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Polynomial transposition

asked 12 years ago

fredrik gravatar image

Let's say I have

R1 = ZZ['y']['x']
R2 = ZZ['x']['y']

Is there a builtin function in Sage for converting an element from R1 to R2 by transposing the coefficients, not just switching the variables? For instance y*x+y should become (x+1)*y, not x*y+x.

I can do it by first coercing to ZZ['x','y'] and then to R2, but that seems clumsy (and possibly inefficient). I could also just write a loop, but that seems ugly (and again possibly inefficient). It would be fine with a function that just transposes the polynomial within R1 (then I can just coerce the result to R2 afterwards).

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answered 12 years ago

updated 12 years ago

I don't know of anything built in to do this, but here's one possible method. I think it requires slightly different definitions of R1 and R2, but maybe you can work something out if you need to keep your definitions:

sage: S1 = ZZ['x,y']
sage: S2 = ZZ['y,x']
sage: a = S1.gen(0) * (S1.gen(1) + 1)
sage: a
x*y + x
sage: a.dict()
{(1, 0): 1, (1, 1): 1}
sage: S2(a.dict())
y*x + y

Edit: sorry, this doesn't do what you want. You should modify the dictionary appropriately:

sage: d = {(p[1], p[0]):a.dict()[p] for p in a.dict()}
sage: S2(d)
y*x + x

So I guess with your original R1 and R2:

sage: R1 = ZZ['y']['x']
sage: R2 = ZZ['x']['y']
sage: a = R1.base_ring().gen(0) * (R1.gen(0) + 1)
sage: a
y*x + y
sage: a.dict()
{0: y, 1: y}
sage: R2({p: a.dict()[p].subs(R2.base_ring().gen(0)) for p in a.dict()})
x*y + x

Edit: There should be a similar way to modify the dictionary here, too. You might also look into a.map_coefficients(...).

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answered 12 years ago

Volker Braun gravatar image

Especially if you are concerned about performance you should never create univariate polynomial rings over univariate polynomial rings. Directly using multivariate polynomial rings will be much faster.

Bonus: its then easy to substitute other variables since all variables are treated at the same level, for example:

sage: S1.<x1,y1> = ZZ[]
sage: S2.<y2,x2> = ZZ[]
sage: p1 = y1*x1 + y1
sage: p1(y2,x2)
y2*x2 + x2
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This is not accurate -- ZZ['x']['y'] is much faster than ZZ['x','y'], especially for large (dense) polynomials.

fredrik gravatar imagefredrik ( 12 years ago )

_Only_ for dense polynomials where the degree in the outer variable isn't too large.

Volker Braun gravatar imageVolker Braun ( 12 years ago )

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Asked: 12 years ago

Seen: 925 times

Last updated: Apr 07 '13