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Overview
Comment:Add a sudoku solver to the recursive query tests in with1.test.
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SHA1: 679eff8759aa25368b977c0d26b78a9fcd9486f5
User & Date: drh 2014-01-18 18:33:44
Context
2014-01-20
14:17
Do not run the tests in with2.test with SQLITE_OMIT_CTE builds. check-in: 8a973912 user: dan tags: trunk
2014-01-18
18:33
Add a sudoku solver to the recursive query tests in with1.test. check-in: 679eff87 user: drh tags: trunk
15:59
Add extra test cases. No changes to code. check-in: d38d485e user: dan tags: trunk
Changes
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Changes to test/with1.test.

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                           ..##+*##########+.++++
                            .......+####....   +
                                 ..+####+.
                                   ..#*..
                                    ....#
                                    +.}}






































finish_test







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                           ..##+*##########+.++++
                            .......+####....   +
                                 ..+####+.
                                   ..#*..
                                    ....#
                                    +.}}

# Solve a sudoku puzzle using a recursive query
#
do_execsql_test 8.2 {
  WITH RECURSIVE
    input(sud) AS (
      VALUES('53..7....6..195....98....6.8...6...34..8.3..17...2...6.6....28....419..5....8..79')
    ),
  
    /* A table filled with digits 1..9, inclusive. */
    digits(z, lp) AS (
      VALUES('1', 1)
      UNION ALL SELECT
      CAST(lp+1 AS TEXT), lp+1 FROM digits WHERE lp<9
    ),
  
    /* The tricky bit. */
    x(s, ind) AS (
      SELECT sud, instr(sud, '.') FROM input
      UNION ALL
      SELECT
        substr(s, 1, ind-1) || z || substr(s, ind+1),
        instr( substr(s, 1, ind-1) || z || substr(s, ind+1), '.' )
       FROM x, digits AS z
      WHERE ind>0
        AND NOT EXISTS (
              SELECT 1
                FROM digits AS lp
               WHERE z.z = substr(s, ((ind-1)/9)*9 + lp, 1)
                  OR z.z = substr(s, ((ind-1)%9) + (lp-1)*9 + 1, 1)
                  OR z.z = substr(s, (((ind-1)/3) % 3) * 3
                          + ((ind-1)/27) * 27 + lp
                          + ((lp-1) / 3) * 6, 1)
           )
    )
  SELECT s FROM x WHERE ind=0;
} {534678912672195348198342567859761423426853791713924856961537284287419635345286179}

finish_test