You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

bob bundy wrote:

I think you need some easier practice first. I'd be happy to go back a few steps and lead you back to this problem, with some easier ones first.

Would you like to try this?

bob

my homework needs to be done lol practice can come latter.I'm more of an IT guy but discrete math comes with my course unfortunately cant wrap my head around math at all.

bob bundy wrote:

How?

The circuit has 6 gates, 4 without the NOTs.

When I partly did the question, I had 4 of the gates done for you.

How have you got it to just 3 gates?

What happened after ( (p' OR ...) OR (...AND...) )

Bob

what exactly do i need to add to it?

bob bundy wrote:

That doesn't seem to require any simplification so what I said before should be fine for (a).

In your earlier thread I made some truth tables. You can do this for your function.

eg

( p ' + q ) AND p

0 1 1 0 0 0Bob

ps. If you post a function first I'll check it for you.

I have come up with, p (p + q ' ) ?

Agnishom wrote:

fourth is eulerian too

thanks so i am correct with the others?

bob bundy wrote:

hi dee93

My picture shows the three logic symbols being used here.

I think I'd work back from the output end, so you get the right brackets.

The last logic gate is an OR so start with

(... OR ...)

The top input to that OR is also an OR and the bottom input is an AND so that makes it

( (...OR...) OR (...AND...))

I'll do one more step for you.

The top input to the OR is from p through a NOT gate so now we have

( (p' OR ...) OR (...AND...) )

See if you can finish it off.

Question. Are you expected to simplify it as well?

Bob

its a two part question I have no idea how to write it.

a) What Boolean function is implemented by the circuit

) The screen shot shows that inputs of p=0 and q=0 give an outputof 1. Confirm that your Boolean function from part (a) evaluates to 1when p and q are set to 0.

**dee93**- Replies: 4

Would I be right in saying that the first and second graphs are Hamiltonian and the third is Eulerian,unsure about the 4th?

**dee93**- Replies: 11

Hi guys can you please help me with this question and explain how i determine which boolean function is in the uploaded circuit?

bob bundy wrote:

hi dee93

It cannot be the same as the other question because the logic statements are different.

Let's go back to first principles.

pq' means p is true and not q is true (or q is false)

A karnaugh map is a way of showing a logic expression diagrammatically. It does a similar job to the venn diagrams I did earlier but copes better when the number of variables goes up.

In the boxes you can put expressions like pq'r or use 0 for false an d 1 for true (101 for this case) or express the logic in words: p true, q false, r true. Below I've taken the earlier diagram and added more labels to show what each box means.

The problem you are trying to do is abc + ac

I would write that as (a AND b AND c) OR (a AND c)

abc is a single box as it represents the case where all three are true.

ac means that a AND c are both true but b can be either. Because of this it will take two boxes.

But that does not mean that you will end up with three boxes shaded because one box is repeated.

Hope that helps.

Bob

i've attempted to do it but unable to draw diagrams on here can you work it out so i can confirm mine is correct?

bob bundy wrote:

see diagram

Then decide which boxes to shade

Bob

2 on far right,and 1 on botton left same as post #17?

bob bundy wrote:

and where will you put c ?

Look at post 17

Bob

some where in the boxes?

bob bundy wrote:

Try it yourself and I'll help out if you get stuck. (after all, I need to know that I've successfully taught you something )

Step 1: Make a set of boxes for three variables. What goes across the top and what down the side?

Bob

a on top b on side?

bob bundy wrote:

?? Did you look at all my recent posts?

Bob

yes but i think you missed one out. abc + ac if you could also simplify that in its dnf form using the map like you have done the others?

bob bundy wrote:

3 variable case:

(b) pq + p'q'r

You have to squeeze up the pq bits to one line.

in the first part of the diagram pq is the green region (two boxes) and p'q'r is the red box.

The DNF version is shown on the right.

Hope that sorts it our for you. I'll leave you to do the third one yourself. Good luck.

Bob

brilliant if you could do the same for the other 2 dnf expressions that would be great

would you know how to simplify these new dnf expression with a karnaugh map?

double post

bob bundy wrote:

But what have you been told it is? You must have been given a rule for it. Or an example.

I'll come clean about this topic.

I did some Boolean algebra and propositional logic back in 1969ish. DNF was not part of it.

So when I saw your question I thought I'd better leave it for someone who knew what it was.

A while later I looked it up on Wiki. It seemed straight forward enough for me to blunder in with a reply.

Wiki says DNF is " it is an OR of ANDs " and all their examples look like this:

(a AND b) OR (c AND d) etc.

If that definition is correct then all your questions have that form already:

(not p AND not q) OR (q)

(p AND q) OR (NOT p AND NOT q AND r)

(a AND b AND c) OR (a AND b)

So I wondered if the questioner wanted FULL DNF or maybe NOT is dis-allowed. But Wiki's examples include NOTs and anyway, some expressions could not be rendered in DNF if NOT was dis-allowed.

If you are able to give the definition according to your course, maybe we can get somewhere.

Bob

This is the definiton I found on a lecture note

Disjunctive Normal Form for a Boolean Expression which means

bob bundy wrote:

I can simplify one:

eg

p'q' + q = (p' + q)(q' + q) = p' + q ( as q' OR q is always true) note: this is now FULL DNF

Here are the truth tables: (apologies if the alignment goes wrong; it's quite hard to do a neat table like this)

p ' AND q '

+q

1 0 0 1 011

1 0 0 0 100

0 1 0 1 011

0 1 1 0 110p '

+q

1 011

1 000

0 111

0 110So both versions give

1

0

1

1How is DNF defined on your course?

Bob

they have called it disjunctive normal function but i assumed its just another word for disjunctive normal form.thanks for that is not possible to convert the others?

bob bundy wrote:

hi dee93

Welcome to the forum.

I thought (pq + r) meant (p AND q OR r) so now I'm confused as all these expressions appear to be in DNF already. Have I got this wrong?

and then post again. I'm fairly confident I can do the manipulations but what am I trying to achieve?

Bob

thanks for the welcome.

I have no idea,this is a homework question and i've been asked to put these into dnf but you think they already are thats very strange.

i have also been asked to simplify them using a kavanaugh map?

**dee93**- Replies: 28

Hi,everyone I am struggling with some homework on discrete mathematics can anyone help me with the following question?

Pages: **1**