At 02\12\27 22:38 -0500 Friday, Elisabeth Varin/Stephane Rouillon wrote:
Craig --I will never understand what you write if I cannot even start from the same place. In the following example, I see 13 "papers" as you say, 4 "candidates" (namely A,B,C,D) and 4 different kinds of ballots or "positions" as said Forest (namely ABCD, BDAC, CDAB , DBCA) Am I right? Steph.>5 ABCD >4 BDAC >3 CDAB >1 DBCA What suppose that there are 5 paper and 6 candidates.
This seems to be only about the default meaning of a term that is using
less words than are needed to get its meaning pinned down. So the thread
is a bad starting point for an argument over the correctness of implying
readers must produce their own wrong restrictions when interpreting bad
English of others.
You seem to have 13 be the number that is found by counting the sum of
all the ballot papers' weights.
The word "papers" is able to refer to more than one idea:
(1) the set of papers, with papers being something or other. Anyway, this
is not a number;
(2) the number of kinds of papers, which is the meaning I had
(3) the number of papers when there is one paper each [that's ill-defined]
(4) the total number of papers, i.e. the sum of their weights
(5) definition (4) but with a constraint that the number is undefined when
some of that ballot papers are negative.
I take meaning of (1) by default and if a number is returned then (2).
I presume it is very unclear to readers where the not-understanding Mr R
is over meanings, (3),(4) and (5).
For example, how many "papers" under your ideas, has this election got ?:
(ABCD) 5 +sqrt(7)/1000
(BDAC) 4
(CDAB) 3
(DBCA) -1
I should not have to respond to questions as simple as this one that Mr R
sent.
Craig Carey
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