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Normal Sudoku rules apply. Numbers on thermometers must increase, starting from the bulb end. Killer clues indicate the sum of three digits in a cage, which cannot include repeated digits. In columns 4,5,6 no two orthogonally adjacent cells contain consecutive digits.

NOTE: As you might have guessed, this was submitted to Cracking The Cryptic, but apparently rejected (I am yet to hear back from CTC after submitting in late August). Assuming my puzzle is sound, my best guess is there is too much "competition" from other setters who are more skilled than me. I would be interested to hear anyone else's opinion on puzzles appearing on CTC.

enter image description here

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  • $\begingroup$ Did you make this puzzle? If yes then that's just amazing. +1 $\endgroup$
    – Anonymous
    Nov 14 '20 at 9:06
  • $\begingroup$ Yes, this is my creation. I googled my name + CTC but could only find a previous "lower-quality puzzle" which somehow got accepted. $\endgroup$
    – happystar
    Nov 14 '20 at 9:22
  • 1
    $\begingroup$ In case the cage totals are too small to read: they are 18+15+17+9+15+22+13+9+17=135 (reading left to right). I have included a check sum since they must sum to 3*45=135, being the last three columns which must include 1-9. If the sum were not 135, we would immediately deduce the puzzle is broken $\endgroup$
    – happystar
    Nov 15 '20 at 0:03
  • 1
    $\begingroup$ @Chris thanks for your feedback. The idea of three columns with different logic is new (unlike e.g. Phistomefel's Theorem!), but the intended break-in actually uses all three parts of the triptych (see accepted solution). I will definitely consider their discord server. $\endgroup$
    – happystar
    Nov 19 '20 at 11:19
  • 1
    $\begingroup$ @oAlt I suspect (but not sure) CTC have attempted the puzzle but failed. Mark and Simon do not successfully solve every puzzle they attempt, and they only cherry-pick the ones they succeed in live-solving. $\endgroup$
    – happystar
    Jan 22 at 21:30
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UPDATE: It's a full solution now.

The first two steps: that the two thermometer tips must be 6,7,8,9 and that the base below the 6 must be 1-5 look fine to me but we can improve the next steps:

Let us put the thermometer base which contains 1-5 on the left:

  123 .tt T??
  ..4 ... t..
  ..5 6t. ttt 

First thing to note is in the top row the 6 can only go to the right-most box. This in turn forces the following placement of 5s:

  123 5tt t..
  ..4 ... 5..
  ..5 6t. ttt 

Now there is only one square left for the center box 4 and that also determines the right thermometer base:

  123 5tt t..
  ..4 ... 5..
  ..5 6t4 321 

Taking advantage of rotational symmetry we now know that both thermometer bases must be 1,2,3. But we can do better than that:

Assuming that the 1-5 base is actually the top leads to a contradiction because

  ..1 ... AAB
  ..2 ... ABB
  543 ... XXX 

XXX must sum to 12 which is only possible via 1,2,9 which leaves only 6-8 for the middle box which cannot be arranged in a legal way. Hence the 1-5 base is the bottom:

  345 ... GGF             GGF = 19 = 2+8+9
  2.. 3.. HHI             HHH = 1+3+5
  1.. ... HII             III = 4+6+7 

GGF must sum to 19 which leaves only 2,8,9. This forces HHH = 1,3,5 and the 3 must be the bottom H.

Next noting III = 4+6+7 and focusing on the right-most column we find that FFF = 22 can only be written 5+8+9 or 6+7+9 but as III cannot keep both 6 and 7 off the right-most column only 5+8+9 is possible

  ..1 ... AAB
  ..2 ... ABB
  t53 ... CCD

  t.4 ... CDD
  t.t ... EEF             FFF = 22 = 5+8+9
  5.6 ... EGF             GGG = 13 = 2+8+3

  345 ... GGF             GGF = 19 = 2+8+9
  2.. 3.. HHI             HH = 1+5
  1.. ... 3II             III = 4+6+7 

Staying in the right-most column we next find that one of the two B's must be 3. As BBB has to sum to 15 the other two B's must be 4+8 or 5+7. In either case 1 and 2 must go to the two D's leaving 6 for the third.

  ..1 ... AAB
  ..2 ... ABB             BBB = 3 + 12
  t53 ... CCD

  t.4 ... CDD             DDD = 6 + 1+2
  t.t ... EE5
  5.6 ... E38

  345 ... GG9             GG = 2+8
  2.. 3.. HHI             HH = 1+5
  1.. ... 3II             III = 4+6+7 

Next, there is only one way left to make EEE = 15 and that is 9+4+2. As one of the thermometer tips in row 5 must be 9, so must be the bottom E.

  ..1 ... AAB
  ..2 ... ABB             BBB = 3 + 12
  t53 ... CCD

  t.4 ... C6D             DD = 1+2
  t.t ... EE5             EEE = 15 = 9+2+4
  5.6 ... E38

  345 ... GG9             GG = 2+8
  2.. 3.. HHI             HH = 1+5
  1.. ... 3II             III = 4+6+7 

Now the bottom tip of CCC can only be 7 and that almost finalizes the thermometers. To finalize CCC let us have a look at row 7. Its center block must accomodate 1+6+7, so 1 has to go in the middle. The two C's in row 3 have to sum to 10, therefore CC = 6+4 or CC = 9+1. But 6+4 would force the last thermometer square to 7 leaving 1,8,9 for the middle block and, again, 1 would have to go in the middle which is not available. Therefore CC = 1+9.

  ..1 ... AAB
  ..2 ... ABB             BBB = 3 + 12
  t53 ... CC2             CC = 1+9

  8.4 ... 761
  9.7 ... EE5             EE = 2+4
  5.6 ... 938

  345 .1. GG9             GG = 2+8
  2.. 3.. HHI             HH = 1+5
  1.. ... 3II             III = 4+6+7 

One of the A's in column 7 has to be 6. As 5 and 7 are already present in column 7, the remaining AA = 12 must split 4+8 and the BB = 12 must split 5+7.

  ..1 ... AAB             AAA = 6 + 12
  ..2 ... ABB             BBB = 3 + 12
  t53 ... 192

  8.4 ... 761
  9.7 ... EE5             EE = 2+4
  5.6 ... 938

  345 .1. GG9             GG = 2+8
  2.. 3.. 51I
  1.. ... 3II             III = 4+6+7 

At this point we can do a number of easy steps leading to

  ..1 ... AAB             AAA = 4+6+8
  ..2 ... A5B             BB = 3+7
  t53 ... 192

  8.4 ... 761
  9.7 ... EE5             EE = 2+4
  5.6 ... 938

  345 .1. GG9             GG = 2+8
  27. 3.4 516
  16. ... 374 

Because of the 6 and 7 in row 7 the 7 in row 6 must be in the center. This forces two more 4's. Now the positions of the 6's in rows 5 and 7 force the 6 in the top center box to be in the middle column. Therefore the 6 in row 3 must be on the thermometer.

  ..1 ... AAB             AAA = 4+6+8
  ..2 ... A5B             BB = 3+7
  653 .4. 192

  8.4 ... 761
  9.7 ... EE5             EE = 2+4
  5.6 47. 938

  345 .1. GG9             GG = 2+8
  27. 3.4 516
  16. ... 374 

The positions of the 8's in rows 3 and 5 force the 9 in row 4 into the middle. As positions left and right of the center square have to be 6 and 8 the center square itself must be 3. This allows a few easy followup moves.

  ..1 ... AAB             AAA = 4+6+8
  ..2 1.. A5B             BB = 3+7
  653 .4. 192

  834 .9. 761
  917 .3. EE5             EE = 2+4
  526 471 938

  345 .1. GG9             GG = 2+8
  279 384 516
  168 ... 374 

Next, the 2 in row 1 is either in column 5 or 6. Either way the 3 in the top center box cannot be in row 1. After placing the 3 at row 2, column 6 everything falls into place.

  781 529 643
  492 163 857
  653 847 192

  834 295 761
  917 638 425
  526 471 938

  345 716 289
  279 384 516
  168 952 374 

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4
  • $\begingroup$ You found the intended break-in, congrats! $\endgroup$
    – happystar
    Nov 19 '20 at 11:17
  • $\begingroup$ There's one thing I don't understand, which had me spinning my wheels for a while. At the point when G+G+F=2+8+9 and F+F+F=5+8+9, how do you deduce that their common F is the 9 and not the 8? The three Gs could be 2+9+2, since the 2s would be in different blocks and columns. $\endgroup$ Nov 19 '20 at 13:29
  • 1
    $\begingroup$ @MishaLavrov Repeats are not allowed within cages, it says so explicitly in them rules. $\endgroup$ Nov 19 '20 at 13:43
  • 1
    $\begingroup$ I guess I should have read the rules more carefully :) Thank you for pointing this out, and for the nice solution! $\endgroup$ Nov 19 '20 at 13:44
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Solution:

enter image description here

Some comments about how I got started (but not how I solved the whole thing, because that would be too long):

The four squares at the ends of the two thermometers must be at least 6 (since there are 5 thermometer squares before them) and are all in the same 3 by 3 block, so they must be a permutation of 6, 7, 8, 9. In particular, the 6 must be the next-to-last square of some thermometer, so one of the two thermometers begins 1, 2, 3, 4, 5, 6. If we rotate the thermometer part of the Sudoku either clockwise or counterclockwise 90 degrees, (we're not sure yet which one) we must get: enter image description here

This lets us fill in 1, 2, 3 at the beginning of the bottom thermometer (since no matter how we rotate the picture, the single given 3 in square (8,4) forces that beginning). From here, the killer Sudoku clues are a natural place to go: the sum of 22 can only be $6+7+9$ or $5+8+9$, and the sum of 9 near the bottom can now only be $1+3+5$ or $1+2+6$.

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1
  • 1
    $\begingroup$ This is correct but there’s a much nicer break in 🙂 $\endgroup$
    – happystar
    Nov 15 '20 at 22:46

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