12
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Let us shy away from the materialistic opulence of 361- cell KenKen layouts (−9 to +9, squared).  Let us contemplate a modest KenKen journey, unburdened by gratuitously extravagant cluenography.*  SerenSerenity may be reached with just a clue or two on a minor board.  Breathe.

Imagine an undelineated 3×3 KenKen puzzle.  All that shows are 2 clue amounts, while others may be hidden.  Imagine the sum of these 2 amounts as small as can be.  Believe that these lead to only one possible completion.  Now . . . how much and in which cells are those 2 amounts?

          Some questions may already enter your consciousness.

Undelineated?    Borders of cages (subregions) are not outlined but can be deduced.

Clue amount?    The number (1− 216 ) before a cage’s arithmetic operator (+, −, ×, / or ÷).

Where?       Each clue amount resides in the leftmost cell of its cage’s top row.

?   Your computer’s level of Kenlightenment already exceeds this search space.

One possible completion?   Even subtleties — such as rotation, reflection and operator substitution — distinguish multiple completions, as do differences in cage outlines.


A path toward clarity

The journey of a thousand undelineated KenKens begins with a single cell.

And without a clue.

     Secret knowledge:  This works well enough as text

   +---------+          +---------+          +---------+
   |    :5   |          |    |5   |          |1 1 |5+ 2|
   |....+....|   -->    |....+....|   -->    |----+    |
   |    :    |          |    :    |          |  2    1 |
   +---------+          +---------+          +---------+

Footnote:
*Cluenography. Noun. Compulsively fetishized depiction of clues.

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  • $\begingroup$ This sets out with an open-ended tag because the minimal solution is not certain at pose time. My first solution had 3 clues with a sum of 27 that result in 4 cages, but I've improved on each of those measures. This cropped up while exploring what is now a subsequent step along this journey. $\endgroup$ – humn Dec 16 '16 at 11:51
  • $\begingroup$ Is it ok if a solution is found but it has rotational solutions? $\endgroup$ – dcfyj Dec 16 '16 at 13:15
  • $\begingroup$ Aw, @dcfyj, rotations count as different. Will clarify that in the statement. $\endgroup$ – humn Dec 16 '16 at 13:16
  • $\begingroup$ Sad, either way what I found wasn't specific enough as the cage borders were too ambiguous. $\endgroup$ – dcfyj Dec 16 '16 at 13:17
  • 1
    $\begingroup$ Word from my Ken master, @ABcDexter - The vast greatness of KenKen can only be found by seeking less within it. We are ready for the next small step. $\endgroup$ – humn Dec 20 '16 at 13:14
5
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UPDATE: A sum 13 solution, as devised by @Neil W! Thanks!

 

Old post, with sum a sum 14 solution in a similar family to Neil's is as follows.

Finally, a puzzle which I think is correct! Maybe you can verify for me, humn?

    _ 7 7
_ _ _
_ _ _

Unique completion:

1 2 3
2 3 1
3 1 2

The right 7 is an L shape going 3 1 2 1.
The left 7 is an L shape going 2 3 2.
The last two numbers are isolated, so their hints are just themselves with no sign marker.

PROOF OF UNIQUENESS:

The leftmost 7's cage must extend downwards by at least one square.
Thus, the rightmost 7's cage must extend down the whole last column.
The rightmost 7's cage must then extend left at least one space across the bottom row.
That cage is complete - 7 is prime, must be a sum, and if it extends any further then the sum is impossible to reach.
Since we know the right hand column is 1, 2, 3, we know that the middle square on the bottom row must be 1 to complete the sum. Now the two known squares of the other 7 cage are [2,3] in some order. To complete the 7 sum, there must be either an extra [1,1] or a [2]. The [1,1] cannot be in the cage, because then two 1s will be in the left hand column. Thus, there is just an extra 2, which must be in the leftmost square of the middle column, completing this cage. Now this cage can be filled out completely since there are two 2s in it. The rest of the board is trivially filled.

The 7 7 can also be reduced even further to 5 8 as pointed out by Neil W, and a similar reasoning will yield a unique solution, depicted by humn (thanks very much!) here.

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  • $\begingroup$ Does anyone know how to work these spoiler things?? xD $\endgroup$ – TheGreatEscaper Dec 16 '16 at 13:47
  • $\begingroup$ @TheGreatEscaper There you go. $\endgroup$ – dcfyj Dec 16 '16 at 13:49
  • $\begingroup$ I play with spoilers a lot, what can I say? :P $\endgroup$ – dcfyj Dec 16 '16 at 13:51
  • 4
    $\begingroup$ The clues must be in the leftmost square of the upper row of the cage they belong to. $\endgroup$ – TheGreatEscaper Dec 16 '16 at 15:03
  • 1
    $\begingroup$ 4 9 unfortunately doesn't, since the 4 can end up being a 212 product or a 121 sum... $\endgroup$ – TheGreatEscaper Dec 17 '16 at 8:51
1
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I'm not sure if I got it correctly, is this a valid solution?

 +--------+
 |05 00 36|
 +  +  +  +
 |00 00 00|
 +  +  +  +
 |00 00 00|
 +--------+  

Turns out the numbers are fixed, but you have playroom with the walls...

 +--------+
 |13 00 00|
 +  +  +  +
 |04 00 00|
 +  +  +  +
 |00 00 00|
 +--------+

 +-----+--+
 | 3  2| 1|
 +--+  +--+
 | 1|3   2|
 |  +--+  |
 | 2  1| 3|
 +-----+--+  

00 is empty field, my ascii art is just bad...

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  • $\begingroup$ Yeah, i don't like it and have to figure out the edits, made some changes and clicked randomly once... $\endgroup$ – Zibelas Dec 16 '16 at 13:23
  • $\begingroup$ I think the second solution is fixed, there are no products in it and you can't shift the numbers around. The fixed lines are just looking ugly in case someone wants to fix them. $\endgroup$ – Zibelas Dec 16 '16 at 13:56
  • $\begingroup$ Aww, think I can get another completion by treating the 4 as 3+1 .. Hope I'm wrong. $\endgroup$ – humn Dec 16 '16 at 14:14
  • $\begingroup$ You can't, else you won't get the 13 anymore since the 1 top right has to be alone, this solution has 3 block and no room for shifting $\endgroup$ – Zibelas Dec 16 '16 at 14:26
  • 1
    $\begingroup$ But that block looks really ugly, not symmetric at all, that's really bad Zen and should be avoided :). $\endgroup$ – Zibelas Dec 16 '16 at 14:56

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