We covered Similarity Graphs today, which are on a Monday handout
that I gave out in class.  See the web site for information
on similarity graphs and pattern recognition and S levels.

Also did the mapping of PANTHERS letters to a hash table

   with f o g (l) where l is a letter from PANTHERS

   and g(l) is the ASCII value of the letter, and
   f(x) = x mod 9

   The successful search value or average should come out
to 11/8 or 1 3/8 and the unsuccessful to 45/9 or exactly 5.

   All you need to know is that the ASCII value of A = 65.
   Since 65 mod 9 = 2
                       or, 65 = 7 * 9 + 2
   you do NOT need
   to know or look up ANY of the other ASCII values, if you 
   proceed like this:

   0   H Q Z      and the range of the function f o g({P,A,N,T,H,E,R,S})
   1   I R               ----- was {0, 1, 2, 3, 6, 8}
   2 A J S
   3 B K T                     which means it was not injective,
   4 C L U                     since the cardinality of the
   5 D M V                     domain = 8 and is greater than
   6 E N W                     cardinality of range = 6.
   7 F O X
   8 G P Y      Of course, it could never be onto (surjective),
                           since cardinality of codomain = 9
                           and PANTHERS is only 8 letters.

   The codomain of the function f o g (l) is {0, 1, 2, 3, 4, 5, 6, 7, 8},
   since the remainder when you divide by 9 will
   always be r >= 0 and r <= 8.

   Since the cardinality of the domain set (8 letters in PANTHERS here),
   is less than the cardinality of the codomain, we know
   that the RANGE of f o g({P, A, N, T, H, E, R, S}) will be 
            -----

   a PROPER SUBSET of the codomain { 0, 1, 2, 3, 4, 5, 6, 7, 8 }.

   Collision resolution is by linear probe method.

   2 A J S

   Easily seen that the function f o g ("A") = f o g ("S"),

   so f o g(letters in Panthers) is NOT injective, i.e. not one-to-one.

   Note that "A" and "S" are synonyms with respect to function f o g()!
                             --------

   Thus ends the partial summary of Wednesday, December 8th class.
                 ---------------

Map the letters in DISCRETE into 8, 9, 10, 11, 12 and 13 locations.

    {D, I, S, C, R, E, T} is 7 distinct letters in the set.

   0   H Q Z     
   1   I R      <------------------ I and R 
   2 A J S      <------------------ S
   3 B K T      <------------------ T             
   4 C L U      <------------------ C             
   5 D M V      <------------------ D             Cardinality of the
   6 E N W      <------------------ E             the RANGE for mod 9
   7 F O X                                        for DISCRET is 6.
   8 G P Y      
   
  Is the cardinality of the RANGE of DISCRET ever = 7 for
                  mod 8, mod 10, mod 11, mod 12 or mod 13?  Try it.

  What is the cardinality of the range for each mapping from mod 8 up to
  the mod 13 mapping?

  Good practice!!!!

  Mark