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The Fugitive
Joined: 28 Aug 2005
Posts: 10
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 Posted: Sun Oct 23, 2005 5:01 am Post subject: Puzzle Grading |
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Can anyone explain how the author grades his puzzles? I ask because I've discovered several fairly obvious inconsistencies.
Example 1:
8th August - Easy
23rd August - Medium
Both contain 26 clues, 44 naked singlets and 11 hidden singlets. Both are easy.
Example 2:
8th April - Hard
3rd October - Very Hard
Both contain 26 clues, 43 naked singlets, 12 hidden singlets, 1 naked couplet and 1 hidden couplet. Both are medium-hard.
Example 3:
25th May - Medium
15th August - Easy
Both contain 26 clues, 53 naked singlets and 2 hidden singlets. Both are easy.
Example 4:
13th May - Medium
Contains 26 clues and 55 naked singlets. Should be rated easy. Compare with 15th August above.
Example 5:
22nd January - Medium
26th January - Medium
27th January - Medium
15th February - Medium
22nd February - Medium
26th February - Medium
1st March - Medium
2nd March - Medium
15th March - Medium
23rd March - Medium
4th April - Medium
15th April - Medium
18th April - Medium
25th April - Medium
29th April - Medium
26th May - Medium
1st June - Medium
23rd June - Medium
All the above have 28 clues and 53 naked singlets. All should be rated easy. Compare with 15th August above.
Example 6:
28th May - Very Hard
24th August - Hard
Both contain 30 clues, 36 naked singlets, 15 hidden singlets, 1 hidden couplet and 1 group interaction. The former has 1 naked couplet while the latter has 2. Both are hard at best.
During my analysis of these puzzles I noticed that every puzzle (including the so called Very Hard ones) can all be solved with a combination of naked subsets (singlets, couplets and triplets), hidden subsets (singlets couplets and triplets) and group interaction (column/row and block interactions). Although many can also be solved using certain combinations of the other logical methods (e.g., block interaction, x-wing, swordfish, xy-wing, colouring, forcing chains and Nishio), they are not required to solve any of the Daily SuDokus (so far at least).
Personally, I think the ratings are far too high for a lot of these puzzles. Indeed, I don't regard any of them as very hard at all. A very hard puzzle should have at the very least an x-wing, a swordfish or an xy-wing -- and preferably any two of them.
Anyhow, here's a "Diabolical" grade puzzle to amuse yourselves with :-)
009006000
750000026
031700008
800070050
000105000
060030007
400007610
520000039
000500800 |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sun Oct 23, 2005 6:41 am Post subject: |
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Hi,
After a nacked pair, X-wing and XW-wing I got to:
| Code: | 2 48 9 38 45 6 1345 7 13
7 5 48 39 19 18 34 2 6
6 3 1 7 45 2 45 9 8
8 14 24 6 7 9 123 5 13
3 79 27 1 8 5 29 6 4
19 6 5 2 3 4 19 8 7
4 89 3 89 2 7 6 1 5
5 2 68 4 16 18 7 3 9
19 179 67 5 69 3 8 4 2 |
And takeing a look at the statistics:
Nr=1 occurs=13 still to set=6 times
Nr=2 occurs=4 still to set=2 times
Nr=3 occurs=7 still to set=3 times
Nr=4 occurs=9 still to set=4 times
Nr=5 occurs=4 still to set=2 times
Nr=6 occurs=4 still to set=2 times
Nr=7 occurs=4 still to set=2 times
Nr=8 occurs=8 still to set=4 times
Nr=9 occurs=11 still to set=5 times
I can see that I have to follow chains of double digits because there are a lot of them.
Asuming: 4 in r4c2 and following the chain I will get to a contradiction
4 in r4c2
2 in r4c3
7 in r5c3
9 in r5c2
8 in r7c2
9 in r7c4
3 in r2c4
8 in r1c4
4 in r1c2
and now setting the 1 in r4c2 will solve it.
We call this technique "Nishio" (if I am not wrong).
see u,
P.S. could be that there is a "shorter" chain that leads to a contradiction. |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sun Oct 23, 2005 6:50 am Post subject: |
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Hi,
Here a shorter contradiction chain:
9 in r9c1
6 in r9c5
7 in r9c3
2 in c5r3
4 in r4c3
1 in r4c2
9 in r6c1
see u, |
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samgj
Site Admin
Joined: 17 Jul 2005
Posts: 106
Location: Cambridge
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| Posted: Sun Oct 23, 2005 9:32 pm Post subject: Re: Puzzle Grading |
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| The Fugitive wrote: | Can anyone explain how the author grades his puzzles? I ask because I've discovered several fairly obvious inconsistencies.
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A dull but straightforward reason -- the grading has evolved a little over time.
| Quote: | Although many can also be solved using certain combinations of the other logical methods (e.g., block interaction, x-wing, swordfish, xy-wing, colouring, forcing chains and Nishio), they are not required to solve any of the Daily SuDokus (so far at least).
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Indeed -- this is semi intentional, but probably won't remain the case. x-wing and the like have not yet been *required*, but various posts here show that they often form part of a route used by the solver.
Sam |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Mon Oct 24, 2005 1:46 am Post subject: Puzzle Grading |
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Fugitive
I share your scepticism of the published gradings. I suspect they are programmed. If so, they are not programmed well.
Prompted by your posting, I ran my own tiny archive. The score is
0 if the puzzle is solved by trivial means
1 is added each time it is necessary to refer to possibilities not entered on the grid
10 is added each time a pair etc needs to be identified
100 is added for each advanced technique.
It is easy to criticise the method but I find programming the way puzzles are attacked by real people very time consuming.
The results are:
Difficult: 1, 1, 1, 1, 0, 0, 0, 0, 0
Fiendish: 8, 6, 16, 25, 2, 1, 4, 2, 1, 4, 1, 11, 3, 3
Ultra Fiendish: 37, 34
Superior: 28, 131, 120, 144, 14
Hard: 31, 18, 3, 1, 4, 10, 1
Very Hard: 17, 30, 16, 19, 11, 14, 16, 13, 60, 5, 4, 18, 40, 37
They suggest to me that:
(a) You are right.
(b) There is a tendency for The Times’ puzzles to get easier (the results are in chronological order of publication).
(c) The target for puzzles rated at about 40 should be 15 – 30 minutes (the winner of the recent Times competition took 13).
Your "diabolical" rated 1, 115, the extra 1,000 representing the need to guess. Incidentally, I think you are right in taking "Nishio" to be synonymous with "guess."
This is not to support the publication of an excess of hard puzzles according to your definition: my preference is for about 20 minutes and no guessing or writing down lots of numbers outside the grid. Call me easy.
Steve R. |
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The Fugitive
Joined: 28 Aug 2005
Posts: 10
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| Posted: Mon Oct 24, 2005 7:30 am Post subject: |
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Thanks to all who responded.
To Sam:
OK, I now understand why some of your earlier puzzles have such odd gradings. But yesterday's medium (23rd October) had 28 clues and 53 naked singlets. How can that possibly be medium? It's quite clearly an easy. How do you arrive at a medium rating for such a trivial puzzle?
Nonetheless, I'm glad to hear you will be employing more advanced techniques as standard in future. Can we expect a "Challenging" rating? 
To someone_somewhere:
Instead of using the XY-Wing (leading to the Naked Couplet and Nishio solutions), you can also use Forced Chains (XY-Wings are really just short Forced Chains). For example:
| Code: |
CHAIN #1: R5C2, R7C2, R1C2 and R4C2
7->9, 9->8, 8->4 and 4->1
CHAIN #2: R5C2, R5C3, R4C3 and R4C2
9->7, 7->2, 2->4 and 4->1
CONCLUSION: R4C2 must be 1
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From there it's Naked Singlets all the way. No need for Nishio.
To Steve:
I agree that it can sometimes be difficult to grade puzzles according to how a human might solve them. And I'm glad you agree with me that the Daily SuDoku gradings are a bit off. But to my mind at least, the grade should reflect the logic that a computer program *will* employ, not the logic that a human *might* employ.
However, we'll have to agree to disagree on Nishio because I don't regard it as guesswork. All techniques -- even Brute Force -- will either prove a value exists in a particular cell, or will prove it does not. In that respect they are all logical. To me, a "guess" is putting a value into a cell without even proving if it can exist in that cell. If it turns out it can, and it is correctly placed, then it is pure "luck" that got you there, not logic. |
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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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| Posted: Mon Oct 24, 2005 11:42 pm Post subject: |
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| The Fugitive wrote: | Thanks to all who responded.
To Sam:
OK, I now understand why some of your earlier puzzles have such odd gradings. But yesterday's medium (23rd October) had 28 clues and 53 naked singlets. How can that possibly be medium? It's quite clearly an easy. How do you arrive at a medium rating for such a trivial puzzle?
Nonetheless, I'm glad to hear you will be employing more advanced techniques as standard in future. Can we expect a "Challenging" rating?
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Hi Fugitive,
It's interesting to see how you categorize the puzzles and grade them. I thought the Oct. 23 puzzle took a long time even though there weren't any difficult techniques required. My paper had a lot of numbers on it! I would be curious as to how you solved it. It must be a challenge for Sam to cater to beginners as well as advanced puzzle solvers. This is a great website. |
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The Fugitive
Joined: 28 Aug 2005
Posts: 10
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| Posted: Wed Oct 26, 2005 4:47 am Post subject: |
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| Quote: | | It's interesting to see how you categorize the puzzles and grade them. I thought the Oct. 23 puzzle took a long time even though there weren't any difficult techniques required. My paper had a lot of numbers on it! I would be curious as to how you solved it. It must be a challenge for Sam to cater to beginners as well as advanced puzzle solvers. This is a great website. |
It's difficult to compare the ways in which humans solve SuDoku. This one in particular has many routes to a the final solution.
For instance, you might start with the X-Wing in R6 and R9, which eliminates the 4 from R1C8, R2C8, R2C9 and R3C8.
Or you might start with the block interaction between B7, C1 and C2, eliminating 7 from R7C2, R8C1, R8C2 and R9C2.
Or the group interaction between B8 and R8, eliminating 2 from R8C1 and R8C2, and 8 from R8C8.
Or the hidden couplet in C3, eliminating 9 from R2C3.
Or the naked couplet in B8 eliminating 6 from R8C4, R8C5 and R8C6, and 7 from R8C4.
Or the hidden singlet in R7, forcing R7C8 to be 8.
Or the naked singlet forcing R2C5 to be c.
These are just examples -- there are certainly more ways to start this puzzle, each of which could lead you to any number of alternate routes.
However the fact remains that, at any one time, there will always be a naked singlet. There are no tricks or breakthroughs that *have* to be performed, so long as you can find those naked singlets -- the only really difficult part of this puzzle. Even if you use alternative methods to reveal other naked singlets they simply lead to a more speedy solution, not a more difficult one.
So, how would I have gone about it? That's difficult to answer because I'm a self-proclaimed expert. I guess you really need to ask yourself why you found it so difficult. However, I'll try and look at the puzzle as a beginner might have looked at it.
Firstly, I try to avoid writing down large lists of candidates in each cell as it makes for a messy solution, and is generally just a waste of time at the start of any game. Had I actually done so, the solution would have been immediately obvious (hence I give it an easy rating).
So I start by first looking at column 5. It's only got 3 unknowns, 268. R2C5 can't be 2 or 8 since they're already in the same block, so it must be 6. Bingo! The first naked singlet. I then pencil mark the remaining 28 in both R5C5 and R8C5 to remind me later.
I then look at column 3 with 4 unknowns, 1789. R7C3 can't be 1 or 8 since they're both in the same block, and 9 is in the same row, so it must be 7. The second naked singlet. This in turn reveals R9C3 to be a 9, leaving 18 in both R2C3 and R5C3, which I pencil in.
Looking at row 7, I can now see it requires 3568. R7C6 must be 6 and R7C9 must be 5. This in turn reveal 3 in R7C2 and 8 in R7C8. Row 7 is now complete.
At this point, block 7 looks the best contender (with only 3 unknowns), but it has no naked singlets. I know there is one in block 8 but it's not immediately obvious unless you happen to look for one there. So at this point it's quite tempting to look at alternative routes. There are certainly lots to choose from, but let's stick to beginner tactics.
If I look at the 6's, I see that the only place a 6 may be placed in block 5 is in R5C4. A hidden singlet. Similarly with the 9's. The only place left in row 5 is R5C6. The 5's also leave R5C7 as yet another hidden singlet, which leads to R6C4. And a 3 in R5C1. At this point I'm left with only 3 unknowns in row 5, so I may as well pencil them in. I already known R5C3 is 18 and R5C5 is 28. R5C9 is 12.
If I turn my attention to column 7, I see it has 3 unknowns, 346. The 3 can't be in B3, so it must be in B9, another hidden singlet. I can pencil in 46 in R1C7 and R3C7.
The 2's show that R9C2 is yet another hidden singlet. This reveals another naked singlet in row 6. R6C2 can't be 234 so it must be 1. Therefore R5C3 is 8, R4C2 is 7 and R4C1 is 2. Block 4 is now complete.
Further naked singlets reveal R5C5 must be 2 and R5C9 a 1. R2C3 is also a 1.
At this stage there are now 4 naked singlets, including the original 7 in R9C4 that I might have missed up until this point. The 3 in R6C6 reveals two more naked singlets and from here on out it's reasonably plain sailing.
If by this time I hadn't found either a naked singlet or a hidden singlet, then I'd have started penciling in the remaining candidates -- of which there would be relatively few. Block 3 would have been a little messy, but the solution would have revealed itself in short time because there was always a naked singlet available.
Now, although I used quite a few hidden singlets in favour of the naked singlet, there was no pressing need to use them. There was always a naked singlet available which would have been found eventually with a more systematic approach. But that alone does not merit a medium rating. Indeed, it should be entirely expected that a complete beginner learn the secrets of the hidden singlet and the naked couplet within their first few easy puzzles -- skills which can be further honed with medium puzzles where the more advanced skills can be honed. |
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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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| Posted: Wed Oct 26, 2005 9:38 pm Post subject: |
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| Thanks Fugitive for a very thorough answer to my question. I've been solving these for about a month and have learned to solve even the very hard ones on this website. But I know I tend to be slow and methodical. The Oct23 puzzle took me a long time because I had too many numbers in the boxes. I need to learn shortcuts now that I know the basics. I will start the puzzle over again and see if I can do it faster. Can you tell me what block interaction is? Thanks. |
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Guest
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| Posted: Thu Oct 27, 2005 3:27 am Post subject: Block Interaction |
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| Quote: | | Thanks Fugitive for a very thorough answer to my question. |
You're welcome.
| Quote: | | Can you tell me what block interaction is? Thanks. |
More commonly referred to as the block/block interaction. You're quite probably familiar with it already as most people discover it without realising it has a name, but let's go over it again for everyone's benefit.
When a candidate is known to reside in only 2 rows (or columns) of 2 interacting blocks, then the candidate MUST appear in the only remaining block in the only remaining row (or column), and can therefore be eliminated from the other two rows (or columns) in that block.
It's much simpler than it sounds. Ultimately, we can eliminate a candidate from up to 6 cells with this method.
At the start of the October 23rd puzzle, I mentioned there was a block interaction between B7, C1 and C2, eliminating candidate 7. Let's explain that in more detail...
If we look at the 7 in block 3, we see that it is on row 2. This means R2C3 can't be 7. And since R1C3 and R3C3 are already known, the 7 must therefore reside in either column 1 or column 2 of block 1.
If we now look at the 7 in block 6, we can see that it is on row 5. This means R5C3 cannot be 7. And since R4C3 and R6C3 are known, the 7 must reside in columns 1 or 2 of block 4.
Since column 3 MUST have a 7 somewhere and we know it isn't in blocks 1 or 4, it must therefore be in block 7. And since we know it must be either R7C3 or R9C3 (R8C3 being known), then it cannot possibly be R7C2, R8C1, R8C2 or R9C2 and can therefore be eliminated from all of these cells (the remaining cells in this block already being known).
All these cells reside in block 7, column 1 and column 2 -- hence the term block interaction between B7, C1 and C2.
Referring to the block/block interaction definition, we'd say that B1 and B4 are interacting with C1 and C2. However, I find this a bit of a mouthful so I prefer to refer directly to the block and columns (or rows) where the eliminations will actually occur, rather than where the interactions occur. The interactive blocks themselves are implied through the column (or row) references, of course.
Note that I also refer to the long-winded "Row/Column and Block Interactions" simply as "Group Interactions". A group is any row, column or block, but the interactions only occur between a block and either a row or a column intersecting the block.
All my other definitions are fairly standard, although I use the terms singlets, couplets, triplets, quadruplets, quintuplets, sextuplets, septuplets and octets when referring to a subset of a group. Obviously, there can never be a subset of 9 (enead) within a group, but it's rare to go beyond a quadruplet (e.g., a hidden octet would have to imply a naked singlet in the ninth cell).
Hope that clarifies things for everyone. |
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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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| Posted: Thu Oct 27, 2005 3:16 pm Post subject: |
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That was a good explanation and I was able to follow it. I didn't notice block interaction with the 7's when I did the puzzle before, although I figured it out in a round about way. It's best to learn how to solve these puzzles by doing them and then asking for help when stuck rather than reading about how to solve them and then trying to solve them. You can learn a lot just by trying. But then you get to the point where I am where you know there must be a faster, easier and cleaner way to solve them. Thanks.  |
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Mars Plastic
Guest
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| Posted: Wed Nov 02, 2005 5:24 pm Post subject: Puzzle Grading |
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Fugitive and Steve R, I agree that the ratings seem inconsistent and higher than they should be, not only on dailysudoku but everywhere.
Steve R, where are the puzzles you rated, I'd like to try some of the higher rated ones.
I've only been doing sudokus for about a month, I don't know all the names of techniques and I haven't been considering myself an expert but I really didn't find Fugitive's "diabolical" one that difficult, it was more challenging than the average one for sure. When I entered it in the draw tool, it said it couldn't solve the puzzle without guessing, Steve R said the same thing. However, I didn't need to guess and only "got stuck" once.
I did encounter one from dailysudoku last month some time that stumped me for a bit, but in general where do you go to get more challenging ones? |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Wed Nov 02, 2005 6:53 pm Post subject: Here's a hard one |
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| Mars Plastic wrote: | | I did encounter one from dailysudoku last month some time that stumped me for a bit, but in general where do you go to get more challenging ones? |
You might try the collection on the KrazyDad blog site -- I've tried a few of his "super tough" puzzles, and they're not easy.
See if you can figure this one out. If you come up with a way to solve it, please post an explanation back in this forum.
| Code: | 6...4...3
.1.....7.
..5...8..
...5.2...
3...9...2
...1.3...
..8...9..
.7.....5.
2...3...4 |
Have a great day! dcb |
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Glassman
Joined: 21 Oct 2005
Posts: 50
Location: England
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| Posted: Thu Nov 03, 2005 2:22 pm Post subject: |
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David — this one has me completely baffled so far. I saw the pair in box 9 immediately, and tried to make headway with r5 and r1 or c1, but nothing was emerging.
Then I reasoned that if I could somehow link one of the last three with box 9, then I could make some progress. No luck there.
I then tried to find a layout of box 5 that would help and connect through to box 9. The most promising were either a 6 or a 7 at r5c6, but I have yet to work these through and look at other possibilities for box 5. So often the most unpromising looking option can prove productive.
Nothing interacts. I believe the key to the puzzle could well lie in choosing one option for box 5 that results in a productive interaction between r1/c1, box 9, and that option for box 5.
Some might consider this guesswork. I don't believe that can apply to a reasonably finite number of possibilities, in this case just four possibilities for this particular cell. I prefer to consider it a logical approach to experimentation.
... and, David, are we looking for a solution, or do we have to go further and prove uniqueness?
Glassman 8)
... and a 7 at r5c6 yields some useful looking possibilities in boxes 3, 6 and 9. This looks to me like xy-wing country, not a technique I either properly understand or have yet mastered, although I must use it as I solve difficult puzzles. A problem (or, perhaps, a benefit?) of learning to solve Sudoku in isolation. |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Nov 04, 2005 12:58 am Post subject: Re: Here's a hard one |
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| Glassman wrote: | | ... and, David, are we looking for a solution, or do we have to go further and prove uniqueness? |
Well, a solution would be good. A demonstration of uniqueness would be better. :)
Incidentally, I didn't introduce this puzzle to the forum -- someone_somewhere did, about a week ago. You can read his post right here. My initial response to that post is right here. So far all we've managed to figure out is that there's a "swordfish" in the "5"s, that there's a hidden pair (the one you spotted in the lower right 3x3 box), and that the "4" in row 5 must appear either at r5c4 or at r5c6.
I did enough work with the "2"s and the "3"s to be sure they won't give one enough information to solve the puzzle. One needs to place a "7" and probably a "4" before having enough information to get to a solution. The most promising line of attack I've been able to identify involves looking for situations where one of these critical digits can only appear in one of two places.
Sorry I'm not more help, but this is a very tough puzzle. I do have good reason to believe the solution is in fact unique -- our friend in Munich doesn't like to waste his time on invalid Sudoku puzzles. dcb |
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Mars Plastic
Guest
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| Posted: Fri Nov 04, 2005 2:10 am Post subject: Super hard one |
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I've got a solution for it. It was pretty tough. I didn't really like it being that hard. I guess it's a case of "be careful what you wish for".
I'm trying some of the super tough ones from that site you pointed out, they're more fun I think.
Anyway, here's the solution I found:
687941523
412358679
935627841
794562318
351894762
826173495
548216937
173489256
269735184
Sorry I can't give a play-by-play on how to solve it. Basically I reduced the 3's and 5's down as far as possible. Then determined that there were only two sets of 3's possible and two sets of 5's possible.
After that I couldn't do anything other than do trial guesses with the 4 possibilities. This one worked.
I don't like having to guess. |
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Guest
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| Posted: Sat Nov 05, 2005 9:10 am Post subject: The Big "X" puzzle |
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Hi,
Back to the Big "X" puzzle:
6 - - - 4 - - - 3
- 1 - - - - - 7 -
- - 5 - - - 8 - -
- - - 5 - 2 - - -
3 - - - 9 - - - 2
- - - 1 - 3 - - -
- - 8 - - - 9 - -
- 7 - - - - - 5 -
2 - - - 3 - - - 4
after finding the Swordfish for digit 5, we can eliminate six digits of 5.
We have than this Candidate table:
6 289 279 2789 4 15789 125 129 3
489 1 2349 23689 2568 689 246 7 569
479 2349 5 23679 1267 1679 8 12469 169
14789 4689 14679 5 678 2 13467 134689 16789
3 568 167 4678 9 4678 1567 168 2
45789 24689 24679 1 678 3 467 4689 56789
145 346 8 2467 12567 1467 9 23 167
149 7 13469 24689 1268 14689 23 5 168
2 569 169 6789 3 156789 167 168 4
Taking a look at the statistics for it:
Nr=1 occurs=28 still to set=7 times
Nr=2 occurs=21 still to set=6 times
Nr=3 occurs=10 still to set=5 times
Nr=4 occurs=26 still to set=7 times
Nr=5 occurs=12 still to set=6 times
Nr=6 occurs=44 still to set=8 times
Nr=7 occurs=28 still to set=7 times
Nr=8 occurs=28 still to set=7 times
Nr=9 occurs=34 still to set=7 times
it is not a bad idea to start working on the Nr=3 and Nr=5.
Yup, there are 2 possible configuration for each.
Only trying the 2x2 possibilities, we can get to the solution.
This is THANKS to Mars Plastic.
I could not find a better way than his one.
see u, |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sat Nov 05, 2005 9:11 am Post subject: |
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Hi,
The previous post was from me, I confess.
see u, |
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