KenKen

[cinelli]

This is a KenKen puzzle which appears in some newspapers. All the digits 1 to 6 must appear in every row and column. In each block the target number in the top left-hand corner is calculated from the digits in all the cells in the block, using the operation indicated by the symbol. For instance with a 2-cell block with 2/, the digits could be (1,2), (2,1), (2,4), (4,2), (3,6) or (6,3).

.---- ---- ---- ---- ---- ----
|6X            |13+ |6   |2-  |
|              |    |    |    |
 ---- ---- ----      ----
|2/  |15X      |         |    |
|    |         |         |    |
      ---- ----      ---- ----
|    |2/  |6   |    |7+  |2-  |
|    |    |    |    |    |    |
 ----      ---- ----
|13+ |    |5-       |    |    |
|    |    |         |    |    |
      ---- ---- ---- ---- ----
|         |11+      |8+  |2/  |
|         |         |    |    |
 ---- ----      ----
|1-       |    |         |    |
|         |    |         |    |
 ---- ---- ---- ---- ---- ----

Cinelli

[jfgw]

Logical deduction without protracted trial and error but it took a little effort.

I would have to go through it again if you want my workings.

.---- ---- ---- ---- ---- ----
|6X            |13+ |6   |2-  |
|  3    1    2 |  5 |  6 |  4 |
 ---- ---- ----      ----
|2/  |15X      |         |    |
|  2 |  5    3 |  4    1 |  6 |
      ---- ----      ---- ----
|    |2/  |6   |    |7+  |2-  |
|  1 |  4 |  6 |  3 |  2 |  5 |
 ----      ---- ----
|13+ |    |5-       |    |    |
|  4 |  2 |  1    6 |  5 |  3 |
      ---- ---- ---- ---- ----
|         |11+      |8+  |2/  |
|  6    3 |  5    2 |  4 |  1 |
 ---- ----      ----
|1-       |    |         |    |
|  5    6 |  4 |  1    3 |  2 |
 ---- ---- ---- ---- ---- ----

Julian F. G. W.

[Gengulphus]

This is a KenKen puzzle which appears in some newspapers. All the digits 1 to 6 must appear in every row and column. In each block the target number in the top left-hand corner is calculated from the digits in all the cells in the block, using the operation indicated by the symbol. For instance with a 2-cell block with 2/, the digits could be (1,2), (2,1), (2,4), (4,2), (3,6) or (6,3).

.---- ---- ---- ---- ---- ----
|6X            |13+ |6   |2-  |
|              |    |    |    |
 ---- ---- ----      ----
|2/  |15X      |         |    |
|    |         |         |    |
      ---- ----      ---- ----
|    |2/  |6   |    |7+  |2-  |
|    |    |    |    |    |    |
 ----      ---- ----
|13+ |    |5-       |    |    |
|    |    |         |    |    |
      ---- ---- ---- ---- ----
|         |11+      |8+  |2/  |
|         |         |    |    |
 ---- ----      ----
|1-       |    |         |    |
|         |    |         |    |
 ---- ---- ---- ---- ---- ----

I believe there is one other rule you haven't stated, namely no duplicate digits in any block. For example, the 13+" block on the left-hand side cannot be 4+5+4, even though those digits can be placed so that the two 4s are neither in the same row nor the same column.

I'm not certain offhand to how great an extent I've used that additional rule in the following spoiler (I've certainly used it at least once), nor whether its use is essential to solving the puzzle, but here goes...

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My general approach with these puzzles is just to chip away at them, looking for cases where I can see only a limited number of possibilities (preferably just 1) for what particular cells or groups of cells contain, and slowly refining my knowledge. In this case, the things I see quickly are:

* The three numbers in the "6X" group at the top left must be 1, 2 and 3 in some order (though I don't know which order yet).

* The two numbers in "15X" group just below it must be 3 and 5 in some order.

* The "5-" group just below the centre must contain 1 and 6, and in this case I know the order because one order places two 6s in the same column.

Moving on to something a bit more complicated, the rightmost column must contain each of 1 to 6, and they must form three pairs, one with ratio 2 and the other two with difference 2. The pair with ratio 2 can be (1,2), (2,4) or (3,6), in each case in some unknown order. Those possibilities leave (3,4,5,6), (1,3,5,6) and (1,2,4,5) respectively for the two pairs with difference 2, and it's easy to see that only the first of those can be split into two pairs with difference 2 and only in one way, namely (3,5) and (4,6). So the pair with ratio 2 is (1,2) and the other two pairs are (3,5) and (4,6). Furthermore, we've already got the fact that the "15X" group contains 3 and 5, which implies that the top one of those two pairs cannot be (3,5), so we know it must be (4,6), and in fact we know the order because one order would imply two 6s in the top row.

Time for a picture showing where we're got to, using what I hope is obvious notation for a group containing particular numbers in an unknown order (note that once I know that, I no longer need to keep the original calculation result and symbol around):

.---- ---- ---- ---- ---- ----
|{123}         |13+ |6   |4   |
|              |    |    |    |
 ---- ---- ----      ---- ----
|2/  |{35}     |         |6   |
|    |         |         |    |
      ---- ----      ---- ----
|    |2/  |6   |    |7+  |{35}|
|    |    |    |    |    |    |
 ----      ---- ----
|13+ |    |1   |6   |    |    |
|    |    |    |    |    |    |
      ---- ---- ---- ---- ----
|         |11+      |8+  |{12}|
|         |         |    |    |
 ---- ----      ----
|1-       |    |         |    |
|         |    |         |    |
 ---- ---- ---- ---- ---- ----

We can now see that the top row cell in the 13+ group must contain 5, reducing the rest of that group to an "8+" group that does not contain 5. That only leaves the possibility that it consists of 1, 3 and 4, and the fact that the second row already contains a 3 allows us to place the 3 on the third row and therefore the 1 and 4 in the two second row cells (though in an unknown order).

The placing of the 3 then allows us to determine the order in the {35} group in the rightmost group. And that in turn allows us to deduce that the "7+" group consists of 2 and 5, since it cannot be 1 and 6 (we've already got 6s on both the third and fourth rows), nor 3 and 4 (ditto for 3s). And already having a 5 on the third row allows us to deduce the order of that 2 and 5.

So we're now at:

.---- ---- ---- ---- ---- ----
|{123}         |5   |6   |4   |
|              |    |    |    |
 ---- ---- ---- ---- ---- ----
|2/  |{35}     |{14}     |6   |
|    |         |         |    |
      ---- ---- ---- ---- ----
|    |2/  |6   |3   |2   |5   |
|    |    |    |    |    |    |
 ----      ---- ---- ---- ----
|13+ |    |1   |6   |5   |3   |
|    |    |    |    |    |    |
      ---- ---- ---- ---- ----
|         |11+      |8+  |{12}|
|         |         |    |    |
 ---- ----      ----
|1-       |    |         |    |
|         |    |         |    |
 ---- ---- ---- ---- ---- ----

Now look at the two "2/" groups. The one in the first column must have a 2 in its upper cell, since that's the only digit remaining in the second row. Its lower cell could contain either 1 or 4, which implies that the upper cell of the "2/" group in the second column contains 4 or 1 respectively - and in either case the lower cell of that group must contain 2. So we know where the 2s are in both the first and second columns, and so the 2 in the "{123}" group must be in its rightmost cell.

Also, the top cell of the "13+" group must be 4, as all the other digits have already been placed in the fourth row. That allows us to determine that the lower cell in the first column "2/" group contains 1 (which in turn says that the leftmost cell of the "123" group contains 3 and its middle cell 1) and the upper cell in the second column "2/" group contains 4 rather than the other way around, and to reduce the rest of the "13+" group to a "9+" group.

So we're now at:

.---- ---- ---- ---- ---- ----
|3   |1   |2   |5   |6   |4   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|2   |{35}     |{14}     |6   |
|    |         |         |    |
 ---- ---- ---- ---- ---- ----
|1   |4   |6   |3   |2   |5   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|4   |2   |1   |6   |5   |3   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|9+       |11+      |8+  |{12}|
|         |         |    |    |
 ---- ----      ----
|1-       |    |         |    |
|         |    |         |    |
 ---- ---- ---- ---- ---- ----

The two remaining cells in the first column must contain 5 and 6, so the "9+" group must contain 6 and 3 or 5 and 4, in each case in that order. The second column already contains a 4, eliminating the 5 and 4 possibility. That allows us to resolve the order of the "{35}" group on the second row and to deduce the numbers in the "1-" group as the remaining numbers in the first and second columns (which fortunately do differ by 1!), to give us:

.---- ---- ---- ---- ---- ----
|3   |1   |2   |5   |6   |4   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|2   |5   |3   |{14}     |6   |
|    |    |    |         |    |
 ---- ---- ---- ---- ---- ----
|1   |4   |6   |3   |2   |5   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|4   |2   |1   |6   |5   |3   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|6   |3   |11+      |8+  |{12}|
|    |    |         |    |    |
 ---- ----      ----
|5   |6   |    |         |    |
|    |    |    |         |    |
 ---- ---- ---- ---- ---- ----

The remaining two numbers in the third column must be 4 and 5, and their order is determined by not having two 5s in the sixth row, and so the remaining number in the "11+" group is 2. That resolves the order of the "{12}" group, and it's then a matter of filling in the one remaining number on the fifth row, which resolves the order of the "{14}" group, and then filling in the one remaining number in each of the fourth and fifth columns to complete the solution:

.---- ---- ---- ---- ---- ----
|3   |1   |2   |5   |6   |4   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|2   |5   |3   |4   |1   |6   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|1   |4   |6   |3   |2   |5   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|4   |2   |1   |6   |5   |3   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|6   |3   |5   |2   |4   |1   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|5   |6   |4   |1   |3   |2   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----

Gengulphus

[jfgw]

I believe there is one other rule you haven't stated, namely no duplicate digits in any block. For example, the 13+" block on the left-hand side cannot be 4+5+4, even though those digits can be placed so that the two 4s are neither in the same row nor the same column.

That was a rule I was unaware of. It would have made the puzzle easier as I could have eliminated some of the assumed options more readily. The solution is unique nonetheless.

Julian F. G. W.

[Gengulphus]

I believe there is one other rule you haven't stated, namely no duplicate digits in any block. For example, the 13+" block on the left-hand side cannot be 4+5+4, even though those digits can be placed so that the two 4s are neither in the same row nor the same column.

That was a rule I was unaware of. It would have made the puzzle easier as I could have eliminated some of the assumed options more readily. The solution is unique nonetheless.

Having examined my solution, I did in fact use it only once, with regard to the four-cell "13+" group near the top of the puzzle. Of the other three groups that aren't all in a single row or column, I never used the fact that the "8+" group summed to 8 at all (it was so close to the end that its contents were all determined by rows and columns having to contain each of 1 to 6), I only used the fact that the "11+" group summed to 11 to deduce that its third cell was 2 after determining that the other two were 4 and 5, and I missed an opportunity to apply it to the three-cell "13+" group on the left side of the puzzle: I got the fact that one cell was 4 and the other two summed to 9, but later got the fact that the other two were 6 and 3, not 5 and 4, from other considerations.

But with regard to that four-cell "13+" group near the top of the puzzle, having deduced that its top cell is 5, I then said:

.---- ---- ---- ---- ---- ----
|{123}         |13+ |6   |4   |
|              |    |    |    |
 ---- ---- ----      ---- ----
|2/  |{35}     |         |6   |
|    |         |         |    |
      ---- ----      ---- ----
|    |2/  |6   |    |7+  |{35}|
|    |    |    |    |    |    |
 ----      ---- ----
|13+ |    |1   |6   |    |    |
|    |    |    |    |    |    |
      ---- ---- ---- ---- ----
|         |11+      |8+  |{12}|
|         |         |    |    |
 ---- ----      ----
|1-       |    |         |    |
|         |    |         |    |
 ---- ---- ---- ---- ---- ----

We can now see that the top row cell in the 13+ group must contain 5, reducing the rest of that group to an "8+" group that does not contain 5. That only leaves the possibility that it consists of 1, 3 and 4, and the fact that the second row already contains a 3 allows us to place the 3 on the third row and therefore the 1 and 4 in the two second row cells (though in an unknown order).

Without the additional rule, there are four additional possibilities for the remaining three cells:

* They contain 3, 3 and 2. That can be eliminated on the same basis that we know the second row already contains a 3.

* They contain 2, 2 and 4. That can be eliminated because it would make the two cells on the second row in the "13+" group be 2 and 4, so that the top cell in the first-column "2/" group be 1, and so require its bottom cell to be 2. But that implies two 2s in the third row.

* They contain 1, 1 and 6. That can be eliminated because it would require a second 6 on the second row.

* They contain 1, 2 and 5. The 5 could not be in either of the second-row cells in the "13+" group, so those two cells would have to contain 1 and 2 and the third-row cell 5, which determines the order in the "{35}" group. It also makes the top cell in the first-column "2/" group a 4, implying in turn that its bottom cell contains 2. That gives us:

.---- ---- ---- ---- ---- ----
|{123}         |5   |6   |4   |
|              |    |    |    |
 ---- ---- ---- ---- ---- ----
|4   |{35}     |{12}     |6   |
|    |         |         |    |
 ---- ---- ---- ---- ---- ----
|2   |2/  |6   |5   |7+  |3   |
|    |    |    |    |    |    |
 ----      ---- ----      ----
|13+ |    |1   |6   |    |5   |
|    |    |    |    |    |    |
      ---- ---- ---- ---- ----
|         |11+      |8+  |{12}|
|         |         |    |    |
 ---- ----      ----
|1-       |    |         |    |
|         |    |         |    |
 ---- ---- ---- ---- ---- ----

Then the only possible position for the third-row 1 is in the top cell of the second-column "2/" group, since a 1 in the top cell of the "7+" group would imply two 6s in the fifth column. In turn, that means that the "7+" cell must have 4 in its top cell and 3 in its bottom cell:

.---- ---- ---- ---- ---- ----
|{123}         |5   |6   |4   |
|              |    |    |    |
 ---- ---- ---- ---- ---- ----
|4   |{35}     |{12}     |6   |
|    |         |         |    |
 ---- ---- ---- ---- ---- ----
|2   |1   |6   |5   |4   |3   |
|    |    |    |    |    |    |
 ---- ---- ---- ---- ---- ----
|13+ |2   |1   |6   |3   |5   |
|    |    |    |    |    |    |
      ---- ---- ---- ---- ----
|         |11+      |8+  |{12}|
|         |         |    |    |
 ---- ----      ----
|1-       |    |         |    |
|         |    |         |    |
 ---- ---- ---- ---- ---- ----

But now the leftmost cell of the fourth row is both required to be 4 to avoid a duplicate digit in the fourth row and required not to be 4 to avoid a duplicate 4 in the first column, eliminating this possibility.

So yes, I agree that the additional rule, which I'm reasonably certain I've seen stated in newspaper KenKens, is not actually needed to solve this one - and that not having it does add a noticeable amount of logical complexity to the solution!

Gengulphus

[cinelli]

Well solved.

The puzzle I set came from The Times, and there you are allowed to have the same digit in a block. Unfortunately I don’t have an example. Perhaps there is another variety of KenKen where there is the additional rule. Like similar puzzles, the challenge is to find a method. You wouldn’t want to do too many of this puzzle, I think.

There are regular puzzles in the I newspaper which I admit I haven’t completely cracked. They are called bridges and, I think, landmines. The trouble with both is that once you have made a mistake (quite easy to do), it is very hard to backtrack.

Cinelli