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Re: Re: "Reduce" wierdness (or too slow?)


I forgot to add the obvious. Why not simply do this:

sols = Reduce[{1000 + t2 == t3, 1200 + w1 == w2, 125 + t1 == t2,
     125 + t5 == t6, k*t3 == 2*(v2 + z2),
     t7*z1 + t2*z2 + z3 + z4 == H + (t2^2)/k + (t7^2)/k + w1,
     t7*z1 + t3*z2 + z3 + z4 == H + (t3^2 + t7^2)/k,
     t7*z1 + z3 + z4 == 2*H + (t7^2)/k, w1 + t6*z1 + z3 == (t6^2)/k +  
w1 + w2,
     w1 == w2 + ((t1 - t2)*(t1 + t2 - k*z2))/k,
     w2 + t5*z1 + z3 == (t5^2)/k + w1 + w2, w2 == (-t1^2)/k + t1*z2 +  
z4,
     z1 == 2*t4/k + v2, z2 == 0}, {H, t1, t2, t3, t4, t5, t6, t7, v2,  
w1, w2,
     z1, z2, z3, z4}, Backsubstitution -> True];

sols1 = Reduce[
    sols, {H, t1, t2, t3, t4, t5, t6, t7, v2, w1, w2, z1, z2, z3, z4},  
Reals];

Now, if you compare sols1 with sols you see that the only difference  
is that the case k!=0 was replaced by two cases k>0 and k<0. These  
two, of course, correspond to the assumption that k is real, which is  
all that is needed in this case to ensure that everything else is  
real. This is in fact exactly the same solution to the same problem as  
I suggested earlier - in effect the inequalities (which are  
automatically added by Reduce if you use the domain restriction to  
Reals) are separated from the equations.

Andrzej Kozlowski


On 25 May 2008, at 19:26, Andrzej Kozlowski wrote:

> The problem is that using Reduce[...,Reals] in this context amounts to
> adding lots of inequalities. To see this, just take a couple of the
> equations:
>
> Reduce[{t7*z1 + t3*z2 + z3 + z4 == H + (t3^2 + t7^2)/k,
>   t7*z1 + z3 + z4 == 2*H + (t7^2)/k}, {H, t1, t2, t3, t4, t5, t6, t7,
> v2, w1,
>   w2, z1, z2, z3, z4}, Reals, Backsubstitution -> True]
>
> You will see all the inequalites that had to be added by Reduce to
> make all expressions that appear in Reduce real. This is actually
> harder rather thans simpler than your original formulation.
>
> Andrzej Kozlowski
>
>
> On 25 May 2008, at 15:02, TuesdayShopping wrote:
>
>> Re: Reduce" wierdness (or too slow?).
>> Thanks for the input from Andrzej Kozlowski, Daniel Lichtblau, Adam
>> Strzebonski and others. OK; No inequations. Below is problem similar
>> (not same) to the earlier one, that does not have any inequations
>> but still Reduce will not solve it in 600 seconds. Even FindInstance
>> will not find a (even partial) solution in 600 seconds. I should
>> have got the solution for H, z2, w1, w2, t1, t2, t3, v2 as below. My
>> goal is that I would like to solve a class of problems (like the one
>> below) for as many variables as I can.
>>
>> Problem:
>> Reduce[{1000 + t2 == t3, 1200 + w1 == w2, 125 + t1 == t2,
>> 125 + t5 == t6, k*t3 == 2*(v2 + z2),
>> t7*z1 + t2*z2 + z3 + z4 == H + (t2^2)/k + (t7^2)/k + w1,
>> t7*z1 + t3*z2 + z3 + z4 == H + (t3^2 + t7^2)/k,
>> t7*z1 + z3 + z4 == 2*H + (t7^2)/k,
>> w1 + t6*z1 + z3 == (t6^2)/k + w1 + w2,
>> w1 == w2 + ((t1 - t2)*(t1 + t2 - k*z2))/k,
>> w2 + t5*z1 + z3 == (t5^2)/k + w1 + w2,
>> w2 == (-t1^2)/k + t1*z2 + z4, z1 == 2*t4/k + v2, z2 == 0}, {H, t1,
>> t2, t3, t4, t5, t6, t7, v2, w1,
>> w2, z1, z2, z3, z4}, Reals, Backsubstitution -> True]
>>
>> (For trying with FindInstance, add variable "k" to list of
>> variables, and of course, remove "Backsubstitution -> True" option)
>>
>> Solutions that must have been found:
>> t1 = ((48 * k - 625)/10),
>> t2 = ((48 * k + 625)/10),
>> t3 = ((48 * k + 10625)/10),
>> v2 = ((48 * k^2 + 10625 * k)/20),
>> w1 = ((9600 * k + 1125000)/k),
>> w2 = ((10800 * k + 1125000)/k),
>> z2 = 0,
>> H= z4 = ((2304 * k^2 + 1020000 * k + 112890625)/(100 * k))
>>
>
>



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