Boole'sche Ausdrücke (Selektionen)
Die Symbole "|" bzw. "&" stehen hier für die logische ODER- bzw. die logische UND-Verknüpfung. "&" wird prioritär vor "|" ausgeführt. Das Symbol "!" bezeichne die logische Umkehrung (Negation).
- Berechnen Sie damit (true | false) & (false | !true).
- Berechnen Sie die folgenden Ausdrücke mit der Bedingung, dass a den Wert 5 und x den Wert (-7) hat. Beachte, dass die logischen Operatoren stärker binden (Klammern) als die Gleichheit.
- 17 < 19
- -3 < -8
- (4 > 3) & (3 > 4)
- (8 % 3 = 0) | (8 % 2 = 0)
- (-13 < -8) | (-1 < 0) & (2 > 3)
- true | false = (5 < 2 * 3) & (false & true)
- a < x
- -x < -a - 2
- Gegeben a = true, b = false, c = true. Berechnen Sie:
- a & (b | c)
- (b & a) | c
- b & (a | c)
- (a & b) | (a & c)
- (( ! a) | (! b)) & ((! a) | (! c))
- Versuchen Sie die folgenden logischen Ausdrücke nur mit den Operatoren AND und NOT darzustellen. Ein ODER (|) darf also nicht mehr vorkommen.
- ! (a | b)
- a | (b & c)
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5 Lösung(en)
def test():
a = 5
x = -7
print 17 < 19
print -3 < -8
print (4 > 3) & (3 > 4)
print (8 % 3 == 0) | (8 % 2 == 0)
print (-13 < -8) | (-1 < 0) & (2 > 3)
print True | False == (5 < 2 * 3) & (False & True)
print a < x
print -x < -a - 2
a = True
b = False
c = True
print a & (b | c)
print (b & a) | c
print b & (a | c)
print (a & b) | (a & c)
print (( not a) | (not b)) & ((not a) | (not c))
test()
*process langlvl(saa2);
*process limits(fixeddec(31));
*process aggregate, attributes;
*process flag(W), source, insource, options, xref,nest, number, offset;
*process gonumber, snap;
*process or ('!') not ('^'); /* module specific */
Boole : proc options(main);
/* ********************************************************** */
/* Autor : philipp gressly freimann (@ sanits-training.ch) */
/* Datum : 10. Nov. 2011 */
/* Aufgabe 3.2 (Programmieraufgaben.ch: Boolesche Ausdruecke. */
/* ********************************************************** */
dcl TRUE bit(1) value('1'b);
dcl FALSE bit(1) value('0'b);
/* Teilaufgabe a) */
put skip list("Teilaufgabe a)");
if(TRUE ! FALSE) & (FALSE ! (^TRUE)) then do;
put skip list("TRUE!");
end;
put skip list("Ist (TRUE!) noch nicht erschienen, so hat");
put skip list(" Teilaufgabe a) FALSE als Loesung.");
/* Teilaufgabe b) */
put skip list("Teilaufgabe b)");
dcl aa fixed(15) init(5);
dcl ax fixed(15) init(-7);
put skip list("17 < 19", 17 < 19);
put skip list("-3 < -8", -3 < -8);
put skip list("(4>3) & (3 > 4)", ((4 > 3) & (3 > 4)));
put skip list(" 8 mod 3 = 0 or 8 MOD 2 = 0",
(0 = MOD(8, 3)) ! (0 = MOD(8,2)));
put skip list("(-13 < -8) ! (-1 < 0) & (2 > 3)" ,
(-13 < -8) ! (-1 < 0) & (2 > 3));
put skip list ("TRUE OR FALSE = (5 < 2*3) & (FALSE & TRUE)",
((TRUE ! FALSE) = ((5 < 2*3) & (FALSE & TRUE))) );
put skip list ("a < x", aa < ax);
put skip list ("x < -a-2", ax < -aa - 2);
/* Teilaufgabe c) */
put skip list("Teilaufgabe c)");
DCL a BIT(1), b BIT(1), c BIT(1);
a = TRUE ;
b = FALSE;
c = TRUE ;
put skip list("a & (b ! c): ",
(a & (b ! c)) );
put skip list("(b & a) ! c: ",
((b & a) ! c));
put skip list("b & (a ! c): ",
b & (a ! c) );
put skip list("(a & b) ! (a & c): ", (a & b) ! (a & c));
put skip list("((^a) ! (^b)) & ((^a) ! (^c)): ",
((^a) ! (^b)) & ((^a) ! (^c)) );
END Boole;
Lösung von: Philipp G. Freimann (BBW (Berufsbildungsschule Winterthur) https://www.bbw.ch)
package ch.programmieraufgaben.selektion;
/**
* Aufgabe 3.2 aus Programmieren lernen
* @author Philipp Gressly (phi AT gressly DOT ch)
*/
public class BoolescheAusdruecke {
public static void main(String[] args) {
new BoolescheAusdruecke().top();
}
void top() {
aufgabeA();
aufgabeB();
aufgabeC();
aufgabeD();
}
void aufgabeA() {
System.out.println("Aufgabe a)");
System.out.println((true | false) & (false | !true));
}
void aufgabeB() {
System.out.println("Aufgabe b)");
int a = 5;
int x = - 7;
System.out.println(17 < 19);
System.out.println(-3 < -8);
System.out.println((4>3) & (3 > 4));
System.out.println((8%3 == 0) | (8%2 == 0));
System.out.println((-13 < -8) | (-1 < 0) & (2 > 3));
System.out.println((true | false) == ((5<2*3) & (false & true)));
System.out.println(a < x);
System.out.println(-x < -a - 2);
}
void aufgabeC() {
System.out.println("Aufgabe c)");
boolean a = true, b = false, c = true;
System.out.println(a & (b | c));
System.out.println((b&a) | c);
System.out.println(b & (a|c));
System.out.println((a&b) | (a&c));
System.out.println(((!a) | (!b)) & ((!a) | (!c)));
}
void aufgabeD() {
boolean a = true, b = false, c = true; // nur ein Beispiel
System.out.println("Aufgabe d)");
System.out.println("(!a) & (!b) : " + ((!a) & (!b)));
System.out.println("!(!(a) & !(b & c)): " + (!(!(a) & !(b & c))));
}
} // end of class BoolescheAusdruecke
Lösung von: Philipp G. Freimann (BBW (Berufsbildungsschule Winterthur) https://www.bbw.ch)
package ch.santis_basis_modul_755.selection_fork;
/**
* @author Gilda.Thode
* @version 1.0
* 2018/02/28
*/
public class Exercise_3_2 {
public Exercise_3_2() {
// TODO Auto-generated constructor stub
}
public static void main(String[] args) {
// TODO Auto-generated method stub
new Exercise_3_2().exercises();
}
void exercises() {
exercise_a();
exercise_b();
exercise_c();
exercise_d();
}
private void exercise_a() {
// TODO Auto-generated method stub
print ("exercise_a\n");
print (true | false & false | !true);
print ("");
}
@SuppressWarnings({ "unused" })
private void exercise_b() {
// TODO Auto-generated method stub
int a = 5;
int x = -7;
print ("exercise_b\n");
if (17 < 19 ) {
print ("1. condition ", true);
}
if (-3 < -8) {
print ("2. condition ", true);
}
if (4 > 3 & 3 > 4) {
print ("3. condition ", true);
}
if ((8 % 3 == 0) || (8 % 2 == 0)) {
print ("4. condition ", true);
}
if ((-13 < -8) || (-1 < 0) && (2 > 3)) {
print ("5. condition ", true);
}
if ((true || false) == ((5 < 2 * 3) && (false && true))) {
print ("6. condition ", true);
}
if (a < x) {
print ("7. condition ", true);
}
if (- x < -a - 2) {
print ("8. condition ", true);
}
print ("");
}
private void exercise_c() {
// TODO Auto-generated method stub
boolean a = true, c = true;
boolean b = false;
print ("exercise_c\n");
if (a && (b || c)) {
print ("1. condition ", true);
}
if ((b && a) || c) {
print ("2. condition ", true);
}
if (b && (a || c)) {
print ("3. condition ", true);
}
if ((a && b) || (a && c)) {
print ("4. condition ", true);
}
if (((! a) || (!b)) && ((! a) || (! c))) {
print ("5. condition ", true);
}
print ("");
}
//exercise d is the secondary part of exercise c --> see the assignments of the booleans
//ATTENTION: the terms are only value for this assignments
private void exercise_d() {
// TODO Auto-generated method stub
boolean a = true, c = true;
boolean b = false;
print ("exercise_d\n");
if (!a & !b) {
print ("1. condition ", true);
}
if (!(!(a) & !((b) && !!(c)))) {
print ("2. condition ", true);
}
print ("");
//the different possibilities for the terms
testab (true, true);
testab (true, false);
testab (false, true);
testab (false, false);
testabc (true, true, true);
testabc (true, true, false);
testabc (true, false, true);
testabc (true, false, false);
testabc (false, true, true);
testabc (false, true, false);
testabc (false, false, true);
testabc (false, false, false);
}
//test methods for both terms
private void testab(boolean a, boolean b) {
// TODO Auto-generated method stub
if ((!(a | b)) == (!a & !b)) {
print ("Okay nice, first term is correct!");
} else {
print ("Error");
}
print ("");
}
private void testabc(boolean a, boolean b, boolean c) {
// TODO Auto-generated method stub
if ((a | (b & c) == !(!(a) & !((b) && !!(c))))) {
print ("Okay nice, second term is correct!");
} else {
print ("Error");
}
print ("");
}
//three print methods
private void print(boolean truefalse) {
// TODO Auto-generated method stub
System.out.println(truefalse);
}
private void print(String string) {
// TODO Auto-generated method stub
System.out.println(string);
}
private void print(String string, boolean b) {
// TODO Auto-generated method stub
System.out.println(string + b);
}
}//end of class
Lösung von: Gilda Thode (SANTIS Training AG)
/* (true | false) & (false | !true)
\----/
<=> (true | false) & (false | false)
\------------/ \-------------/
<=> true & false
\--------------------/
<=> false */
console.log( (true || false) && (false || !true) ); // false
let a = 5, x = -7;
console.log( 17 < 19 ); // true
console.log( -3 < -8 ); // false
/* (4 > 3) & (3 > 4)
\-----/ \-----/
<=> true & false
\-------------/
<=> false */
console.log( (4 > 3) && (3 > 4) ); // false
/* (8 % 3 = 0) | (8 % 2 = 0)
\---------/ \---------/
<=> false | true
\----------------/
<=> true */
console.log( (8 % 3 == 0) || (8 % 2 == 0) ); // true
/* (-13 < -8) | (-1 < 0) & (2 > 3)
\--------/ \------/ \-----/
<=> true | true & false
\-------------/ (& vor |)
<=> true | false
\--------------------/
<=> true */
console.log( (-13 < -8) || (-1 < 0) && (2 > 3) ); // true
/* true | false = (5 < 2 * 3) & (false & true)
\----------/ \----/ \------------/
<=> true = (5 < 6 ) & false
\---------/
<=> true = true & false
\-------------------/
<=> true = false
\------------------------/
<=> false */
console.log( true || false == (5 < 2 * 3) && (false && true) ); // true (!)
// hier muss noch eine klammer gesetzt werden:
console.log( (true || false) == (5 < 2 * 3) && (false && true) ); // false
/* a < x
\---/
<=> 5 < -7 <=> false */
console.log( a < x ); // false
/* -x < -a - 2
\/ \----/
<=> +7 < -5 - 2
\----/
<=> 7 < -7 <=> false */
console.log( -x < -a - 2 ); // false
a = true;
let b = false, c = true;
/* a & (b | c)
| \--------------/
<=> true & (false | true)
\------------/
<=> true & true
\--------------/
true */
console.log( a && (b || c) ); // true
/* (b & a ) | c
\------------/ |
<=> (false & true) | true
\------------/
<=> false | true
\--------------/
<=> true */
console.log( (b && a) || c ); // true
/* b & (a | c)
| \-----------/
<=> false & (true | true)
\-----------/
<=> false & true
\--------------/
false */
console.log( b && (a || c) ); // false
/* ( a & b ) | ( a & c )
| | | |
<=> (true & false) | (true & true)
\------------/ \-----------/
<=> false | true
\-------------------/
<=> true */
console.log( (a && b) || (a && c) ); // true
/* ((!a) | (!b)) & ((!a) | (!c))
| | | |
<=> ((!true) | (!false)) & ((!true) | (!true))
\-----/ \------/ \-----/ \------/
<=> (false | true) & (false | false)
\---------------/ \----------------/
<=> true & false
\--------------------------/
<=> false */
console.log( ((!a) || (!b)) && ((!a) || (!c)) ); // false
// anwendung des gesetzes nach de Morgan
// https://de.wikipedia.org/wiki/De-morgansche_Gesetze
// in kürze: diskunktion kann man durch negation und konjunktion ersetzen
/*
I.
a b
0 0: !(a | b) <=> !a & !b
\-----/ \/ \/
!0 <=> 1 & 1
\/ \----/
1 <=> 1 (1)
0 1: !(a | b) <=> !a & !b
\-----/ \/ \/
!1 <=> 1 & 0
\/ \----/
0 <=> 0 (1)
1 0: !(a | b) <=> !a & !b
\-----/ \/ \/
!1 <=> 0 & 1
\/ \----/
0 <=> 0 (1)
1 1: !(a | b) <=> !a & !b
\-----/ \/ \/
!1 <=> 0 & 0
\/ \----/
0 <=> 0 (1) */
console.log( !(a || b) == !a && !b );
/*
II.
a b c
0 0 0: a | (b & c) <=> !(!a & !(b & c)
| \-----/ | \-----/
0 | 0 <=> !(1 & ! 0)
\-------/ \----/
0 <=> !(1 & 1)
\---------/
0 <=> ! 1
\-----/
0 <=> 0 (1)
0 0 1: a | (b & c) <=> !(!a & !(b & c)
| \-----/ | \-----/
0 | 0 <=> !(1 & ! 0)
\-------/ \----/
0 <=> !(1 & 1)
\---------/
0 <=> ! 1
\-----/
0 <=> 0 (1)
0 1 0: a | (b & c) <=> !(!a & !(b & c)
| \-----/ | \-----/
0 | 0 <=> !(1 & ! 0)
\-------/ \----/
0 <=> !(1 & 1)
\---------/
0 <=> ! 1
\-----/
0 <=> 0 (1)
0 1 1: a | (b & c) <=> !(!a & !(b & c)
| \-----/ | \-----/
0 | 1 <=> !(1 & ! 1)
\-------/ \----/
1 <=> !(1 & 0)
\--------/
1 <=> ! 0
\----/
1 <=> 1 (1)
1 0 0: a | (b & c) <=> !(!a & !(b & c)
| \-----/ | \-----/
1 | 0 <=> !(0 & ! 0)
\-------/ \----/
1 <=> !(0 & 1)
\--------/
1 <=> ! 0
\----/
1 <=> 1 (1)
1 0 1: a | (b & c) <=> !(!a & !(b & c)
| \-----/ | \-----/
1 | 0 <=> !(0 & ! 0)
\------/ \----/
1 <=> !(0 & 1)
\--------/
1 <=> ! 0
\----/
1 <=> 1 (1)
1 1 0: a | (b & c) <=> !(!a & !(b & c)
| \-----/ | \-----/
1 | 0 <=> !(0 & ! 0)
\------/ \----/
1 <=> !(0 & 1)
\--------/
1 <=> ! 0
\----/
1 <=> 1 (1)
1 1 1: a | (b & c) <=> !(!a & !(b & c)
| \-----/ | \-----/
1 | 1 <=> !(0 & ! 1)
\------/ \---/
1 <=> !(0 & 0)
\--------/
1 <=> ! 0
\----/
1 <=> 1 (1) */
console.log( a || (b && c) == !(!a && !(b && c)) ); // lissalanda@gamx.at
Lösung von: Lisa Salander (Heidi-Klum-Gymnasium Bottrop)
Verifikation/Checksumme:
- false
- Antworten:
- 17 < 19 -> true
- -3 < -8 -> false
- (4 > 3) & (3 > 4) -> false
- (8 % 3 = 0) | (8 % 2 = 0) -> true
- (-13 < -8) | (-1 < 0) & (2 > 3) -> true
- true | false = (5 < 2*3) & (false & true) -> false
- a < x -> false
- -x < -a - 2 -> false
- true, true, false, true, false
-
- (! a) & (! b)
- !(!(a) & !(b & c))
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Meta
| Zeit: | 0.5 |
| Schwierigkeit: | Mittel |
| Webcode: | tby9-5gzh |
| Autor: | Philipp G. Freimann (BBW (Berufsbildungsschule Winterthur) https://www.bbw.ch) |