## 1/27 usa today

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Here is my way
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`Almost locked sets xy wing rule+----------------+----------------+----------------+| 478 ^57   579  | 3689 89   1    | 2    3478 3469 | | 178  3    179  | 689  2    4    | 5    178  69   | | 148  6    2    | 389  5    7    | 348  1348 349  | +----------------+----------------+----------------+| 5    48   13   | 189  6    289  | 349  234  7    | | 2    17   1367 | 179  4    59   | 369  35   8    | | 9    48  %67   | 78   3    25   |%46   25   1    | +----------------+----------------+----------------+| 367 ^257  57   | 4    78   368  | 1    9   *235  | | 137  1279 8    | 5    179  39   |*347  6    234  | | 1367 1579 4    | 2    179  369  |*378 *38  *35   | +----------------+----------------+----------------+A=^B=%C=*x=7y=4z=2and we getR5C2<>7`
bennys

Posts: 156
Joined: 28 September 2005

Hi Tso.

Thanks for your excelent explanation of the loop.

a) Yes, it is right.
b) Once I have looked for the puzzle I almost immediatly spoted the AUR, constructed mentaly the link from "2" to "9", and then I started the construction of links from the "9" (and not from the "2", because there is in the same unit of r1c3 a bivalue node with "9", whereas that doesn't happen for "2").
c) The nice loop notation its just that: a notation. More important is to visualize the implications of the loop in the puzzle and see by ourselves that r7c2 must really be "2". Also, it is very important to confirm the deduction of a loop by looking to the puzzle.

Hope this help.

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

This one was certainly much tougher than I thought it was going to be and I think nigh on impossible on paper i.e. without a programme to highlight particular candidate numbers and colouring facility. This was how I managed it eventually:

After basic eliminations from locked candidates etc. I got to here:

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`{478}  {57}   {579}  {3689} {89}   {1}    {2}    {3478} {3469} {178}  {3}    {179}  {689}  {2}    {4}    {5}    {178}  {69}   {148}  {6}    {2}    {389}  {5}    {7}    {348}  {1348} {349}  {5}    {48}   {13}   {189}  {6}    {289}  {349}  {234}  {7}    {2}    {17}   {1367} {179}  {4}    {59}   {369}  {35}   {8}    {9}    {48}   {67}   {78}   {3}    {25}   {46}   {25}   {1}    {367}  {257}  {57}   {4}    {78}   {368}  {1}    {9}    {235}  {137}  {1279} {8}    {5}    {179}  {39}   {347}  {6}    {234}  {1367} {1579} {4}    {2}    {179}  {369}  {378}  {38}   {35}`

Colouring conjugate 4s permits elimination of 4s at r3c7 and r4c7.
Colouring conjugate 5s permits elimination of 5 at r7c2.
Colouring conjugate 8s permits elimination of 8s at r3c4 and r4c4.

Grid now as follows:
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`{478}  {57}   {579}  {3689} {89}   {1}    {2}    {3478} {3469} {178}  {3}    {179}  {689}  {2}    {4}    {5}    {178}  {69}   {148}  {6}    {2}    {39}   {5}    {7}    {38}   {1348} {349}  {5}    {48}   {13}   {19}   {6}    {289}  {39}   {234}  {7}    {2}    {17}   {1367} {179}  {4}    {59}   {369}  {35}   {8}    {9}    {48}   {67}   {78}   {3}    {25}   {46}   {25}   {1}    {367}  {27}   {57}   {4}    {78}   {368}  {1}    {9}    {235}  {137}  {1279} {8}    {5}    {179}  {39}   {347}  {6}    {234}  {1367} {1579} {4}    {2}    {179}  {369}  {378}  {38}   {35}   `

Now there's a triple in row 4, so you can exclude 9 from r4c6 and 3 from r4c8.
An xy-wing (r1c5, r3c4, r3c7) - eliminate 8 from r1c8.
Another xy-wing (r4c7, r5c8, r5c6) - eliminate 9 from r5c7.
Singles in row 4!

Grid now at this point:
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` {478}  {57}   {579}  {3689} {89}   {1}    {2}    {347}  {3469} {178}  {3}    {179}  {689}  {2}    {4}    {5}    {178}  {69}   {148}  {6}    {2}    {39}   {5}    {7}    {38}   {1348} {349}  {5}    {48}   {3}    {1}    {6}    {28}   {9}    {24}   {7}    {2}    {17}   {167}  {79}   {4}    {59}   {36}   {35}   {8}    {9}    {48}   {67}   {78}   {3}    {25}   {46}   {25}   {1}    {367}  {27}   {57}   {4}    {78}   {368}  {1}    {9}    {235}  {137}  {1279} {8}    {5}    {179}  {39}   {347}  {6}    {234}  {1367} {1579} {4}    {2}    {179}  {369}  {378}  {38}   {35}   `

Colouring conjugate 1s - eliminate 1 from r2c1.
xy-wing (r3c4,7; r1c5) - eliminate 9 from r1c4.
xy-chain (r7c2) -7- (r7c5) -8- (r1c5) -9- (r3c4) -3- (r3c7) -3- (r5c7) -6- (r6c7) -6- (r6c3) -7- (r7c3) -7- (r7c2) Loop is closed thus eliminate 3s in r8,9c7; r3c8,9, eliminate 7s in r7c1; r1,2,5c3.
Colouring conjugate 7s - eliminate 7s from r8,9c1.

Grid is now:
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`{478}  {57}   {59}   {368}  {89}   {1}    {2}    {347}  {3469} {78}   {3}    {19}   {689}  {2}    {4}    {5}    {178}  {69}   {148}  {6}    {2}    {39}   {5}    {7}    {38}   {148}  {49}   {5}    {48}   {3}    {1}    {6}    {28}   {9}    {24}   {7}    {2}    {17}   {16}   {79}   {4}    {59}   {36}   {35}   {8}    {9}    {48}   {67}   {78}   {3}    {25}   {46}   {25}   {1}    {36}   {27}   {57}   {4}    {78}   {368}  {1}    {9}    {235}  {13}   {1279} {8}    {5}    {179}  {39}   {47}   {6}    {234}  {136}  {1579} {4}    {2}    {179}  {369}  {78}   {38}   {35}`

Getting there!
Now we have an x-wing of 7s so r1c2 = 5
And the rest is singles.
CathyW

Posts: 316
Joined: 20 June 2005

Hi Tarek.

First, thanks for the puzzle you have posted. Second, I want to congratulate you for this very good and interesting puzzle, where some advanced techniques can be applied. My solution have 9 steps, and, personaly, I rate it just a little easier than exercise #5. Clearly, your generator is getting better and better. Keep the good work.

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Starting with candidate grid Cathy posted
CathyW wrote:
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`{478}   {45789} {579}   {3689}  {89}    {1}     {2}     {3478}  {3469}  {178}   {3}     {179}   {689}   {2}     {4}     {5}     {178}   {69}    {148}   {6}     {2}     {389}   {5}     {7}     {3489}  {1348}  {349}   {5}     {148}   {13}    {189}   {6}     {289}   {349}   {234}   {7}     {2}     {17}    {1367}  {179}   {4}     {59}    {369}   {35}    {8}     {9}     {478}   {67}    {78}    {3}     {258}   {46}    {245}   {1}     {367}   {257}   {3567}  {4}     {78}    {368}   {1}     {9}     {235}   {137}   {1279}  {8}     {5}     {179}   {39}    {347}   {6}     {234}   {1367}  {1579}  {4}     {2}     {1789}  {3689}  {378}   {3578}  {35}    `

Placing a 1 or 7 in r5c2=>r1c2=5

If r5c2=7=>r1c2=5
If r5c2=1=>r4c3=3=>r5c3=67and r6c3=67=>r7c3=5=>r1c3<>5=>r1c2=5
Therefore r1c2=5

Therefore r7c3=5 and r9c9=5

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`{478}   {5}     {79}    {3689}  {89}    {1}     {2}     {3478}  {3469}  {178}   {3}     {179}   {689}   {2}     {4}     {5}     {178}   {69}    {148}   {6}     {2}     {389}   {5}     {7}     {3489}  {1348}  {349}   {5}     {148}   {13}    {189}   {6}     {289}   {349}   {234}   {7}     {2}     {17}    {1367}  {179}   {4}     {59}    {369}   {35}    {8}     {9}     {478}   {67}    {78}    {3}     {258}   {46}    {245}   {1}     {367}   {27}    {5}     {4}     {78}    {368}   {1}     {9}     {235}   {137}   {1279}  {8}     {5}     {179}   {39}    {347}   {6}     {234}   {1367}  {179}   {4}     {2}     {1789}  {3689}  {378}   {3578}  {5}    `

Using the xy-wing of 79 in r1c3, 89 in r1c5 and 78 in r7c5

Placing 8 or 9 in r1c5 enables you to eliminate the 7 in r7c2 and place the 2 in r7c2

If r1c5=8=>r7c5=7=>r7c2=2
If r1c5=9=>r1c3=7=>(r5c3<>7 and r6c3<>7)=>r5c2=7=>r7c2<>7=>r7c2=2
Therefore r7c2=2

This implies r7c9=3

Code: Select all
`{478}   {5}     {79}    {3689}  {89}    {1}     {2}     {478}   {3469}  {178}   {3}     {179}   {689}   {2}     {4}     {5}     {178}   {69}    {148}   {6}     {2}     {389}   {5}     {7}     {3489}  {1348}  {49}   {5}     {148}   {13}    {189}   {6}     {289}   {349}   {234}   {7}     {2}     {17}    {1367}  {179}   {4}     {59}    {369}   {35}    {8}     {9}     {478}   {67}    {78}    {3}     {258}   {46}    {245}   {1}     {367}   {2}     {5}     {4}     {78}    {368}   {1}     {9}     {3}   {137}   {1279}  {8}     {5}     {179}   {39}    {47}    {6}     {24}   {1367}  {179}   {4}     {2}     {1789}  {3689}  {78}    {78}    {5}    `

I had overlooked the locked 7's in boxes 3 and 9 so had a naked pair in r9

The naked pair of 78's in r9 lead to the exposure of two more sets of naked pairs on r9, also from here, it's singles all the way.

MCC
MCC

Posts: 1275
Joined: 08 June 2005

Carcul & Tso,
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`[2 = (5&7)] - [7 = 8] - [8 = 9] - [9 =AUR= 2][r7c23    ] - [r7c5 ] - [r1c5 ] - [r17c23   ] => r7c2 = 2`

Does this personalized cycle notation correctly describe Carcul's loop? I like including the links inside the cells/nodes so that I can better keep track of what is going on. The (5&7) is a grouping that is true only when both 5 and 7 are true in the node. Since the 2 in r7c2 is a discontinuity connected by strong links, it must be true!?

Interesting aside, I don't think you can count an AUR strong link as a nominal link, because r7c2 = 2 would not in general imply that r1c3 <> 9
Myth Jellies

Posts: 593
Joined: 19 September 2005

Hi Myth Jellies.

Myth Jellies wrote:Does this personalized cycle notation correctly describe Carcul's loop?

Sorry, but I don't understand a little of your notation.

Myth Jellies wrote:I don't think you can count an AUR strong link as a nominal link, because r7c2 = 2 would not in general imply that r1c3 <> 9

First, I don't know what a "nominal link" is. Perhaps you could first explain your own terms when they differ from the commonly used ones. Second, I have never said that the link in an AUR is a strong one: if you read carefully the post about AURs, you will read that the "link is similar but not completely equivalent to the type of link that exists in an ALS having two of his values in just one cell each." So, I don't understand why r7c2=2 would imply r1c3<>9.

Myth Jellies wrote:Since the 2 in r7c2 is a discontinuity connected by strong links, it must be true!?

Instead of trying to understand your notation, or mine, why don't you make r7c2<>2 and see what happen? Also, in terms of the traditional nice loop rules, r7c2 is a discontinuity with one strong link and two weak links with different labels from the one of the strong link, and so the labels of the weak links can be eliminated from this node, leaving "2" as the only possible value for r7c2.

Hope this help.

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

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