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  1. May 27, 2024
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    • Hague Matthew UXAC009's avatar
      add(parameters): basic parameter implementation · 7886989c
      Hague Matthew UXAC009 authored and Hague Matthew UXAC009's avatar Hague Matthew UXAC009 committed
      Detects any constant that is not a string in a string expression as a
      parameter.
      
      A better implementation will check that all param variables are strings
      fixed to be length 1.
      7886989c
    • Hague Matthew UXAC009's avatar
    • Hague Matthew UXAC009's avatar
      bug(parikh_image): repair overrestrictive optimisation · fbcc3817
      Hague Matthew UXAC009 authored
      Optimisation was requiring that, if a label may have n different
      characters, then each of the n characters are different in two ways:
      
      1. char(i) < char(i+1) (for some ordering of char vars)
      2. there was some difference in predicate satisfaction between char(i)
         and char(i+1).
      
      This fails when there is only one possible satisfying character for the
      label. Then we cannot have two characters that are the same but also
      different.
      
      Relax optimisation so that
      
      1. char(i) < char(i+1) OR the count of characters of char(i+1) is 0,
      2. same for predicate satisfaction (only differ if count > 0),
      3. if count(i) is 0, then so is count(i+1).
      fbcc3817
    • Hague Matthew UXAC009's avatar
      bug(parikh_image): repair overrestrictive optimisation · ae642748
      Hague Matthew UXAC009 authored
      Optimisation was requiring that, if a label may have n different
      characters, then each of the n characters are different in two ways:
      
      1. char(i) < char(i+1) (for some ordering of char vars)
      2. there was some difference in predicate satisfaction between char(i)
         and char(i+1).
      
      This fails when there is only one possible satisfying character for the
      label. Then we cannot have two characters that are the same but also
      different.
      
      Relax optimisation so that
      
      1. char(i) < char(i+1) OR the count of characters of char(i+1) is 0,
      2. same for predicate satisfaction (only differ if count > 0),
      3. if count(i) is 0, then so is count(i+1).
      ae642748
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