[sv-ec] Alternatives for errata 172

As per my action item from the last meeting here are alternatives for assignment of values to a fixed number
of bins when the range_list contains duplicate values.

There seem to be 3 alternatives.

A) Eliminate duplicate values, sort values
   - Determine the set of unique values. Let V be the cardinality of this set.
   - Sort the values in increasing order.
   - Evenly allocate the values to the bins. The first bin gets the first N values, the second bin gets the next N values, and so on. The last bin gets the remaining values. N is the truncated integer value of V/n, where n is the size of the bin as specified by the user using "bins b[n]" notation. Note that if V < n, the bins 0 to V-1 will get one value each starting from the smallest value from the sorted set; rest of the bins do not get any value.

B) Eliminate duplicate values, retain user order
   - Determine the ordered set of unique values. If a value occurs more than once, the 1st occurance of the value is retained and each subsequenct repetition of the value is ignored. Let V be the cardinality of this set.
   - Evenly allocate the values to the bins. The first bin ...<same text as in A>.

C) Retain duplicate values, retain user order
   - If a fixed number of bins is specified and the 'range_list' contains duplicate values, the duplicate values are retained for the purpose of allocating values to bins.
   - Determine the set of values (including duplicates). Let V be the cardinality of this set.
   - Evenly allocate the values to the bins. The first bin ...<same text as in A>.

My preference is for option B. The users order of values should be preserved and the elimination of duplicate values is consistent with the '[]' form where a bin is created for each (unique) value.

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For illustration of the alternatives, here each of the alternatives is applied to:
  bins bns1[3] = { 9, 1, 3, 1, 7, 1, 5, 2, 1, 4, 8, 6 };
  bins bns2[3] = {5,6, [8:11], [1:9] };

A) bns1; the sorted unique values are [1:9]
        1st bin: [1:3]
      2nd bin: [4:6]
      3rd bin: [7:9]
   bns2; the sorted unique values are [1:11]
      1st bin: [1:3]
      2nd bin: [4:6]
      3nd bin: [7:11]

B) bns1; the unique ordered values are {9,1,3,7,5,2,4,8,6}
        1st bin: {9,1,3}
      2nd bin: {7,5,2}
      3rd bin: {4,8,6}
   bns2; the sorted unique values are {5,6,8,9,10,11,1,2,3,4,7}
      1st bin: {5,6,8}
      2nd bin: {9,10,11}
      3nd bin: {1,2,3,4,7}

C) bns1; there are 12 values
        1st bin: {9,1,3,1}
      2nd bin: {7,1,5,2}
      3rd bin: {1,4,8,6}
   bns2; there are 15 values
      1st bin: {5,6,8,9,10}
      2nd bin: {11,1,2,3,4}
      3nd bin: {5,6,7,8,9}

- Ray
Received on Wed Sep 29 14:29:49 2004