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<Paper uid="C67-1008">
  <Title>SLANT GRAMMAR CALCULUS</Title>
  <Section position="7" start_page="0" end_page="4" type="concl">
    <SectionTitle>
6. Wanted: Non-Intersecting Vaults
</SectionTitle>
    <Paragraph position="0"> The two above-mentioned cancellation rules are, rewritten in the modified (simpler) notation, fused into one: If in a string of characters, an underlined character and a non-underlined character appear, separated only by a space character and denoting the same atom, erase all three characters.</Paragraph>
    <Paragraph position="1"> Our task may also be formulated thus: Draw a connecting line from each denominator in the string to a numerator with the same name, in the correct direction - i.e., not over the numerator of the own symbol - like vaults over the strings, in such a way that no two such vaults intersect and that one numerator - say \[ - remains untouched by any vault nor roofed by one. If this succeeds, the string is a grammatical sequence of type t.</Paragraph>
    <Paragraph position="2"> Example'.</Paragraph>
    <Paragraph position="3"> ct 7. Identification of the Correct Numerator If all numerator symbols are different the task is trivial. If several numerator atoms carry the same name, we have to decide which one is intended by a given denominator. How do we do this ? The problem is in this form a purely computational one - and no easy such problem.</Paragraph>
    <Paragraph position="4"> 8. Algorithm with Stacking One algorithm can be summarized as follows: We have a given string of words and want to ascertain whether its type is one of the set T = Is, t .... \]. For each word we have a given set of alternative category symbols.</Paragraph>
    <Paragraph position="5"> We first consider the simpler problem of analyzing a given string of category symbols and see if it can be reduced to s.</Paragraph>
    <Paragraph position="6"> We join the (not underlined symbol) s followed by space to the beginning of the given symbol string. When the reduction rules are applied the resultant string should vanish; we say it is reduced to unity.</Paragraph>
    <Paragraph position="7"> The string now contains exactly as many, say n, numerator and denominator atoms. Every numerator should be paired with one other denominator; the whole probhefn is to decide which denominator.</Paragraph>
    <Paragraph position="8"> Whenever we thus pair two atoms, the intervening string of atoms should vanish under reduction, be reducible to unity. The same holds for the rest of the original string, after the two paired atoms and everything between them has been removed:  s a_ ab_c'd'- d e ~-c_fg g i In general, then, when one wants to test whether a pairing is a good match or not, one is left with two simpler, isomorphic problems, those of reducing two strings to unity: s a abfg g f_; d d e e Normally we cannot decide beforehand which pairing to test; we must try alternatives. For each arbitrary choice of procedure, we store the alternatives in an OR-stack, For each pairing we place the bracketed-out string in an ANDstack, to be handled later when and if the string on our working table vanishes. If we have emptied the table and the AND-stack, the given string was reducible to s. If we have an irreducible residue on the table, we clear the table and try the next alternative if there is one in the OR-stack, deleting what has been stored in the AND-stack since the last arbitrary choice; if we cannot reduce to unity the string on the table and the OR-stack is empty, we must give the task up.</Paragraph>
    <Paragraph position="9"> To minimize the number of pairings to test for each string we analyze, we draw up a binary matrix M of potential matches, with m.. = 1, if the denominator No. i and the numerator No. j are the same character and are placed in proper order and reasonably wide apart in the string. (E.g., if i is a &amp;quot;non-recursive&amp;quot; type of syntagm, &amp;quot;reasonably&amp;quot; placed means that,-is the nearest numerator written with the same character as i and placed on the proper side of i.) If any row or column in M is empty, the task is hopeless. Otherwise, we select the atom whose row or column contains the least number of 1 ~s. To analyze a string of words we assign to each word one symbol of the type ABCDE, where C is the set of numerators in all the word ts category symbols, B is the set of nominator atoms appearing immediately before the numerator in any one of the word's category symbols, etc. On the string of these new symbols the above procedure, mutatis mutandis, is applied. Thus, instead of the condition in (I) above that two atoms should be the &amp;quot;same&amp;quot; characters, it is required that one underlined atom A and a non-underlined atom B appearing separated only by space should fulfill the condition A n B ~ 0.</Paragraph>
  </Section>
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