Having these two bitsets as an example:
1 1 0 0
1 0 0 1
How can I get this following result:
0 1 0 0
From my point of view this kind of logic operation must be a subtraction but omitting the borrow bit.
Is it possible to have a simple logical operation to get that result?
Thanks in advance.
it is only true, if A is true and B is false.
So it can be solved as A & ~B
Related
So this is the truth table:
In_1 In_2 In_3 Out
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 1
How do I create a circuit that matches this truth table?
Here is what I have tried, but this did not work:
Please help would be appreciated.
You just need to connect the output to a constant logic 1, which is usually just the power supply voltage. This is a trivial logic function and requires no gates to implement.
The solution is to apply Kmap. We will be ending up with an equation that will be similar to
but this is not the reduced form. If we try to reduce this equation we will end up with 1.
Check this tool for better understanding.
https://www.boolean-algebra.com/kmap/
So this is the truth table
In_1 In_2 In_3 Out
0 0 0 0
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 0
I would like to create a circuit based on this truth table.
This is what I have tried, but failed
A Karnaugh map as suggested by paddy, will give you a set of minterms which fulfil the expression. That is the classical way to tackle such problems.
By inspection of the truth-table you can convince yourself, that the output is true whenever In_1 is unequal In_3 or In_1 is unequal In_2:
f = (In_1 xor In_2) or (In_1 xor In_3)
I'm trying to construct a multiplex gate. It has two inputs, and one selector. I got as far as
the truth table.
A | B | Sel | Out
0 0 1 0
0 1 1 0
1 0 1 1
1 1 1 1
0 0 0 0
0 1 0 1
1 0 0 0
1 1 0 1
And this is where my method fails. I've constructed simpler gates such as AND, and OR. Those were so simple I didn't need an articulate method. I went to wikipedia to see if I could get
a method. Instead I only discovered which gates I need to construct the circuit. For my goals, this misses the point. More important to me is the method that arrives at the answer, rather than the answer itself. I know I need to use DeMorgan's Laws, but fall down when trying to come up with specifics. Any hints would be most welcome.
Just to elaborate on Keith's answer, here's the Karnaugh map for your truth table:
AB
00 01 11 10
___________
sel 0 | 0 1 1 0
1 | 0 0 1 1
This is created by grouping A and B, and then making a matrix of the outputs for any given input. Note the column headings do not count in binary, rather they are more like a grey code, having only one transition between each column.
Now that's done, you can write an equation that ORs together terms that cover all the 1s in the Karnaugh map.
On the Karnaugh map, it's pretty easy to see terms that cover multiple 1s. For example, the term B.sel' (B and not sel) covers both the 1's in the top row.
That combined with A.sel for the 1's in the bottom row gives the equation
output = B.sel' + A.sel
This works out at 4 gates, including the NOT.
You can make a Karnaugh Map, which will help you pick the gates you need to implement your function.
I have a table with different sorts of things. The thing is that there are columns that just some of the rows fill with data. Because the other rows doesn't need that. Is it bad to have some cells empty? Should I store all of then in different tables and have a main table where columns that all objects need are? And then when I want to select something I do joins instead?
It is not bad to have some empty cells. Mostly you need only multiple tables when you can have a 1 on more relation between the tables.
From the little information you have provided, this might be a case of storing information about different types of objects in one table, like cars, people, houses etc.
And forgive me in advance for my overly simplistic example, could not think of a better one.
People have attributes like name, family name, birth date. Cars have attributes like brand, color, number of doors, engine volume etc. Houses have area, number of bedrooms, garden (yes or no) etc.
When such a table would be populated with entries, it would look like this:
id name attr1 atttr2 attr3 attr4 attr5 attr6
1 car1 1 1 0 0 0 0
2 car2 1 1 0 0 0 0
3 man1 0 0 1 1 0 0
4 man2 0 0 1 1 0 0
5 house1 0 0 0 0 1 1
6 house2 0 0 0 0 1 1
1 means cell has some data, 0 means cell is empty.
If your table looks like this, you might be better off splitting your table into more tables, as database normalization suggests.
I'm needing to create a truth table, and I really need to find a resource to explain how it works. I'll give an example of a problem.
I have to create a truth table based on this: A*(B+AB)=AB
So the truth table looks something like:
0 0
0 1
1 0
1 1 for A*(B+AB)=AB
How do I even begin to solve this? Are there any good resources that give a good explanation on what to do?
Ok So I then did one more complicated that involves a NOT.
! indicates not
!(A*!B+!AB) = AB+!(A+B)
So I did C = A*!B D=!A*B then !(C+D) for the left side. My final answer for that side is
0 0 1
0 1 0
1 0 0
0 0 1
So the right side is this
C = A * B D = A + B then C + !D
so that looked like this
0 0 1
0 1 0
0 1 0
1 1 1
I think I'm getting it? :)
Edit: I put in some extra explanation given your comment (which is now deleted).
A and B are two boolean variables. For example, in a program, A might be firstTestOK and B might be secondTestOK. Each of A and B can be either true (1) or false (0).
A+B means A or B which is true if either A or B is true. A*B means A and B is is true only if both A and B are true.
All of the combinations for A, B are:
A is false and B is false
A is false and B is true
A is true and B is false
A is true and B is true
This can be written more compactly as a truth table as follows:
A B
0 0
0 1
1 0
1 1
What you've been asked to do is show A*(B+AB) is the same as AB. So, for each combination, we work out the left-hand-side, which is A*(B+AB) and the right-hand-side, which is AB:
A B C=A*B D=B+C A*D = A*B
0 0 0 0 0 0
0 1 0 1 0 0
1 0 0 0 0 0
1 1 1 1 1 1
so, looking at all of the combinations in the last two columns, we see that the results are the same, so AD=A(B+AB) is AB.
Since the left-hand-side is a little complicated, I did it in steps by breaking it up into pieces, by introducing C and D.