I am attempting to work a problem from a textbook in sympy, but sympy fails to find a solution which appears valid. For interest, it is the design of a PID controller using direct synthesis with a second order plus dead time model.
The whole problem can be reduced to finding K_C, tau_I and tau_D which will make
K_C*(s**2*tau_D*tau_I + s*tau_I + 1)/(s*tau_I)
= (s**2*tau_1*tau_2 + s*tau_1 + s*tau_2 + 1)/(K*s*(-phi + tau_c))
for given tau_1, tau_2, K and phi.
I have tried to solve this by matching coefficients:
import sympy
s, tau_c, tau_1, tau_2, phi, K = sympy.symbols('s, tau_c, tau_1, tau_2, phi, K')
target = (s**2*tau_1*tau_2 + s*tau_1 + s*tau_2 + 1)/(K*s*(-phi + tau_c))
K_C, tau_I, tau_D = sympy.symbols('K_C, tau_I, tau_D', real=True)
PID = K_C*(1 + 1/(tau_I*s) + tau_D*s)
eq = (target - PID).together()
eq *= sympy.denom(eq).simplify()
eq = sympy.poly(eq, s)
sympy.solve(eq.coeffs(), [K_C, tau_I, tau_D])
This returns an empty list. However, the textbook provides the following solution:
booksolution = {K_C: 1/K*(tau_1 + tau_2)/(tau_c - phi),
tau_I: tau_1 + tau_2,a
tau_D: tau_1*tau_2/(tau_1 + tau_2)}
Which appears to satisfy the equations I'm trying to solve:
[c.subs(booksolution).simplify() for c in eq.coeffs()]
returns
[0, 0, 0]
Can I massage this into a form which sympy can solve? What am I doing wong?
Edit: This finds the correct solution, but requires a little too much thought from my side to order the equations:
eqs = eq.coeffs()
solution = {}
solution[K_C] = sympy.solve(eqs[1], K_C)[0]
solution[tau_D] = sympy.solve(eqs[0], tau_D)[0].subs(solution)
solution[tau_I] = sympy.solve(eqs[2], tau_I)[0].subs(solution).simplify()
In SymPy 1.0 (to be released soon) I get this answer
In [25]: sympy.solve(eq.coeffs(), [K_C, tau_I, tau_D])
Out[25]:
⎡ ⎧ -(τ₁ + τ₂) τ₁⋅τ₂ ⎫⎤
⎢{K_C: 0, τ_I: 0}, ⎨K_C: ───────────, τ_D: ───────, τ_I: τ₁ + τ₂⎬⎥
⎣ ⎩ K⋅(φ - τ_c) τ₁ + τ₂ ⎭⎦
which looks like your textbook's solution.
Related
I have a list of polynomial expressions, (in my case obtained as the output of a Groebner basis computation), that I would like to view. I am using Jupyter, and I have started off with
import sympy as sy
sy.init_printing()
This causes an individual expression to be given nicely typeset. For a non-Groebner example:
sy.var('x')
fs = sy.factor_list(x**99-1)
fs2 = [x[0] for x in fs[1]]
fs2
The result is a nice list of LaTeX-typeset expressions. But how do I print these expressions one at a time; or rather; one per line? I've tried:
for f in fs2:
sy.pprint(f)
but this produces ascii pretty printing, not LaTeX. In general the expressions I have tend to be long, and I really do want to look at them individually. I can of course do
fs2[0]
fs2[1]
fs2[2]
and so on, but this is tiresome, and hardly useful for a long list. Any ideas or advice? Thanks!
Jupyter (through IPython) has a convenience function called display which works well with SymPy:
import sympy as sy
sy.init_printing()
sy.var('x')
fs = sy.factor_list(x**99-1)
fs2 = [x[0] for x in fs[1]]
for f in fs2:
display(f)
Output:
You can also get the latex code for each of these polynomials by using the latex function:
import sympy as sy
from sympy.printing.latex import latex
sy.init_printing()
sy.var('x')
fs = sy.factor_list(x**99-1)
fs2 = [x[0] for x in fs[1]]
for f in fs2:
print(latex(f))
Output:
x - 1
x^{2} + x + 1
x^{6} + x^{3} + 1
x^{10} + x^{9} + x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1
x^{20} - x^{19} + x^{17} - x^{16} + x^{14} - x^{13} + x^{11} - x^{10} + x^{9} - x^{7} + x^{6} - x^{4} + x^{3} - x + 1
x^{60} - x^{57} + x^{51} - x^{48} + x^{42} - x^{39} + x^{33} - x^{30} + x^{27} - x^{21} + x^{18} - x^{12} + x^{9} - x^{3} + 1
I want to achieve this kind of simplification:
sqrt(2)*sqrt(pi)*(p**2 + (-p**2 + 4*pi**2)*exp(2*I*p) - 4*pi**2)*exp(-I*p)/(p**2 - 4*pi**2)**2=-2*I*sqrt(2*pi)*sin(p)/(p**2 - 4*pi**2)
However, sympy.simplify can't simplify this expression:
f=sqrt(2)*sqrt(pi)*(p**2 + (-p**2 + 4*pi**2)*exp(2*I*p) -\
4*pi**2)*exp(-I*p)/(p**2 - 4*pi**2)**2
print(sympy.simplify(f))
#sqrt(2)*sqrt(pi)*(p**2 + (-p**2 + 4*pi**2)*exp(2*I*p) - 4*pi**2)*exp(-I*p)/(p**2 - 4*pi**2)**2
How to simplify this expression with SymPy?
Besides, I don't want to use Piecewise((sqrt(2)*I/(2*sqrt(pi)), Eq(p, -2*pi))...)
Just massage the expression a bit. You know that with fractions, you generally factorize them and then you cancel like-terms. Then you simplify after that:
from sympy import *
p = Symbol("p", real=True)
f = sqrt(2)*sqrt(pi)*(p**2 + (-p**2 + 4*pi**2)*exp(2*I*p) - 4*pi**2)*exp(-I*p)/(p**2 - 4*pi**2)**2
f = simplify(expand(cancel(factor(f))))
print(f)
Gives
-2*sqrt(2)*I*sqrt(pi)*sin(p)/(p**2 - 4*pi**2)
I have been working on some integrations and even though the system is working, it takes much more time to work than it should.
The problem is that the expressions are many pages, and even though they are 3 variables only, sy.simplify just crashes the Kernel after 4 hours or so.
Is there a way to make such lengthy expressions more compact?
EDIT:
Trying to recreate a test expression, using cse. I can't really substitute the symbols to make a final expression, equal to the 1st one
sy.var('a:c x')
testexp = sp.log(x)+a*(0.5*x)**2+(b*(0.5*x)**2+b+sp.log(x))/c
r, e = sy.cse(testexp)
FinalFunction = sy.lambdify(r[0:][0]+(a,b,c,x),e[0])
Points = sy.lambdify((a,b,c,x),r[0:][1])
FinalFunction(Points(1,1,1,1),1,1,1,1)
>>>NameError: name 'x1' is not defined
cse(expr) is sometimes a way to get a more compact representation since repeated subexpressions can be replaced with a single symbol. cse returns a list of repeated expressions and a list of expressions (a singleton if you only passed a single expression):
>>> from sympy import solve
>>> var('a:c x');solve(a*x**2+b*x+c, x)
(a, b, c, x)
[(-b + sqrt(-4*a*c + b**2))/(2*a), -(b + sqrt(-4*a*c + b**2))/(2*a)]
>>> r, e = cse(_)
>>> for i in r: pprint(Eq(*i))
...
_____________
╱ 2
x₀ = ╲╱ -4⋅a⋅c + b
1
x₁ = ───
2⋅a
>>> for i in e: pprint(i)
...
x₁⋅(-b + x₀)
-x₁⋅(b + x₀)
You are still going to have long expressions but they will be represented more compactly (and more efficiently for computatation) if cse is able to identify repeated subexpressions.
To use this in SymPy you can create two Lambdas: one to translate the variables into the replacement values and the other to use those values:
>>> v = (a,b,c,x)
>>> Pts = Lambda(v, tuple([i[1] for i in r]+list(v)))
>>> Pts(1,2,3,4)
(2*sqrt(2)*I, 1/2, 1, 2, 3, 4)
>>> Func = Lambda(tuple([i[0] for i in r]+list(v)), tuple(e))
>>> Func(*Pts(1,2,3,4))
(-1 + sqrt(2)*I, -1 - sqrt(2)*I)
I have an expression (expr, see below) that I am unable to simplify in SymPy. For real and positive x, expr is equivalent to x**3 + 2*x, but simplify and refine do not simplify the expression at all. (Mathematica does the simplication without any effort).
How to simplify this expression with SymPy?
from sympy import *
x = var('x')
expr = 16*x**3/(-x**2 + sqrt(8*x**2 + (x**2 - 2)**2) + 2)**2 - 2*2**(S(4)/5)*x*(-x**2 + sqrt(8*x**2 + (x**2 - 2)**2) + 2)**(S(3)/5) + 10*x
expr1 = simplify(expr) # does nothing
expr2 = refine(expr, Q.positive(x)) # does nothing
It can be done!
I rescind my earlier answer. Your expression can be simplified using Sympy. Here's how:
import sympy as sym
x = sym.symbols('x', positive=True)
expr = 16*x**3/(-x**2 + sym.sqrt(8*x**2 + (x**2 - 2)**2) + 2)**2 - 2*2**(sym.S(4)/5)*x*(-x**2 + sym.sqrt(8*x**2 + (x**2 - 2)**2) + 2)**(sym.S(3)/5) + 10*x
sym.simplify(sym.factor(sym.factor(sym.expand(sym.radsimp(expr))), deep=True))
Output:
x*(x**2 + 2)
Basically, I dug through all of the docs on sympy.simplify until I found that magic combination. Also, you have to define x as positive when you create the symbol, just as I did in the code above.
Comment on Mathematica
"Mathematica does the simplication without any effort"
I don't think you should ever underestimate the quantity of time and money that has gone into making the heuristic nightmare that is Mathematica's Simplify seem like it "just works". Sadly, in a lot of ways Sympy is still in it's infancy in comparison. sympy.simplify is one of those ways.
I need to write a regular expression that can detect a string that contains only the characters x,y, and z, but where the characters are different from their neighbors.
Here is an example
xyzxzyz = Pass
xyxyxyx = Pass
xxyzxz = Fail (repeated x)
zzzxxzz = Fail (adjacent characters are repeated)
I thought that this would work ((x|y|z)?)*, but it does not seem to work. Any suggestions?
EDIT
Please note, I am looking for an answer that does not allow for look ahead or look behind operations. The only operations allowed are alternation, concatenation, grouping, and closure
Usually for this type of question, if the regex is not simple enough to be derived directly, you can start from drawing a DFA and derive a regex from there.
You should be able to derive the following DFA. q1, q2, q3, q4 are end states, with q1 also being the start state. q5 is the failed/trap state.
There are several methods to find Regular Expression for a DFA. I am going to use Brzozowski Algebraic Method as explained in section 5 of this paper:
For each state qi, the equation Ri is a union of terms: for a transition a from qi to qj, the term is aRj. Basically, you will look at all the outgoing edges from a state. If Ri is a final state, λ is also one of the terms.
Let me quote the identities from the definition section of the paper, since they will come in handy later (λ is the empty string and ∅ is the empty set):
(ab)c = a(bc) = abc
λx = xλ = x
∅x = x∅ = ∅
∅ + x = x
λ + x* = x*
(λ + x)* = x*
Since q5 is a trap state, the formula will end up an infinite recursion, so you can drop it in the equations. It will end up as empty set and disappear if you include it in the equation anyway (explained in the appendix).
You will come up with:
R1 = xR2 + yR3 + zR4 + λ
R2 = + yR3 + zR4 + λ
R3 = xR2 + + zR4 + λ
R4 = xR2 + yR3 + λ
Solve the equation above with substitution and Arden's theorem, which states:
Given an equation of the form X = AX + B where λ ∉ A, the equation has the solution X = A*B.
You will get to the answer.
I don't have time and confidence to derive the whole thing, but I will show the first few steps of derivation.
Remove R4 by substitution, note that zλ becomes z due to the identity:
R1 = xR2 + yR3 + (zxR2 + zyR3 + z) + λ
R2 = + yR3 + (zxR2 + zyR3 + z) + λ
R3 = xR2 + + (zxR2 + zyR3 + z) + λ
Regroup them:
R1 = (x + zx)R2 + (y + zy)R3 + z + λ
R2 = zxR2 + (y + zy)R3 + z + λ
R3 = (x + zx)R2 + zyR3 + z + λ
Apply Arden's theorem to R3:
R3 = (zy)*((x + zx)R2 + z + λ)
= (zy)*(x + zx)R2 + (zy)*z + (zy)*
You can substitute R3 back to R2 and R1 and remove R3. I leave the rest as exercise. Continue ahead and you should reach the answer.
Appendix
We will explain why trap states can be discarded from the equations, since they will just disappear anyway. Let us use the state q5 in the DFA as an example here.
R5 = (x + y + z)R5
Use identity ∅ + x = x:
R5 = (x + y + z)R5 + ∅
Apply Arden's theorem to R5:
R5 = (x + y + z)*∅
Use identity ∅x = x∅ = ∅:
R5 = ∅
The identity ∅x = x∅ = ∅ will also take effect when R5 is substituted into other equations, causing the term with R5 to disappear.
This should do what you want:
^(?!.*(.)\1)[xyz]*$
(Obviously, only on engines with lookahead)
The content itself is handled by the second part: [xyz]* (any number of x, y, or z characters). The anchors ^...$ are here to say that it has to be the entirety of the string. And the special condition (no adjacent pairs) is handled by a negative lookahead (?!.*(.)\1), which says that there must not be a character followed by the same character anywhere in the string.
I've had an idea while I was walking today and put it on regex and I have yet to find a pattern that it doesn't match correctly. So here is the regex :
^((y|z)|((yz)*y?|(zy)*z?))?(xy|xz|(xyz(yz|yx|yxz)*y?)|(xzy(zy|zx|zxy)*z?))*x?$
Here is a fiddle to go with it!
If you find a pattern mismatch tell me I'll try to modify it! I know it's a bit late but I was really bothered by the fact that I couldn't solve it.
I understand this is quite an old question and has an approved solution as well. But then I am posting 1 more possible and quick solution for the same case, where you want to check your regular expression that contains consecutive characters.
Use below regular expression:
String regex = "\\b\\w*(\\w)\\1\\1\\w*";
Listing possible cases that above expression returning the result.
Case 1: abcdddd or 123444
Result: Matched
Case 2: abcd or 1234
Result: Unmatched
Case 3: &*%$$$ (Special characters)
Result: Unmatched
Hope this will be helpful...
Thanks:)