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migrating from traditional regexp cardinalities to shexj abstract syntax

From: Eric Prud'hommeaux <eric@w3.org>
Date: Thu, 3 Mar 2016 17:38:23 -0500
To: Iovka Boneva <iovka.boneva@univ-lille1.fr>
Cc: public-shex-dev@w3.org
Message-ID: <20160303223820.GJ24478@w3.org>
This email lays out the current cardinality tests in preparation for
adapting them to ShExJ's abstract syntax.

The triple expressions in shexj are one of:

  triple constraint: predicate, inverse, value expression, cardinality
  allOf: sequence of triple expressions, cardinality
  someOf: sequence of triple expressions, cardinality
  inclusion: ref to a shape's triple expression.

Currently the permissible cardinality on any of these is {0,∞}.  We
may restrict the permissible cardinalities on allOfs and containers of
allOfs in the future in order to simplify the 

The ICDT paper gave the membership formula like:

Σ: edge labels
∆: set of symbols, possibly in Σ
ω: ∆→N: map symbols in ∆ to number of occurrences
   e.g. w₀ = ⦃a, a, a, c, c⦄ means {a:3, c:2, *:0}

      I(ε) = [0;∞]
      I(a[n,m]) = [⌈ω(a)/m⌉;⌊ω(a)/n⌋]
        0/∞ = 0
        i/∞ = 1 for i ≥ 1
  or: I(E₁ ∣ E₂) = I(E₁) ⊕ I(E₂)
        ⊕: pairwise addition: [n₁;m₁]⊕[n₂;m₂]=[n₁+n₂; m₁+m₂]
        m+∞ = ∞+m = ∞
 and: I(E₁ ∥ E₂) = I(E₁) ∩ I(E₂)
        [n₁;m₁]∩[n₂;m₂]=[max(n₁,n₂);min(m₁,m₂)]
              ⎧         [0;∞] if ω∆(E) = ε
      I(E*) = ⎨         [1;∞] if ω∆(E) ≠ ε and I(E) ≠ 0
              ⎩             ∅ otherwise
              ⎧         [0;0] if ω∆(E) = ε
      I(E⁺) = ⎨[1;max(I(E))⌉] if ω∆(E) ≠ ε and I(E) ≠ 0
              ⎩             ∅ otherwise
        all ε∥E* are expressed as E⁺


Example:
 I(a[n,m]) tests
  <S> { :p . } / { <s> :p true }
    ∆={:p}, ω={:p→1}
    I(:p[1,1]) = [⌈1/1⌉;⌊1/1⌋] = [1;1] pass
  <S> { :p . } / { <s> :q true }
    ∆={:p}, ω={:p→0}
    I(:p[1,1]) = [⌈0/1⌉;⌊0/1⌋] = [0;0] fail
?⎧<S> { :p .? } / { <s> :q true }
/⎨  ∆={:p}, ω={:p→0}
0⎩  I(:p[0,1]) = [⌈0/1⌉;⌊0/0⌋] = [0;1] pass (adding rule for 0/0=∞)
  <S> { :p .{2} } / { <s> :p true }
    ∆={:p}, ω={:p→1}
    I(:p[2,2]) = [⌈1/2⌉;⌊1/2⌋] = [1;0] fail
⅔⎧<S> { :p .{2,3} } / { <s> :p 1 }
/⎨  ∆={:p}, ω={:p→1}
1⎩  I(:p[2,3]) = [⌈1/3⌉;⌊1/2⌋] = [1;0] fail
⅔⎧<S> { :p .{2,3} } / { <s> :p 1,2 }
/⎨  ∆={:p}, ω={:p→2}
2⎩  I(:p[2,3]) = [⌈2/3⌉;⌊2/2⌋] = [1;1] pass
⅔⎧<S> { :p .{2,3} } / { <s> :p 1,2 }
/⎨  ∆={:p}, ω={:p→3}
3⎩  I(:p[2,3]) = [⌈3/3⌉;⌊3/2⌋] = [1;2] pass
⅔⎧<S> { :p .{2,3} } / { <s> :p 1,2,3,4 }
/⎨  ∆={:p}, ω={:p→4}
4⎩  I(:p[2,3]) = [⌈4/3⌉;⌊4/2⌋] = [2;2] fail
?/0 is really modeled as I(I(:p[1,1]) ∣ I(ε)) below

  I(ε) tests
  <S> { } / { <s> :p true }
    ∆={:p}, ω={}
    I(ε) = [0;∞] pass
?⎧<S> { :p .? } / { <s> :q true }
/⎨  ∆={:p}, ω={:p→0}
0⎩  I(I(:p[1,1]) ∣ I(ε)) = I([0;0] ∣ [0;∞]) = [0;∞]
⅔⎧<S> { :p .{2,3} } / { <s> :p 1 }
/⎨  ∆={:p}, ω={:p→1}
1⎩  I(:p[2,3]) = [⌈1/3⌉;⌊1/2⌋] = [1;0] fail
⅔⎧<S> { :p .{2,3} } / { <s> :p 1 }
/⎨  ∆={:p}, ω={:p→1}
1⎩  I(I(I(:p[2,2]) ∥ I(I(:p[1,1]) ∣ I(ε)))) =
    I(I([⌈1/2⌉;⌊1/2⌋] ∥ I([⌈0/1⌉;⌊0/1⌋] ∣ [0;∞]))) =
    I(I([1;0] ∥ I([0;0] ∣ [0;∞]))) = I(I([1;0] ∥ [0;∞])) = I([1;0]) fail
⅔⎧<S> { :p .{2,3} } / { <s> :p 1,2 }
/⎨  ∆={:p}, ω={:p→2}
2⎩  I(I(I(:p[2,2]) ∥ I(I(:p[1,1]) ∣ I(ε)))) =
    I(I([⌈2/2⌉;⌊2/2⌋] ∥ I([⌈0/1⌉;⌊0/1⌋] ∣ [0;∞]))) =
    I(I([1;1] ∥ I([0;0] ∣ [0;∞]))) = I(I([1;1] ∥ [0;∞])) = I([1;1]) pass
⅔⎧<S> { :p .{2,3} } / { <s> :p 1,2,3 }
/⎨  ∆={:p}, ω={:p→3}
3⎩  I(I(I(:p[2,2]) ∥ I(I(:p[1,1]) ∣ I(ε)))) =
    I(I([⌈3/2⌉;⌊3/2⌋] ∥ I([⌈0/1⌉;⌊0/1⌋] ∣ [0;∞]))) =
    I(I([2;1] ∥ I([0;0] ∣ [0;∞]))) = I(I([2;1] ∥ [0;∞])) = I([2;1]) fail
     -- overruns if <s> :p 3 assigned to [2,2] group
    I(I([⌈2/2⌉;⌊2/2⌋] ∥ I([⌈1/1⌉;⌊1/1⌋] ∣ [0;∞]))) =
    I(I([1;1] ∥ I([1;1] ∣ [0;∞]))) = I(I([1;1] ∥ [1;∞])) = I([1;1]) pass
     -- accepts if <s> :p 3 assigned to I(I(:p[1,1]) ∣ I(ε))
⅔⎧<S> { :p .{2,3} } / { <s> :p 1,2,3,4 }
/⎨  ∆={:p}, ω={:p→3}
4⎩  I(I(I(:p[2,2]) ∥ I(I(:p[1,1]) ∣ I(ε)))) =
    I(I([⌈4/2⌉;⌊4/2⌋] ∥ I([⌈0/1⌉;⌊0/1⌋] ∣ [0;∞]))) =
    I(I([2;2] ∥ I([0;0] ∣ [0;∞]))) = I(I([2;2] ∥ [0;∞])) = I([2;2]) fail
     -- overruns if <s> :p 4 assigned to [2,2] group
    I(I([⌈3/2⌉;⌊3/2⌋] ∥ I([⌈1/1⌉;⌊1/1⌋] ∣ [0;∞]))) =
    I(I([2;1] ∥ I([1;1] ∣ [0;∞]))) = I(I([2;1] ∥ [1;∞])) = I([2;1]) pass
     -- overruns if <s> :p 4 assigned to I(I(:p[1,1]) ∣ I(ε))

⅔/1, ⅔/2, ⅔/3, ⅔/4 really modeled as I(:p[2,3]) above

I'd continue here but I need help understanding ω∆(E) ≠ ε and I(E) ≠ 0.

-- 
-ericP

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Received on Thursday, 3 March 2016 22:38:39 UTC

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