So we found the two prime implicants T'and R'. Both of those are of size 4 and if we look at the Carnot map we can clearly see there are no more prime implicants of size 4. And so then we go to size 2 and indeed there are... groupings of two that are present but no grouping of two is present which is not already entirely contained with one of the previously identified prime members.
And likewise this thing can be said for any groupings of one. So the two prime implicants, T prime and R prime, are the only prime implicants. And so now our next step, as usual after we have found all of our prime implicants, is to identify which of the prime implicants are essential. So we'll say essential prime implicants. And of course the way to identify these is by identifying the essential one cells.
I encourage you to take a moment to try to do this yourself. But the essential one cells will be here, here, here, and here. The only two one cells that are not essential are the one in the upper left-hand corner and the one in the upper right-hand corner. Both of those one cells are common to both of the prime implicants. So now that we have identified the essential one cells, we ask ourselves, which prime implicants are essential.
T prime is essential because it has this essential one cell and this essential one cell. And the R prime prime implicant is also essential because it has these two essential one cells. Therefore, both of the prime implicants that we have found are essential. And now we look for minimal sums.
And this function is F3. So F3. Of R, S, and T, we write down the two essential prime applicants, T prime or R prime, and by including T prime, we have accounted for those terms and these terms, and then by including R prime in this sum, we have also so accounted for those terms and again as in all our previous examples so far by merely including the essential prime implicants we have accounted for the entire function and therefore this expression is sufficient. It is the only minimal sum for this problem.
And note that here we have two literals. So we have gone from, by using the Carnot map technique, we've gone from an expression with six literals to an expression with two literals. Two easy ways that you could show that these expressions are equivalent would be to find the truth table for each expression and make sure that the two truth tables are the same, or by algebraic methods you could fairly easily show that the second expression is equal to the first. So, that concludes all of the examples for Lecture 11. And now we'll look at some problems. So, the two problems for Lecture 11 are shown here.
Lecture 11.1 says... Find all minimal sums of f of x, y, z equals the sum of the min terms 1, 3, 5, and 7. And your choices are these four choices. 11.2 says find all minimal sums of g of x, y, z equals the product of the max terms 2 and 4. And here you have these four choices. So that concludes lecture 11. Good luck.