F = ABC + AC + C'D'
is there a way to minimise this function even further because i want to make the circuit diagram with only 2 input nand gates
any suggestions ?
thanks
First, simplify:
F = ABC + AC + C'D'
F = AC(B + 1) + C'D'
F = AC + C'D'
Now, put in terms of ANDs and NOTs only:
F = (AC + C'D')'' [double negation]
F = ( (AC)'(C'D')' )' [DeMorgan's]
Then noting that:
NOT can be implemented via 2-input NAND by tying its inputs together.
AND can be implemented via 2-input NAND by combining NAND and NOT.
You should be able to implement F in this form directly using only 2-input NANDs.
This is the solution for the minimalization using a Karnaugh Table.
Related
Using Octave's symbolic package, I define a symbolic function of t like this:
>> syms a b c d t real;
>> f = poly2sym([a b c], t) + d * exp(t)
f = (sym)
2 t
a⋅t + b⋅t + c + d⋅ℯ
I also have another function with known coefficients:
>> g = poly2sym([2 3 5], t) + 7 * exp(t)
g = (sym)
2 t
2⋅t + 3⋅t + 7⋅ℯ + 5
I would like to solve f == g for the coefficients a, b, c, d such that the equation holds for all values of t. That is, I simply want to equate the coefficients of t^2 in both equations, and the coefficients of exp(t), etc. I am looking for this solution:
a = 2
b = 3
c = 5
d = 7
When I try to solve the equation using solve, this is what I get:
>> solve(f == g, a, b, c, d)
ans = (sym)
t 2 t
-b⋅t - c - d⋅ℯ + 2⋅t + 3⋅t + 7⋅ℯ + 5
───────────────────────────────────────
2
t
It solves for a in terms of b, c, d, t. This is understandable since in essence there is no difference between the variables b, c and t. But I was wondering if there was a method to somehow separate the terms (using their symbolic form w. r. t. the variable t) and solve the resulting system of linear equations on a, b, c, d.
Note: The function I wrote here is a minimal example. What I am really trying to do is to solve a linear ordinary differential equation using the method of undetermined coefficients. For example, I define something like y = a*exp(-t) + b*t*exp(-t), and solve for diff(y, t, t) + diff(y,t) + y == t*exp(-t). But I believe solving the problem with simpler functions will lead me to the right direction.
I have found a terribly slow and dirty method to get the job done. The coefficients have to be linear in a, b, ... though.
The idea is to follow these steps:
Write the equation in f - g form (which equals zero)
Use expand() to separate the terms
Use children() to get the terms in the equation as a symbolic vector
Now that we have the terms in a vector, we can find those that are the same function of t and add their coefficients together. The way I checked this was by checking if the division of two terms had t as a symbolic variable
For each term, find other terms with the same function of t, add all these coefficients together, save the obtained equation in a vector
Pass the vector of created equations to solve()
This code solves the equation I wrote in the note at the end of my question:
pkg load symbolic
syms t a b real;
y = a * exp(-t) + b * t * exp(-t);
lhs = diff(y, t, t) + diff(y, t) + y;
rhs = t * exp(-t);
expr = expand(lhs - rhs);
chd = children(expr);
used = false(size(chd));
equations = [];
for z = 1:length(chd)
if used(z)
continue
endif
coefficients = 0;
for zz = z + 1:length(chd)
if used(zz)
continue
endif
division = chd(zz) / chd(z);
vars = findsymbols(division);
if sum(has(vars, t)) == 0 # division result has no t
used(zz) = true;
coefficients += division;
endif
endfor
coefficients += 1; # for chd(z)
vars = findsymbols(chd(z));
nott = vars(!has(vars, t));
if length(nott)
coefficients *= nott;
endif
equations = [equations, expand(coefficients)];
endfor
solution = solve(equations == 0);
I am working on simplifying the expression f = x'yz + xy'z + xyz' + xyz. Actually, it may not be this expression. The question is: simplify the boolean expression for a voting system, the system being: three people vote on multiple candidates, and two or more people should agree(true) on the candidate in order to pass. So I think the answer would be xy + yz + xz, but I can't figure out the process between. Can anyone explain?
From the idempotent/identity law, we have x + x = x, and so xyz + xyz = xyz. Applying this principle, we can rewrite your expression as:
f = x'yz + xy'z + xyz' + xyz
=> f = x'yz + xy'z + xyz' + xyz + xyz + xyz --OR with xyz twice without affecting the value
=> f = x'yz + xyz + xy'z + xyz + xyz' + xyz --Rearrange
=> f = yz (x + x') + xz (y + y') + xy(z' + z) --Group
=> f = yz + xz + xy --Since x+x' = 1
That said, as the diagram clearly shows, you can simply take AND together each pair of inputs, and OR them together to get the same result. By doing this, you ensure that:
If any 2 of the 3 inputs are true, your overall result is true
When all 3 are true, the result is still true
The advantage of expressing it in this way is that you can just focus on each pair of inputs at one time, without worrying about the impact of the third one.
A simple way without involved logical reasoning
Write a truth table. For three inputs, there are 2^3 = 8 rows.
Four rows correspond to the given terms in your sum-of-products expression.
Enter the eight values of your expression into a Karnaugh map:
Group adjacent 1-terms to blocks as shown.
A pair of cells can be merged into a bigger block, if they just differ in one input. This way, the blocks double their cell-count and reduce their input-count by one in every merge step.
Each of the resulting blocks corresponds to one implicant term in the minimized expression.
Drawing the map and finding the blocks can be done automatically using a nice online tool of Marburg University.
I want to simplify the following expression:
F = (A+B+C)(A+B'+C)(A'+B+C)
I have simplified it accordingly.
F = (A+B+C)(A+B'+C)(A'+B+C)
F = (A+C)(A'+B+C)
F = AA' + AB + AC + A'C + BC + C
F = AB + C(A + A' + B + 1) = AB + C
However, the correct answer is (A+C)(B+C).
Where in my "current" proof am I going wrong? I have seen the solution, but I want to know why my current approach is wrong.
Nothing wrong - it's just two different ways of expressing the same thing.
If the goal is minimization I would argue that your solution is "better" since it only references each term once.
Wolfram Alpha is your friend in cases like this.
My own equation is a bit longer, but the following example shows perfectly where I struggle at the moment.
So far I have been using the let() and letsimp() function
to substitute longer terms in an equation,
but in this example they have no effect:
(%i1) eq: ((2*u+a^2+d) * y+x)/2*a = x;
2
a ((2 u + d + a ) y + x)
(%o1) ------------------------ = x
2
(%i2) let(2*u+a^2+d, %beta);
2
(%o2) 2 u + d + a --> %beta
(%i3) letsimp(eq);
2
a ((2 u + d + a ) y + x)
(%o3) ------------------------ = x
2
What is the preferred way to replace 2*u+a^2+d with %beta in this sample equation?
And why has letsimp() no effect?
Thank you very much!
letsimp applies only to "*" expressions. You could try subst.
I have the following function to be reduced/simplified.
F(A,B,C,D) = BC + (A + C'D') where ' denotes the complement
Here's my solution:
= BC + (A + C'D')'
= BC + (A + (C+D)
= BC + (A + C + D)
= BC + C + A + D
= C(B + 1) + A + D
= C*1 + A + D
= C + A + D
Is this correct?
As in traditional algebra, if you do something to one side of the equation, you must do it to the other side, including complementing. Here we state the original equation:
F'(A,B,C,D) = BC + (A + (CD)')
Since we have F' instead of F, my intuition tells me to complement both sides, but first I distribute the complement in the term (CD)' to make life easier in the long run:
F' = BC + (A + (C'+ D'))
Now we can complement both sides of the equation:
1: F = '(BC)'(A + (C'+ D')) The OR becomes AND after distributing complement
Now let's distribute the complements inside just to see what we get:
2: F = (B'+ C')(A'(CD))
Now we can just distribute the right term (A'(CD)) over the two terms being OR'ed:
3: F = B' (A'(CD)) + C' (A'(CD))
We see that the right term goes away since we have a CC' and thus we are left with:
4: F = A'B'CD
Hopefully I didn't make a mistake. I know you've found the answer, but others reading this might have a similar question and so I did it out to save duplicate questions from being asked. Good Luck!