From: Xan Gregg <xan.gregg@jmp.com>

Date: Thu, 21 Jul 2005 13:33:46 -0400

Message-Id: <0fddad752d5e1e48f0c11763476dd68b@jmp.com>

Cc: <xmlschema-dev@w3.org>

To: "Roger L. Costello" <costello@mitre.org>

Date: Thu, 21 Jul 2005 13:33:46 -0400

Message-Id: <0fddad752d5e1e48f0c11763476dd68b@jmp.com>

Cc: <xmlschema-dev@w3.org>

To: "Roger L. Costello" <costello@mitre.org>

> Note that I am assuming that the enumeration facet takes precedence > over all > other facets. I believe that this is true, isn't it? I don't think it is. Just because a value satisfies one facet doesn't give a free ride past the others. I like your organization into three parts (pattern, enumerations, and range), but unfortunately you need the *intersection* of each part's value space. Consider this variation of your exanmple: > <simpleType name="foo"> > <restriction base="byte"> > <pattern value=".*[13579]"/> <!-- odd numbers --> > <minInclusive value="0"/> > <maxInclusive value="35"/> > <enumeration value="11"/> > <enumeration value="22"/> > <enumeration value="33"/> > <enumeration value="44"/> > <enumeration value="55"/> > </restriction> > </simpleType> The value space is {11, 33} with cardinality 2. Algorithm becomes: 1. Compute range [lo..hi], as before. 2. If enumeration present, cardinality is number of enumeration values that are in [lo..hi] and satisfy any pattern values; done. 3. If pattern present, cardinality is number of values in [lo..hi] that satisfy pattern values; done. 3. Otherwise, cardinality is hi - lo + 1. xan On Jul 21, 2005, at 12:06 PM, Roger L. Costello wrote: > Hi Xan, > > I like your algorithm! I wonder if checking for enumerations > shouldn't be > done first, and then patterns, and then the 7 steps you described? > > Algorithm for Computing the Cardinality of a byte simpleType (with > Xan's > suggestions) > > 1. If there are enumeration facets then > cardinality = count the number of enumeration facets > Done. > > 2. If there are pattern facets then > cardinality = the maximum cardinality of all the pattern facets. > Done. > > 3. Otherwise: > 3.1. lo = -128, hi = 127 > 3.2. minInclusive present => lo = max(lo, minInclusive) > 3.3. maxInclusive present => hi = min(hi, maxInclusive) > 3.4. minExclusive present => lo = max(lo, minExclusive + 1) > 3.5. maxExclusive present => hi = min(hi, maxExclusive - 1) > 3.6. totalDigits present => lo = max(lo, -10^^totalDigits + 1); > hi = min(hi, 10^^totalDigits - 1) > 3.7. cardinality = hi - lo + 1 > > Note that I am assuming that the enumeration facet takes precedence > over all > other facets. I believe that this is true, isn't it? > > Does this algorithm seem correct? /RogerReceived on Thursday, 21 July 2005 17:33:50 GMT

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