The New Sudoku Players' Forum

[morl]

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...|7..|..5
..4|.2.|8..
3..|..1|.6.
---+---+---
.8.|...|4..
..9|..6|..2
5..|3..|.7.
---+---+---
.7.|.8.|...
6..|..4|...
...|5..|.1.

WARNING: This one is very hard. SudokuWiki didn't solve it.

[Mauriès Robert]

Hi,

WARNING: This one is very hard. SudokuWiki didn't solve it.

I, for one, don't find this grid difficult. It has a TDP level =1, which means that with a set of two conjugated tracks, it is solved completely. For example by tracing the tracks from pair 8b2 and using their interactions.
Robert

[Cenoman]

WARNING: This one is very hard. SudokuWiki didn't solve it.

Hi morl,
one must be aware that Andew's solver (sudokuwiki.org) does not use embedded ALS's in the chains used in krakens (called unit forcing nets or cell forcing nets on his site), and does not calculate any multi-kraken (complex net) I needed these techniques to solve the puzzle.

I, for one, don't find this grid difficult. It has a TDP level =1, which means that with a set of two conjugated tracks, it is solved completely. For example by tracing the tracks from pair 8b2 and using their interactions.

I quite don't agree with Robert's evaluation concerning the puzzle difficulty. With a solver running only basic techniques, you see immediately that 8r3c4 is a backdoor and that 8r1c6 leads to a contradiction with singles only. But it is pure T&E and it says nothing, how difficult buildings the tracks might be.
The puzzle is rated S.E. 8.9, and to me it not over-rated. I have tried solutions with AIC's and krakens only. They need very long sequences. So, here is a one-stepper, with a four kraken net (computer generated...)

Code: Select all

 +-------------------------+-----------------------+---------------------------+
 |  1289   1269    1268    |  7      3469   389    |  1239    2349    5        |
 |  179    1569    4       |  69     2      359    |  8       39      1379     |
 |  3      259     257-8   |  48     459    1      |  279     6       479      |
 +-------------------------+-----------------------+---------------------------+
 |  127    8       12367   |  129    1579   2579   |  4       359     1369     |
 |  147    134     9       |  48     1457   6      |  135     358     2        |
 |  5      1246    126     |  3      149    289    |  169     7       1689     |
 +-------------------------+-----------------------+---------------------------+
 |  1249   7       1235    |  1269   8      239    |  23569   23459   3469     |
 |  6      12359   12358   |  129    1379   4      |  23579   23589   3789     |
 |  2489   2349    238     |  5      3679   2379   |  23679   1       346789   |
 +-------------------------+-----------------------+---------------------------+

Kraken row (6)r7c479 demonstrating -6r7c7 => -7r4c3
(6)r7c4 - (6=93)r2c48 - r4c8 = (36-7)r4c39
(6)r7c7
(6)r7c9 - r4c9=(6-7)r4c3

Kraken row (7)r8c579 demonstrating -7r8c7 => -7r3c9
(7)r8c5 - r9c6 = r4c6 - r4c3 = r3c3 - (7)r3c9
(7)r8c7
(7)r8c9 - (7)r3c9

Kraken cell (479)r3c9 demonstrating -7r3c9 => -8r3c3
(4)r3c9 - (4=8)r3c4 - (8)r3c3
(7)r3c9
(9)r3c9-(9=3)r2c8 - r4c8 = (36-7)r4c39 = (7-8)r3c3

Kraken column (5)r578c7
(5)r5c7 - (5=398)r245c8 - r5c4 = (8)r3c4 -(8)r3c3
(5-6)r7c7 => -(7)r4c3 = (7-8)r3c3
(5-7)r8c7 => -(7)r3c9 => -(8)r3c3
=>-8r3c3; ste

[Mauriès Robert]

Hi Cenoman,
You write:

I quite don't agree with Robert's evaluation concerning the puzzle difficulty. With a solver running only basic techniques, you see immediately that 8r3c4 is a backdoor and that 8r1c6 leads to a contradiction with singles only. But it is pure T&E and it says nothing, how difficult buildings the tracks might be.

You can't say that and reduce TDP to T&E. TDP can be used in different ways: in the manner of AICs, in the manner of colouring, etc...
In this example I used it in the way of colouring, that is to say that starting from a pair of candidates whose tracks I see (observation of the puzzle) that the tracks from these candidates had a good potential for development, I gradually exploited the interactions of the two tracks.
A first development is the following one which allows to eliminate 2r4c3 and 1r5c4 see puzzle1.

puzzlz1: Show

We then continue the development of the two tracks, which gives two other eliminations 1r2C1 and 4r5c5 (see puzzle2)

puzzlz1: Show

And so on, we get numerous eliminations and validations that allow us to end the grid with the basics.

Obviously by developing the two tracks a little more each time you get more eliminations at once.

This way of using TDP is not T&E and if the interactions of the two tracks were not enough to conclude, I could use another set of conjugated tracks (or extensions) to make more eliminations or validations. The advantage of the TDP colouring scheme over the AIC scheme is that it allows several eliminations to be made at the same time.

So I maintain that this grid is not very difficult if one practices TDP well.
An element in evaluating the difficulty level of a puzzle is also the number of backdoors that the puzzle contains. This one contains 23 of them!

Sincerely
Robert

[eleven]

My personal definition of T&E is: all which needs a rubber.
If i have to put in auxiliary digits to find eliminations (or a solution), which i have to rubber out again in case of no success, then it's T&E.

[totuan]

Hi All,
Nice to see most of you still here !

My path for this one - ER8.9. After basic SSTS:

Code: Select all

 *-----------------------------------------------------------------------------*
 | 1289    1269    1268    | 7       3469    389     | 1239    2349    5       |
 | 179     1569    4       | 69      2       359     | 8       39      1379    |
 | 3       259     2578    | 48      459     1       | 279     6       479     |
 |-------------------------+-------------------------+-------------------------|
 | 127     8       12367   | 129     1579    2579    | 4       359     1369    |
 | 147     134     9       | 48      1457    6       | 135     358     2       |
 | 5       1246    126     | 3       149     289     | 169     7       1689    |
 |-------------------------+-------------------------+-------------------------|
 | 1249    7       1235    | 1269    8       239     | 23569   23459   3469    |
 | 6       12359   12358   | 129     1379    4       | 23579   23589   3789    |
 | 2489    2349    238     | 5       3679    2379    | 23679   1       346789  |
 *-----------------------------------------------------------------------------*

01- Present as diagram: => r2c1<>1

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DP(17)r245c1
 ||                               
(2)r4c1-r6c23=(2-8)r6c6=r5c4-r3c4=(8-7)r3c3=r2c1*
 ||                         |
(4)r5c1-(4=8)r5c4-----------
 ||
(9)r2c1*

02- (3)r4c8=(36-7)r4c39=r3c3-(7=9)r2c1-(9=3)r2c8 => loop: r4c3<>12, r4c9<>19, r2c2469<>9, r1578c8<>3, some singles
03- => r5c2<>14, r5c2=3

Code: Select all

                            -(8)r3c3=r3c4-(8=4)r5c4*       
                          /                             
(3)r5c2=(3-7)r4c3=(7)r3c3—------r2c1=(7-1)r2c9=r2c2*

04- (5=1)r5c7-r6c9=(1-7)r2c9=r2c1-r3c3=(7-6)r4c3=r4c9-r6c7=r7c7 => r7c7<>5
05- Present as diagram: => r3c4<>4, stte

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(7)r7c5-r5c5=r5c1-r4c3=(7-8)r3c3=r3c4*
 ||                       
(7)r7c9-(7)r3c9
 ||      || 
 ||     (9)r3c9-r2c8=(9-7)r2c1=(7-8)r3c3=(8)r3c4*
 ||      ||
 ||     (4)r3c9*
 ||
(7)r7c7-(5)r7c7=r5c7-(5=8)r5c8-(8=4)r5c4*

Ps: for me, easy puzzles are puzzles I have solved by myself then puzzles that I have not solved yet are the difficult ones so most of puzzles on “daily puzzles” here are very difficult

totuan

[Mauriès Robert]

Hi,

My personal definition of T&E is: all which needs a rubber.
If i have to put in auxiliary digits to find eliminations (or a solution), which i have to rubber out again in case of no success, then it's T&E.

So in this case Cenoman's resolution is T&E, because before finding the right Krakens, he has necessarily made tests that give nothing.
Sincerly
Robert

[eleven]

There is a slight difference.

All i have to remember from (eventually useless) AIC's or Kraken is the outcome, in the case i can use it later. I would not write it into the grid.
Normally I start with a plan to eliminate this or that, and look, if i can split that into digestible steps (by far not being in totuan's league for that).

In the rubber case i just follow this or that assumption (from possibly the same elimination plan) for a hopefully long run, which i cannot do in my head, because very soon it is a complex net. Only after that i would look, if there are common outcomes of different paths. No question, that for hard puzzles this is a very effective way.

You also might remember SpAce's markup method. I wouldn't say, that it looked very clear, when it became complex, but he stated, that no mark became ever wrong (just unnnecesary).

As i said, this is just my personal opinion, and i will not discuss it any further, unless you find a basic mistake in it.

[denis_berthier]

Code: Select all

...|7..|..5
..4|.2.|8..
3..|..1|.6.
---+---+---
.8.|...|4..
..9|..6|..2
5..|3..|.7.
---+---+---
.7.|.8.|...
6..|..4|...
...|5..|.1.

WARNING: This one is very hard. SudokuWiki didn't solve it.

I will not even comment eleven's "definition" of T&E, by which everything you don't like can be called T&E. This is just the standard means used by some to reject the work of others without studying it.
There is a standard definition of T&E and it was given by Ruud, more than fifteen years ago.
In my books, I formalised it into a precise procedure and this leads to a clear broad classification of puzzles according to the depth of T&E their solution requires. The T&E procedure is neither DFS nor BFS and it has nothing to do with backdoors or guessing.

That being said, I don't like tracks (yes, I explicitly write "I don't like"; I don't feel the need to hide this behind pseudo-definitions). The reason I don't like tracks is very clear: they rely on OR branching (from the very start). It may be hard to develop a chain, but tracks require that you develop two chains in parallel; that's enough to settle their case for me.

This puzzle is in T&E(1) and we can therefore be sure a priori that it is solvable by braids (as often in such a case, it is indeed solvable by whips).
It is rated 8.9 by SER and it has W rating 6 (see resolution path below. So, by the two classifications, it is very far from the hardest known ones (SER 11.9) but also very far from being easy. I can easily accept that, for a human player, it is very hard.

Hidden Text: Show

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***********************************************************************************************
***  SudoRules 20.1.s based on CSP-Rules 2.1.s, config = W+SFin
***  using CLIPS 6.31-r761
***********************************************************************************************
...7....5..4.2.8..3....1.6..8....4....9..6..25..3...7..7..8....6....4......5...1.
219 candidates, 1368 csp-links and 1368 links. Density = 5.73080306648234%
hidden-pairs-in-a-column: c4{n4 n8}{r3 r5} ==> r5c4 ≠ 1, r3c4 ≠ 9
finned-swordfish-in-columns: n7{c3 c6 c7}{r3 r4 r9} ==> r9c9 ≠ 7
biv-chain[4]: r5n7{c5 c1} - b1n7{r2c1 r3c3} - r3n8{c3 c4} - c4n4{r3 r5} ==> r5c5 ≠ 4
biv-chain[4]: r5n8{c8 c4} - r3n8{c4 c3} - c3n7{r3 r4} - b4n3{r4c3 r5c2} ==> r5c8 ≠ 3
biv-chain[4]: c4n4{r5 r3} - r3n8{c4 c3} - c3n7{r3 r4} - b4n3{r4c3 r5c2} ==> r5c2 ≠ 4
whip[4]: r6n2{c3 c6} - c6n8{r6 r1} - r3n8{c4 c3} - c3n7{r3 .} ==> r4c3 ≠ 2
whip[4]: r2n3{c9 c6} - r2c8{n3 n9} - r4c8{n9 n5} - c6n5{r4 .} ==> r1c8 ≠ 3
whip[5]: r3n8{c3 c4} - c6n8{r1 r6} - r5c4{n8 n4} - r6n4{c5 c2} - r6n2{c2 .} ==> r3c3 ≠ 2
whip[5]: r2n7{c9 c1} - c3n7{r3 r4} - r4n6{c3 c9} - r4n3{c9 c8} - r2c8{n3 .} ==> r2c9 ≠ 9
whip[6]: c1n8{r9 r1} - c6n8{r1 r6} - r5c4{n8 n4} - c1n4{r5 r7} - c1n2{r7 r4} - b5n2{r4c4 .} ==> r9c1 ≠ 9
whip[6]: c6n8{r6 r1} - r3n8{c4 c3} - c3n7{r3 r4} - r4n6{c3 c9} - r4n9{c9 c8} - r4n3{c8 .} ==> r6c6 ≠ 9
biv-chain[3]: r6c6{n2 n8} - r5c4{n8 n4} - b4n4{r5c1 r6c2} ==> r6c2 ≠ 2
whip[6]: r6n2{c3 c6} - r9n2{c6 c7} - b3n2{r1c7 r1c8} - b3n4{r1c8 r3c9} - r3c4{n4 n8} - b5n8{r5c4 .} ==> r7c3 ≠ 2, r8c3 ≠ 2
whip[6]: r4n6{c9 c3} - c3n7{r4 r3} - r2n7{c1 c9} - b3n1{r2c9 r1c7} - b3n3{r1c7 r2c8} - r4n3{c8 .} ==> r4c9 ≠ 9, r4c9 ≠ 1
biv-chain[4]: c9n1{r6 r2} - r2n7{c9 c1} - c3n7{r3 r4} - r4n6{c3 c9} ==> r6c9 ≠ 6
whip[5]: b6n1{r6c9 r5c7} - b3n1{r1c7 r2c9} - r2n7{c9 c1} - r5c1{n7 n4} - b5n4{r5c4 .} ==> r6c5 ≠ 1
biv-chain[3]: r6c5{n9 n4} - r5c4{n4 n8} - b6n8{r5c8 r6c9} ==> r6c9 ≠ 9
biv-chain[3]: b6n9{r4c8 r6c7} - r6c5{n9 n4} - r1n4{c5 c8} ==> r1c8 ≠ 9
biv-chain[4]: r5c8{n5 n8} - r5c4{n8 n4} - r6c5{n4 n9} - b6n9{r6c7 r4c8} ==> r4c8 ≠ 5
whip[1]: b6n5{r5c8 .} ==> r5c5 ≠ 5
naked-pairs-in-a-column: c8{r2 r4}{n3 n9} ==> r8c8 ≠ 9, r8c8 ≠ 3, r7c8 ≠ 9, r7c8 ≠ 3
biv-chain[4]: c5n6{r9 r1} - r2c4{n6 n9} - c8n9{r2 r4} - r6n9{c7 c5} ==> r9c5 ≠ 9
biv-chain[4]: r6n9{c7 c5} - b5n4{r6c5 r5c4} - r5n8{c4 c8} - r6c9{n8 n1} ==> r6c7 ≠ 1
naked-triplets-in-a-block: b6{r4c8 r4c9 r6c7}{n9 n3 n6} ==> r5c7 ≠ 3
hidden-single-in-a-row ==> r5c2 = 3
biv-chain[4]: r6c7{n6 n9} - c8n9{r4 r2} - r2c4{n9 n6} - b8n6{r7c4 r9c5} ==> r9c7 ≠ 6
whip[4]: r4c9{n3 n6} - r9n6{c9 c5} - c5n3{r9 r1} - c7n3{r1 .} ==> r8c9 ≠ 3
biv-chain[5]: b3n1{r1c7 r2c9} - r6c9{n1 n8} - b5n8{r6c6 r5c4} - b5n4{r5c4 r6c5} - r6n9{c5 c7} ==> r1c7 ≠ 9
biv-chain[5]: c7n6{r7 r6} - r6n9{c7 c5} - b5n4{r6c5 r5c4} - r5n8{c4 c8} - b6n5{r5c8 r5c7} ==> r7c7 ≠ 5
whip[5]: c8n9{r4 r2} - b2n9{r2c4 r1c6} - b2n8{r1c6 r3c4} - c4n4{r3 r5} - r6c5{n4 .} ==> r4c5 ≠ 9
biv-chain[6]: c8n3{r2 r4} - b6n9{r4c8 r6c7} - r6c5{n9 n4} - r5c4{n4 n8} - r6n8{c6 c9} - c9n1{r6 r2} ==> r2c9 ≠ 3
biv-chain[3]: c6n5{r4 r2} - r2n3{c6 c8} - c8n9{r2 r4} ==> r4c6 ≠ 9
whip[3]: b3n9{r3c9 r2c8} - r2n3{c8 c6} - b2n5{r2c6 .} ==> r3c5 ≠ 9
biv-chain[5]: r7n5{c3 c8} - c7n5{r8 r5} - c7n1{r5 r1} - r2c9{n1 n7} - b1n7{r2c1 r3c3} ==> r3c3 ≠ 5
whip[1]: c3n5{r8 .} ==> r8c2 ≠ 5
whip[5]: b5n1{r4c5 r5c5} - c7n1{r5 r1} - b3n3{r1c7 r2c8} - r4n3{c8 c9} - r4n6{c9 .} ==> r4c3 ≠ 1
whip[2]: c9n1{r2 r6} - b4n1{r6c3 .} ==> r2c1 ≠ 1
biv-chain[4]: c5n6{r9 r1} - r2c4{n6 n9} - r2c1{n9 n7} - r5n7{c1 c5} ==> r9c5 ≠ 7
biv-chain[5]: c7n6{r7 r6} - r4n6{c9 c3} - c3n7{r4 r3} - r2c1{n7 n9} - r2c4{n9 n6} ==> r7c4 ≠ 6
hidden-single-in-a-block ==> r9c5 = 6
hidden-single-in-a-block ==> r2c4 = 6
whip[3]: b2n9{r2c6 r1c5} - c5n3{r1 r8} - b8n7{r8c5 .} ==> r9c6 ≠ 9
biv-chain[4]: c5n3{r8 r1} - b3n3{r1c7 r2c8} - r4c8{n3 n9} - b5n9{r4c4 r6c5} ==> r8c5 ≠ 9
biv-chain[3]: c5n9{r1 r6} - r6c7{n9 n6} - c2n6{r6 r1} ==> r1c2 ≠ 9
finned-jellyfish-in-columns: n9{c1 c6 c5 c8}{r2 r7 r1 r6} ==> r6c7 ≠ 9
stte
GRID 0 SOLVED. rating-type = W+SFin, MOST COMPLEX RULE TRIED = W[6]
268739145
914625837
357841269
186257493
739416582
542398671
475183926
621974358
893562714

[Mauriès Robert]

Hi Denis,
Thank you for your insight into the concept of T&E and at the same time as an answer to Eleven's point of view.
I understand that you don't like tracks, but I would like to point out that biv-chains and whips, which I understand correspond to patterns, are equivalent to sets of two conjugated tracks (see definition here) of which only one track is developed (so only one chain).
For example on the resolution you give :

biv-chain[4]: r5n7{c5 c1} - b1n7{r2c1 r3c3} - r3n8{c3 c4} - c4n4{r3 r5} ==> r5c5 ≠ 4
is equivalent to
P(7r5c1) ={7r5c1, 7r3c3, 8r3c4, 4r13c5,...} et P(7r5c5) =(7r5c5, ..} => -4r5c5

whip[4]: r6n2{c3 c6} - c6n8{r6 r1} - r3n8{c4 c3} - c3n7{r3 .} ==> r4c3 ≠ 2
is equivalent to
P(2r6c3) ={2r6c6, 8r1c6, 8r3c3, 7r4c3,...} et P(2r6c23) =(2r6c23, ..} => -2r4c3

I can solve this in several stages without ever developing the second track.
Cordially
Robert

[denis_berthier]

I understand that you don't like tracks, but I would like to point out that biv-chains and whips, which I understand correspond to patterns, are equivalent to sets of two conjugated tracks [...] of which only one track is developed (so only one chain).

Hi Robert
Let's put bivalue chains aside; they are a very specific and elementary reversible pattern (and a special case of whips); they are just the super-symmetric view of the basic AICs (i.e. with no included sub-patterns - only bivalue/bilocal 2D-cells).

A whip is not equivalent to a set of anything, a whip is a single pattern (no more no less than a bivalue chain).
A whip doesn't correspond to patterns, a whip is a pattern.
A whip is a clearly defined logical pattern with no or-branching, contrary to tracks.
Some tracks may not have or-branching, but the possibility of or-branching is inherent in your definition. This possibility is excluded for all the types of chains I've ever introduced.

In what follows, I use T&E in my precise sense, not as a disguised insult.
We already had this discussion. A single track corresponds to T&E starting with a single candidate. A pair of conjugated tracks correspond to two interacting T&E procedures .
It is trivially true that any whip or braid can be seen as a track, because any whip or braid elimination can be obtained by T&E and a single track is a way of doing T&E. In the same way, any forcing whip or braid can be seen as a pair of conjugated tracks or as two interacting T&E procedures.
Of course, something essential to whips or braids is lost in the translation - in particular their inherent notion of length.

In PBCS, I've shown that forcing braids don't have any noticeable additional power wrt braids. And in some previous discussion, you admitted you didn't have any example showing any advantage of a pair of conjugated tracks over a unique track.
This example illustrates this result once more. There is a simple solution with whips; no need of forcing braids or conjugated tracks.

To be honest, your presentation of tracks is so obscure that it's difficult to say when you develop one track or two at the same time.
In the solution of this puzzle, it seems clear you developed two conjugated tracks in parallel.
Only after I gave my whip solution, you transposed it to a notation with tracks. Can you confirm?

[Mauriès Robert]

To be honest, your presentation of tracks is so obscure that it's difficult to say when you develop one track or two at the same time.
In the solution of this puzzle, it seems clear you developed two conjugated tracks in parallel.
Only after I gave my whip solution, you transposed it to a notation with tracks. Can you confirm?

Indeed in my conception of the TDP I do not give precise indications on the possibilities offered (a single track, two conjugated tracks, an anti-track, etc...), to solve a puzzle. I leave that to the choice of each one according to what one likes to use.
What I can say is that the choice of a resolution with conjugated tracks allows a faster resolution than the choice of a resolution with only one track or anti-track. In the same way, one can impose limited track lengths on oneself, which lengthens the number of resolution steps. An Skyscraper is the same as using a short track, but if you want to lengthen the track you can find other eliminations in general, etc...
I try to give various examples of different types of resolutions to those interested in TDP in order to show them its use in different forms and that it is sufficient for solve.
In the case of this puzzle, I have chosen the fastest method (question of time spent on everything I do), that is to say, that of two conjugated tracks that I have developed to the maximum by looking for interactions (eliminations and validations). But I also know how to do with a resolution that uses only anti-tracks of limited length.
Robert

[denis_berthier]

Indeed in my conception of the TDP I do not give precise indications on the possibilities offered (a single track, two conjugated tracks, an anti-track, etc...), to solve a puzzle. I leave that to the choice of each one according to what one likes to use.

Sorry for my late answer, but I've been busy.
In that case, I can't see what's different between TDP and mere colouring.

[Mauriès Robert]

Hi Denis,

In that case, I can't see what's different between TDP and mere colouring.

To say this is to fail to see that TDP includes several methods based on the notions of track and anti-track, i.e. sets of candidates that are built with the TB starting from generator. It's like saying I don't see any difference between your whips theory and the AIC chain theory. It's as if I was saying that there's no point in defining whips when you can define g-whips !
- Among these methods, there is the one consisting in building the solution with a set of two conjugated tracks, by developing the two tracks until we get the solution, even if we use extensions. This is what you call DFS, and I call the "coloring" method. Nothing to do with simple coloring for what I know, but rather with Bernard Borrelly's virtual coloring or GEM.
- There is the "step by step" method using successive conjugated tracks that exploit the different pairs available on the puzzle. We treat an X-wing, a Skyscraper, and other simple models this way, except that we don't need to know which model it is.
- There's also the method of using only one track or anti-track to do step-by-step eliminations, much like you do with your chains and others with AICs. A target attached to a whip is an invalid track generator. It is no more complicated to detect these targets with tracks than with whips. I have also shown in different examples on this forum how to use an anti-track that does the same job as AICs.
Sincerely
Robert

[denis_berthier]

Hi Robert,
Sure, one can use different colouring schemes and when you start from two candidates there's even no reason to consider only bivalue/bilocal ones.

Among these methods, there is the one consisting in building the solution with a set of two conjugated tracks, by developing the two tracks until we get the solution, even if we use extensions. This is what you call DFS,

Just a point of vocabulary. There nothing I call DFS. DFS has been known in Computer Science for decennies. And it doesn't start with a conjugated pair.