Integer Variables - Which Three Logical Expressions Are Equivalent To Each Other
Apr 12, 2014
Assuming that x, y, and z are integer variables, which of the following three logical expressions are equivalent to each other, that is, have equal values for all possible values of x, y, and z?
(x == y && x != z) || (x != y && x == z)
(x == y || x == z) && (x != y || x != z)
(x == y) != (x == z)
None of the three
A. I and II only
B. II and III only
C. I and III only
D. I, II, and III
I selected B, but got it wrong. I really think I need understanding boolean logic. The correct answer says something else but I don't get the logic. Here is the correct answer:
Answer Key : The following model answer has been provided to you by the grader. Carefully compare your answer with the one provided here.
Expression III is the key to the answer: all three expressions state the fact that exactly one out of two equalities, x == y or x == z, is true. Expression I states that either the first and not the second or the second and not the first is true. Expression II states that one of the two is true and one of the two is false. Expression III simply states that they have different values. All three boil down to the same thing. The answer is E.
In exercise 4, I get the same problem:
The expression !((x <= y) && (y > 5)) is equivalent to which of the following?
A. (x <= y) && (y > 5)
B. (x <= y) || (y > 5)
C. (x >= y) || (y < 5)
D. (x > y) || (y <= 5)
E. (x > y) && (y <= 5)
Exercise 4
ABCDE
Incorrect
Score: 0 / 1
Submitted: 2/10/2014 8:21pm
Your answer is incorrect.
Answer Key
The following model answer has been provided to you by the grader. Carefully compare your answer with the one provided here. The given expression is pretty long, so if you try to plug in specific numbers you may lose a lot of time. Use De Morgan's Laws instead:
!((x <= y) && (y > 5))
!(x <= y) || !(y > 5)
When ! is distributed,
&& changes into ||, and vice-versa
I write a small program. I want post my integer variables in the next class. Here a code(I wanna change answer color):
Main class(pagrindine.java): import java.awt.*; import javax.swing.*; import java.awt.event.*; import javax.swing.event.*; //Mokinio rinkinukas su skaiciofke kur gali keisti atsakymo spalva, pabandyti sukurti uzrasu knygute.
[code]....
So, I want to post variables r,g,b from class slideris.java in the pagrindiness.java.
a. Create an application named ArithmeticMethods whose main() method holds two integer variables. Assign values to the variables. In turn, pass each value to methods named displayNumberPlus10(), displayNumberPlus100(), and displayNumberPlus1000(). Create each method to perform the task its name implies.
b. Modify the ArithmeticMethods class to accept the values of the two integers from a user at the keyboard.
I was almost finished with my program when I noticed I had a big logical error. At the very beginning of my else if statements I would jump down to my else statement.
Java Code:
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The title says already where my difficulties are. I forgot to say, the "S" printing part works, but why the others doesn't. To make it more clear:
java Monoton 1 3 3 4 & java Monoton 1 3 4 1
=> nothing (my output)
I forgot to notice, with 1 3 3 4 as parameters it jumps out at if (b < c), which is expected. but it jumps out of the whole if instead just run the else part. that's the essence of my problem.
The Exercise:
=> The program should print the following letters;
S, if every number is true bigger than the before,
M, if every number is bigger or equal than the before,
N in the other cases
Examples (from the exercise)
=> java Monoton 0 1 2 4
S
=> java Monoton 1 3 3 4
M
=> java Monoton 1 3 4 1
N
The Exercise should done without the use of logical operators or other combined requirements.
Java Code:
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[Code] ....
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