
Hi,
i've started to read the ppl documentation, and i've found a doubt reading section at page 8:
1.4.5 Removing Dimensions from the vector space.
and in particular, how the remove_space_dimension works.
In fact from the example i can not understand why applying the operator on the singleton set {(3,1,0,2)^T} in R^4 with the variable set {x_1, x_2} we obtain just {(3,2)^T}. Namely, i can not understand which is the connection between the set of variables gived to the operator and the two dimensions removed from P (i.e the second and the third).
Instead, the remove_higher_space_dimension operator is simpler because is easy to understand that it removes all the dimensions i >= m.
Thanks,
Alessandro Vincenzi

On Wed, 1 Nov 2006, Alessandro Vincenzi wrote:
Hi,
i've started to read the ppl documentation, and i've found a doubt reading section at page 8:
1.4.5 Removing Dimensions from the vector space.
and in particular, how the remove_space_dimension works.
In fact from the example i can not understand why applying the operator on the singleton set {(3,1,0,2)^T} in R^4 with the variable set {x_1, x_2} we obtain just {(3,2)^T}. Namely, i can not understand which is the connection between the set of variables gived to the operator and the two dimensions removed from P (i.e the second and the third).
The variables x_1 and x_2 are the second and third dimensions. After removing these dimensions, the remaining dimensions (the 1st and 4th) are renumbered to be the 1st and 2nd.
If it is not clear, can you suggest a better explanation?
Pat
Instead, the remove_higher_space_dimension operator is simpler because is easy to understand that it removes all the dimensions i >= m.
Thanks,
Alessandro Vincenzi
PPL-devel mailing list PPL-devel@cs.unipr.it http://www.cs.unipr.it/mailman/listinfo/ppl-devel

Alessandro Vincenzi wrote:
Hi,
i've started to read the ppl documentation, and i've found a doubt reading section at page 8:
1.4.5 Removing Dimensions from the vector space.
and in particular, how the remove_space_dimension works.
In fact from the example i can not understand why applying the operator on the singleton set {(3,1,0,2)^T} in R^4 with the variable set {x_1, x_2} we obtain just {(3,2)^T}. Namely, i can not understand which is the connection between the set of variables gived to the operator and the two dimensions removed from P (i.e the second and the third).
In the manual (and in the library itself), the space dimensions are indexed starting from zero. That is why removing x_1 and x_2 gives the result above. When coding the example above in C++, you will write something like
Variable x1(1); Variable x2(2); Variables_Set vs(x1, x2);
At a first sight, the manual is clear enough on this respect. Anyway, let us know if you find any place where we number the dimensions using an inconsistent or ambiguous indexing scheme.
Cheers, Enea.
Instead, the remove_higher_space_dimension operator is simpler because is easy to understand that it removes all the dimensions i >= m.
Thanks,
Alessandro Vincenzi
PPL-devel mailing list PPL-devel@cs.unipr.it http://www.cs.unipr.it/mailman/listinfo/ppl-devel
participants (3)
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Alessandro Vincenzi
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Enea Zaffanella
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P M Hill