Hello,

I intend create a table truth with 20 variables (input) and one output. Is this possible?
I am using the software, to windows, Logic Friday, but this software only can handle maximum 16 inputs.

Which the total is 1 048 576, because 2^20 equal this great number, I try create this table: https://snag.gy/VjxE47.jpg

Everything else is despicable, so I can disregard, right?
Thanks

 

I have no idea what might be "despicable" about everything else and, even if it were, why should you be able to disregard something just because you think that it is somehow abhorrent?

Hopefully the nature of the relationships in your logic table will allow you to use a highly condensed number of rows in which many of the entries are "don't care". But you have to be careful about not including the same condition in multiple rows unless the output for all such rows are the same.

 

Ok, maybe I should change word "despicable" to "don't care". It is?

Assuming the principle which this table it is right, how can I developer solutions in logic gate from this table? Because on Logic Friday I can't handle this total of variables.

Thanks

 

Is that your entire table? Or does it continue the same pattern for A4?

From your table you seem to be indicating that the output is always a 1.

Under what conditions is the output ever a 0?

Reading between the lines, you seem to be saying that the output should be a 1 if any of the P signals are HI at the same time that any of the A signals are HI (and presumably LO for all other cases). Is that a correct statement?

 

If only tables of 16 logical variables are possible, then you can use four tables and a combining function.
F(d1,...d20)=F1d17!d18!d19!d20+F21d17d18!d19!d20+F3!d17!d18!d19!d20+F4!d17!d18!d19d20

 

If only tables of 16 logical variables are possible, then you can use four tables and a combining function.
F(d1,...d20)=F1d17!d18!d19!d20+F21d17d18!d19!d20+F3!d17!d18!d19!d20+F4!d17!d18!d19d20

Huh?....

F(d1,...d20)=F1(d17)(!d18)(!d19)(!d20)+F21(d17)(d18)(!d19)(!d20)+F3(!d17)(!d18)(!d19)(!d20)+F4(!d17)(!d18)(!d19)(d20)

I assume your second term has a typo in it and this should be

F(d1,...d20)=F1(d17)(!d18)(!d19)(!d20)+F2(!d17)(d18)(!d19)(!d20)+F3(!d17)(!d18)(!d19)(!d20)+F4(!d17)(!d18)(!d19)(d20)

So when any two terms of d17 through d20 were TRUE, F(d1,...d20) would be identically FALSE.

Why wouldn't you need 16 tables, one for each possible combination of d17...d20?

 

I am sorry.
I made a mistake and greatly simplified the situation. I have been studying the truth table for a long time. But your comment, my brain cleared up. I remembered everything. It is necessary not four functions, but as many functions, as many units in the main table. If they are few, then lucky. If not, then alas is bad.

 

Is that your entire table? Or does it continue the same pattern for A4?

From your table you seem to be indicating that the output is always a 1.

Under what conditions is the output ever a 0?

Reading between the lines, you seem to be saying that the output should be a 1 if any of the P signals are HI at the same time that any of the A signals are HI (and presumably LO for all other cases). Is that a correct statement?

No, this is only rows which I intend consider to my analyze. Because the entire table will have 1 048 576 rows/scenarios.

Reading between the lines, you seem to be saying that the output should be a 1 if any of the P signals are HI at the same time that any of the A signals are HI (and presumably LO for all other cases). Is that a correct statement?

Yes, you are correct.

As I said I intend also like this:

(P1 !P2!P3!P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A1
+
(P1! P2 !P3!P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A1
+
(P1!P2! P3 !P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A1
+
....
(P1 !P2!P3!P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A2
+
(P1! P2 !P3!P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A2
+
(P1!P2! P3 !P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A2
+
....
(P1 !P2!P3!P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A3
+
(P1! P2 !P3!P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A3
+
(P1!P2! P3 !P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A3
+
....
(P1 !P2!P3!P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A4
+
(P1! P2 !P3!P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A4
+
(P1!P2! P3 !P4!P5!P6!P7!P8!P9!P10!P11!P12!P13!P14!P15!P16)*A4
+
....

Thanks

 

No, this is only rows which I intend consider to my analyze. Because the entire table will have 1 048 576 rows/scenarios.

Yes, you are correct.

If this IS the description of your problem: the output should be a 1 if any of the P signals are HI at the same time that any of the A signals are HI (and presumably LO for all other cases).

The just read that sentence and think about what it is saying. Let's make it even more apparent.

The output should be 1 if (X is HI) and (Y is HI).

If you had signals X and Y, what would your circuit look like?

Now let's consider what X is.

(X is HI) = (any of the P signals are HI)
(Y is HI) = (any of the A signals are HI)

 

I think I'm figuring out where you're headed.

Yes, if we consider all P which variable X and Y all variable Y, your though is correct.