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Re: Re: Re: Resolve/Reduce is taking forever
On 15 Jun 2006, at 16:27, Andrzej Kozlowski wrote:
>
> Select, of course, is of a completely different kind of creature; it
> obviously makes use of some kind of search algorithm and symbolic
> algebra is used only to the extent that it is used by the function
> specified as the second argument to Select. So the question about the
> "complexity with respect to the number of variables" reduces to the
> same question for the selection functions used.
Of course not "some kind of serarch" but the obvious one: it just
goes through the list in order and applies the selection criterion to
each element in turn. The time complexity will depend on the
criterion and on what is contained in the list.
Andrzej Kozlowski
>
>
> On 14 Jun 2006, at 19:29, Bonny Banerjee wrote:
>
>> That was very helpful. Thanks.
>>
>> Is there a text where I can find the complexity of Mathematica
>> functions
>> such as Reduce, Resolve, NMinimize, Select, etc. with respect to
>> the number
>> of variables? Please let me know.
>>
>> --Bonny.
>>
>>
>>
>> ----- Original Message -----
>> From: "Andrzej Kozlowski" <akoz at mimuw.edu.pl>
To: mathgroup at smc.vnet.net
>> Subject: [mg67298] [mg67289] [mg67259] Re: [mg67217] Resolve/Reduce is
>> taking forever
>>
>>
>>> On 13 Jun 2006, at 14:07, Bonny Banerjee wrote:
>>>
>>>> I am trying to write a simple function that determines the
>>>> conditions
>>>> for a
>>>> curve to be on the left of a straight line. A curve is to the
>>>> left of a
>>>> straight line if each point on the curve is to the left of the
>>>> straight
>>>> line. The curve is specified using parametric equations:
>>>>
>>>> x -> a3Ã?t^3 + a2Ã?t^2 + a1Ã?t + a0
>>>> y -> b1Ã?t + b0
>>>>
>>>> where t is the parameter, 0<=t<=1, and {a0, a1, a2, a3, b0, b1}
>>>> are real
>>>> coefficients. The straight line is specified using two points
>>>> {x1,y1}
>>>> and
>>>> {x2,y2}.
>>>>
>>>> Here is the function:
>>>>
>>>> isLeftofLine[{x1_, y1_}, {x2_, y2_}, {a0_, a1_, a2_, a3_, b0_,
>>>> b1_}] =
>>>> Resolve[
>>>> ForAll[t, 0 <= t <= 1 => fxy[{x1, y1}, {x2, y2}, {a3Ã?t^3 +
>>>> a2Ã?t^2
>>>> +
>>>> a1Ã?t + a0, b1Ã?t + b0}] <= 0],
>>>> {a0, a1, a2, a3, b0, b1}, Reals]
>>>>
>>>> where
>>>>
>>>> fxy[{x1_, y1_}, {x2_, y2_}, {x_, y_}] =
>>>> (x - x1)Ã?(y2 - y1) - (y - y1)Ã?(x2 - x1)
>>>>
>>>> I tried Resolve and Reduce but both are taking forever. I waited
>>>> for
>>>> more
>>>> than 4 hours but could not get any result from any of them.
>>>> Considering
>>>> this
>>>> is a simple logical expression with only one universal
>>>> quantifier, I am
>>>> surprised at what might be taking so long.
>>>>
>>>> Any insights would be very helpful. Also, any alternative method
>>>> for
>>>> solving
>>>> the same problem, such as using any other function in place of
>>>> Reduce/Resolve or using a different representation for the curve or
>>>> straight
>>>> line, would be nice to know. I preferred using parametric
>>>> equations for
>>>> representing the curve as the curve is finite.
>>>>
>>>> Thanks,
>>>> Bonny.
>>>>
>>>
>>>
>>>
>>> First of all, your "function" will not work even, in this form, in a
>>> thousand years, because you are demanding Mathematica to solve a
>>> purely
>>> symbolic problem on the right hand side, before any numerical
>>> values have
>>> been substituted for x1, x2,y1,y2. This is obviously impossible
>>> (see the
>>> comments below) . So as a minimum you need to replace your
>>> immediate
>>> assignment = by Delayed assignment :=, and substitute numerical
>>> values
>>> for the x's and the y's. Your function should then look like this:
>>>
>>> isLeftofLine[{x1_, y1_}, {x2_, y2_},{a0_, a1_, a2_, a3_, b0_,
>>> b1_}] :=
>>> Resolve[
>>> ForAll[t, 0 <= t <= 1, fxy[{x1, y1}, {x2, y2}, {a3Ã?t^3 +
>>> a2Ã?t^2 +
>>> a1Ã?t + a0, b1Ã?t + b0}] <= 0],
>>> {a0, a1, a2, a3, b0, b1}, Reals]
>>>
>>> When you evaluate this with numerical values for the coordinates
>>> of the
>>> points you will get a relationship between the a's and b's .
>>> However,
>>> even if the problem is now correctly formulated it will take a
>>> very long
>>> time to solve it. The algorithm that is needed, called
>>> CylindricalAlgebraicDecomposition, is doubly exponential in the
>>> number of
>>> variables (it is the number of variables and not the number of
>>> quantifiers that counts). So let's try first solving a version of
>>> your
>>> problem with a smaller number of unknowns, say 4. Here is the
>>> reduced
>>> problem.
>>>
>>> fxy[{x1_, y1_}, {x2_, y2_}, {x_, y_}] =
>>> (x - x1)Ã?(y2 - y1) - (y - y1)Ã?(x2 - x1)
>>>
>>>
>>> isLeftofLine[{x1_, y1_}, {x2_, y2_}, {a_, b_, c_, d_}] :=
>>> Resolve[
>>> ForAll[t, 0 <= t <= 1 , fxy[{x1, y1}, {x2, y2}, {t^3 +
>>> a*t^2 + b, c* t + d}] <= 0],
>>> {a, b, c, d}, Reals]
>>>
>>> Now there are only four variables. Let's try some concrete values of
>>> {x1,y1} and {x2,y2}.
>>>
>>> In[27]:=
>>> cond=isLeftofLine[{0, 0}, {3, 4}, {a, b, c, d}]
>>>
>>> Out[27]=
>>> b â?? Reals && ((a < -6 && ((c <= (1/3)*(8*a + 12) &&
>>> d >= (1/3)*(4*a + 4*b - 3*c + 4)) ||
>>> (Inequality[(1/3)*(8*a + 12), Less, c, LessEqual,
>>> 0] && d >= (1/81)*(8*a^3 + 27*c*a + 108*b) +
>>> (1/81)*Sqrt[64*a^6 + 432*c*a^4 + 972*c^2*a^2 +
>>> 729*c^3]) || (c > 0 && d >= (4*b)/3))) ||
>>> (-6 <= a <= -3 && ((c <= (1/3)*(8*a + 12) &&
>>> d >= (1/3)*(4*a + 4*b - 3*c + 4)) ||
>>> ((1/3)*(8*a + 12) < c < 0 &&
>>> d >= (1/81)*(8*a^3 + 27*c*a + 108*b) +
>>> (1/81)*Sqrt[64*a^6 + 432*c*a^4 + 972*c^2*a^2 +
>>> 729*c^3]) || (c >= 0 && d >= (4*b)/3))) ||
>>> (-3 < a < -1 && ((c <= (1/3)*(-a^2 + 2*a + 3) &&
>>> d >= (1/3)*(4*a + 4*b - 3*c + 4)) ||
>>> ((1/3)*(-a^2 + 2*a + 3) < c < 0 &&
>>> d >= (1/81)*(8*a^3 + 27*c*a + 108*b) +
>>> (1/81)*Sqrt[64*a^6 + 432*c*a^4 + 972*c^2*a^2 +
>>> 729*c^3]) || (c >= 0 && d >= (4*b)/3))) ||
>>> (a >= -1 && ((c <= (1/3)*(4*a + 4) &&
>>> d >= (1/3)*(4*a + 4*b - 3*c + 4)) ||
>>> (c > (1/3)*(4*a + 4) && d >= (4*b)/3))))
>>>
>>>
>>> We can now use it to check the condition cond for various choices of
>>> {a,b,c,d}, for example
>>>
>>> cond/.Thread[{a, b,c,d}->{2,3,4,5}]
>>>
>>> True
>>>
>>> So here is the good news and the bad news. I think that if I have
>>> understood you correctly, your problem is solvable in theory. I
>>> strongly
>>> doubt, however, that it is solvable in reasonable time. You can
>>> try out
>>> the above on your computer and see how long it takes. Then
>>> remember: the
>>> algorithm is double exponential in the number of variables so 4
>>> hours
>>> seems like a very optimistic estimate.
>>>
>>> Andrzej Kozlowski
>>> Tokyo, Japan
>>
>
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