0
1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1



Question 2

Complete
Marked out of 3.00

Question text

Tabelo vsebovalnikov lahko naložite z izbirnikom datotek ali jo vnesete v spodnjo tabelo. Če se odločite za slednje, lahko negacije zapisujete s klicajem kot v naslednjem primeru: $m_1 =$ !x1 !x2 !x3 x4.

Upload your table of implicants or fill out the table below. In case of the latter, you can write down the negations using an exclamation mark as in the following example: $m_1 = $ !x1 !x2 !x3 x4.

Answer text Question 2


4 3 2 1
(1)
(1)
(1)
(1)
(2)
(2)
(2)
(2)
(3)
(3)
(3)
(3)
(4)
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(8)
(8)
(9)
(9)
(9)
(9)


Answer files Question 2

Question 3

Complete
Marked out of 3.00

Question text

Z uporabo izbirnika datotek naložite tabelo pokritij ali izpolnite spodnjo tabelo.

Upload the essential prime implicant table or fill out the table below.

Answer text Question 3




Answer files Question 3

Question 4

Complete
Marked out of 2.00

Question text

Zapišite MDNO funkcije $f(x_1, x_2, x_3, x_4)$. Negacijo lahko zapišete s klicajem, disjunkcijo pa s črko v, npr. $m_0 \vee m_1 =$ !x1 !x2 !x3 !x4 v !x1 !x2 !x3 x4..

Write down the MDNF of $f(x_1, x_2, x_3, x_4)$. You can write down the negations with exclamation marks and disjunctions using letter v, e.g. $m_0 \vee m_1 =$ !x1 !x2 !x3 !x4 v !x1 !x2 !x3 x4.