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.
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 want to ask if it is possible to do a query and have a result like this? The more difficult part is to have a half-split column within the same row.
M stands for Male and F stands for Female for those who are wondering.
Teacher Class Level Name 7 8 9 10 11
M F M F M F M F M F
Mr A Class 1 Ant 0 0 3 9 0 1 0 0 0 0
Ms B Class 1 Bumble Bee 0 0 11 5 1 0 0 0 0 0
Mrs C Class 2 Cat 0 0 0 1 7 1 0 0 0 0
Mdm D Class 3 Dog 0 0 0 0 0 0 7 0 1 0
I have attached the same table but in excel so it's easier to see. It will be good if this can be done in pure sql query.
Thank you in advance to those who help in any way.
The closest result you could get solely using SQL would be to have a table returned with the following column headers:
Teacher Name, Class Level, Name, 7_M, 7_F,8_M, 8_F,9_M, 9_F,10_M, 10_F,11_M, 11_F.
Now, to display those X_M and X_F column as "half-split", you would have to do it outside SQL (EDIT: For the reasons Radim mentioned in his comment).
I'm trying to write a query to process a single table that looks like this:
record_id item_id part_id part_length
----------- ------- -------- ------------
1 0 0 123.12
2 0 0 123.09
3 0 1 231.24
4 0 1 239.14
5 1 0 45.91
6 1 0 46.12
7 1 1 62.24
8 1 1 59.40
which is basically a table of inaccurate length measurements of some parts of some items recorded multiple times (not twice, actually each part has 100s of measurements). With a single select, I want to get a result like this:
record_id item_id part_id unit part_length_ratio
----------- ------- -------- ----- ----------------
1 0 0 1 123.12 / 231.24
2 0 0 1 123.09 / 239.14
3 0 1 0 231.24 / 123.12
4 0 1 0 239.14 / 123.09
5 1 0 1 45.91 / 62.24
6 1 0 1 46.12 / 59.40
7 1 1 0 62.24 / 45.91
8 1 1 0 59.40 / 46.12
which is basically selecting each part of an item as the unit and calculates the ratio of the length of other parts of the same item to this unit while matching the measurement times. I wrote a script which computes this kind of table but would like to do it with sql. I can understand if you fail to understand the question :)
for each item i
for each part unit of i
for each part other of i
if unit != other
print i.id other.part_id unit.part_id other.length / unit.length
As I said in a comment, tables are unordered sets: there is no first or second row...
... unless if you want to use the id column to explicitly order the rows.
However, can you guarantee that there will always be (exactly) two samples for each case and that the "lower ID" always match the first sample? This appears to be quite fragile as in real-life, there will probably have cases where a test will be performed twice or a test will be missing or done "late". Not mentioning concurrent access to your DB.
Can't you simply add a "sample number" column?
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