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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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 Posted: Sun Feb 19, 2006 4:20 pm Post subject: A "4-star constellation" example |
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Here's a pretty good example of a "4-star constellation." This is from the David Bodycombe puzzle for Saturday, Feb 18, rated six stars.
After doing the obvious stuff I arrived at this position.
| Code: | 36 7 4 5 38 136 18 2 9
9 8 356 16 7 2 145 134 36
2 35 1 9 348 346 458 7 368
368 34 36 2 5 16 9 134 7
7 1 9 8 34 34 6 5 2
3568 345 2 16 9 7 148 134 38
35 6 35 7 1 9 2 8 4
4 2 8 3 6 5 7 9 1
1 9 7 4 2 8 3 6 5 |
Notice that the "6" in row 6 must either occur at r6c1 or at r6c4. Suppose that r6c1 = 6. Then we have a nice double-implication chain, or "4-star constellation":
A. r6c1 = 6 ==> r6c4 = 1
B. r6c1 = 6 ==> r2c3 = 6 ==> r2c4 = 1
Since we can't have two "1"s in column 4 the "6" can't go at r6c1, and we must have r6c4 = 6. This isn't enough to break the puzzle wide open, but it will allow one to make quite a bit of progress, leading to an easily resolved "BUG" pattern. dcb
PS There's more than one way to look at this pattern. For instance, consider this chain of reasoning:
r6c1 = 6 ==> r6c4 = 1 ==> r2c4 = 6 ==> {3, 5} pair in column 3 ==> r4c3 = 6
and now we have two "6"s in the middle left 3x3 box. I've noticed this in a lot of Sudoku puzzles -- with a chain like this there are usually two or three different ways to "see" the contradiction. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Feb 20, 2006 2:40 am Post subject: Bodycombe Puzzles |
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This Saturday puzzle is very similar to the Friday puzzle I posted previously. The starting point is:
| Code: |
Puzzle: DB021806
+-------+-------+-------+
| . . 4 | 5 . . | . . 9 |
| . 8 . | . 7 2 | . . . |
| 2 . 1 | 9 . . | . 7 . |
+-------+-------+-------+
| . . . | 2 5 . | 9 . . |
| . 1 9 | . . . | 6 5 . |
| . . 2 | . 9 7 | . . . |
+-------+-------+-------+
| . 6 . | . . 9 | 2 . 4 |
| . . . | 3 6 . | . 9 . |
| 1 . . | . . 8 | 3 . . |
+-------+-------+-------+
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In the position given by David, may I point out the Unique rectangle in C5 and C6? It is
The argument goes as follows: The pattern of possibilities
34 34
34 34
does not allow a unique solution. However, one of the squares in R3 must be <4>, for there are no other possible values <4> in their block. Therefore neither R3C5 or R3C6 can be <3>.
More to come,
Keith |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Feb 20, 2006 2:51 am Post subject: |
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The Sunday David Bodycombe puzzle is hidden in the TV book that comes with the Sunday newspaper (in my case, the Detroit Free Press). Like the Friday puzzle, it has become much more interesting and challenging in the last few weeks. Here is today's puzzle (which is rated 5 stars):
| Code: |
Puzzle: DB021906
+-------+-------+-------+
| 3 . . | . . . | 6 7 . |
| . 5 . | 6 . . | 9 3 . |
| . . . | 5 7 . | . . 4 |
+-------+-------+-------+
| . . . | 9 . 4 | . . . |
| . . 6 | . . . | 8 . . |
| . . . | 3 . 7 | . . . |
+-------+-------+-------+
| 8 . . | . 9 6 | . . . |
| . 6 4 | . . 5 | . 2 . |
| . 1 3 | . . . | . . 6 |
+-------+-------+-------+
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The only clue I'll give, for now, is that there is a very good reason I have posted this puzzle in this thread!!
Best wishes,
Keith |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Mon Feb 20, 2006 4:21 pm Post subject: DB021806 again |
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Thanks for the Sunday puzzle, Keith. I only get these in the paper on Saturdays.
On the Saturday puzzle, there's also a nice "coloring" solution available. To get to this (slightly different) position one must recognize that the "8" in r4c1 is unique in row 4 (starting from the position posted previously).
| Code: | 36+ 7 4 5 38 136- 18 2 9
9 8 356 16 7 2 145 134 36
2 35 1 9 348 346 458 7 368
8 34 36= 2 5 16~ 9 134 7
7 1 9 8 34 34 6 5 2
356 345 2 16 9 7 148 134 38
35 6 35 7 1 9 2 8 4
4 2 8 3 6 5 7 9 1
1 9 7 4 2 8 3 6 5 |
There are only two ways to place a "6" in rows 1 & 4.
r1c6 = 6 ==> r4c6 <> 6 ==> r4c3 = 6
r1c6 <> 6 ==> r1c1 = 6
So either r1c1 = 6, or else r4c3 = 6 ... either way there cannot be a "6" at r2c3. So the "6" in column 3 must appear at r4c3 -- this is essentially the same conclusion forced by the "4-star constellation" in my previous post. dcb
PS The "BUG" pattern that eventually emerges from this puzzle can also be resolved by "coloring" on the digit "3". |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Mon Feb 20, 2006 6:21 pm Post subject: Cloudy Skies |
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If I remember correctly someone_somwhere’s four-star constellation took the form of a rectangle containing four cells:
…..XY…………..XZetc…
………………………….......
…..Yetc………..YZetc…
………………………….......
with only two places for Y in the bottom row and only two places for Z in the right hand column. The alpha star (top left) admits only two possibilities, X and Y. X can then be eliminated from the cell containing XZetc.
I can’t spot this pattern in David’s example of yesterday, 19 February, which started this thread. Perhaps I am looking in the wrong part of the sky. Perhaps the constellation has developed beyond its original form. If the latter, would it not be better to qualify the term “four-star constellation” when using the development so as to minimise confusion?
Steve |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Mon Feb 20, 2006 10:50 pm Post subject: What's in a name? |
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| Steve R wrote: | | I can’t spot this pattern in David’s example of yesterday, 19 February, which started this thread. |
Well, I'm not real sure myself, Steve. 
| Someone_somewhere wrote: | Here the definition of a "4 stars constellation" similar to the "5 ...":
1. it involves 4 stars (meaning 4 cells)
2. it has 2 starting digits, in the so called starting cell. We always find this alpha star, the bright one to start with.
3. we follow 2 chains, or pathes, to other stars/cells, forcing a digit TO BE or NOT TO BE in this cell.
4. the overall result is a chain of 4 stars, with an ending in some result, as to set or exclude a digit in one of the stars, or a contradiction. |
I guess, to comply with #2, I should start the argument from r6c4:
A. r6c4 = 1 ==> r2c4 = 6 ==> r4c3 = 6
B. r6c4 = 1 ==> r6c1 = 6
So there I have the chains involving four cells and ending in a contradiction. It sort of looks like what s_s was talking about , to me. But hey, a rose by any other name would smell as sweet.
BTW, where is someone_somewhere, anyway? We haven't heard from him in quite a while. dcb |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Feb 20, 2006 11:39 pm Post subject: What pattern is this? |
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I have found this pattern only once or twice:
| Code: |
AB . . . AC
. . . . .
. . . . .
BD . . . CD
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Values in other squares do not matter. If the top left square is A, follow a clockwise cycle to give
| Code: |
A . . . C
. . . . .
. . . . .
B . . . D
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If the top left square is B, follow a counter-clockwise cycle to give
| Code: |
B . . . A
. . . . .
. . . . .
D . . . C
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Either way, you can eliminate all other possibilities of A in the top row, C in the right column, D in the bottom row, and B in the left column.
These situations can involve more than four squares. Personally, I have been calling them "bidirectional cycles", where "cycle" is a closed chain. If you find one, the situation is similar to an X-wing, where one of two adjacent "corners" must be some particular value.
You can try to find one of these in the Sunday puzzle posted above! (It has 5 squares.)
Keith |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Tue Feb 21, 2006 5:39 pm Post subject: What's in a name? |
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David
Thanks for the response. I had forgotten the wise words you quote from someone_somewhere, taking them as a general description of what was going on rather than a definition. Perhaps he will say more when he surfaces again.
By coincidence both approaches crop up at the same position in the six stars puzzle you posted on Friday, 17th.
| Code: | 236 59 23 8 129 7 159 156 4
1 47 24 6 29 5 79 3 8
67 8 59 3 19 4 2 156 179
259 6 7 1 4 3 59 8 29
59 15 8 7 6 2 13459 45* 139
24 3 124 5 8 9 6 7 12
3457 145 6 2 357 8 14 9 37
347 2 134 9 37 6 8 14 5
8 579 59 4 35 1 37 2 6 |
As well as the widely defined variety of constellation which you used to put 9 in r5c1, there is a narrowly defined one with alpha star r5c2, using only two 1s in row 7 and only two 4s in column 7. This narrow version eliminates 5 from r5c7.
Making both these moves gives a (finned) X-wing for 5 in rows 1 and 5. So r3c9 is not 5 and the only place for 5 in row 3 is r3c3.
The solution of the puzzle is then relatively straightforward, with the emphasis on “relatively.”
Steve |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Mar 03, 2006 4:36 pm Post subject: Another "5 Star Constellation" example |
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Here's a good example of a "5-star constellation." This is the "Nightmare" puzzle for Friday, 3 March, 2006.
| Code: | . 1 . . . . . 7 4
. 5 . . 8 . 3 . .
2 . . . . 9 . . .
. . 6 5 . . . . .
7 4 . . 1 . . 3 8
. . . . . 4 1 . .
. . . 2 . . . . 3
. . 4 . 7 . . 8 .
6 3 . . . . . 9 . |
After resolving 18 cells and making a few eliminations via simple coloring I found a "5-Star Constellation" that allowed me to set the value at r6c8. The rest of the puzzle was duck soup. dcb |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Mar 03, 2006 9:14 pm Post subject: |
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| Quote: | After doing the obvious stuff I arrived at this position.
| Code: |
36 7 4 5 38 136 18 2 9
9 8 356 16 7 2 145 134 36
2 35 1 9 348 346 458 7 368
368 34 36 2 5 16 9 134 7
7 1 9 8 34 34 6 5 2
3568 345 2 16 9 7 148 134 38
35 6 35 7 1 9 2 8 4
4 2 8 3 6 5 7 9 1
1 9 7 4 2 8 3 6 5 |
Notice that the "6" in row 6 must either occur at r6c1 or at r6c4. Suppose that r6c1 = 6. Then we have a nice double-implication chain, or "4-star constellation":
A. r6c1 = 6 ==> r6c4 = 1
B. r6c1 = 6 ==> r2c3 = 6 ==> r2c4 = 1 |
David (or anyone),
Forgive me, but I just can't understand "B", above. How does one get to r2c3 being a "6"? It seems to me that it has to be a "5" because the "6" in r6c1 forces a "3" in r1c1 and a "1" in r6c4, which forces a "6" in r2c4, thus leaving a "5" as the only choice for r2c3. I'm obviously missing something here.  |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Mar 03, 2006 11:41 pm Post subject: |
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Marty,
I agree with you. David must have mistyped something. However, his conclusion is valid:
R6C1 = 6 ==> R1C1 = 3 and R4C3 = 3 ==> R7C1 = R7C3 = 5.
The mystery to me is how you decide to test the value <6> in R6C1?
Keith |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sat Mar 04, 2006 12:56 am Post subject: "Constellations" are mysterious ... |
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| Marty R wrote: | David (or anyone),
Forgive me, but I just can't understand "B", above. How does one get to r2c3 being a "6"? It seems to me that it has to be a "5" because the "6" in r6c1 forces a "3" in r1c1 and a "1" in r6c4, which forces a "6" in r2c4, thus leaving a "5" as the only choice for r2c3. I'm obviously missing something here. |
Hi, Marty! Thanks for asking the question. It's a good one.
The really interesting thing about the "n-star constellations" or "double-implication chains" is that they depend on both positional logic and on value-oriented logic.
In the case you're asking about, I've used positional logic.
| Code: | 36 7 4 5 38 136 18 2 9
9 8 356 16 7 2 145 134 36
2 35 1 9 348 346 458 7 368
368 34 36 2 5 16 9 134 7
7 1 9 8 34 34 6 5 2
3568 345 2 16 9 7 148 134 38
35 6 35 7 1 9 2 8 4
4 2 8 3 6 5 7 9 1
1 9 7 4 2 8 3 6 5 |
-- There has to be a "6" in the top left 3x3 box somewhere.
-- Only two slots are open for a "6" in that 3x3 box -- at r1c1 & at r2c3.
-- If r6c1 = 6 there can't be a "6" at r1c1, so then there would have to be a "6" at r2c3.
I abbreviated this logic by writing "r6c1 = 6 ==> r2c3 = 6" ... I thought that would be clear enough, but I'm happy to provide this clarification. The notion of "positional logic" took me a little while to catch onto.
You'll notice also that this "positional logic" conclusion is at variance with the result you obtained using "value-oriented logic." That's not very surprising, since I was reasoning toward a contradiction. In other words, one might just as well argue that assuming "r5c1 = 6" forces one to place two different values at r2c3 -- a "5" using value-oriented logic around the loop r6c1 - r6c4 - r2c4, and a "6" by tracing a different path, from r6c1 - r1c1 - r3c2 - r2c3. It's all the same -- a contradiction is a contradiction. Either approach yields the result r6c1 <> 6.
| Keith wrote: | I agree with you. David must have mistyped something. However, his conclusion is valid:
R6C1 = 6 ==> R1C1 = 3 and R4C3 = 3 ==> R7C1 = R7C3 = 5.y
The mystery to me is how you decide to test the value <6> in R6C1? |
I don't think I mistyped anything, Keith -- please see the reply to Marty R, above.
Once you've spotted a "constellation" you can usually start arguing from any point inside it. The conclusion "r6c1 <> 6" is what caught my eye, so I started the argument from that cell. Perhaps the {1, 6} pair in column 4 is a more natural place to start.
-- There has to be a "6" in column 4 somewhere.
-- Since there are only two spots for a "6" in column 3, we can conclude that if r2c4 = 6, then r4c3 = 6. But then r6c1 <> 6, because there can't be two "6"s in the middle left 3x3 box.
-- If r6c4 = 6 the conclusion r6c1 <> 6 is immediate, because there can't be two "6"s in row 6.
Viewed from this perspective we can see this "4-star constellation" as an exercise in multi-coloring.
| Code: | 36 7 4 5 38 136 18 2 9
9 8 356+ 16= 7 2 145 134 36
2 35 1 9 348 346 458 7 368
368 34 36- 2 5 16 9 134 7
7 1 9 8 34 34 6 5 2
3568 345 2 16~ 9 7 148 134 38
35 6 35 7 1 9 2 8 4
4 2 8 3 6 5 7 9 1
1 9 7 4 2 8 3 6 5 |
That's another cute thing about the "constellations" -- sometimes they can be analyzed as two sets of binary chains involving a single digit. Other times they involve more than one digit, and can't be worked out in terms of simple coloring. dcb |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sat Mar 04, 2006 4:07 pm Post subject: someone_somewhere is still alive |
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Sorry for not beeing active in this forum for a longer time ...
I am active now in a project in UK where I have similar excitement as solving the "almost imposible" Sudoku's.
In the airport, I noticed a lot of small Sudoku books ... thinking not to buy one, in order not to be dissapointed if the games are to easy.
Maybe I will get a couple of "Nightmaire" from Sudo Cue or "Super Tough" from Krazydad because the planes started having delays as soon as the first snow storm is approching.
Not sure if from the airport I can see the 4 and 5 star constellation.
On the weekend I take a look here and note that everthing is on the right way. People are solving and enjoying.
BTY, I have keep one or two Sudoku's from the minimal 17, that I have found out to be as difficult for human, as beeing from an different galaxy. The 4, 5, etc stars all all in our galaxy. The solving of this particular ones, means looking far outside the stars of our galaxy.
If you got not tired, reading so many words (less than what Alan's 
and are still interested, please ask David, whom I will send the exceptional ones that I have found.
see u, |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 04, 2006 6:40 pm Post subject: |
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| Quote: | -- There has to be a "6" in the top left 3x3 box somewhere.
-- Only two slots are open for a "6" in that 3x3 box -- at r1c1 & at r2c3.
-- If r6c1 = 6 there can't be a "6" at r1c1, so then there would have to be a "6" at r2c3.
I abbreviated this logic by writing "r6c1 = 6 ==> r2c3 = 6" ... I thought that would be clear enough, but I'm happy to provide this clarification. The notion of "positional logic" took me a little while to catch onto.
You'll notice also that this "positional logic" conclusion is at variance with the result you obtained using "value-oriented logic." That's not very surprising, since I was reasoning toward a contradiction. In other words, one might just as well argue that assuming "r5c1 = 6" forces one to place two different values at r2c3 -- a "5" using value-oriented logic around the loop r6c1 - r6c4 - r2c4, and a "6" by tracing a different path, from r6c1 - r1c1 - r3c2 - r2c3. It's all the same -- a contradiction is a contradiction. Either approach yields the result r6c1 <> 6. |
Thank you for the explanation. I can see that either your positional or my value-oriented leads to different contradictions which both result in precluding a "6" in r6c1.
However, I don't know how I could have known that "r6c1=6==>r2c3=6" equated to the logic described above it.
Being a newbie to a sudoku forum, I'd like to ask if the notation you use is considered standard on forums, i.e., "==>" representing an arrow and "<>" meaning unequal. |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sat Mar 04, 2006 7:14 pm Post subject: "Standards" for notation |
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| Marty R wrote: | | Being a newbie to a sudoku forum, I'd like to ask if the notation you use is considered standard on forums, i.e., "==>" representing an arrow and "<>" meaning unequal. |
I don't know if it's "standard" for fora, or not, Marty. I'm not even sure that any such "standards" exist.
By "==>" I intend to represent an implication arrow, which is a more or less standard symbol in most textbooks on mathematics and/or symbolic logic. So
A ==> B
means "if A, then B," and
A <==> B
means "A, if and only if B" (or "A and B are logically equivalent").
I learned to use this notational shortcut about 30 years ago, when there was a lot less control over computer-generated fonts than there is today. Old habits die hard.
For the other trick, "<>" meaning "is not equal to," I can actually claim a meaningful, or semi-official, precedent. This is the notation used in C (a widely used computer language) and also in JavaScript (a widely used scripting language for internet browsers). So I would expect most computer programmers, at least, to recognize what "<>" stands for. dcb |
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