Greenplum DB changing table column length

Using your current method you might use a query such as the one below. This example uses derived tables to treat each backup date range as if it where a table. Then joins two backup date derived tables (named Today and Previous) on the emp# (I have used EmpNo in my example) and uses the WHERE clause to get only records with at least on difference between the two. Note that the syntax for derived tables may vary depending on what database product you are using. Adding the database product to the question might help the question be answered more accurately.

SELECT 
    Today.EmpNo
    , Today.First AS TodayFirst
    , Today.Last AS TodayLast
    , Today.DOB AS TodayDOB
    , Previous.First AS PreviousFirst
    , Previous.Last AS PreviousLast
    , Previous.DOB AS PreviousDOB
FROM
    (
        SELECT *
        FROM employeehist
        WHERE BackupDate = '3/10/2014'
    ) AS Today


    INNER JOIN 
    (
        SELECT *
        FROM employeehist
        WHERE BackupDate = '3/9/2014'
    ) AS Previous
        ON ( Today.EmpNo = Previous.EmpNo )


WHERE  ( Today.First != Previous.First OR Today.Last != Previous.Last OR Today.DOB != Previous.DOB );

To find all versions of a particular employee record use DISTINCT. This will return unique versions of each record.

SELECT DISTINCT First, Last, DOB
FROM employeehist
WHERE EmpNo = 1;

You might also consider creating a trigger on the employee table which will only write records to the history table when changes are made. This would allow you to track all edits to an employee without having to compare each backup version to every other backup version.

Greenplum doesn't (as at version 4.3.8) maintain a record in the catalogue linking indexes on partitions to indexes on base tables. The best option is to follow the partitions and find indexes on the partitions matching the definition of the base index.

CREATE OR REPLACE FUNCTION drop_child_indexes (index_name varchar)
RETURNS VOID
AS
$functionBody$
DECLARE
  child_index_name varchar;
BEGIN

  FOR child_index_name IN
    SELECT child_index.indexrelid::regclass
      FROM pg_index AS parent_index
        -- Find the partitioning scheme for the table the index is on
        INNER JOIN pg_partition ON pg_partition.parrelid = parent_index.indrelid
        -- Follow the links through to the individual partitions
        INNER JOIN pg_partition_rule ON pg_partition_rule.paroid = pg_partition.oid
        -- Find the indexes on each partition
        INNER JOIN pg_index AS child_index ON child_index.indrelid = pg_partition_rule.parchildrelid
          -- Which are on the same field as the named index
          AND child_index.indkey = parent_index.indkey
          -- Using the same comparison operator
          AND child_index.indclass = parent_index.indclass
      -- Filtered for the index we're trying to drop
      WHERE parent_index.indexrelid = $1::regclass::oid
      -- Drop leaves first, even if it doesn't really matter in this case
      ORDER BY pg_partition.parlevel DESC

LOOP
  RAISE NOTICE '%', child_index_name||' ';
  EXECUTE 'DROP INDEX '||child_index_name||';';
END LOOP;

END
$functionBody$
LANGUAGE plpgsql;

If there another index with the same definition (fields, comparison operators) it's partition indexes will also be dropped. Not a perfect answer but better than substring matching.