### Solving a difficult KenKen puzzle

That right there is the hardest KenKen puzzle I’ve ever come across. Maybe I’m just missing a strategy and there’s something obvious I overlooked, but I found the levels of ambiguity in this one to be very, very difficult to unravel. I’ve been coming back to this puzzle for months. Having finally cracked it, I wanted to document the solution. Partly to remember what happened, but mostly because it just felt so darn good to finally figure out.

With KenKen puzzles, there is no guessing. The puzzles are logical, and after doing hundreds of them over the past several years, I’ve only come across one or two that seemed to have equally forked possiibilities. And I hated those.

KenKen is similar to Sudoku, rows and columns contain n-digits with no repeats (a Latin Square). With KenKen, it can be helpful to know the sum of all digits in the puzzle’s dimension since every row and column must add up to that Triangular number. For a six digit puzzle this is 21. For this particular puzzle, it’s also worth remembering that there are only three paths to 11: `{1, 4, 6}`

, `{2, 3, 6}`

and `{2, 4, 5}`

. Note the digits unique to each pathway.

### Starting Out

The key to breaking into this one was column 6. Since column 6’s top cage sums to 11, and **F6** is 5, the sum of **D6** and **E6** must equal 5. Since the only possibilities are combinations of 1, 2, 3 and 4, and **E6** can’t be 1 because the sum of **E5** and **F5** can’t be 12, there are only three possibilities, **D6**: `{1, 2, 3}`

and **E6**: `{2, 3, 4}`

. Since the sum of both cages is 20 and we know 5 is accounted for, the sum of **C3** through **F6** must be 15. That means **A5** and **B5** have to add up to 6. The remaining half of that cage, **B4** and **C4** must add up to 7.

We now know the sum of **B4** through **F6**, so **A4** must be 3. This also reveals **F3** is also 3 because the sums of all cells in the top two cages of column 3 are known.

Having a 3 in **F3** means the remaining slots of that cage, **F1** and **F2** can only be 2 and 4. This means **F4** must be 1 and **F5** must be 6 because, like **E6**, **F5** can’t be 1. Additionally, **D4** and **E4** must contain 4 and 6 because there’s no other way to get to 11 using a 1. The last two cells of column 4 can only be 2 and 5.

The upper left cage, **A2**, **B1** and **B2**, has limited choices. Since the largest possible number is 6, the sum of any two adjacent sides can’t be less than 9. Likewise, since the largest adjacent sum is 11, neither **A2** or **B1** can be less than 4. The only exception is **B2** which could be 3 since both adjacent cells could be 6.

But the **D2**, **E2** cage can’t contain 3 and 4. If it did, then **A2** and **B2** could only be 5 and 6, which would mean **B1** would have to be 4. The problem is that **F1** and **F2** can only be 2 or 4 and column 2’s 4 would already be taken by the 3 and 4 in **D2** and **E2**. So, the **D2**, **E2** cage can only be a combination of 2 and 5 or 1 and 6, the sum of **A2**, **B2** must be 9 and **B1** has to be 6.

At this point, things get hairy. I couldn’t see any obvious connection and essentially started trying alternate solutions.

### A dead end

There are only two remaining combinations for reaching 11 in the **A6**, **B6**, **C6** cage, but if **A2** was 6 and **B2** was 3, the only remaining combination would be `{1, 4, 6}`

because the top row would already have 6, 1, and 3, leaving only a 2 or 4 as candidates in **A6** except the 2 would be taken by any combination of **A3**, **B3**, **B4**, **A5** and **B5**. So, from here, the top row could only be filled out one way:

That cascades down to **E6**. Now the **D2**, **E2** cage has would have to be 2 or 5 because the 6 was taken by **A2**. But **E6** couldn’t be 2 because then **E5** would have to be 5 which would make **E2** impossible. And here’s where this tangent falls apart all over the board. That was a long way to go to prove that **A2** and **B2** have to be 4 and 5.

### It all comes together

Disproving that was painful, but now we’re starting to get somewhere. Confirming **A2**‘s and **B2**‘s candidates reveals **C2** is 3, **F2** is 2 and **F1** is 4. This also limits **C1**‘s candidates to 2 and 5. **C4** is also limited to 2 and 5, so no other cells in row C can contain a 2 or 5.

Even better, there’s now only one way to fill in the top two rows:

That pretty much gives it all away. Columns 2, 4 and 6 are quickly revealed and the remaining values are easily discovered.

The solved puzzle:

### Solvers

It was interesting and humbling to run this puzzle through two KenKen solvers after finishing it by hand.

Michael Heyeck’s NekNek solver solved the puzzle instantly. Michael documented the creation of his KenKen solver, including the complete python source code (less than 150 lines with comments).

Below is the puzzle in JSON notation for the Poison Meatball solver. It was slightly reassuring that this took a few seconds to get to the answer.

```
{
"width": 6,
"height": 6,
"values": [1, 2, 3, 4, 5, 6],
"rules": [{
"op": "=",
"value": 1,
"cells": [{"x": 0, "y": 0 }]
}, {
"op": "+",
"value": 15,
"cells": [{"x": 1, "y": 0 }, {"x": 0, "y": 1 }, {"x": 1, "y": 1 }]
}, {
"op": "+",
"value": 10,
"cells": [{"x": 2, "y": 0 }, {"x": 3, "y": 0 }, {"x": 2, "y": 1 }]
}, {
"op": "+",
"value": 13,
"cells": [{"x": 4, "y": 0 }, {"x": 3, "y": 1 }, {"x": 4, "y": 1 }, {"x": 3, "y": 2 }]
}, {
"op": "+",
"value": 11,
"cells": [{"x": 5, "y": 0 }, {"x": 5, "y": 1 }, {"x": 5, "y": 2 }]
}, {
"op": "+",
"value": 13,
"cells": [{"x": 0, "y": 2 }, {"x": 1, "y": 2 }, {"x": 0, "y": 3 }, {"x": 0, "y": 4 }]
}, {
"op": "+",
"value": 11,
"cells": [{"x": 2, "y": 2 }, {"x": 2, "y": 3 }, {"x": 2, "y": 4 }]
}, {
"op": "+",
"value": 7,
"cells": [{"x": 4, "y": 2 }, {"x": 4, "y": 3 }, {"x": 5, "y": 3 }]
}, {
"op": "+",
"value": 7,
"cells": [{"x": 1, "y": 3 }, {"x": 1, "y": 4 }]
}, {
"op": "+",
"value": 11,
"cells": [{"x": 3, "y": 3 }, {"x": 3, "y": 4 }, {"x": 3, "y": 5 }]
}, {
"op": "+",
"value": 13,
"cells": [{"x": 4, "y": 4 }, {"x": 5, "y": 4 }, {"x": 4, "y": 5 }]
}, {
"op": "+",
"value": 9,
"cells": [{"x": 0, "y": 5}, {"x": 1, "y": 5}, {"x": 2, "y": 5}]
}, {
"op": "=",
"value": 5,
"cells": [{"x": 5, "y": 5}]
}]
}
```

Gil Kessler8:32am on Sunday, June 2nd, 2013 #Thank you for a truly hard puzzle. I had begun to think they could not really be difficult (and certainly not a 6×6!). Your help in getting B1 was invaluable. And even after, it was very hard (but I was able to finish on my own). Many thanks again.

Rick Samuelson1:25am on Monday, June 17th, 2013 #Thanks for posting that very difficult puzzle. Like Gil Kessler, I thought I could solve all kenkens but I couldn’t solve this one without reading some of your steps. You’re certainly right that it “was a long way to go to prove that A2 and B2 have to be 4 and 5.” I’ve gone through some of those but this one I just couldn’t see … so I learned something. Thanks again.

a contributor7:49pm on Wednesday, July 24th, 2013 #{“width”:6,”height”:6,”values”:[1,2,3,4,5,6],”rules”:[{“op”:”=”,”value”:1,”cells”:[{“x”:0,”y”:0}]},{“op”:”+”,”value”:15,”cells”:[{“x”:1,”y”:0},{“x”:0,”y”:1},{“x”:1,”y”:1}]},{“op”:”+”,”value”:10,”cells”:[{“x”:2,”y”:0},{“x”:3,”y”:0},{“x”:2,”y”:1}]},{“op”:”+”,”value”:13,”cells”:[{“x”:4,”y”:0},{“x”:3,”y”:1},{“x”:4,”y”:1},{“x”:3,”y”:2}]},{“op”:”+”,”value”:11,”cells”:[{“x”:5,”y”:0},{“x”:5,”y”:1},{“x”:5,”y”:2}]},{“op”:”+”,”value”:13,”cells”:[{“x”:0,”y”:2},{“x”:1,”y”:2},{“x”:0,”y”:3},{“x”:0,”y”:4}]},{“op”:”+”,”value”:11,”cells”:[{“x”:2,”y”:2},{“x”:2,”y”:3},{“x”:2,”y”:4}]},{“op”:”+”,”value”:7,”cells”:[{“x”:4,”y”:2},{“x”:4,”y”:3},{“x”:5,”y”:3}]},{“op”:”+”,”value”:7,”cells”:[{“x”:1,”y”:3},{“x”:1,”y”:4}]},{“op”:”+”,”value”:11,”cells”:[{“x”:3,”y”:3},{“x”:3,”y”:4},{“x”:3,”y”:5}]},{“op”:”+”,”value”:13,”cells”:[{“x”:4,”y”:4},{“x”:5,”y”:4},{“x”:4,”y”:5}]},{“op”:”+”,”value”:9,”cells”:[{“x”:0,”y”:5},{“x”:1,”y”:5},{“x”:2,”y”:5}]},{“op”:”=”,”value”:5,”cells”:[{“x”:5,”y”:5}]}]}

tabishkhan5:40am on Saturday, July 19th, 2014 #its such a hard puzzle i am not able to solve it but i am definitely gonna solve it either it will take time http://travelwithsmile.com/TourPackages.aspx

Steppenwolf7:14am on Sunday, April 19th, 2015 #Interesting puzzle. However, I didn’t think it was particularly difficult. It took about 5 mins to get the 3 in the top row and 3, 1 and 6 in the bottom row. After that I got stuck for 15 mins, until I realized that the 11 in the middle two columns has to be 6,4 and 1. It then took about 10 mins to focus on the top 2 rows and try out the 2 possibilities (2,5 and 5,2) for the top 2 elements of the third column. The rest took about 10 mins to complete.

Mark Jones9:34am on Saturday, May 21st, 2016 #A new trick in my play book is to extend the Rule of 21 across multiple rows or columns. That started me out with a 3 at F3, which led directly to a 3 at A4. The combos of 11+ were also very helpful, as you note. Thanks for a fun puzzle :)

Le big Maque9:10pm on Wednesday, August 31st, 2016 #Thanks for posting. Trying a “+ only”-puzzle was a new experience for me! I managed showing uniqueness of the solution only by striking out two dead ends. (As I’ve read your solution now, we went similar tracks, just that the minor information {3,4}={E5,E6} can be derived independently from solving the upper two rows.)

The sum rules are pretty useful most of the time. Unfortunately, somewhat lengthy indirect arguments,or lets call it “trial-and-error” approach to pin it down, seem inevitable if one wants to solve a puzzle with only the + operation. But nevertheless, it was a fun one.

h5:06am on Wednesday, May 23rd, 2018 #thanks

h5:06am on Wednesday, May 23rd, 2018 #thank you

bruce kamolnick6:37pm on Wednesday, August 1st, 2018 #Never could see any of the puzzles (the graphics were broken). Would you possibly email me a PDF of this column?

Discovered KenKen just a few weeks ago, and good enough for the Sunday Times 5x5s and 7x7s, but still pretty much in rookie territory, and using tons of if this then that trials.

Thanks Joe!

Bruce