Very impressive/difficult 9x9 today. I'll be interested in successful solvers' strategies.

I narrowed things down as much as was obviously possible to begin with. Then I divided the puzzle into sections - rows 1, 2 and 3; rows 4, 5 and 6; rows 7, 8 and 9; columns a, b and c... to see if any section had enough cells and cages filled so that the remaining cells had only a limited number of possibilities. For instance, if you can put some limits on d1e1f1e2, b2c2, and/or f2f3 then there are only certain numbers that can be used in b3 and h3, because b3 + h3 = 7 or 9 or whatever.

One of those did, so I started checking whether some of those limited possibilities had obvious conflicts. At some point one of these _didn't_ have any conflicts after several steps assuming that it was true so (I apologize, clm, knowing that you always stick to reason and wouldn't follow a hunch just because it hadn't caused an obvious conflict) I just plugged that one in and the solution fell quickly into place.

So, a semi-analytical solution...

Thanks for that jaek, very interesting.

I tried to solve this one myself as well (I usually only do
the timed puzzles), and haven't succeeded yet

So far I'm thinking that (after putting all the obvious stuff) the 60x
and 378x cages are key (a handful of combinations, which somewhat
determine the center 25+, maybe something possible with the top/bottom/left/right sums...)

This structure (very little times appeared) in the 9's is probably the most difficult due to the number (6 today) and position of the subtraction cages. To show the full solution requires time (I hope to have it in a not far future to fully develop one of these). I agree with jaek with respect to the strategy by isolating specific cells every three lines. It's clear (addition rule for the three righmost columns) that g4 + g6 = 10 >>> [2,8] or [4,6] ... but the interesting thing is that f4 + f6 = even; since (parity rule for the three leftmost columns) c4 + c6 = odd >>> d4 + d6 = 26 - odd = odd then (parity rule for the three central columns) "60x" is odd, since "378x", which is [1,6,7,9] or [2,3,7,9] (excluding the [3,3,6,7] due to the double 3), is always odd. So we have concluded, Patrick, that "60x" = [1,3,4,5] being invalid the combination [1,2,5,6] (and obviously excluding [2,2,3,5] due to the double 2).

But this is only to start, much more analysis is required, I have first eliminated [4,6] for the pair g4-g6 then arriving to [2,8] and then necessarily f4 = 5, etc. ... Apologize because now it's prior to 24:00 CET.

I found the puzzle solvable (although difficult) after I had eliminated [4,6] for the pair g4-g6 like clm did. What irked me was that I found no real "analytical" way of doing so, meaning that I had to do some deep trial and error (combining [4,6] for the pair g4-g6 first with [4,7] and then with [5,6] for the 11+ box in column d) before I was satisfied that I could indeed rule [4,6] out.

Hi, Bram, to eliminate the pair [4,6] we can do this other “analysis”:
Case g4 = 4 >>> f4 = 1. If we observe the graphic we quickly arrive to b4 = 3. In these conditions, considering the addition of the three central rows (total of 135): a + b + c = 19 (the sum of cells b6, f6 and h6); but if b = 5 >>> a + c = 14 cann’t be accomplished and if b = 7 >>> a + c = 12 and, as c6-d6 = [4,9], this condition cann’t be accomplished.



Case g4 = 6 >>> f4 = 3 (c4-d4 = [4,9]) >>> f6 = 5 >>> f7-f8 = [4,8] >>> f2 = 2; f3 = 6. The three 2’s in columns d, e and f (the three central columns) force “378x” = [1,6,7,9] with a sum of 23 and now, considering the addition of the three central columns (total of 135): a + b = 17 (the sum of cells d4 and d6) which is impossible due to the 8 in cage “16x”.

clm, you have restored my faith in analysis!

I had found no better way to proceed with the puzzle than having [4,6] as candidates in g4 and g6 and then considering the special cases of [4,7] and [5,6] in d2 and d3, which lead to a complicated process of trial and error. Your solution, analyzing the two possible orderings of [4,6] within g4 and g6 in turn, is lean and clear.

(Maybe there even exists an analytical way of going about my initial approach, too, but if so I failed to find it, most probably due to my taking the three central rows (as opposed to columns) into consideration only at a very late point, long after I had painstakingly eliminated [4,6] by trial and error. I am not going to revisit it since I am actually on a calcudoku hiatus – but got intrigued when I logged in to submit a solution from my cache of solved book puzzles and then noticed this thread.)

Thank you for the crystal-clear demonstration

On a side note, I think patterned 9x9s are generally rather difficult and underrated compared to 12x12s and 15x15s. The latter can usually be solved by a series of not-too-difficult mini-analyses whereas in the former you often reach a critical point by which you need to find an opportunity for doing a bit of higher-level analysis like the one you demonstrated. I suspect this is due to the solving/rating algorithm more often resorting to brute force, i e trial and error (which a computer should be able to conduct more easily than a person, and without the risk of miscalculation) rather than doing the most intricate kinds of analysis.

(And on a side note to the side note, some 10x10s, although not patterned, have also been more difficult than the 12x12s and 15x15s. But this is not generally the case, and of course the one from 2 April is much easier than the patterned 9x9 and, fittingly, also has a lower difficulty rating.)

Absolutely, I am of the same opinion: it's true that we have 3 days for the 12's and 30 days for the 15's but, for instance, today's 7x7 difficult (the subtractions only is always rated 9 points) is quite easy as compared with the yesterday's 9x9 (though sometimes the 7x7 subtractions only is quite difficult ). Another clearer case: the 10x10 on fridays, 14 points, is generally easy and does not take more than 15-20 minutes (I suppose in the other hand that this 10x10 makes happy most puzzlers with those 14 points ).

A possibility (suggestion): To assign 11 (or 12) points to the tuesday's 9x9's (perhaps this would motivate more puzzlers to fight against this puzzle.

I'm glad everyone who posted found this one so hard.
I gave up after making a mathematical error at some point, having used some or most of the techniques described here!
Glad I didnt miss anything blindingly obvious.
This was the hardest 9x9 I've ever done though I am relatively new to the site as a subscriber.
I've been doing them for a year or so and would welcome this degree of difficulty on a regular basis.

I just noticed that this puzzle only has a rating of 130. It's a little bit funny as it was the most difficult puzzle I've ever done.