Sudoku Players' Forum (clone)

Righto. So it seems that on the (pretty vast by now) preponderance of evidence, you can't have the same number twice in an enclosure (how's that for a term?). Any ideas on whether or not it's necessary to know this, however? I think all the ones in The Times so far can be solved without recourse to this rule, but then again there haven't been that many. As for the one you posted in the other thread, I'm having trouble even working out where to start with it!

It helped me knowing you could not double the numbers in 'enclosures'
When I thought you could I had trouble solving them.

At last I finished yesterdays moderate, 20 minutes took me 20 hours!!
Todays was a piece of P### compared to yesterdays did it in about 20 minutes, which was only twice the target time..

The 45 rule was obvious I thought.

I have found it useful to print of a sheet with all the options of
2, 3 and 4 enclosures and find it useful as a quick reference when eliminating or proving numbers
example of the 3 below.. the four is a bit bigger and I havnt thought about the 5 yet as I think it would be too big and will just have to work out on the fly...

6= 1+2+3
7= 1+2+4
8= 1+2+5, 1+3+4
9= 1+2+6, 1+3+5, 2+3+4
10= 1+2+7, 1+3+6, 1+4+5, 2+3+5
11= 1+2+8, 1+3+7, 1+4+6, 2+3+6, 2+4+5
12= 1+2+9, 1+3+8, 1+4+7, 1+5+6, 2+3+7, 2+4+6, 3+4+5
13= 1+3+9, 1+4+8, 1+5+7, 2+3+8, 2+4+7, 2+5+6, 3+4+6
14= 1+4+9, 1+5+8, 1+6+7, 3+5+6, 2+3+9, 2+4+8, 2+5+7, 3+4+7
15= 1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6
16= 1+6+9, 1+7+8, 2+5+9, 2+6+8, 3+4+9, 3+5+8, 3+6+7, 4+5+7
17= 1+7+9, 2+6+9, 2+7+8, 3+5+9, 3+6+8, 4+5+8, 4+6+7
18= 1+8+9, 2+7+9, 3+7+8, 3+6+9, 4+5+9, 4+6+8, 5+6+7
19= 2+8+9, 3+7+9, 4+6+9, 4+7+8, 5+6+8,
20= 3+8+9, 4+7+9, 5+7+8, 5+6+9
21= 4+8+9, 5+7+9, 6+7+8
22= 5+8+9, 6+7+9
23= 6+8+9
24= 7+8+9

Thanks for that it will be so useful. Anyone done one for 4 boxes?

*Need to remember that if repeats are allowed the number of combinations increase*

here are the 4's

please let me know if you spot any mistakes

10 1+2+3+4
11 1+2+3+5
12 1+2+3+6 1+2+4+5
13 1+2+3+7 1+2+4+6 1+3+4+5
14 1+2+3+8 1+2+4+7 1+2+5+6 1+3+4+6 2+3+4+5
15 1+2+3+9 1+2+4+8 1+2+5+7 1+3+4+7 1+3+5+6 2+3+4+6
16 1+2+4+9 1+2+5+8 1+2+6+7 1+4+5+6 1+3+4+8 1+3+5+7 2+3+4+7 2+3+5+6
17 1+2+5+9 1+2+6+8 1+4+5+7 1+3+4+9 1+3+5+8 1+3+6+7 2+3+4+8 2+3+5+7 2+4+5+6
18 1+2+6+9 1+2+7+8 1+4+5+8 1+4+6+7 1+3+5+9 1+3+6+8 2+3+4+9 2+3+5+8 2+3+6+7 2+4+5+7 3+4+5+6
19 1+2+7+9 1+4+5+9 1+4+6+8 1+3+6+9 1+3+7+8 1+5+6+7 2+3+5+9 2+3+6+8 2+4+5+8 2+4+6+7 3+4+5+7
20 1+2+8+9 1+3+7+9 1+4+6+9 1+4+7+8 1+5+6+8 2+3+6+9 2+3+7+8 2+4+5+9 2+4+6+8 2+5+6+7 3+4+5+8 3+4+6+7
21 1+3+8+9 1+4+7+9 1+5+6+9 1+5+7+8 2+3+7+9 2+4+6+9 2+4+7+8 2+5+6+8 3+4+8+9 3+4+6+8 3+5+6+7
22 1+4+8+9 1+5+7+9 1+6+7+8 2+3+8+9 2+5+6+9 2+5+7+8 3+4+6+9 3+4+7+8 3+4+6+8 3+4+7+8 3+5+6+8 4+5+6+7
23 1+5+8+9 1+6+7+9 2+4+8+9 2+5+7+9 2+6+7+8 3+4+7+9 3+4+6+9 3+4+7+9 3+5+6+9 3+5+7+8 4+5+6+8
24 1+6+8+9 2+5+8+9 2+6+7+9 3+4+8+9 3+5+7+9 3+6+7+8 4+5+6+9 4+5+7+8
25 1+7+8+9 2+6+8+9 3+5+8+9 3+6+7+9 4+5+7+9 4+6+7+8
26 2+7+8+9 3+6+8+9 4+5+8+9 4+6+7+9 5+6+7+8
27 3+7+8+9 4+6+8+9 5+6+7+9
28 5+6+8+9 4+7+8+9
29 5+7+8+9
30 6+7+8+9

Has anyone begun to work out a theory and strategy for these yet? Apart from the first 'gentle', the ones I've finished so far all seem to involve a significant amount of T&E (or, in honour of their creator, at the minimum a lot of nishio).

Is there always a pure logic path to a solution? If not, is there a way we can tell when T&E is necesary?

The 45 rule is one logical foundation: when the totals in a row, column or box exceed 45, then any 'spillover' cells have to contain the difference. Then there are the '6' rule (=1,2,3), the '17' rule (=9,8), the '24' rule (=9,8,7) etc. But these are isolated instances. Does anyone have more generalised approaches?

A lot of the time I found it quite difficult to tell if what I was doing was T&E or not I think the puzzle is a lot more ambiguous in this respect than regular sudoku. A few things I learnt, however, were:

• A 15 total in a two-cell enclosure can only be made with 7&8 or 6&9, and for some reason they seem to come up a lot.
• Whatever you do, don't get so involved in the maths that you forget to apply the usual sudoku logic as well.
• Don't forget about 'underspill' as well as overspill: you have complete enclosures using up eight of the nine cells in a unit, the remaining cell takes the value of 45 minus the sum of the enclosures.

roger888 wrote:I'm not sure the criticism is fair,

Sure it is.

roger888 wrote:or that the alleged ambiguity exists.

Sure it does.


Multiple people found the instructions ambiguous at best -- including you:

roger888 wrote:Has anyone begun to work out a theory and strategy for these yet? Apart from the first 'gentle', the ones I've finished so far all seem to involve a significant amount of T&E (or, in honour of their creator, at the minimum a lot of nishio).

Is there always a pure logic path to a solution? If not, is there a way we can tell when T&E is necesary?

No T&E is required if you know all the rules -- no duplicate digits in ANY group -- row, column, box or numbered enclosure. Ignore this rule at your own peril.

roger888 wrote:The 45 rule is one logical foundation: when the totals in a row, column or box exceed 45, then any 'spillover' cells have to contain the difference.

??? The rows, columns and boxes cannot exceed 45.

roger888 wrote:Then there are the '6' rule (=1,2,3), the '17' rule (=9,8), the '24' rule (=9,8,7) etc. But these are isolated instances. Does anyone have more generalised approaches?

Red herring. There is no '6' rule, nor is there a '24' rule, only a rule that no digit may appear twice in an enclosure.


This puzzle is a hybrid of Sudoku and Cross Sums. Search the web for "cross sums" and "stategy" or "tactics", etc. Some (but not all) of what you find may be helpful. For example, here and here and here.

Cross Sums are (is?) one of the most popular graphic logic puzzles in Japan. (Nikoi has published 22 books of Cross Sums aka Kakro and 24 books of Sudoku.) No other puzzles in this genre are nearly as popular.

All three of these puzzles have share the common solving method of making pencil marks within cells and eliminating candidates. Each cell in the solution has 9 possible states. Only a very tiny percentage of all graphic logic puzzles share this charactoristic. Just my opinion -- I think it has something to do with their popularity, as it allows a larger range of difficulty and an obvious 'foothold' to attack the harder ones. I've had the experience while solving other types of puzzles, of having made some sort of logical deduction, but having no simple way to store that information within the solving area.

tso wrote:

roger888 wrote:The 45 rule is one logical foundation: when the totals in a row, column or box exceed 45, then any 'spillover' cells have to contain the difference.

??? The rows, columns and boxes cannot exceed 45.

I think what he means is that, where you have a set of enclosures that together make up an entire box (or row or column) plus one additional cell (the 'spillover'), you can work out the content of that cell by adding up the totals of the enclosures and subtracting 45 from it.

roger888 wrote:Then there are the '6' rule (=1,2,3), the '17' rule (=9,8), the '24' rule (=9,8,7) etc. But these are isolated instances. Does anyone have more generalised approaches?

Red herring. There is no '6' rule, nor is there a '24' rule, only a rule that no digit may appear twice in an enclosure.

Well no, they aren't given rules of the puzzle, but they're logical inferences from the rules, so they can be treated as if they are rules, and they are useful to emply, so are hardly red herrings.

has the no repeats within an enclosure rule been verified yet?

With hindsight, my use of the term 'rule' was ill-advised. I meant it in the sense of an algorithmic rule for solving, not a constitutive rule of the puzzle. Sorry.

More significant to me is the assertion:

tso wrote:No T&E is required if you know all the rules -- no duplicate digits in ANY group -- row, column, box or numbered enclosure. Ignore this rule at your own peril.

Is this true? Presumably not in all cases. How do we evalute when T&E is required (allowing for the fuzziness of what constitutes T&E, at the extreme I mean brute force and trying all possibilities in turn and following them through to a solution or a contradiction)?

This is probably equivalent to asking whether anyone yet has the foundation for a solving algorithm. My guess at present is that this would need to be based on intersecting sets of possible permutations. But I'm a recreational solver, not a mathematician or programmer.

I managed the Killer Su Dokus Nos. 1 & 2 without difficulty, but I simply cannot see a way to get going on No.3 (Moderate) in the Times today.

Hint please.

PaulIQ164 wrote:Whatever you do, don't get so involved in the maths that you forget to apply the usual sudoku logic as well.

Ooohh, how true this little gem is. I've almost disappeared up my own backside in the last two days applying maths when sudoku logic was more than enough. Of course, some people may feel that happened some time ago... But the point is, these things should come with a caveat reading IT'S A SUDOKU.

And regarding today's, the best place to start is the middle. There's only five numbers which can add up to 35, so what does that tell you about the four corner cells in the middle box? Following from this, look at column 5. This has two separate pairs which total 5, so these cells must contain 1,2,3 and 4.

Was that too much of a hint?

I liked the way todays went.
The start in the middle was a good use of the 45 rule and then just working out.
Most of the time it was symmetrical.

And regarding today's, the best place to start is the middle. There's only five numbers which can add up to 35, so what does that tell you about the four corner cells in the middle box? Following from this, look at column 5. This has two separate pairs which total 5, so these cells must contain 1,2,3 and 4.

Was that too much of a hint?

No, just right, thanks. I've now struggled through. But too much maths. I graduated in maths 50 years ago. I've done my share of maths for this lifetime!

Wow - I've finally done the five from Wednesday's paper. Overall, I think the "Moderate" actually took the longest although I haven't timed myself properly. Definitely a case of practice helps if you've got the patience and hours to spare

None of the enclosures had repeated numbers but I'd still like confirmation as to whether that is a rule or not.