A’B(D’ + C’D)’ + B( A + A’CD)’
Trying to simplify the above boolean expression having problems with the apostrophe outside the bracket.
I applied the compliment De Morgans rule and got: A’B(DC+D’) + B(A’A+C’+D’) is this correct?
So looking at the left side we have A'B(D'+C'D)'
A'B(D(C'D)') after applying De Morgan's
A'B(D(C+D')) after applying De Morgan's again
From here you can take back over. You just forgot the parenthesis
Related
I've been stuck on this for quite some time. Is there a command in R that will create a Box Cox Linear transformation that has a different a parameter for my independent variable and dependent variable?
chicago.reg <- lm(data = Chichagodata, sprice ~ nrooms + lvarea + hage + lsize + ptaxes + sspend + mspend +
medinc + dfcl + particle + sulfur+pctwht+ dfni +aircon+ garage+ nbath + cook+ohare)
Note that I do not want to transform my variables pctwht, dfni, aircon, garage, nbath, cook,and ohare as they are dummy variables.
Thank you if you can provide me insight on this issue.
You should have a look into coxBox and boxcox functions. The former allow to transform individual variables/lists, while the latter works on lm (or aov) objects. In addition you might want to read this and this posts...
So, I have a really long symbolic exrpression in Matlab that I want to copy/paste into a JavaScript-code to animate a numerical solution for it. The problem is that some places in my code i get exponents (mostly ^2), when I'd rather have Matlab express it as A*A.
I have mulitple (and different) expressions like
cos(th2t)^2
That I would rather have expressed as
cos(th2t)*cos(th2t)
Any way I can do this? The alternative is to use a text editor afterward and search for powers of 2 and replace it, but there are multiple different expressions, so that would take some time...
This is an example of one of the exrpressions I end up with:
(J31*(2*ddth2t*cos(th1t) - 2*w12*w13 - 4*dth1t*dth2t*sin(th1t) - 4*dth2t*w13*sin(th1t) + dth1t^2*sin(2*th2t)*cos(th1t) - w12^2*sin(2*th2t)*cos(th1t) + w13^2*sin(2*th2t)*cos(th1t) + 2*dth2t*w11*sin(2*th2t) + 2*w12*w13*cos(th2t)^2 + ddth1t*sin(2*th2t)*sin(th1t) + 4*dth1t*dth2t*cos(th2t)^2*sin(th1t) + 2*dth1t*w13*sin(2*th2t)*cos(th1t) + 4*dth2t*w13*cos(th2t)^2*sin(th1t) + w11*w12*sin(2*th2t)*sin(th1t) + 4*dth1t*w12*cos(th1t)^2*cos(th2t)^2 - 2*dth1t*w11*sin(2*th1t)*cos(th2t)^2 - 2*dth2t*w11*sin(2*th2t)*cos(th1t)^2 + 2*w12*w13*cos(th1t)^2*cos(th2t)^2 - w11*w13*sin(2*th1t)*cos(th2t)^2 - 4*dth2t*w12*cos(th1t)*cos(th2t)*sin(th1t)*sin(th2t)))/(2*(J11 + J31))
Steve's answer should work fine if you want to rely on the ** operator. However, since that operator is not officially supported, and that solution doesn't directly answer the OP's question, here is a function that can expand out the exponents in a symbolic expression.
function [ text ] = remove_powers( exp )
%Cleans the powers from T, and return expanded text representation
% Define functions
enclose =#(t) ['(' t ')'];
expand_pow=#(b,p) strjoin(strcat(cell(1,p),enclose(char(b))),'*');
count_pow=#(s) arrayfun(#(k) count(char(s(k)),'^'), 1:length(s));
sym2str = #(s) strrep(char(s), ' ', '');
% Check for fractions
[N,D]=numden(exp);
if ~isequal(D,sym(1))
% pass up the num and den
text = [remove_powers(N) '/' enclose(remove_powers(D))];
else
% Split a into subterms
Ts = children(exp);
% Clean children
has_pow=count_pow(Ts)>0;
text = sym2str(exp);
if sum(count_pow(Ts))<count_pow(exp)
% We have removed a power, clean it, expand it, and pass up
text = expand_pow(remove_powers(Ts(1)),Ts(2));
else
% Just clean the unclean children and pass up
for t=Ts(has_pow)
text = strrep(text,sym2str(t),remove_powers(t));
end
end
end
end
The function uses the children function in Matlab to recursively clean each subexpression and replace it (as text) in the parent. This method is better than using regex because it avoids the issue of parsing the syntax, as Steve mentioned in the comment with respect to cos(x^2+2*y)^2.
Which makes for a good example:
syms x y real
exp = cos(x^2+2*y)^2;
cleaned_exp = remove_powers(exp)
Outputs: (cos(2*y+(x)*(x)))*(cos(2*y+(x)*(x)))
Notice that since Matlab is doing the parsing, there was no need to parse the order of precedence for the '^' operators, which could be difficult to accomplish with regex.
To test the OP's example:
syms ddth1t dth1t th1t ddth2t dth2t th2t w11 w12 w13 J11 J31 real
exp = (J31*(2*ddth2t*cos(th1t) - 2*w12*w13 - 4*dth1t*dth2t*sin(th1t) - 4*dth2t*w13*sin(th1t) + dth1t^2*sin(2*th2t)*cos(th1t) - w12^2*sin(2*th2t)*cos(th1t) + w13^2*sin(2*th2t)*cos(th1t) + 2*dth2t*w11*sin(2*th2t) + 2*w12*w13*cos(th2t)^2 + ddth1t*sin(2*th2t)*sin(th1t) + 4*dth1t*dth2t*cos(th2t)^2*sin(th1t) + 2*dth1t*w13*sin(2*th2t)*cos(th1t) + 4*dth2t*w13*cos(th2t)^2*sin(th1t) + w11*w12*sin(2*th2t)*sin(th1t) + 4*dth1t*w12*cos(th1t)^2*cos(th2t)^2 - 2*dth1t*w11*sin(2*th1t)*cos(th2t)^2 - 2*dth2t*w11*sin(2*th2t)*cos(th1t)^2 + 2*w12*w13*cos(th1t)^2*cos(th2t)^2 - w11*w13*sin(2*th1t)*cos(th2t)^2 - 4*dth2t*w12*cos(th1t)*cos(th2t)*sin(th1t)*sin(th2t)))/(2*(J11 + J31));
cleaned_exp = remove_powers(exp);
isequal(sym(cleaned_exp),exp) % This should be 1
count(cleaned_exp,'^') % This should be 0
As expected, the new expression is equivalent to the original, but has no '^' symbols.
This answer suggests that you could call the exponential operator in Javascript with e.g. A**2. As such you could replace all instances of ^ with **.
(Solution from comments)
My question is quite similar to this one, but with a difference. I want to create a macro (or whatever) that behaves this way:
julia> #my-macro x + 2
:(x + 2)
(note that x + 2 is not enclosed in quotes). Is there something like that in Julia? And if there is not, how do I do it? (Please, give a detailed explanation about why it works.)
The input expression to the macro needs to be quoted because a macro returns an expression, which are evaluated, while you would like to get the expression itself, hence you need an extra quoting. The quoting can be done as:
macro mymacro(ex)
Expr(:quote,ex) # this creates an expression that looks like :(:(x + 2))
end
e=#mymacro x + 2 #returns :(x + 2)
Another shorter possibility:
macro mymacro(ex)
QuoteNode(ex)
end
e = #mymacro x + 2 #returns :(x + 2)
Hi I have this expression XY'Z+YY'+X+XY'+XZ ( ' means not) Can anyone please show me the simplification of this expression? Thanks in advance.
This is what i got so far:
XY'Z+YY'+XY'+X(1+Z)
XY'Z+YY'+XY'+XY'+X
Y'(XZ+Y)+X(1+Y')
Y'(XZ+Y)+X
I know 1+Y=1 but 1+Y'=1 too?
Y'(XZ+Y)+X
change this back to:
XY'Z+YY'+X
Then simplify a bit:
XY'Z+YY'+X
YY' + X(1+Y'Z)
0 + X
X
I was asked this question in an interview.
Write a c# program which will print out all the errors in a statement.
(a + b == 3 and x == y or b / c == d)
Allowed keywords are and/or, braces are allowed and have to match. The statement has to be logically correct. Print out all the errors.
Something like a compiler or sql analyzer.
Any idea how to go about syntax checking?
You're being asked to build a parser.
The computer science theory behind compilers is extensive and complex. To keep it simple, I recommend reading Jack Crenshaw's compiler tutorial. It will show you in the first few chapters how to do exactly this by building a top-down recursive descent parser. His example is almost exactly what you're trying to do. It's in Pascal, but it's still easy enough to follow and the concept still applies.
Google for "jack crenshaw let's build a compiler", or browse to:
http://compilers.iecc.com/crenshaw/
That's actually pretty complicated if you want to get it 100% right. The direction to research would have to do with the keywords "finite state automata", "regular expressions" (not to be confused with Regex) and "context-free grammars". Also there's an approach called "recursive descent parser" but you'll need to understand the above concepts first.
Matching brackets and parenthesis is easy.
string expression = "(a + b + (c + d) )";
int brackets = 0;
int parenthesis = 0;
foreach(char c in expression)
{
if(c == '(') ) parenthesis++;
if(c == ')') ) parenthesis--;
if(parenthesis < 0) { // ERROR!!! }
// Same logic with brackets
}
if (parenthesis != 0) { ERROR!!! }
Also, I would use a similar approach with "statements mergers" ( +, -, *, / ) and "statements comparators" ( ==, !=, <, <=, ... )
For each word in your expression, only a few valid keywords can follow.
As it was mentionned before, writing parsers is a difficult task, planning every exception case right away is hardly possible. This should be the general idea instead of a definitive solution.
On a final note, it is perfectly ok to have multiple passes to scan for different things. First pass is brackts/parenthesis only, then check for something else in another loop.