Paul,
<<No particular significance - they are just integers representing a
column "cut point">>
IIUC, after the cut point the matrix becomes all zeros (which would surely
negate any benefit of its being a WIBM, wouldn't it?); but I just can't
see
how this would be any better than having two intervals, one which is
governed by factorisation and the other which is not, yet does have
positive
column sums. Recall, I'm doing it for the sake of notation.
<<
> The reason it was always a problem to me was that the two matrices, the
> one
> whose distribution of nonzero entries is determined by factorisation and
> the
> other whose is not, are distinguished by two different arguments for the
> height of each column entry. I had been told that if you set out your
> definitions with all your notation etc. for one set of arguments, you
> don't
> do it for a different set of arguments.
Just the opposite - practice parsimony with symbolism. That's why I
defined R(M,r,s) with _no_ more restrictions on r and s than were
needed to make the notation make sense.>>
I wonder, then, if you could advise me on notation for the two different
matrices, for, a) column sum in n; b) number of columns for which column
sum
is >0 [o(x,y,.), I think I'll be able to infer the rest.
<<If y - x = d then your pairs _must_ look like
(x, _), (x + 1, _), ..., (x + d, _). You have never said whether you
are requiring y - x = y' - x'.>>
Yes, I need y - x to be equal to y' - x'.
<<
> And of all
> possible sets of such pairs, I want the set that yields the lowest
> possible
> sum of all values r-s.
There's more than one set? What are they?>>
If you have any set of numbers, say 3,7,1, and pair them with another set,
say 6,4,1, you get different results according to the pairing. For (3,6),
(7,4), (1,1) only (7,4) will give a set R; 7-4 =3, so sum(R) = 3. Now we
try
a different pairing: (3,4), (7,1),(3,1). Here, (7,1) and (3,1) both give
sets R, and sum(R)=6+2=8. As I say, I want the set of pairs that gives the
lowest sum(R). I am not considering pairs (n,n') for which
rank(M,n)=rank(M,n'), so the number of pairs considered might very well be
less than y-x+1.
Recall also that I stipulate that if |R|=r-s, s>1; and if |S|=t-u, u>1.
And
if rank(M,n)=0 and rank(M,n')=1, then (n,n') is not a member of R or S,
but
a member of V.
<<Are you allowing your intervals to overlap?>>
Yes.
<<
> <<rank(M,z) = 1 says z is a prime - in fact if
> supp(M.z) = {i}, z is the i-th prime.>>
>
> Not by my definition of P(m).
Your definition of P(m) has nothing to do with rank(M,n) and supp(M,n).>>
But rank(M,n) is column sum. For all but m columns in the whole
'factorisation' matrix, only a column sum of zero indicates that n is
prime.
<<
> You're not
> rewriting the whole matrix, from start to finish, such as to discard all
> access to the old sup****ts, are you?
_I'm_ not rewriting the matrix at all - that's your job I reckon. >>
But at what point does the matrix become, in effect, a histogram? The
original rows, in each of which the nonzero entries are separated by an
interval of a prime in P(m), are still there, aren't they, before the
initial defining criteria for
R(M,r,s) have been laid down?
<<
> Why don't we just state that any column
> contains h members of P(m), and tag the entries for that column as
> {1,2...h}?
I don't see what you are trying to do. For WIBM T, T(i,j) very nicely
"tags" the i-th entry of column j; for that matter, so does (i,j). >>
If you have a column with some zero entries interspersed with nonzero
entries, then the i-th prime (if 2 is the first prime, 3 the second, 5 the
third etc.) is, of course, not necessarily on the i-th row in that column.
So I deduce from what you say above that our matrix, in its basic
structure,
takes the form of a histogram, with no interspersed zeroes. This dashes
the
hopes I expressed in my previous paragraph. Can't we retain the
interspersed
zeroes, and tag the nonzero entries as I suggest?
<<We don't know -or, at least you haven't said - what r and
s are.>>
In the example I gave, I said r = 5 and s = 3.
<<R(M,r,s) has _no_ relevant content except its size. It does not contain
"tags" for _anything_ unless you explicitly associate elements of
R(M,r,s) with whatever it is you are trying to tag.>>
I have always said that the very reason I am creating the set R, for a
given
(n,n'), is so that I can pair its members with members of another set S
(which pertains to a different (n,n') in [x,y]x[x',y']). The |R| (= r-s)
entries with the highest tag numbers in a column n are all the members of
R.
The |S| (= t-u) entries with the highest tag numbers in a column n' are
all
the members of S. I'll need notation so that I can speak of the tag number
(which, incidentally, is never 1) of a member of R in relation to the tag
number of a member of S (which is also never 1). The tag number of a
column
entry, the highest of which in n is rank(M,n), is fixed, irrespective of
whether or not it is referenced as a member of R or of S, and irrespective
of |R| and |S|. Aren't we now in a position to define R and S - and, for
that matter, V?
<<I don't know if this is of any use to you at all but there is an
operation XOR sometimes denoted by \otimes, I'll use (x). It works like
this: 1 (x) 1 = 0 (x) 0 = 0; 1 (x) 0 = 0 (x) 1 = 1. If you XOR one
column term by term with another you'll get 0's where the both were 1's
or both 0's and 1's where one or the other, but not both, were 1.>>
Hmm. I gave this some thought but couldn't see how it would help. Will
keep
it in the back of my mind, though.
Many thanks for your continuing assistance.
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