It is possible without guessing. Most of the corner areas are solvable to a certain extent, independently of each other. The least solvable is the lower left corner, but here you can find out that the only way the vertical 33 can work is going to be 1,2,6,7,8,9. With this info, you can find out that the 18 adjacent to it can only be composed of the lowest possible value for each of its three elements. That allows the bottom left corner to be solved. This then determines the combination making the vertical 26 in the middle of the puzzle, and once this is solved (based on finding the only value which works for the 19 crossing it) the rest of the puzzle is straightforward.

Posted 10th Feb 2019 at 16:01

carlotes Daily subscriber Rated puzzle: Hard Completion time: 8:37 Used 'show wrong moves'

Thanks a lot Chessmama. I'll try again with these suggestions.

Posted 11th Feb 2019 at 01:18

Pseudo Daily subscriber Rated puzzle: Moderate Best completion time: 8:50 Time on first attempt: 10:44 Used 'remove' Used 'show wrong moves'

Another way you can do it is to first solve most of the top half, then add up the rows which sum to 7, 20, 6, 19 and 13, and subtract the columns which intersect those rows. You can then figure out that the bottom squares in the 26 and 33 columns must sum to 18, and therefore they must both be 9.

Posted 11th Feb 2019 at 04:20

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

I mean, I am not saying such difficult games should not be posted. Of course different difficulty level puzzles must be posted every now and then. I see 90% of puzzles rated Easy by members' average, and only few Mild, even less rated moderate. I have hardly seen and DIFFICULT, or HARD or Extreme, and I used to wonder that when learned members here find these puzzles easy that take me hours to solve, what kind of puzzle would they find hard.

Posted 11th Feb 2019 at 04:21

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

My suggestion is - The site should tell the difficulty level beforehand as judged by the site, This way we can see that site is also posting difficult, hard, extreme puzzles, it is the wise members who are finding even those level to be Easy. :-)

Posted 11th Feb 2019 at 05:46

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

You wrote "vertical 33 can work is going to be 1,2,6,7,8,9."

Good enough, 33 is pair of 12 and 6789

"With this info, you can find out that the 18 adjacent to it can only be composed of the lowest possible value for each of its three elements."

How? 17 across is 89, if 17 has 8 then remaining digits of 18 should total to 10 giving

if 17 has 8 in 18 column, then remaining digits of 18 should total to 10 giving 19 37 46 combinations. if 17 has 9 in 18 column, then remaining digits of 18 should total to 9 giving 18 27 36 45 combinations.

why should it give lowest values for each of 3 elements of 18?

Posted 11th Feb 2019 at 05:50

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

I had solved it, but I could find reasonable clues upto this point only.

@vsrawat: You are almost there. The square to the right of the blue highlighted square cannot be 6 or more, because you cannot then make 24 vertically, so it is 1, 2, 3 or 5. But then the square to the left of the blue square cannot be 1, 2, 3 or 5, because then the maximum of those three squares (the highlighted plus the ones on each side) plus the 4 is 5+4+3+2 = 14, leaving no solution for the last square. So the one to the left of the blue must be minimum 6, but then the only solution for the 18 is 4+8+6.

Posted 11th Feb 2019 at 08:49

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

yes, that is it.

I can now see that,

Let me understand it in my slow way -

in the cell to the right of blue, the column of 24 has 9 already taken, so other three digits should total up to 15

In these if the lower-most of the three cells (above 9) has the only possibility of 68,

if it is a 6 at lower-most (above 9), top two should total up to 9, that is 1+8,2+7, (6+3 out as 6 is below), 4+5, and if it is an 8 at lower-most (above 9), top two should total up to 7, that is 1+6,2+5,4+3,

now as top cell has only possibilities of 45678, then these 123 have to be in the cell below (the cell right to the blue cell) and in case of 4+5, 4 is not a possibility in this lower cell, it has to be in the top cell and 5 will come below, so it will be 1235.

Yes, as you said, it can't be 6 or more in the second cell. :-)

Posted 11th Feb 2019 at 09:05 Last edited by vsrawat 11th Feb 2019 at 09:28

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

however, to the second part of logic, (left of the blue square), is still not clear

Now, as the 24 across has: 4 (given) and possibilities 6789, 12356789, 13, 1235,

total of three cells, 4 (given) + 3 (max of 1-3) + 5 (max of 1235) = 12 that leaves 12 to come from total of two cells 6789, 12356789

12 can come from 7+5, (Edited: 8+4, no 4 is already there), 9+3 combo

so, you said "12356789 cell cannot be 1, 2, 3 or 5, but I am finding that it can be (Edited: 5/3)

however it can surely not be 1-2.

Posted 11th Feb 2019 at 09:55 Last edited by Elisabeth 11th Feb 2019 at 09:58

Elisabeth Daily subscriber Rated puzzle: Hard Completion time: 27:50 Used 'valid marks' Used 'auto remove' Used 'show wrong moves'

Full marks for perseverance, vsrawat!! I'm afraid I gave up, used all the aids and did a bit of guessing:)

I used to worry about all the "Easy" results. Now I look at the lower end of the first column (no aids) which is where I hope to be, to see what standard is set. I also note how long/short the no aids column is compared with the other one; thiis can be an indication. It might help if we had more than the 3 options, perhaps a "Mild" after "Easy"? If Gareth chose the level of difficulty we might still not agree! (Maybe people don't like to admit they found the puzzle hard?)

Posted 11th Feb 2019 at 10:07

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

however, with 1&2 removed from 12356789 cell,

combo for 18: should come from 356789 where 8 or 9 is a must (from 17 across possibilities)

18 possibilities are: x189 x279 369 378 459 468 x567 x indicates not possible due to above conditions

that leaves 18's possibilities 369 378 459 468

in case of 4 (459 468), 4 can only be in the top cell, so lower two cells can have combo of 9+5 or 8+6, making it 56 in lower most cell

or in case of 3 (369 378), 3 can only be in this lower most cell, so the combo for 24 in cell left of blue get reduced from 356789 to 356

and if 3 is in the lower most cell, the combo for upper two could be 6+9 or 7+8, meaning the top cell can only have 6 or 7 and of course 4 for the other possibilities mentioned above.

possibilities for 28 across are 4789 5689, now 5 is nowhere there so only 4789 can be there for 28 across, so 6 cannot be in 28 across, so 18 down combo in case of 3 (369 378) cannot now be 369, and can only be 378

thus 18 down, can only have 47 in top, 89 in the middle and 356 in the bottom.

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

further, possibilities for 24, are (where 4 is already given): x4-1289 x4-1379 4-1568 4-2369 x4-2378 4-2567

as 789 are only in a single cell, the possibilities above that mention more than 789 together are out. marked with x above

that leaves 24 = 4-1568 4-2369 x4-2567

also 2 is at only one place, putting 2 there will leave 5 at only one place removing 5 from there will also remove 6 from there, that will leave 67 in a single cell, so that is also not possible.

that leaves 24 = 4-1568 4-2369

as 8 and 9 in only one cell, if they go there, that leaves that will leave 156 and 236, and 6 is left at only 1 place where it will go,

so, in 24 across: 356 cell must have 6 leaving 158 and 239 for remaining 3 cells of 24 across

now 18 down has 6 fixed at bottom, that leaves 8+4 or 9+3, but 3 is not there, so it has to be 4 8 6 for 18 down

that brings 17 across to be 9 and 8, that removes 9 from 33 down, that removes 24 = 4-1568 x46-239 and leaves 24 = 46-158

that removes 7 and from 78 cell and leaves 8 there, and also fixes 1 and 5 as they are at one place only.

it goes a long way filling several cells.

Posted 11th Feb 2019 at 11:18 Last edited by vsrawat 11th Feb 2019 at 13:00

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

19 across: 136-27 136-45 (just paired to show common)

13 across: 148 238 247

33 down (2-9 given): 1678 3478 3568

I can do iteration, no big deal in this small set, still would like to learn if any method is there.

Posted 11th Feb 2019 at 13:06

Pseudo Daily subscriber Rated puzzle: Moderate Best completion time: 8:50 Time on first attempt: 10:44 Used 'remove' Used 'show wrong moves'

In the 19 across, notice that the 2nd, 4th and 5th squares are "almost" 1234. So it's worth checking what happens if you "force" them to be 1234: * If you put 1, 2 or 3 in the first square, those four squares will be 1234 (i.e. sum to 10), which would require the 3rd square to be 9, which is not possible. * Similarly, if the 3rd square is 1, 2 or 3 then the 1st square must be 9, which is also not possible.

Posted 11th Feb 2019 at 13:16 Last edited by Pseudo 11th Feb 2019 at 13:17

Pseudo Daily subscriber Rated puzzle: Moderate Best completion time: 8:50 Time on first attempt: 10:44 Used 'remove' Used 'show wrong moves'

Regarding difficulty, I suspect most (if not all) types of puzzles could be objectively rated based on the types of techniques needed to solve them. But it would be misleading to label these "easy" or "hard" etc. as those are pretty subjective terms. You could use something like 1-5 stars though.

For me, "easy" means I didn't get stuck for very long, "moderate" means I got stuck for a moderate about of time and "hard" means I got stuck for a significant amount of time.

Posted 11th Feb 2019 at 14:18

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

to start iteration, I think starting with 3 down (1 2) or 31 down (3 4) iteration will be much more beneficial because they would quickly fix one another's values and would ripple through to remove 1234 across the rows and column, then any mismatch can be quickly identified.

but yeah, previous "stuck" that I thought could be resolved by sudoku logic without going through iteration, but this one seems to have no clue and iteration is the only solution unless any member could find some way.

Posted 11th Feb 2019 at 15:21

Pseudo Daily subscriber Rated puzzle: Moderate Best completion time: 8:50 Time on first attempt: 10:44 Used 'remove' Used 'show wrong moves'

Here's another approach that doesn't use "educated guessing": Look at the 19 across. There are 5 ways to make 19 from 5 squares: * 12349 * 12358 * 12367 * 12457 * 13456

The first two aren't possible (no 8 or 9). The fourth one also isn't possible (can't have both 5 and 7). Of the remaining two combinations, the 567 must go in the 1st and 3rd squares, while the other three squares must be 1-5. In fact, with either 12367 or 13456 the 1st square must be 6 and the 3rd square must be 5 or 7.

With either combination there must be a 3, and the only place for the 3 is the last square in that row.

Posted 11th Feb 2019 at 16:59

Penelope Daily subscriber Has started but not yet finished this puzzle

Going back to average times and easy versus hard etc. I think, as Elisabeth says, that the only way to get a clear idea of how long a puzzle might take, or how difficult it is, is to look at the results table. Times and levels of difficulty are so skewed by people using aids and finishing a puzzle really quickly, then calling it easy.

Posted 11th Feb 2019 at 17:14

vsrawat Rated puzzle: Hard Best completion time: 3:09:01 Time on first attempt: 3:11:15 Used 'show wrong moves' Used 'check puzzle' when incorrect

That is it.

be it 12367 or 13456, in both cases, 5 or 7 are going to come in a single cell. that leaves 5-1236, 7-1346, and 6 can only be in a single cell.

filling that up gives 12 12 123 so 12 pairs out filling 3rd cell with 3.

going on, that leaves 33 down with only possibility of 678. that gives it the final push to crumble.

yes. this is also logic, no iteration required.

Now I can sleep happily otherwise it would have come in my nightmare.

Wonder where is Gareth. Tomorrow, seeing it, (someone) is going to be very angry with me for having wasted so many site resources. :-)

thanks.

Posted 11th Feb 2019 at 17:25

gareth Administrator Daily subscriber Has started but not yet finished this puzzle

Also, don't forget that if you tick 'Show current clue fits' then the player will show you which possible number combinations fit for a clue. It also crosses out those that won't fit based on existing placed numbers, although it doesn't consider what will fit compared to your pencilmarks since it doesn't know for sure that you have placed all possible pencilmarks.

Posted 11th Feb 2019 at 17:26

gareth Administrator Daily subscriber Has started but not yet finished this puzzle

On difficulty ratings, I actually have very accurate ratings for almost all of the puzzles on the site, but as you know it only shows ratings averaged from player scores. It's been interesting how most people don't rate the puzzles accurately - it must be a psychological thing.

Posted 11th Feb 2019 at 21:25 Last edited by Chessmama 11th Feb 2019 at 21:36

@vsrawat: In response to: "however, to the second part of logic, (left of the blue square), is still not clear Now, as the 24 across has: 4 (given) and possibilities 6789, 12356789, 13, 1235"

You can solve it other ways (as Pseudo suggests), but my logic is as follows:

*IF* the second of the cell possibilities you give (12356789) had any of the values 1, 2, 3 or 5, then that cell plus the two following cells could *maximally* be 2 + 3 + 5. Add that to the 4 you already have gives 14, and then the last cell (6789) would need to be 10 to make 24.

*Therefore* this cell *cannot* be any of 1, 2, 3 or 5. It is a more advanced logic than simple exclusion: it's exclusion based on a maximum sum of possible values. But you don't have to iterate.

EDIT: You can also see it another way with your logic:

"total of three cells, 4 (given) + 3 (max of 1-3) + 5 (max of 1235) = 12 that leaves 12 to come from total of two cells 6789, 12356789

12 can come from 7+5, (Edited: 8+4, no 4 is already there), 9+3 combo"

In order to get yourself in a position to make a two cell 12-combo, you had to "assign" 3 and 5 to two unknown cells. That then stops both of your possible combinations of 7+5 and 9+3 for the other two cells. Therefore some part of the assumption that the two cells you "guessed" are 3 and 5 must be wrong.

Posted 12th Feb 2019 at 17:57

pianobookspottery Daily subscriber Rated puzzle: Hard Completion time: 26:26:17 Used 'show wrong moves'

This was not a moderate puzzle in my opinion. This was fairly difficult.

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Rating scores out of 10.0 show the average difficulty rating chosen by users, where 1.0 is "Easy" and 10.0 is "Hard".

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Minor aid is defined as no more than one use of 'Check solution' when incomplete and/or no more than one use of 'Check solution' when wrong; and/or using highlighting aids (show repeated digits, show broken inequalities and show valid/invalid placements [slitherlink] only). Major aid is any and all other use of the solving aids except for 'show wrong'.

Add new comment)HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectHardCompletion time:8:37Used 'show wrong moves'EasyCompletion time:9:11HardCompletion time:8:37Used 'show wrong moves'ModerateBest completion time:8:50Time on first attempt:10:44Used 'remove' Used 'show wrong moves'HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectHardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectHardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectGood enough, 33 is pair of 12 and 6789

"With this info, you can find out that the 18 adjacent to it can only be composed of the lowest possible value for each of its three elements."

How?

17 across is 89,

if 17 has 8 then remaining digits of 18 should total to 10 giving

if 17 has 8 in 18 column, then remaining digits of 18 should total to 10 giving 19 37 46 combinations.

if 17 has 9 in 18 column, then remaining digits of 18 should total to 9 giving 18 27 36 45 combinations.

why should it give lowest values for each of 3 elements of 18?

HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectscreenshot

After this, I had to go though complex iterations and rough-shod trial and error.

It is not a complain, just trying to learn how to identify simpler clues that I had missed.

Last edited by vsrawat 11th Feb 2019 at 06:23HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectno obvious clue, but a simplest iteration tells that two 4s can't be there vertically so it can't be both 4.

Those who are afraid of iterations can't go past this without iteration, so it would prompt and encourage them to learn iterations.

Full marks for this.

EasyCompletion time:9:11You are almost there. The square to the right of the blue highlighted square cannot be 6 or more, because you cannot then make 24 vertically, so it is 1, 2, 3 or 5.

But then the square to the left of the blue square cannot be 1, 2, 3 or 5, because then the maximum of those three squares (the highlighted plus the ones on each side) plus the 4 is 5+4+3+2 = 14, leaving no solution for the last square.

So the one to the left of the blue must be minimum 6, but then the only solution for the 18 is 4+8+6.

HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectI can now see that,

Let me understand it in my slow way -

in the cell to the right of blue, the column of 24 has 9 already taken, so other three digits should total up to 15

In these if the lower-most of the three cells (above 9) has the only possibility of 68,

if it is a 6 at lower-most (above 9), top two should total up to 9, that is 1+8,2+7, (6+3 out as 6 is below), 4+5, and

if it is an 8 at lower-most (above 9), top two should total up to 7, that is 1+6,2+5,4+3,

now as top cell has only possibilities of 45678, then these 123 have to be in the cell below (the cell right to the blue cell) and in case of 4+5, 4 is not a possibility in this lower cell, it has to be in the top cell and 5 will come below, so it will be 1235.

Yes, as you said, it can't be 6 or more in the second cell. :-)

Last edited by vsrawat 11th Feb 2019 at 09:28HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectNow, as the 24 across has: 4 (given) and possibilities 6789, 12356789, 13, 1235,

total of three cells, 4 (given) + 3 (max of 1-3) + 5 (max of 1235) = 12

that leaves 12 to come from total of two cells 6789, 12356789

12 can come from 7+5, (Edited: 8+4, no 4 is already there), 9+3 combo

so, you said "12356789 cell cannot be 1, 2, 3 or 5, but I am finding that it can be (Edited: 5/3)

however it can surely not be 1-2.

Last edited by Elisabeth 11th Feb 2019 at 09:58HardCompletion time:27:50Used 'valid marks' Used 'auto remove' Used 'show wrong moves'I used to worry about all the "Easy" results. Now I look at the lower end of the first column (no aids) which is where I hope to be, to see what standard is set. I also note how long/short the no aids column is compared with the other one; thiis can be an indication. It might help if we had more than the 3 options, perhaps a "Mild" after "Easy"? If Gareth chose the level of difficulty we might still not agree! (Maybe people don't like to admit they found the puzzle hard?)

HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectcombo for 18: should come from 356789

where 8 or 9 is a must (from 17 across possibilities)

18 possibilities are: x189 x279 369 378 459 468 x567 x indicates not possible due to above conditions

that leaves 18's possibilities 369 378 459 468

in case of 4 (459 468), 4 can only be in the top cell, so lower two cells can have combo of 9+5 or 8+6, making it 56 in lower most cell

or in case of 3 (369 378), 3 can only be in this lower most cell,

so the combo for 24 in cell left of blue get reduced from 356789 to 356

and if 3 is in the lower most cell, the combo for upper two could be 6+9 or 7+8, meaning the top cell can only have 6 or 7

and of course 4 for the other possibilities mentioned above.

possibilities for 28 across are 4789 5689, now 5 is nowhere there so only 4789 can be there for 28 across, so 6 cannot be in 28 across,

so 18 down combo in case of 3 (369 378) cannot now be 369, and can only be 378

thus 18 down, can only have 47 in top, 89 in the middle and 356 in the bottom.

stuck again.

revised screenshot

HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectas 789 are only in a single cell, the possibilities above that mention more than 789 together are out. marked with x above

that leaves 24 = 4-1568 4-2369 x4-2567

also 2 is at only one place, putting 2 there will leave 5 at only one place

removing 5 from there will also remove 6 from there, that will leave 67 in a single cell, so that is also not possible.

that leaves 24 = 4-1568 4-2369

as 8 and 9 in only one cell, if they go there, that leaves that will leave 156 and 236, and 6 is left at only 1 place where it will go,

so, in 24 across: 356 cell must have 6

leaving 158 and 239 for remaining 3 cells of 24 across

now 18 down has 6 fixed at bottom, that leaves 8+4 or 9+3, but 3 is not there,

so it has to be 4 8 6 for 18 down

that brings 17 across to be 9 and 8,

that removes 9 from 33 down,

that removes 24 = 4-1568 x46-239 and leaves 24 = 46-158

that removes 7 and from 78 cell and leaves 8 there, and also fixes 1 and 5 as they are at one place only.

it goes a long way filling several cells.

Last edited by vsrawat 11th Feb 2019 at 13:00HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrect20 across: 3-1268 3-1457 3-2456 (3 is given)

19 across: 136-27 136-45 (just paired to show common)

13 across: 148 238 247

33 down (2-9 given): 1678 3478 3568

I can do iteration, no big deal in this small set, still would like to learn if any method is there.

ModerateBest completion time:8:50Time on first attempt:10:44Used 'remove' Used 'show wrong moves'* If you put 1, 2 or 3 in the first square, those four squares will be 1234 (i.e. sum to 10), which would require the 3rd square to be 9, which is not possible.

* Similarly, if the 3rd square is 1, 2 or 3 then the 1st square must be 9, which is also not possible.

Last edited by Pseudo 11th Feb 2019 at 13:17ModerateBest completion time:8:50Time on first attempt:10:44Used 'remove' Used 'show wrong moves'For me, "easy" means I didn't get stuck for very long, "moderate" means I got stuck for a moderate about of time and "hard" means I got stuck for a significant amount of time.

HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectbut yeah, previous "stuck" that I thought could be resolved by sudoku logic without going through iteration, but this one seems to have no clue and iteration is the only solution unless any member could find some way.

ModerateBest completion time:8:50Time on first attempt:10:44Used 'remove' Used 'show wrong moves'* 12349

* 12358

* 12367

* 12457

* 13456

The first two aren't possible (no 8 or 9). The fourth one also isn't possible (can't have both 5 and 7). Of the remaining two combinations, the 567 must go in the 1st and 3rd squares, while the other three squares must be 1-5. In fact, with either 12367 or 13456 the 1st square must be 6 and the 3rd square must be 5 or 7.

With either combination there must be a 3, and the only place for the 3 is the last square in that row.

HardBest completion time:3:09:01Time on first attempt:3:11:15Used 'show wrong moves' Used 'check puzzle' when incorrectbe it 12367 or 13456, in both cases, 5 or 7 are going to come in a single cell. that leaves 5-1236, 7-1346, and 6 can only be in a single cell.

filling that up gives 12 12 123 so 12 pairs out filling 3rd cell with 3.

going on, that leaves 33 down with only possibility of 678. that gives it the final push to crumble.

yes. this is also logic, no iteration required.

Now I can sleep happily otherwise it would have come in my nightmare.

Wonder where is Gareth. Tomorrow, seeing it, (someone) is going to be very angry with me for having wasted so many site resources. :-)

thanks.

Last edited by Chessmama 11th Feb 2019 at 21:36EasyCompletion time:9:11In response to:

"however, to the second part of logic, (left of the blue square), is still not clear

Now, as the 24 across has: 4 (given) and possibilities 6789, 12356789, 13, 1235"

You can solve it other ways (as Pseudo suggests), but my logic is as follows:

*IF* the second of the cell possibilities you give (12356789) had any of the values 1, 2, 3 or 5, then that cell plus the two following cells could *maximally* be 2 + 3 + 5. Add that to the 4 you already have gives 14, and then the last cell (6789) would need to be 10 to make 24.

*Therefore* this cell *cannot* be any of 1, 2, 3 or 5.

It is a more advanced logic than simple exclusion: it's exclusion based on a maximum sum of possible values. But you don't have to iterate.

EDIT:

You can also see it another way with your logic:

"total of three cells, 4 (given) + 3 (max of 1-3) + 5 (max of 1235) = 12

that leaves 12 to come from total of two cells 6789, 12356789

12 can come from 7+5, (Edited: 8+4, no 4 is already there), 9+3 combo"

In order to get yourself in a position to make a two cell 12-combo, you had to "assign" 3 and 5 to two unknown cells. That then stops both of your possible combinations of 7+5 and 9+3 for the other two cells. Therefore some part of the assumption that the two cells you "guessed" are 3 and 5 must be wrong.

HardCompletion time:26:26:17Used 'show wrong moves'Add new commentAdd a commentYour comment:You can however view other players' statistics and comments in the tables above.

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