The expression is: x' + x + xz + yz - which reads x not or x or x and z or y and z. This expression is the reduced form of a very large expression which I am trying to prove equal to 1.
I know the above expression is equivalent to 1 according to Wolfram and another expression reducer. Any explanation would be very helpful. Thank you in advance.
We'll assume that Wolfram uses 1 as a synonym for true. The first two terms of the expression are x' and x. One or the other of them must be 1 (true), so the logical OR of the two of them is 1. Then the logical OR of true and anything else is going to be true. That's why the entire expression reduces to 1.
The key to understanding this expression is idea of or operator. If you have true or false or smth there is no matter what is smth, the expression still will be true. So in your case xz + yz is redundant here because x' + x covers all the set of possible values
Related
Sorry for the basic question, but it's quite hard to find too much discussion on Maxima specifics.
I'm trying to learn some Maxima and wanted to use something like
x:2
x+=2
which as far as I can tell doesn't exist in Maxima. Then I discovered that I can define my own operators as infix operators, so I tried doing
infix("+=");
"+=" (a,b):= a:(a+b);
However this doesn't work, as if I first set x:1 then try calling x+=2, the function returns 3, but if I check the value of x I see it hasn't changed.
Is there a way to achieve what I was trying to do in Maxima? Could anyone explain why the definition I gave fails?
Thanks!
The problem with your implementation is that there is too much and too little evaluation -- the += function doesn't see the symbol x so it doesn't know to what variable to assign the result, and the left-hand side of an assignment isn't evaluated, so += thinks it is assigning to a, not x.
Here's one way to get the right amount of evaluation. ::= defines a macro, which is just a function which quotes its arguments, and for which the return value is evaluated again. buildq is a substitution function which quotes the expression into which you are substituting. So the combination of ::= and buildq here is to construct the x: x + 2 expression and then evaluate it.
(%i1) infix ("+=") $
(%i2) "+="(a, b) ::= buildq ([a, b], a: a + b) $
(%i3) x: 100 $
(%i4) macroexpand (x += 1);
(%o4) x : x + 1
(%i5) x += 1;
(%o5) 101
(%i6) x;
(%o6) 101
(%i7) x += 1;
(%o7) 102
(%i8) x;
(%o8) 102
So it is certainly possible to do so, if you want to do that. But may I suggest maybe you don't need it? Modifying a variable makes it harder to keep track, mentally, what is going on. A programming policy such as one-time assignment can make it easier for the programmer to understand the program. This is part of a general approach called functional programming; perhaps you can take a look at that. Maxima has various features which make it possible to use functional programming, although you are not required to use them.
Example:
11: 01011
18: 10010
2: 00010
Ans: 00010
At position zero (right most), I have more 0s than the number of 1s. Hence my answer should be 0. Similarly, at position one, I have 1 as the dominant bit, and so on.
Is there a way to find this using basic bitwise operators or should I use any other bitmasks that could achieve this result? I am looking for a O(1) solution for the same.
If you have exactly three inputs x, y, z, you can use
(x & y) | (z & (x | y))
Similar formulas could probably be found for any other constant number of inputs.
Given that a boolean expression is in conjunctive normal form: is there a "simple" algorithm to simplify it while keeping it in CNF?
In particular, what property of the following expression causes this simplifiation?
(~a+b+c)(a+~b+c)(a+~c)
simplifies to ...
(~a+b+c)(a+~b)(a+~c)
The Karnaugh map of your example is:
To get a simplified DNF, '1' cells are grouped to get a cover with the minimum number of minterms.
Similarly, one can group the '0' cells to get an inverse cover with the minimum number of terms.
The inverse map:
The literals of the resulting terms have to be inverted to arrive at the desired minimum CNF
(a + ~b) (a + ~c) (~a + b + c)
The procedure makes use of the fact that the inverse of a minterm is a maxterm (commonly called CNF clause) with inverted literals.
I am trying to solve this problem :
(Write a program to compute the real roots of a quadratic equation (ax2 + bx + c = 0). The roots can be calculated using the following formulae:
x1 = (-b + sqrt(b2 - 4ac))/2a
and
x2 = (-b - sqrt(b2 - 4ac))/2a
I wrote the following code, but its not correct:
program week7_lab2_a1;
var a,b,c,i:integer;
x,x1,x2:real;
begin
write('Enter the value of a :');
readln(a);
write('Enter the value of b :');
readln(b);
write('Enter the value of c :');
readln(c);
if (sqr(b)-4*a*c)>=0 then
begin
if ((a>0) and (b>0)) then
begin
x1:=(-1*b+sqrt(sqr(b)-4*a*c))/2*a;
x2:=(-1*b-sqrt(sqr(b)-4*a*c))/2*a;
writeln('x1=',x1:0:2);
writeln('x2=',x2:0:2);
end
else
if ((a=0) and (b=0)) then
write('The is no solution')
else
if ((a=0) and (b<>0)) then
begin
x:=-1*c/b;
write('The only root :',x:0:2);
end;
end
else
if (sqr(b)-4*a*c)<0 then
write('The is no real root');
readln;
end.
do you know why?
and taking a=-6,b=7,c=8 .. can you desk-check it after writing the pesudocode?
You have an operator precedence error here:
x1:=(-1*b+sqrt(sqr(b)-4*a*c))/2*a;
x2:=(-1*b-sqrt(sqr(b)-4*a*c))/2*a;
See at the end, the 2 * a doesn't do what you think it does. It does divide the expression by 2, but then multiplies it by a, because of precedence rules. This is what you want:
x1:=(-1*b+sqrt(sqr(b)-4*a*c))/(2*a);
x2:=(-1*b-sqrt(sqr(b)-4*a*c))/(2*a);
In fact, this is because the expression is evaluated left-to-right wrt brackets and that multiplication and division have the same priority. So basically, once it's divided by 2, it says "I'm done with division, I will multiply what I have now with a as told".
As it doesn't really seem clear from the formula you were given, this is the quadratic formula:
As you can see you need to divide by 2a, so you must use brackets here to make it work properly, just as the correct text-only expression for this equation is x = (-b +- sqrt(b^2 - 4ac)) / (2a).
Otherwise the code looks fine, if somewhat convoluted (for instance, you could discard cases where (a = 0) and (b = 0) right after input, which would simplify the logic a bit later on). Did you really mean to exclude negative coefficients though, or just zero coefficients? You should check that.
Also be careful with floating-point equality comparison - it works fine with 0, but will usually not work with most constants, so use an epsilon instead if you need to check if one value is equal to another (like such: abs(a - b) < 1e-6)
Completely agree with what Thomas said in his answer. Just want to add some optimization marks:
You check the discriminant value in if-statement, and then use it again:
if (sqr(b)-4*a*c)>=0 then
...
x1:=(-1*b+sqrt(sqr(b)-4*a*c))/2*a;
x2:=(-1*b-sqrt(sqr(b)-4*a*c))/2*a;
This not quite efficient - instead of evaluating discriminant value at once you compute it multiple times. You should first compute discriminant value and store it into some variable:
D := sqr(b)-4*a*c;
and after that you can use your evaluated value in all expressions, like this:
if (D >= 0) then
...
x1:=(-b+sqrt(D)/(2*a);
x2:=(-b-sqrt(D)/(2*a);
and so on.
Also, I wouldn't write -1*b... Instead of this just use -b or 0-b in worst case, but not multiplication. Multiplication here is not needed.
EDIT:
One more note:
Your code:
if (sqr(b)-4*a*c)>=0 then
begin
...
end
else
if (sqr(b)-4*a*c)<0 then
write('The is no real root');
You here double check the if-condition. I simplify this:
if (a) then
begin ... end
else
if (not a)
...
Where you check for not a (in your code it corresponds to (sqr(b)-4*a*c)<0) - in this case condition can be only false (for a) and there is no need to double check it. You should just throw it out.
I am unsure how to use the Distributive property on the following function:
F = B'D + A'D + BD
I understand that F = xy + x'z would become (xy + x')(xy + z) but I'm not sure how to do this with three terms with two variables.
Also another small question:
I was wondering how to know what number a minterm is without having to consult (or memorise) the table of minterms.
For example how can I tell that xy'z' is m4?
When you're trying to use the distributive property there, what you're doing is converting minterms to maxterms. This is actually very related to your second question.
To tell that xy'z' is m4, think of function as binary where false is 0 and true is 1. xy'z' then becomes 100, binary for the decimal 4. That's really what a k-map/minterm table is doing for you to give a number.
Now an important extension of this: the number of possible combinations is 2^number of different variables. If you have 3 variables, there are 2^3 or 8 different combinations. That means you have min/maxterm possible numbers from 0-7. Here's the cool part: anything that isn't a minterm is a maxterm, and vice versa.
So, if you have variables x and y, and you have the expression xy', you can see that as 10, or m2. Because the numbers go from 0-3 with 2 variables, m2 implies M0, M1, and M3. Therefore, xy'=(x+y)(x+y')(x'+y').
In other words, the easiest way to do the distributive property in either direction is to note what minterm or maxterm you're dealing with, and just switch it to the other.
For more info/different wording.