## An xyt Color Wrap

In the spirit of Valentine’s Day, I come across a coloring resource in Berthier’s xyt chain.

The elaboration of Royle 17-20565 in Chapter XVII of The Hidden Logic of Sudoku is a challenge in itself. By the theory of looking for stuff where the light is better, I  look for line subsets as the line is marked, and the lines in closure don’t get examined until – you guessed it – closure.

So with this one, I was in closure on column 5 when I spotted this naked quad. Of course the hidden triple is just as available at this point in Sysudoku basic.

Even if you were too busy sending Valentine cards, and declined the homework, you can appreciate how tough the line marking was, and can imagine how much worse number scanning and hidden logic transformations would be.

So with some relief, I followed up, and got the next cleanup bonus.

If you haven’t gone further already, avoid looking at the grid below. Instead, mark the follow up on the naked triple on your own grid, and contemplate where you would start xyt-chains.

The ticket to ride is a pair of linked bv, with a potential victim within sight of the starting candidate.

The SudoRules first xyt (black) makes a single cell repair in an XY eliminating 4r3c8.

This enables an xyt making two XY repairs and rounding two corners, for another elimination, and a hidden single in r6.

The follow up celebration ends with a skyscraper in the 7’s that you can easily spot (slink-wink-slink), then Berthier reports a hidden xy chain for the collapse.

Instead, we choose to invest the bv dividends in Medusa coloring, and manage to color most of the bv with two clusters.

The bridging logic is

Not(orange and green) =>     Red or blue.

And also,

Not( red and green) =>

Orange or blue.

Oops, that means blue is true, because

(red or blue)& (orange or blue) => (red or orange) and blue => blue.

In trying the obvious, we stumble upon an insight: The xyt version of forcing chain logic is a recourse for coloring.

Starting on one of the few remaining uncolored bv, the xyt assumption “if 9 is false” then includes that red is false and orange is true, and in two ways, 9r5c3 is false. No need to go further. 9r5c3 is false, regardless of 9r5c7.  That wraps red, with two red 4’s in c3.

Remember the shortcut wink? Coloring applies in all the bv clover examples the Royle 17 series exhibits in THLS.

We dig a bit deeper into the xyt chain next post. This time, our Royle foil is Royle 17-33442. Its another “solved in half the numbers” wonder.

I'm John Welch, a retired engineering professor, father of 3 wonderful daughters and granddad to 7 fabulous grandchildren. Sudoku analysis and illustration is a great hobby and a healthy mental challenge.
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### 4 Responses to An xyt Color Wrap

1. GUENTER TODT says:

Hello John,
You took 1,5,7 for the green and blue coloring and got no information about the truth of red and orange. More interesting for green and blue are 4 and 9 in W. Take 4r6c2 blue and 9r6c2 green and color a little more. Then you get blue and orange as true colors, and only singles are left.
Be well
Guenter

• Sudent says:

No, Guenter, your alternative is not more interesting than the role of the red/orange cluster in the blue wrap, followed by the shortcut wrap by xyt chain.

• GUENTER TODT says:

John, I did not want to discuss whether solutions are more or less interesting. In some respect every solution is interesting. My comment should give a focus on coloring strategies. Let me explain it more detailed. When you use simple coloring, you hope that you get a color trap or a color wrap. If not, you choose a second coloring. A computer may try all possibilities, but a human solver needs a strategy. Which starting cell should be chosen? Which candidates should be colored? You color first 1, 4, 7, 9 in red and orange – only the red 1r9c6 is missing. At least two of these numbers should be colored in blue and green. You choose 1, 7 and in addition 5 as a new number. With 1, 7 you get a wrap with 5 clues. I notice that 4 and 9 in box W see already red and orange candidates and promise more success. This is my coloring: 4r6c2 blue, 9r6c2 green, 9r6c7 blue, 9r5c7 green, 4r5c7 blue, 4r5c3 green, 4r9c2 green. Blue and red is not possible because of 4r6c2 blue and 4r6c9 red. Green and red is not possible because of 4r5c3 green and 4r8c3 red. Green and orange is not possible because of 9r6c2 green and 9r2c2 orange. Therefore blue and orange are true, and my wrap has altogether 15 clous. If you prefer a wrap with 5 clous, it is your choice.

• Sudent says:

Guenter, I hate to dispute such a diligent reader, but you did say your alternative coloring was more interesting than the one I published, to illustrate a capability of coloring that I had not published before. You made a judgement. I understood your alternative. You could have presented it as an alternative, rather than a superior one. It’s rather pointless to search through alternative colorings for the shortest or most decisive one.