jco from Brazil

to the Sudoku site, jco. We hope to see you commenting on the puzzle pages. I have briefly visited Brazil. Once when I visited the Iguazu Falls after trekking in Peru and the second time for an overnight stopover on my way to Antarctica.

Hello Anne,
Thank you for the welcome! I am glad to know that you visited Brazil and saw a wonderful place that I myself still need to see sometime. I've never been in Australia, but surely it is in my plans for the future.
I am happy to have found this wonderful place made of people that love Sudoku!
It will take me some time to learn how things work.
Regards,
jco

Nice to see you received my comment and replied. I often wonder if new members realise they have a page of their own. I have been on this site just after it started up fifteen years ago, before we even had our own pages. Perhaps you'd like to visit my page and see how we have set them up. Also, if you leave a comment on my page, or anyone else's page, they are likely to see them and reply to you on your page.
Regards,
Anne

Hello jco, thank you so much for visiting my page and for your lovely comments.
I must admit I don't understand how the chaps on 'tough' Sudoku work it out and the comments they make about how to solve it. I just work it as I go but rarely do the 'tough' puzzle, just the easy, medium and hard.
Regards,
Anne

Hi jco,

You wrote 'I know of your expertise on the subject!'
Many thanks! It is just the result of 13 years, practicing sudoku...
I feel uneasy to help you, because I don't know what you have already learnt and what you are missing.

Have you seen my comment to rwm's solution ? In this comment, I give a chain (AIC) solving the puzzle in one step, using the UR(56)gh35:
(1)gh5==(4)gh5-g4=a4-a9=(4-1)g9=(1)g45 =>-1i56.h6; singles to 81

First, a side remark: in your pencilmarks, you have no candidate 4a8, nor 4g5.
It means you have already considered the row X-Wing (4)ag49, while rwm and myself have not. These are valid eliminations, but to me, X-wing, XY-Wing, simple coloring, Swordfish, and others are not part of basic techniques. They have to be counted as separates steps. You may have noticed that some players disagree on that point.

Your own logic is sound, and I would write the following AIC:
(1)gh5 == (4)h5 - h8 = (4-1)g9 = h7 - h2 = (1)i2 => -1 i56

Detailed information:
AIC is the acronym of Alternate Inference Chain.
Such name comes from the alternance Strong Link / Weak link (SL / WL)
Two candidates (or group of candidates) are in a strong link, if, according to sudoku rules, at least one out of these must be True. In Eureka writing convention, strong links use the '=' symbol. Example in the above AIC: 4h8 = 4g9
When you write h8<>4 => g9=4, you prove that 4h8 and 4g9 are in a strong link: at least one is true, in other words they can't all be False. Not that the SL 4h8 = 4g9 is equivalent to the boolean equation 4h8 OR 4g9 = True (with '=' meaning 'equals' here)
Likewise, two candidates (or group of candidates) are in a weak link, if, according to sudoku rules, at most one out of these can be True. In Eureka writing convention, weak links use the '-' symbol. Example in the above AIC: 4h5 - 4h8
When you write h5=4 => h8<>4, you prove that 4h5 and 4h8 are in a weak link: at most one is true, in other words they can't all be True. Not that the WL 4h5 = 4h8 is equivalent to the boolean equation 4h5 NAND 4h8 = True (with '=' meaning 'equals' here)
An AIC starts with a SL and ends with a SL.
Using the alternance, you can prove that either the first term is True OR the last term is True. They can't both be False. (Actually, either the first term is True OR all terms on the rightside of the '=' SL symbol are True, including the last one of course). Any candidate in sight of both endterms is False (otherwise, contradiction)
'In sight of...' is synonym for 'in a weak link with...'
Last detail. I have used a '=='symbol in the first (strong) link of my AIC. This is just to draw reader's attention: this strong link is not inferred straight away from sudoku rules, but from the UR pattern.
Some players would write (1)gh5 =UR= (4)h5. Acceptable too, matter of taste.
Note that (1==4)gh5 or (1=UR=4)gh5 would also be valid.

Now a last suggestion:
You can see that my AIC eliminates three candidat

Hi jco,
You wrote 'I think in your solution today one part is missing (in the typing)
i146 before loop, right?'
Definitely right. Thank you. Correction now posted in the Tough page.
On the New Sudoku Players forum, I noticed that you use Hodoku, known to be a good software. I'm not Hodoku user myself. SpAce knows Hodoku very well.
Have a nice week too!
Cenoman

Thank you for your Christmas Wishes, jco!
Wishing you a very Merry Christmas and a Happy & Healthy and better 2021! Have a wonderful time over the Festive Season.
Best wishes, Anne

Hi jco,
Nice to hear from you. Like some of the others on this site, you are an expert on using Eureka notation.
I am usually able to 'read' Eureka, but can't 'write' it myself.

All the best for the Holiday period,

Alfred.

1. (5)i9 = (5-6*)e9 = (6-4)e6 = (4)gi6 - (4=8)g5 - (8&)g2 = [(5)i9 = (5)e9 - (5=4)e4 - (4)c4 = (4)c2 - (4 =& 2)g2 - (2=7)g9]
\
(4=6*)i2

1. (5)i9 = (5-6*)e9 = (6-4)e6 = (4)gi6 - (4=8)g5 - (8&)g2 =
[(5)i9 = (5)e9 - (5=4)e4 - (4)c4 = (4)c2 - (4 =& 2)g2 - (2=7)g9]
\
(4=6*)i2

1. (5)i9 = (5-6*)e9 = (6-4)e6 = (4)gi6 - (4=8)g5 - (8&)g2 =
[(5)i9 = (5)e9 - (5=4)e4 - (4)c4 = (4)c2 - (4 =& 2)g2 - (2=7)g9]
\
(4=6*)i2