SOLO-013 · Advanced Mastermind

Six Slots Without an Echo

Find the only orderly code of 6 digits.

This page contains the solution.

The answer

Code: 331037.

The reasoning

1. Separates two constraints for each line: the number of equal positions per position, and then the total of common occurrences. This total is well placed + wrong placed; for each number, the minimum of the two multiplicities is taken.
2. To cut down the work, ignore the positions for now. Let n₀,…,n₇ be the counts of the eight digits. Their sum is 6.
3. Proposal 345125 requires min(n1, 1) + min(n2, 1) + min(n3, 1) + min(n4, 1) + min(n5, 2) = 2.
4. Proposal 073421 requires min(n0, 1) + min(n1, 1) + min(n2, 1) + min(n3, 1) + min(n4, 1) + min(n7, 1) = 4.
5. Proposal 573237 requires min(n2, 1) + min(n3, 2) + min(n5, 1) + min(n7, 2) = 3.
6. Proposal 531562 requires min(n1, 1) + min(n2, 1) + min(n3, 1) + min(n5, 2) + min(n6, 1) = 2.
7. These limitations of occurrences alone leave 16 multi-group(s).Here they are, written in ascending order: 000237; 001277; 001337; 001377; 002237; 011277; 011337; 011377; 012277; 012777; 013337; 013777; 022237; 024677; 033467; 034677.
8. Multi ensemble 000237: after successive application of the numbers of property placed, there remains 32 → 13 → 1 → 0 arrangement(s).
9. Multi ensemble 001277 : after successive application of the numbers of property placed, there remains 48 → 10 → 0 → 0 arrangement(s).
10. Multi ensemble 001337 : after successive application of the numbers of property placed, there remains 66 → 28 → 2 → 0 arrangement(s).
11. Multi ensemble 001377 : after successive application of the numbers of property placed, there remains 48 → 19 → 1 → 0 arrangement(s).
12. Multi ensemble 002237 : after successive application of the numbers of property placed, there remains 66 → 15 → 2 → 0 arrangement(s).
13. Multi ensemble 011277 : after successive application of the numbers of property placed, there remains 66 → 21 → 0 → 0 arrangement(s).
14. Multi ensemble 011337 : after successive application of the numbers of property placed, there remains 72 → 24 → 2 → 0 arrangement(s).
15. Multi ensemble 011377 : after successive application of the numbers of property placed, there remains 66 → 23 → 3 → 0 arrangement(s).
16. Multi ensemble 012277 : after successive application of the numbers of property placed, there remains 66 → 10 → 0 → 0 arrangement(s).
17. Multi ensemble 012777 : after successive application of the numbers of property placed, there is left 32 → 6 → 0 → 0 arrangement(s).
18. Multi ensemble 013337 : after successive application of the numbers of property placed, there is left 56 → 21 → 3 → 1 arrangement(s).
19. Multi ensemble 013777 : after successive application of the numbers of property placed, there is left 32 → 11 → 1 → 0 arrangement(s).
20. Multi ensemble 022237 : after successive application of the numbers of property placed, there is left 56 → 7 → 1 → 0 arrangement(s).
21. Multi ensemble 024677 : after successive application of the numbers of property placed, there remains 96 → 39 → 4 → 0 arrangement(s).
22. Multi ensemble 033467 : after successive application of the numbers of property placed, there remains 132 → 58 → 3 → 0 arrangement(s).
23. Multi ensemble 034677 : after successive application of the numbers of property placed, there remains 96 → 45 → 4 → 0 arrangement(s).
24. So we can make two separate tables, occurrences and then places. Only one line in the second table keeps an arrangement; that's the code announced. This separation avoids blindly browsing all the ordered words.
25. Starts with the 262.144 ordered words of length 6 out of 0 to 7. The initial zeros or words with repetition are not removed.
26. Index 1, proposal 345125: retains exactly the configurations with 1 position equal(s) and 2 common occurrence(s). There are 26,034 candidates left.
27. Index 2, proposal 073421: retains exactly the configurations with 0 position level(s) and 4 common occurrence(s). There are 2,490 candidates left.
28. Index 3, proposal 573237: retains exactly the configurations with 2 positions and 3 common occurrence(s). There are 65 candidates left.
29. Index 4, proposal 531562: retains exactly the configurations with 2 positions and 2 common occurrence(s). There are 1 candidate(s) left.
30. In index 1, the candidate 000000 would give 0 well placed and 0 badly placed, instead of 1 and 1: it is eliminated.
31. In index 2, candidate 000015 would give 1 well placed and 1 ill placed, instead of 0 and 4: it is eliminated.
32. In index 3, candidate 100137 would give 2 well placed and 0 badly placed, instead of 2 and 1: it is eliminated.
33. This filtering can be done on paper in branches: first fixes the possible occurrences, then affects their places and cuts a branch as soon as a line exceeds its number of places or can no longer reach its common total. You must not choose a candidate only because it satisfies the last line.
34. Final check against 345125: 1 well placed, 1 wrong placed.
35. Final check against 073421 : 0 well placed, 4 wrong placed.
36. Final check against 573237: 2 well placed, 1 wrong placed.
37. Final check against 531562: 2 well placed, 0 wrong placed.
38. The exhaustive check takes all possible words and has exactly one solution. The numbers are therefore determined at all places; it is not just a compatible code found at random.

Common mistakes and false leads

• Consolidate wrongly placed and figure absent.
• Count twice an occurrence of the code when the proposal repeats a number.
• Prohibit repetitions or the initial zero without the rules requiring it.

{
  "multisets_and_position_counts": [
    {
      "bag": "000237",
      "remaining_after_position_constraints": [
        32,
        13,
        1,
        0
      ]
    },
    {
      "bag": "001277",
      "remaining_after_position_constraints": [
        48,
        10,
        0,
        0
      ]
    },
    {
      "bag": "001337",
      "remaining_after_position_constraints": [
        66,
        28,
        2,
        0
      ]
    },
    {
      "bag": "001377",
      "remaining_after_position_constraints": [
        48,
        19,
        1,
        0
      ]
    },
    {
      "bag": "002237",
      "remaining_after_position_constraints": [
        66,
        15,
        2,
        0
      ]
    },
    {
      "bag": "011277",
      "remaining_after_position_constraints": [
        66,
        21,
        0,
        0
      ]
    },
    {
      "bag": "011337",
      "remaining_after_position_constraints": [
        72,
        24,
        2,
        0
      ]
    },
    {
      "bag": "011377",
      "remaining_after_position_constraints": [
        66,
        23,
        3,
        0
      ]
    },
    {
      "bag": "012277",
      "remaining_after_position_constraints": [
        66,
        10,
        0,
        0
      ]
    },
    {
      "bag": "012777",
      "remaining_after_position_constraints": [
        32,
        6,
        0,
        0
      ]
    },
    {
      "bag": "013337",
      "remaining_after_position_constraints": [
        56,
        21,
        3,
        1
      ]
    },
    {
      "bag": "013777",
      "remaining_after_position_constraints": [
        32,
        11,
        1,
        0
      ]
    },
    {
      "bag": "022237",
      "remaining_after_position_constraints": [
        56,
        7,
        1,
        0
      ]
    },
    {
      "bag": "024677",
      "remaining_after_position_constraints": [
        96,
        39,
        4,
        0
      ]
    },
    {
      "bag": "033467",
      "remaining_after_position_constraints": [
        132,
        58,
        3,
        0
      ]
    },
    {
      "bag": "034677",
      "remaining_after_position_constraints": [
        96,
        45,
        4,
        0
      ]
    }
  ]
}

{
  "candidates": 262144,
  "solutions": 1,
  "clues": 4,
  "survivors_after_clues": [
    26034,
    2490,
    65,
    1
  ]
}

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