## Backdoor, T&E or Colouring?

Everything about Sudoku that doesn't fit in one of the other sections

### Backdoor, T&E or Colouring?

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After singles the Box Analysis suggested possible single digit eliminations in 489, but nothing came of it. So what next? Simple speculation on the flow-on effects of nominating a particular value in a certain cell quickly showed that this would force another of that value in the same column, immediately proving that the initial assumption could not be true, and the only other possibility quickly solved it. So I virtually fell over the solution to a puzzle which the Daily Sudoku Draw/Play rated as ‘Too Hard,’ whatever that means.
So is this Colouring in action, a Backdoor, or just dumb luck?
Yogi
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Location: New Zealand

### Re: Backdoor, T&E or Colouring?

You proved that a value must go in a cell, then found that value was a backdoor. Dumb luck? No way ... an inspired choice.

Mathimagics
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Location: Canberra

### Re: Backdoor, T&E or Colouring?

Hi Yogi,

Yogi wrote:After singles the Box Analysis suggested possible single digit eliminations in 489, but nothing came of it.

Didn't you see the Skyscaper on 8s? It doesn't solve any cells but it helps. After that there's an interesting double-W-Wing that works like a Remote Pair, eliminating both digits from the same cell (only one needed, though). Then you can finish the puzzle with a third W-Wing or a Grouped Turbot Crane (i.e. ER).

There are also plenty of more complicated single-digit options that solve it in one move. Here's one:

Code: Select all
`.--------------------.-------------.-----------------.|   6     2     7    | 5  89    3  | 489   1     489 || af489   5    a489  | 6  7     1  | 2    b89    3   ||   3     1    b8-9  | 4  2    c89 | 6     7     5   |:--------------------+-------------+-----------------:|   1     7     3    | 8  49    5  | 49    6     2   ||  e4589  489   4589 | 2  6    d49 | 7     3     1   ||   2     49    6    | 3  1     7  | 489   5     489 |:--------------------+-------------+-----------------:|   7     489   4589 | 1  458   2  | 3    c489   6   ||   48    6     1    | 9  3     48 | 5     2     7   ||  e4589  3     2    | 7  458   6  | 1    c489  d89  |'--------------------'-------------'-----------------'`

(9)r2c13 = r2c8&r3c3 - r3c6|r79c8 = r5c6&r9c9 - r59c1 = (9)r2c1 => -9 r3c3; stte

or:

Mutant 4x6-Obifish: (9)R2C16B9\r59c8[r3b11] => -9 r3c3; stte

So is this Colouring in action

Did you use actual coloring and follow both paths without presumptions on which would lead to what? If not, then it was T&E. If you only test a single candidate, even with coloring, it's T&E. Real coloring investigates two strongly linked options at the same time, trying to find verities (where both parities agree) instead of contradictions (but they can happen too, solving the coloring immediately).

a Backdoor, or just dumb luck?

Well, you definitely found a backdoor since it solved the puzzle. If you'd tested that first, instead of the other option that resulted in a contradiction, it would have led to the solution immediately. Had you accepted that without proving the other option false, it would have been a mere guess (edit: sorry, I meant an inspired choice! ). That wouldn't count for anything in my books (edit: but it's not the only opinion on the matter). Now you actually solved the puzzle, even if via T&E, so that definitely counts.

Btw, here's a recent one where I used honest T&E. In that case it doesn't really count, though, because I knew there was a backdoor to be revealed if I killed those candidates. So, it wasn't real solving at all. (Choosing that particular backdoor as a target was an inspired choice, though, as I had a feeling the others would be harder to crack. I haven't really investigated if I was right or not.)
-SpAce-: Show
Code: Select all
`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

SpAce

Posts: 2353
Joined: 22 May 2017

### Re: Backdoor, T&E or Colouring?

SpAce wrote:That wouldn't count for anything in my books (edit: but it's not the only opinion on the matter).

It's certainly not mine!

Mathimagics
2017 Supporter

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Joined: 27 May 2015
Location: Canberra

### Re: Backdoor, T&E or Colouring?

Thank you ALL for your comments. It’s good to have a diversity of opinion. That's why we have a forum, right? Coming from a sales background I tend to be driven more by results. The catch-cry is ‘Keep It Simple!’ I just want to find the one unique solution to a puzzle and know that I solved it by myself. This means that a lot of the time it’s more important to know WHAT works rather than why it works or what its name is, in my humble opinion. However, if there are approaches that will get there sooner or more simply, then I would try to learn them if I could find a source which will help me understand them and teach me.
I get the point that true Colouring involves investigating two options and looking for a verity elsewhere, much like BUG+2. I think I did note the 8skyscraper, but the elimination did not solve a cell, only confined 8 to column3 in box1. So I went somewhere else.
Yogi
2017 Supporter

Posts: 174
Joined: 05 December 2015
Location: New Zealand

### Re: Backdoor, T&E or Colouring?

Hi Yogi,
If I have understood your point of view correctly, you are looking for simplicity and efficiency in resolution.
I suggest that you take an interest in the TDP (track's technique) which allows you to solve without having to know all the expert techniques (wing, fish, ALS, etc...).
To put it simply, you build two chains P(A) and P(B) from a pair of strongly linked candidates A and B, and you look for the candidates common to these two chains, they are solutions.
P(A) is constructed by assuming (hypothesis) that A is the solution and using basic techniques to place the other candidates resulting from this hypothesis. Ditto for P(B).
Here is an example of how to solve the puzzle you are proposing, which I think is rather easy.
A=9r1c5 => P(A) = {9r1c5, 9r5c6, 9r6c2, 9r9c9, 9r7c3,8r3c3, ...}
B=9r3c6 => P(B) = {9r3c6,8r3c3, ... }
=> r3c3=8 and the puzzle ends with the basic techniques.

puzzle: Show
P(A) is marked in blue, P(B) is marked in yellow.

You can also reason with a single anti-track P'(B) obtained by assuming that B is eliminated, this gives :
P'(9r3c6) = {9r1c5, 9r5c6, 9r6c2, 9r9c9, 9r7c3, ...} which eliminates any candidate who sees the starting candidate B=9r3c6 and the finishing candidate 9r7c3, here the 9r3c3.

I published on this forum a theoretical document on TDP that suggests that the technique is complicated. In practice it is very easy to apply.
I give on this forum examples of resolution with TDP.
Robert
Mauriès Robert

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Location: France