The New Sudoku Players' Forum

[ronk]

In this case a or b is part of the n-tuple. Then you also can safely eliminate a or b from the 2 UR cells.
I dont know, if this variant already has a name.

"Cannibalistic UR Type 3?" Seriously, I view that as a continuous chain.

[RW]

I dont know, if this variant already has a name.

If there was other candidates '5' in the row (and the box) but still the two bivalue cells '18' and '38', then you could eliminate the 8s from the UR cells as an UR+2kx. However, the bivalue cells and the two UR cells '158' and '358' would anyway form a naked quad to eliminate the rest of the 5s from the box and then the 5s from the row could be eliminated as locked candidates, and we would arrive at the UR type 4. To keep some 5s in the row there would have to be a 5 in one of the bivalue cells as well:

Code: Select all

.   18  .
.   358 .
158 .   358Y
------------
58  .   58

Now you can use the UR+2k to eliminate '8' from '358Y'.

RW

[ravel]

I see, the UR+2kx is more general than the type 3 variant, because it only depends on an ALS, that locks the extra candidates of one of the two UR cells, when the other one is set to an UR number. And when you have an ALS for all extra candidates you always have one for both groups of extra candidates.

[Sheldon Wu]

[quote="ravel"]I see, the UR+2kx is more general than the type 3 variant, because it only depends on an ALS, that locks the extra candidates of one of the two UR cells, when the other one is set to an UR number. And when you have an ALS for all extra candidates you always have one for both groups of extra candidates.[/quote]

[Sheldon Wu]

In this case a or b is part of the n-tuple. Then you also can safely eliminate a or b from the 2 UR cells.
I dont know, if this variant already has a name.

"Cannibalistic UR Type 3?" Seriously, I view that as a continuous chain.

[Cec]

Hi Sheldon Wu,

You seem to be having trouble posting a message. Click on Here for suggested procedure to post to the Forum.

Cec

[Myth Jellies]

Ruud's Daily Nightmare for 8/4/06 provided some opportunities for some advanced UR deductions.

Code: Select all

 *-----------*
 |.4.|5.1|.8.|
 |..2|.6.|9..|
 |...|9.7|...|
 |---+---+---|
 |.17|...|69.|
 |28.|...|.51|
 |.3.|...|.4.|
 |---+---+---|
 |...|789|...|
 |...|...|...|
 |..1|2.6|8..|
 *-----------*

basic methods take you to here...

Code: Select all

 *---------------------------------------------------------------------*
 |B3679    4     B369    | 5      23     1      | 237    8      2367   |
 | 1357   d57     2      | 348    6      348    | 9     -137    3457   |
 |A13568   56    A3568   | 9      234    7      | 145    1236   23456  |
 |-----------------------+----------------------+----------------------|
 | 45      1      7      | 348    345   C23458  | 6      9     C238    |
 | 2       8      49     | 6      79     34     | 37     5      1      |
 |B569     3     B569    | 1      79    C258    | 27     4     C278    |
 |-----------------------+----------------------+----------------------|
 | 3456    256    3456   | 7      8      9      | 145    1236   23456  |
 |A678    D2679  A68     | 34     1      345    | 45   d'267   D2679   |
 | 3457   D579    1      | 2      45     6      | 8    d'37    D34579  |
 *---------------------------------------------------------------------*

...and pickings are kind of slim while you search for that
(4=3)r5c6 - (3=5&4)r4c15 xyz-wing to kill the 4's in r4c46.

In the meantime, there are a few advanced uniqueness rectangle tricks that you can take advantage of. Three of them have extra candidates in all four cells. Uppercase letters mark out the four rectangles.

A) 68-UR in r38c13. Two strong 8-links form a corner at r3c1, and the opposite corner contains just the base candidates, 68. Therefore r3c1 <> 6. [edited to fix conclusion cell]

B) 69-UR in r16c13. Two strong 9-links form a corner at r1c1. There is also a strong 6-link that does not include that corner, therefore r1c1 <> 6.

C) 28-UR in r46c69. Two strong 2-links form a corner at r4c6. Two strong 8-links also form a corner at r6c9. For the same reason as in B, we know that r4c6 <> 8, and r6c9 <> 2.

D) 79-UR in r89c29. Here the 9's are locked into the UR, so if the deadly pattern is to be avoided, it will have to be the 7's that move out. If the move occurs out of r89c2, then the only other spot in c2 is r2c2. If the move occurs out of r89c9, then the only other spots in box 9 are r89c8. Since one or both of these alternate landing places has to contain a 7, any cell which sees both of them cannot contain a 7. Therefore r2c8 <> 7.


Geeky AIC stuff, just for grins and giggles...

If you'd like to express these uniqueness tricks as AICs, you need to note that A, B, and C use a special URCorner avoidance weak link that is defined as follows: if you have two strong links forming a corner for one of your base candidates, then you have two URC weak links that exist between the other base candidate in that corner and each of the base candidates in the opposite corner. The URC weak link is represented by "-URC-", and I've emphasized the strong UR corner links with a "= (N&N)[cell1,cell2]" structure. Thus if you'd like to express these tricks as AIC's you could write something like the following....

A: 8[r3c1] = (8&8)[r3c3,r8c1] - 8[r8c3]=6[r8c3] -URC- 6[r3c1]=(1358)[r3c1] => r3c1 <> 6
B: 9[r1c1] = (9&9)[r1c3,r6c1] - 6[r6c1] = 6[r6c3] -URC- 6[r1c1]=(379)[r1c1] => r1c1 <> 6
C1: 2[r4c6] = (2&2)[r4c9,r6c6]-8[r4c9] = 8[r6c9] -URC- 8[r4c6]=(2345)[r4c6] => r4c6 <> 8
C2: 8[r6c9] = (8&8)[r4c9,r6c6]-2[r4c9] = 2[r4c6] -URC- 2[r6c9]=(78)[r6c9] => r6c9 <> 2
D: 7[r2c2] = (7&9)[r89c2] -UR- (7&9)[r89c9] = 7[r89c8] => r2c8 <> 7

Proof of the URC weak link

Say you have

Code: Select all

abW  |  abX
     |
abY  |  abZ

in cells w, x, y, and z; with strong links a[w]=a[x] and a[w]=a[y] forming a strong corner on w. Opposite cell is z.

deadly pattern avoidance tells you that ab[z] - (ab&ab&ab)[w,x,y]

continuing we get...

Code: Select all

                                                b[w]
                                              /
ab[z] - (ab&ab&ab)[w,x,y]=(W or X or Y)[w,x,y] - a[x] = a[w] - b[w]
                                              \
                                                a[y] = a[w] - b[w]

since all these chains have alternating inferences, we know that we can rewrite
A - B = ... = C - D
as
A - D
thus for the W or X or Y cases we always get ab[z] -...- b[w] same as ab[z] - b[w] which is the UR Corner weak link.

[ronk]

Proof of the URC weak link

Say you have

Code: Select all

abW  |  abX
     |
abY  |  abZ

in cells w, x, y, and z; with strong links a[w]=a[x] and a[w]=a[y] forming a strong corner on w. Opposite cell is z.

deadly pattern avoidance tells you that ab[z] - (ab&ab&ab)[w,x,y]

continuing we get...

Code: Select all

                                                b[w]
                                              /
ab[z] - (ab&ab&ab)[w,x,y]=(W or X or Y)[w,x,y] - a[x] = a[w] - b[w]
                                              \
                                                a[y] = a[w] - b[w]

The exclusion of b from the w cell depends upon b[x]=b[z] and/or b[y]=b[z]. I don' t see that requrement showing up in the above notation ... probably because I don't understand the ab[z] - (ab&ab&ab)[w,x,y] thingie.

Would you please put that weak link into words?

TIA, Ron

[keith]

MJ,

Question:
You said:

D) 79-UR in r89c29. Here the 9's are locked into the UR, so if the deadly pattern is to be avoided, it will have to be the 7's that move out. If the move occurs out of r89c2, then the only other spot in c2 is r2c2. If the move occurs out of r89c9, then the only other spots in box 9 are r89c8. Since one or both of these alternate landing places has to contain a 7, any cell which sees both of them cannot contain a 7. Therefore r2c8 <> 7.

I agree that the 7's have to move out, and I think I agree that two 7's have to move out. Also, the two that move out cannot be on a diagonal, since that leaves the UR as still possible with 7's on the other diagonal.

You have chosen the 7's to move out of one or other of the columns.

Why do you not consider the 7's to possibly move out of one of the rows?

Observation:
The hunt for UR reductions may lead you to look at <34> in R28C46.

There is a strong link on <4> in C4 and one on <3> in R8. Not enough for a UR reduction, I think.

However, there is also a strong link on <8> in R2. Thus, R2C6 cannot be <3>. {R2C6=3; R2C4=8; R8C4=4; R8C6=3, a contradiction in C6.}

If you make this non-UR reduction and continue, you get to a point where basic techniques run out. However, UR's A and D survive, and your reductions solve the puzzle.

Thank you, a most interesting post!

Keith

[ronk]

D) 79-UR in r89c29. Here the 9's are locked into the UR, so if the deadly pattern is to be avoided, it will have to be the 7's that move out. If the move occurs out of r89c2, then the only other spot in c2 is r2c2. If the move occurs out of r89c9, then the only other spots in box 9 are r89c8. Since one or both of these alternate landing places has to contain a 7, any cell which sees both of them cannot contain a 7. Therefore r2c8 <> 7.

Due to the four UR strong links for digit 9 we effectively have two weak links for UR digit 7, i.e., r89c2-7-r89c9 and r8c29-7-r9c29. Combining the first of these with the grouped strong links for digit 7 in c2 and c8 we can say ...

r89c9-7-r89c2=7=r2c2-7-r2c8=7=r89c8-7-r89c9, implies r89c9<>7

... a grouped turbot fish with a UR as one of the weak links.

If you make this non-UR reduction and continue, you get to a point where basic techniques run out. However, UR's A and D survive, and your reductions solve the puzzle.

Following the xyz-wing without using any URs, a finned swordfish in r249c159 solves the puzzle too ... but then we wouldn't have had any fun with URs.

[daj95376]

Following the xyz-wing without using any URs, a finned swordfish in r249c159 solves the puzzle too ... but then we wouldn't have had any fun with URs.

Does this result in [r2c9]<>4 and [r3c5]<>4 ? That's what I get using Templates. This eventually leads to a rare Hidden Triple.

[ronk]

Does this result in [r2c9]<>4 and [r3c5]<>4 ? That's what I get using Templates.

Sorry, I should have said the exclusion was [r3c5]<>4. The exclusion [r2c9]<>4 is then due to line-box interaction.

[Myth Jellies]

Proof of the URC weak link

Say you have

Code: Select all

abW  |  abX
     |
abY  |  abZ

in cells w, x, y, and z; with strong links a[w]=a[x] and a[w]=a[y] forming a strong corner on w. Opposite cell is z.

deadly pattern avoidance tells you that ab[z] - (ab&ab&ab)[w,x,y]

continuing we get...

Code: Select all

                                                b[w]
                                              /
ab[z] - (ab&ab&ab)[w,x,y]=(W or X or Y)[w,x,y] - a[x] = a[w] - b[w]
                                              \
                                                a[y] = a[w] - b[w]

The exclusion of b from the w cell depends upon b[x]=b[z] and/or b[y]=b[z]. I don' t see that requrement showing up in the above notation ...

Here I am not exactly trying to show that b is excluded from the w cell. I am trying prove the -URC- link, i.e. to show that b in the w cell excludes a or b in the z cell. Once you establish that, then as you say, strong links involving b[z] easily exclude b from cell w. I showed that in the A, B, and C chains above.

probably because I don't understand the ab[z] - (ab&ab&ab)[w,x,y] thingie.

Would you please put that weak link into words?

TIA, Ron

ab[z] - (ab&ab&ab)[w,x,y] (better would be ab[z] -UR- (ab&ab&ab)[w,x,y])
If cell z contains ab, then cells w, x, & y cannot contain ab & ab & ab, and vice versa. If they did, then we would have a deadly pattern UR.

(ab&ab&ab)[w,x,y]=(W or X or Y)[w,x,y]
Either cells w, x, and y contain ab & ab & ab; or w, x, and y must contain a W or X or Y

At this point I split into three simple chains, one for each possibility (W or X or Y) and show that they all result in excluding b from cell w. Combining it all together we have a or b in cell z excluding b in cell w. Thus we can use
b[w] -URC- b[z], and b[w] -URC- a[z] as well as b[w] -URC- ab[z]

[Myth Jellies]

MJ,

Question:
You said:

D) 79-UR in r89c29. Here the 9's are locked into the UR, so if the deadly pattern is to be avoided, it will have to be the 7's that move out. If the move occurs out of r89c2, then the only other spot in c2 is r2c2. If the move occurs out of r89c9, then the only other spots in box 9 are r89c8. Since one or both of these alternate landing places has to contain a 7, any cell which sees both of them cannot contain a 7. Therefore r2c8 <> 7.

I agree that the 7's have to move out, and I think I agree that two 7's have to move out. Also, the two that move out cannot be on a diagonal, since that leaves the UR as still possible with 7's on the other diagonal.

I agree it sounds like I said all 7's move out, but that is not what I meant. Just to be clear, without uniqueness considerations the rectangle could contain two 7's, one in each pairing. With uniqueness and the locked 9's, only one 7 can be allowed in the rectangle. Therefore one of the pairings must lose its 7. We do not assume which pairing loses the 7, however, any cell which sees all possible non-UR landing zones for both pairings, cannot contain a 7.

You have chosen the 7's to move out of one or other of the columns.

Why do you not consider the 7's to possibly move out of one of the rows?

Since the pairings in columns are also constrained by a box, you are free to consider alternate non-UR locations for the 7 in either that box or that column.

[daj95376]

Code: Select all

+--------------+
|  -   -    #  |
| 12   -   123 |
|  -   -    #  |
+--------------+
| 12   -   124 |
|  -   -    #  |
|  -   -   34  |
+--------------+
|  -   -    #  |
|  -   -    #  |
|  -   -    #  |
+--------------+


There are obvious variations when the extra candidates are in the same row and/or the same box. This is usually called a Type 3 Unique Rectangle. There is also a diagonal variant.

Here is a diagonal UR Type 3 variant -- I think. ronk searched for it in the top1495 and Ruud's Top 10000 without success.

Does anyone have a real world example?

Code: Select all

 [cell #]<>1234
 +---------------+
 |  12   . 123   |
 |   #   .   .   |
 |1234   .   .   |
 |---------------+
 |   .   .   .   |
 |   .   .   .   |
 | 124   .  12   |
 +---------------+


Note: Explaining the elimination of <3> in the upper-left cell of the UR seems to be a stumbling point. (Thanks ronk!)

[Edited]: I should have originally stated that digits <1> and <2> are optional in the cell with <1234>. If either digit is missing, then the eliminations in [cell #] are reduced accordingly. I'm sorry for posting the most complex variant!