Dear list members,

I have turned to Singular to calculate the intersections of
polynomials that Maple 11 has had problems with. (My solution sets
are infinite, and Maple's polynomial arithmetic apparently is
particular poor in such cases.) After defining the ring, polynomials
and ideal, I use facstd. I would like to factorise my results further
than at present.

For example, the appended code produces the following output:

[1]:
_[1]=v
_[2]=b3-b2v+3bv2-2bv-v3
[2]:
_[1]=9v-4
_[2]=b-2v
[3]:
_[1]=v-1
_[2]=b-v

Thus, [1,2] simplifies to "=b" and [3,2] to "=b-1". How can I ask
Singular to perform these simplifications?

Thank you,

Colin Rowat, Department of Economics, University of Birmingham

APPENDED CODE

LIB "primdec.lib";
ring r=0,(b,s,v),lp;
poly v11h= b^3 + v^3 - 2*b*v;
poly v21h= v^3 - 3*b*v^2 + b^2*v + 2*b*v -b^3;
ideal i1121h = v11h,v21h;
list fi1121h = facstd(i1121h);
fi1121h;

Now you have three solutions. Any system that knows about Groebner
bases should be able to do that. What's the problem with Maple?

On May 8, 9:01?am, cliclic... at (no spam) freenet.de wrote:
Now you have three solutions. Any system that knows about Groebner
bases should be able to do that. What's the problem with Maple?

Thank you Martin. With the example posted, Maple can get me the
result that I want. My problem arises for more complicated examples,
when I run into memory allocation difficulties. I do not have first
hand knowledge of this, but am told that Maple's polynomial arithmetic
is poor, and particularly so for systems with infinite solutions - as
here.

Thus, I turned to Singular as a possibly more efficient package for
harder problems.

In the simple example that I first posted, including

option(redSB)

before the facstd() call performs the simplifications that I wanted.
I am now trying this on more complicated examples.

On May 8, 9:01 am, cliclic... at (no spam) freenet.de wrote:
Now you have three solutions. Any system that knows
about Groebner
bases should be able to do that. What's the problem
with Maple?

Thank you Martin. With the example posted, Maple can
get me the
result that I want. My problem arises for more
complicated examples,
when I run into memory allocation difficulties. I do
not have first
hand knowledge of this, but am told that Maple's
polynomial arithmetic
is poor, and particularly so for systems with
infinite solutions - as
here......

I am also not exactly sure what you want done. Can you post a "more
complicated example"? It may be that resultant techniques give the
best answer. (They usually are best for solving systems of polynomial
equations.)

Thank you Robert. Appended is the "more complicated
example" that I
have just tried running. While it solves in just
under two hours with
Maple 11 (running on an IBM x3655 with 8GB RAM), it
Killed out of
Singular on my linux laptop (Pentium M at 1.5GHz and
512MB RAM).
Thus, the example involves polynomials of the order
mentioned as
"record-hunter" by Martin.

Is it reasonable to expect solutions to such systems?
If so, to what
extent will I need to tailor my solution finding
strategy to the ideal
at hand?

Best,

Colin

LIB "primdec.lib";
ring r=0,(b,s,v),lp;
poly v22l = (2*s-1)^2 * ((5*s^2-2*s-1)*b^2 -
(8*s^2-5*s-1)*b - 1)*v^5
+ (2*s-1) * ((4*s-1)*(s-1)^2 * b^3 -
2*(7*s^3-6*s^2+1)*b^2 + (32*s^3 -
36*s^2 + 9*s + 2)*b + 2*(s-1))*v^4 +
((1-2*s)*(s-1)*(11*s^2-10*s
+2)*b^3 + (9*s^4 - 12*s^3 - s^2 + 4*s - 1)*b^2 + (5*s
- 48*s^4 -
41*s^2 + 81*s^3 + 1)*b - (1-s)^2)*v^3 + ((1 + 21*s^4
- 10*s + 34*s^2 -
45*s^3)*b^3 + 2*s*(s-1)*(s^2-s+1)*b^2 +
s*(s-1)*(16*s^2-13*s+1)*b)*v^2
+ (2*s*(1-s)*(2*s-1)^2 * b^3 - s^2 * (1-s)^2 * b^2 -
2*s^2 * (1-s)^2 *
b)*v + b^3 * s^2 * (1-s)^2;
poly v21l = 3*(s-1)*(2*s-1)^3 * b * v^6 - (2*s-1)^2 *
((s^2 - 4*s +
1)*b^2 + (23*s^2 - 29*s + *b + 1)*v^5 +
(2*s-1)*((s-1)*(7*s^2 - 5*s
+ 1)*b^3 + 2*(5*s^3 - 12*s^2 + 6*s - 1)*b^2 + (59*s^3
- 96*s^2 + 51*s
- 7)*b + 2*(s-1))*v^4 + ((1-s)*(31*s^3 - 34*s^2 +
14*s - 2)*b^3 +
(-37*s^2 + 48*s^3 + 10*s - 21*s^4 - 1)*b^2 + (29*s +
141*s^3 - 2 -
69*s^4 - 101*s^2)*b - (1-s)^2)*v^3 + ((6*s^2 - 6*s
+1)*(2*s-1)^2 * b^2
+ 2*s*(s-1)*(2*s-1)^2 * b + s*(s-1)*(19*s^2 - 19*s +
4))*b*v^2 + (1-
s)*(2*s*(2*s-1)^2 * b^2 - s^2 * (1-s)*b - 2*s^2 *
(1-s))*b*v + s^2 *
b^3 * (1-s)^2;

Thank you Robert. Appended is the "more complicated
example" that I
have just tried running. While it solves in just
under two hours with
Maple 11 (running on an IBM x3655 with 8GB RAM), it
Killed out of
Singular on my linux laptop (Pentium M at 1.5GHz and
512MB RAM).
Thus, the example involves polynomials of the order
mentioned as
"record-hunter" by Martin.

Is it reasonable to expect solutions to such systems?
If so, to what
extent will I need to tailor my solution finding
strategy to the ideal
at hand?

Best,

Colin

LIB "primdec.lib";
ring r=0,(b,s,v),lp;
poly v22l = (2*s-1)^2 * ((5*s^2-2*s-1)*b^2 -
(8*s^2-5*s-1)*b - 1)*v^5
+ (2*s-1) * ((4*s-1)*(s-1)^2 * b^3 -
2*(7*s^3-6*s^2+1)*b^2 + (32*s^3 -
36*s^2 + 9*s + 2)*b + 2*(s-1))*v^4 +
((1-2*s)*(s-1)*(11*s^2-10*s
+2)*b^3 + (9*s^4 - 12*s^3 - s^2 + 4*s - 1)*b^2 + (5*s
- 48*s^4 -
41*s^2 + 81*s^3 + 1)*b - (1-s)^2)*v^3 + ((1 + 21*s^4
- 10*s + 34*s^2 -
45*s^3)*b^3 + 2*s*(s-1)*(s^2-s+1)*b^2 +
s*(s-1)*(16*s^2-13*s+1)*b)*v^2
+ (2*s*(1-s)*(2*s-1)^2 * b^3 - s^2 * (1-s)^2 * b^2 -
2*s^2 * (1-s)^2 *
b)*v + b^3 * s^2 * (1-s)^2;
poly v21l = 3*(s-1)*(2*s-1)^3 * b * v^6 - (2*s-1)^2 *
((s^2 - 4*s +
1)*b^2 + (23*s^2 - 29*s + *b + 1)*v^5 +
(2*s-1)*((s-1)*(7*s^2 - 5*s
+ 1)*b^3 + 2*(5*s^3 - 12*s^2 + 6*s - 1)*b^2 + (59*s^3
- 96*s^2 + 51*s
- 7)*b + 2*(s-1))*v^4 + ((1-s)*(31*s^3 - 34*s^2 +
14*s - 2)*b^3 +
(-37*s^2 + 48*s^3 + 10*s - 21*s^4 - 1)*b^2 + (29*s +
141*s^3 - 2 -
69*s^4 - 101*s^2)*b - (1-s)^2)*v^3 + ((6*s^2 - 6*s
+1)*(2*s-1)^2 * b^2
+ 2*s*(s-1)*(2*s-1)^2 * b + s*(s-1)*(19*s^2 - 19*s +
4))*b*v^2 + (1-
s)*(2*s*(2*s-1)^2 * b^2 - s^2 * (1-s)*b - 2*s^2 *
(1-s))*b*v + s^2 *
b^3 * (1-s)^2;

I eliminated s from these two equations, yielding one equation in b and v. If this is not what you want, let me know.

It takes 0.17 seconds and around 350K of RAM. The resultant has these four factors:

3*v^6*b - 9*v^5*b + 9*v^4*b - 3*v^3*b

2*v - 1

v^2*b^5 - v^3*b^4 - 4*v^2*b^4 + 2*v*b^4 + 3*v^4*b^3 + 2*v^3*b^3 + 2*v^2*b^3 - 4*v*b^3 + b^3 - v^5*b^2 - 12*v^4*b^2 + 10*v^3*b^2 - v*b^2 + 4*v^5*b + 10*v^4*b - 20*v^3*b + 11*v^2*b - 2*v*b - 4*v^5 + 4*v^4 - v^3

b^3 + 2*v^3*b^2 - 4*v^2*b^2 - v*b^2 - 2*v^3*b + 7*v^2*b - 2*v*b - v^3

I did not check that each of these is irreducible.

Robert H. Lewis wrote:

I eliminated s from these two equations, yielding one equation in b and v. If this is not what you want, let me know.

It takes 0.17 seconds and around 350K of RAM. The resultant has these four factors:

3*v^6*b - 9*v^5*b + 9*v^4*b - 3*v^3*b

2*v - 1

v^2*b^5 - v^3*b^4 - 4*v^2*b^4 + 2*v*b^4 + 3*v^4*b^3 + 2*v^3*b^3 + 2*v^2*b^3 - 4*v*b^3 + b^3 - v^5*b^2 - 12*v^4*b^2 + 10*v^3*b^2 - v*b^2 + 4*v^5*b + 10*v^4*b - 20*v^3*b + 11*v^2*b - 2*v*b - 4*v^5 + 4*v^4 - v^3

b^3 + 2*v^3*b^2 - 4*v^2*b^2 - v*b^2 - 2*v^3*b + 7*v^2*b - 2*v*b - v^3

I did not check that each of these is irreducible.

The first and third factor reduce to 3·b·v^3·(v - 1)^3 and (b^3 -
b^2·v + 3·b·v^2 - 2·b·v - v^3)·(b·v - 2·v + 1)^2, respectively. The
timing is really impressive.

Thank you for this. That speed does seem amazing (my Singular code has hit
16 hours and is still running), and gives me new hope about the results that
are obtainable.

Robert - did you use Fermat or Singular to obtain the above? If the latter,
that's just the resultant() command?

Finally, a basic interpretational question, as my algebra is strained here:
if I am interested in all intersections of the two polynomials, v21l and
v22l, the resultant above gives me all combinations of b and v in the affine
variety <v21l,v22l>?
I then need to see what further restrictions are imposed by s?

Finally, a basic interpretational question, as my algebra is strained
here: if I am interested in all intersections of the two polynomials,
v21l and v22l, the resultant above gives me all combinations of b and
v in the affine variety <v21l,v22l>? I then need to see what further
restrictions are imposed by s?

Here is the invocation in FriCAS (or Axiom or OpenAxiom if you prefer): ...
(3) -> factors factor(resultant(v22l, v21l, s))

(3)
[[factor= 81,exponent= 1], [factor= b,exponent= 8],
[factor= v - 1,exponent= 8], [factor= v,exponent= 13],
[factor= 2v - 1,exponent= 1], [factor= (b - 2)v + 1,exponent= 7],
3 2 2 3
[factor= v - 3b v + (b + 2b)v - b ,exponent= 2],
2 3 2 2 2 3
[factor= (2b - 2b - 1)v + (- 4b + 7b)v + (- b - 2b)v + b ,exponent= 1]]
Type: List Record(factor: Polynomial Integer,exponent: Integer)
Time: 0.11 (EV) + 0.004 (OT) = 0.12 sec

I do not know about affine varieties, but the resultant vanishes if and only if
v22l and v21l have a common root as polynomials in s. Hm, I'm afraid I do not
understand your question, sorry. Maybe you could expand?

For my simple example (q.v. my initial post), the
factorize(resultant()) command in Singular yields:

-2 [multiplicity 1]
b [multiplicity 7]
b-1 [multiplicity 1]
9b-8 [multiplicity 1]

All except the first of these therefore correspond to a partial
solution - thus the first step in an elimination and extension
strategy?

If so, is there a generally optimal way of automating the whole procedure
(ideally in Singular) or must one tailor this to each pair of polynomials?

On May 9, 11:10 pm, cliclic... at (no spam) freenet.de wrote:
Robert H. Lewis wrote:

I eliminated s from these two equations, yielding
one equation in b and v. If this is not what you
want, let me know.

It takes 0.17 seconds and around 350K of RAM.
The resultant has these four factors:

.... snip ....


The first and third factor reduce to 3·b·v^3·(v -
1)^3 and (b^3 -
b^2·v + 3·b·v^2 - 2·b·v - v^3)·(b·v - 2·v + 1)^2,
respectively. The
timing is really impressive.

Dear Robert and Martin,

Thank you for this. That speed does seem amazing (my
Singular code
has hit 16 hours and is still running), and gives me
new hope about
the results that are obtainable.

Robert - did you use Fermat or Singular to obtain the
above? If the
latter, that's just the resultant() command?....

If what you want are the solutions to the single equation v22l = v21l
and not those to the pair of simultaneous equations [v22l = 0, v21l =
0]
I am looking for the simultaneous solutions.