B

]]>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?

]]>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

]]>see diagram

Then decide which boxes to shade

Bob

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

]]>Then decide which boxes to shade

Bob

]]>and where will you put c ?

Look at post 17

Bob

some where in the boxes?

]]>Look at post 17

Bob

]]>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?

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

Bob

]]>?? 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

]]>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

]]>(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

]]>(a) p'q' + q

Diagram shows the p'q' box in a yucky yellow colour (if p' and q' are both 1s then p and q are zeros) and all the boxes for q in green.

then I've identified in the second part the three boxes in DNF form.

Now to sort out three variables.

Bob

]]>