From: <petsa@us.ibm.com>

Date: Thu, 20 May 1999 14:46:04 -0400

To: Marina_Nikitina@infoimage.com

cc: Www-Xml-Schema-Comments@w3.org

Message-ID: <85256777.00671795.00@D51MTA03.pok.ibm.com>

Received on Thursday, 20 May 1999 14:46:18 UTC

Date: Thu, 20 May 1999 14:46:04 -0400

To: Marina_Nikitina@infoimage.com

cc: Www-Xml-Schema-Comments@w3.org

Message-ID: <85256777.00671795.00@D51MTA03.pok.ibm.com>

Marina: Thanks for your comments. I'm glad someone is reading the draft with care! You raise 3 points: 1. Yes, h must >= 0. We will fix this. 2. For some specific values x = v. 3. Since values in C are approximations for values in M, many values in M map to a single value in C. Perhaps this needs to be made more specific. Regards, Ashok (Embedded image moved to Marina_Nikitina@infoimage.com file: 05/18/99 02:22 PM pic03455.pcx) To: www-xml-schema-comments@w3.org cc: (bcc: Ashok Malhotra/Watson/IBM) Subject: exact and approximate In 2.5.1.4 you are defining approximate datatypes as follows: Then for every value v' in C, there is a corresponding value v in M and a real value h such that P(x) = v' for all X in M such that |v - x| < h. First of all, because of absolute value, h should be >0 (or >=0). Then x in M must be different from v in M. So, you better say , for example, ... and a real value h >0 such that P(x) = v' for all x != v in M such that |v - x| < h Next: Furthermore, for at least one value v' in C, there is more than one value y in M such that P(y) = v' Question: is it possible for v'' from C and v'' !=v' that P(y) =v' and P(y)=v''? I am just trying to understand your concept of "approximate datatype". Thanks

- application/octet-stream attachment: pic03455.pcx

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