## August 5, 2015

Post puzzles for others to solve here.

### Re: August 5, 2015

daj95376 wrote:
blue wrote:I hoped to generate some discussion about the idea of using a strong link between locked sets that overlap in a cell (or cells?), but that don't share any candidates. I don't recall seeing anything like that before ... which isn't saying a lot.

Good luck. After SteveG48's solution to the July 25th puzzle, I've given up on trying to discuss most notation. However, I normally do really well following your notation, so I knew that I wanted to discuss your solution when I got lost following the notation.

If you do get a successful discussion going on notation, then I hope you manage to include the various possible interpretations of "nm". I think it has progressed past being a pair in two cells.

BTW: I updated my chain's notation in hopes of generating a discussion (as well).

Code: Select all
` Legend: -( 4 | 6 ) --> ( -4 & -6 )         -( 4 & 6 ) --> ( -4 | -6 ) discontinuous loop: 9r9c3 =UR= 3r7c8 - (3=4&6)r147c7 - (4|6=137)r2c279 - (1=9)r279c3`

Bottom Line: Nothing replaces a proper network diagram (using multiple lines) at times.

_

Thanks again for the help Danny.

9r9c3=3r7c8 a strong link -- both can be true! One of these days I will learn that.

If 3r7c8 is true then 9r9c3 is also true.
if 3r7c8 is not true then 9r9c3 has to be to break the ur. The discontinuous loop is two strong links at 9r9c3.

I believe this to be true
Candidates (-aaa) have an "or" | relationship.
Candidates (=bbb) have an "and" & relationship.

It does not hurt to put the | or & in but I still don't see why it is needed.

I appreciate your patience with me.
dan

ArkieTech

Posts: 3236
Joined: 29 May 2006
Location: NW Arkansas USA

### Re: August 5, 2015

daj95376 wrote:If you do get a successful discussion going on notation, then I hope you manage to include the various possible interpretations of "nm". I think it has progressed past being a pair in two cells.

BTW: I updated my chain's notation in hopes of generating a discussion (as well).

Code: Select all
` Legend: -( 4 | 6 ) --> ( -4 & -6 )         -( 4 & 6 ) --> ( -4 | -6 ) discontinuous loop: 9r9c3 =UR= 3r7c8 - (3=4&6)r147c7 - (4|6=137)r2c279 - (1=9)r279c3`

I see that you've dropped the '5' from 456r147c7, and the '45' from 459r279c3.
I'm slightly curious why you didn't drop the '37' from 137r2c279 as well:
9r9c3 =UR= 3r7c8 - (3=4&6)r147c7 - (4|6=1)r2c279 - (1=9)r279c3
This bit that's indented here, is a bunch of "stream of thought" stuff, that I wrote, and was about to delete.
I decided to leave it in, in case it has an "entertainment value" for anyone.
That way you get something where the left/right reversal is unquestionably:
(9=1)r279c3 - (1=4|6)r2c279 - (4&6=3)r147c7 - 3r7c8 =UR= 9r9c3
From there, if you like, you can fill in details that make it easier to follow (in the PM), for a left to right reading:
(9=145)r279c3 - (1=37(4|6))r2c279 - (4&6=35(4|6))r147c7 - 3r7c8 =UR= 9r9c3
What a mess, right ? It's technically correct, though ... unless I've made a stupid mistake.

At that point, you'ld probably want to look back at the PM grid, and try to find a different ALS grouping for the same cells -- one that gives a more readable chain.
(9=45)r79c3 - (4=137)r2c239 - (1|7=3456)r1247c7 - 3r7c8 =UR= 9r9c3
Then if you like, you can drop some things out, to give:
(9=4)r79c3 - (4=1&7)r2c239 - (1|7=3)r1247c7 - 3r7c8 =UR= 9r9c3
BTW: I wasn't trying to make a point in any of that, other than maybe that when it comes to chains incorpoating ALS's (or worse), the left/right reversals aren't always easy to follow, even if they're easy to write down.

About 'mn' notation, I don't have much to say, except:
1) I don't really like seeing (ab=cd) for a 2-cell ALS, since it looks symmetric on its face, but the logic can only be (a|b = c&d) or (a&b = c|d), neither of which is symmetric. On the other hand, usually it's clear fromt he context, what is meant, and it's usually (a|b = c&d). In that case, I'ld probably still rather see (a|b=cd). [ That's all just personal opinion/preference, of course. ]
2) If you use 'mn' with an ALS of 3 or more cells, then you should probably include an '&' or '|' symbol. Again though, if it's used like (m|n=??), then it's usually clear from the context, what (mn=??) would mean -- clear at least, as long as the '??' part has as many digits as there are ALS cells.
3) I like what Dan just wrote:
I believe this to be true
Candidates (-aaa) have an "or" | relationship.
Candidates (=bbb) have an "and" & relationship.

It seems like a reasonable rule of thumb for left to right reading, and so a reasonable choice for writing as well.
[ The intended meaning, would be clear from the context, at least. ]
blue

Posts: 838
Joined: 11 March 2013

### Re: August 5, 2015

blue wrote:
daj95376 wrote:If you do get a successful discussion going on notation, then I hope you manage to include the various possible interpretations of "nm". I think it has progressed past being a pair in two cells.

BTW: I updated my chain's notation in hopes of generating a discussion (as well).

Code: Select all
` Legend: -( 4 | 6 ) --> ( -4 & -6 )         -( 4 & 6 ) --> ( -4 | -6 ) discontinuous loop: 9r9c3 =UR= 3r7c8 - (3=4&6)r147c7 - (4|6=137)r2c279 - (1=9)r279c3`

I see that you've dropped the '5' from 456r147c7, and the '45' from 459r279c3.
I'm slightly curious why you didn't drop the '37' from 137r2c279 as well:
9r9c3 =UR= 3r7c8 - (3=4&6)r147c7 - (4|6=1)r2c279 - (1=9)r279c3

I'm still trying to adjust to using minimal linking information. So, I missed reducing the '137' to '1'. I'm not sure if reducing complex ALS terms will ever catch on (or should), but everyone does it without thinking in less complex scenarios. Consider this chain segment:

... - 7r3c8 = (7-6)r2c7 = (6)r2c1 = ...

Here the candidates <14> are ignored in r2c7 ... unless ... the chain segment is part of a continuous loop. Even then, people pick up the eliminations -14r2c7 without correctly representing the bidirectional notation in r2c7. Here, the notation is minimal and no one thinks about it.

This bit that's indented here, is a bunch of "stream of thought" stuff, that I wrote, and was about to delete.
I decided to leave it in, in case it has an "entertainment value" for anyone.
That way you get something where the left/right reversal is unquestionably:
(9=1)r279c3 - (1=4|6)r2c279 - (4&6=3)r147c7 - 3r7c8 =UR= 9r9c3
From there, if you like, you can fill in details that make it easier to follow (in the PM), for a left to right reading:
(9=145)r279c3 - (1=37(4|6))r2c279 - (4&6=35(4|6))r147c7 - 3r7c8 =UR= 9r9c3
What a mess, right ? It's technically correct, though ... unless I've made a stupid mistake.

At that point, you'ld probably want to look back at the PM grid, and try to find a different ALS grouping for the same cells -- one that gives a more readable chain.
(9=45)r79c3 - (4=137)r2c239 - (1|7=3456)r1247c7 - 3r7c8 =UR= 9r9c3
Then if you like, you can drop some things out, to give:
(9=4)r79c3 - (4=1&7)r2c239 - (1|7=3)r1247c7 - 3r7c8 =UR= 9r9c3
BTW: I wasn't trying to make a point in any of that, other than maybe that when it comes to chains incorpoating ALS's (or worse), the left/right reversals aren't always easy to follow, even if they're easy to write down.

Yes, I encountered several scenarios for the notation as well. I'm a firm believer that a coherent l-to-r representation is sufficient. I believe that the creator of a chain should have reasonable expectations that the reader can fill in "and/or logic" for multiple candidates without the need for special symbols. However, creating cumbersome notation for the express purpose of reducing the number of terms, is not acceptable.

Here is the fundamental sequence that I see before "gathering terms":

(9=45)r79c3 - (4=1*)r2c3 - (1=3)r2c9 - (3=7)r2c2 - [(*17=46)r2c7 + 456r14c7] - (45=3)r7c7 - 3r7c8 =UR= 9r9c3

There are many permutations for gathering terms in this sequence. Some are more confusing than others.

About 'mn' notation, I don't have much to say, except:
1) I don't really like seeing (ab=cd) for a 2-cell ALS, since it looks symmetric on its face, but the logic can only be (a|b = c&d) or (a&b = c|d), neither of which is symmetric. On the other hand, usually it's clear from the context, what is meant, and it's usually (a|b = c&d). In that case, I'ld probably still rather see (a|b=cd). [ That's all just personal opinion/preference, of course. ]
2) If you use 'mn' with an ALS of 3 or more cells, then you should probably include an '&' or '|' symbol. Again though, if it's used like (m|n=??), then it's usually clear from the context, what (mn=??) would mean -- clear at least, as long as the '??' part has as many digits as there are ALS cells.
3) I like what Dan just wrote:
I believe this to be true
Candidates (-aaa) have an "or" | relationship.
Candidates (=bbb) have an "and" & relationship.

It seems like a reasonable rule of thumb for left to right reading, and so a reasonable choice for writing as well.
[ The intended meaning, would be clear from the context, at least. ]

As for your (1) issue, I have an expanded notation for it. I didn't include it earlier because it wasn't used.

Code: Select all
` Legend: -( 4 | 6 ) --> ( -4 & -6 )         -( 4 & 6 ) --> ( -4 | -6 )         -( 4 + 6 ) --> ( -4 & -6 )         =( 4 + 6 ) --> ( =4 & =6 )`

Now, (ab=cd) for a 2-cell ALS, exists as a fully qualified (a+b=c+d) in a 2-cell ALS.

At this point, I'll repeat:

I believe that the creator of a chain should have reasonable expectations that the reader can fill in "and/or logic" for multiple candidates without the need for special symbols. Thus, I would consider (ab=cd) to be sufficient for a 2-cell ALS.

_
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

### Re: August 5, 2015

Experience shows that the more things that are implied rather than properly expressed in a notation the more likely it will be that a chain a) will contain errors leading to withdrawn posts or b) cause misunderstandings leading to fairly trivial discussions. If every argument in a chain is a clear Boolean it might add a few characters but the writing discipline involved would strangle many logical errors at birth and the unambiguous chain would be easier for readers at all levels to follow and validate.

Things would be different if there were source references available for each person's notation style, but as it is, this discussion will be soon lost in the jungle. How then will a newcomer be able to understand anything? For example it took me some time to find <this thread from Jan 2013> and I knew what I was looking for. Those posts appeared to sort some of these issues out, but now they're all forgotten and the efforts taken wasted.

I've lost count of how many times I've lost count of the number of times these recurring points have been raised by some experienced players.
David P Bird
2010 Supporter

Posts: 1040
Joined: 16 September 2008
Location: Middle England

### Re: August 5, 2015

David P Bird wrote:... How then will a newcomer be able to understand anything? ...

Sorry, but i can't hear this newcomer argument anymore.
Newcomers don't know, what an AIC is. It takes them weeks to learn it. And it is not needed to understand an elimination.
They could just follow a normal implication stream like this:
Code: Select all
`+---------------------+----------+----------------------+| 4568  9     2458    | 1  36  7 | (456)   2456    235  || 46    (37)  (14)    | 5  9   2 | (1467)  8       (13) || 56    37    125     | 8  36  4 | 9       2567    1235 |+---------------------+----------+----------------------+| 3     2     89      | 4  7   1 | (56)    569     58   || 19    5     7       | 6  2   8 | 13      39      4    || 18    4     6       | 3  5   9 | 17      27      128  |+---------------------+----------+----------------------+| 7     1     (45)*   | 2  8   6 | (345)   (3)45*  9    || 2     6     3       | 9  4   5 | 8       1       7    || 459   8     (9)-45* | 7  1   3 | 2       45*     6    |+---------------------+----------+----------------------+`

UR 45r79c48: 9r9c3 or 3r7c8 (at least one must be true)
3r7c8 (-> r7c7=45) -> triple 456r147c7 (-> r2c7=17) -> triple 137r2c279 (-> r2c4=4) -> pair 45r27c3 -> r9c3=9
=> r9c3=9 (in both cases)
eleven

Posts: 2075
Joined: 10 February 2008

### Re: August 5, 2015

eleven wrote:Sorry, but i can't hear this newcomer argument anymore.
Newcomers don't know, what an AIC is. It takes them weeks to learn it. And it is not needed to understand an elimination.
My point is simple, this forum must attract new players as old ones drop out otherwise it will die.

Unlike you, many quite experienced players are unable to apply Boolean logic beyond the most basic patterns because they lack a discipline to follow. In my eyes that discipline is to define the Booleans that must be true or false and to ensure every weak and strong link is sound with no possible exceptions.

UR 45r79c48: 9r9c3 or 3r7c8 (at least one must be true)
3r7c8 (-> r7c7=45) -> triple 456r147c7 (-> r2c7=17) -> triple 137r2c279 (-> r2c4=4) -> pair 45r27c3 -> r9c3=9
=> r9c3=9 (in both cases)

I find you leave a lot of gaps for the reader to bridge, eg r9c3 & r9c8 don't appear anywhere in the main stream.

Applying a disciplined approach there are no gaps
(9=45)r79c3 -UR- (45=3)r79c8 - (3=456)r147c7 - (46#1=17#1)r2c7 - (137=4)r2c239 - (45=9)r79c3 => r9c3 = 9
or shorter
(9=45)r79c3 -UR- (45=3)r79c8 - (3=456)r147c7 - (46#1)r2c7 = (6,4)r2c1,r2c3 - (45=9)r79c3 => r9c3 = 9

The reader can take these in smaller bites and validate every link far more easily. Although it doesn't concern you, they're also clearly bidirectional. OK, the notation must be learnt (as would yours) but it enables a longer journey.

[Edit typos in chains pointed out by ronk]
Last edited by David P Bird on Fri Aug 07, 2015 7:06 pm, edited 2 times in total.
David P Bird
2010 Supporter

Posts: 1040
Joined: 16 September 2008
Location: Middle England

### Re: August 5, 2015

blue wrote:I hoped to generate some discussion about the idea of using a strong link between locked sets that overlap in a cell (or cells?), but that don't share any candidates. I don't recall seeing anything like that before ... which isn't saying a lot.

I'm happy to discuss this topic with you. We could shorten the number of links in this chain considerably by using an AMSLS to provide the link that you suggest. Consider the native sets: (1)r2, (3)r2, (7)r2, (4)c3, (5)c3, (3)c7, (4)c7, (5)c7, (6)c7. There is a maximum of 9 possible truths in the 8 cells 2N2379, 7N3, 147N7. So if either 4 or 5 is false in column 3, then 3 must be true in column 7. The other direction reads that if 3 is false in column 7, then both 4 and 5 are true in column 3. Hence:

(45)r27c3 = (3)r7c7[AMSLS133445567:2n2379,7n3,147n7] - (3)r7c8 = (9)r9c3[AUR45:r79c38] => r9c3 <> 45.

The link is possible, but a larger number links of lower complexity is much easier to follow IMHO.

blue wrote:9r9c3 =UR= 3r7c8 - (3=4&6)r147c7 - (4|6=1)r2c279 - (1=9)r279c3

Don't you mean: 9r9c3 =UR= 3r7c8 - (3=4&6)r147c7 - (4|6=1)r2c279 - (1=45)r27c3 => r9c3 <> 45.
FWIW, I much prefer this form where superfluous ALS candidates that play no part in the elimination are dropped from the notation.

I agree. a&b means a and b. a|b means a or b or both (DPB uses ab#1 for this). In typical mathematical shorthand a&b can be contracted to just ab. & and | work fine when there are only two candidates, but they can't match DPBs abc#2 notation when at least 2 of abc are true for example.

P.S. DPB beat me in the race to post. What he said.
sultan vinegar

Posts: 81
Joined: 27 August 2013

### Re: August 5, 2015

David P Bird wrote:OK, the notation must be learnt (as would yours) but it enables a longer journey.

Unfortunately i can't ask a newcomer, what is easier to understand for him - because they see all the AIC's and run away
eleven

Posts: 2075
Joined: 10 February 2008

### Re: August 5, 2015

eleven wrote:
David P Bird wrote:OK, the notation must be learnt (as would yours) but it enables a longer journey.

Unfortunately i can't ask a newcomer, what is easier to understand for him - because they see all the AIC's and run away

Did you ask me? I'm a newcomer and I read MJs AIC guide. I can categorically say that MJs rigid disciplined water-tight logical approach works.
sultan vinegar

Posts: 81
Joined: 27 August 2013

### Re: August 5, 2015

David P Bird wrote:Applying a disciplined approach there are no gaps
(9=45)r79c3 -UR- (45=3)r79c8 - (3=456)r147c7 - (46#1=17#1)r2c2 = (137=4)r2c239 - (45=9)r79c3 => r9c3 = 9

Oops, three strong links in a row, r2c7 not mentioned, ...
ronk
2012 Supporter

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Location: Southeastern USA

### Re: August 5, 2015

ronk wrote:Oops, three strong links in a row, r2c7 not mentioned, ...
Thanks ronk, corrected now.
David P Bird
2010 Supporter

Posts: 1040
Joined: 16 September 2008
Location: Middle England

### Re: August 5, 2015

sultan vinegar wrote:Did you ask me? I'm a newcomer ...

Simply because you have learned that stuff on the Eureka forum. So you are no real newcomer, but biased.
You can notate your things as you want, so far i understood each notation. But i insist, that knowing AIC's is no requirement (and only a very limited help) for being a good solver.
Letting others know, that they would need to learn it, is all but newcomers friendly.
eleven

Posts: 2075
Joined: 10 February 2008

### Re: August 5, 2015

eleven wrote:
sultan vinegar wrote:Did you ask me? I'm a newcomer ...

Simply because you have learned that stuff on the Eureka forum. So you are no real newcomer, but biased.
You can notate your things as you want, so far i understood each notation. But i insist, that knowing AIC's is no requirement (and only a very limited help) for being a good solver.
Letting others know, that they would need to learn it, is all but newcomers friendly.

That premise might be true when solving puzzles as easy as these are. In other words, you can get away with implying that your apparent preferred non-Eureka method (over the years) of transmitting solutions is just as good as with AICs when the solutions are relatively short. But the difficulty of expressing an overall solution in a way that others can understand is proportional to the number of steps and complexity required. It's the same reason that we have common spoken languages- otherwise understanding and/or transmitting anything but the most simple ideas would be impossible.

Point being, that maybe a 'several-steps' solution using your non-AIC 'notation' would be ingenious, but how would most of us know since the logic becomes harder to follow the more 'ifs', 'thens' and arrows there are. For instance, try expressing a solution to Gurth's December 23, 2014 ER=8.3 puzzle using your preferred, non-Eureka (AIC) method and let's see how it looks:

Code: Select all
`*-----------------------------------------------------------------------------* | 1456    356     9       | 4567    4567    8       | 1357    2       3457    | | 7       3568    12348   | 24569   2456    23459   | 1359    3459    345     | | 245     35      234     | 24579   1       234579  | 8       34579   6       | |-------------------------+-------------------------+-------------------------| | 259     3579    6       | 8       2457    24579   | 357     1       3457    | | 1258    578     1278    | 12457   3       2457    | 567     4567    9       | | 159     4       137     | 1579    57      6       | 2       357     8       | |-------------------------+-------------------------+-------------------------| | 3       1       5       | 467     9       47      | 67      8       2       | | 4689    6789    478     | 2567    25678   257     | 35679   35679   1       | | 689     2       78      | 3       5678    1       | 4       5679    57      | *-----------------------------------------------------------------------------*`
DonM
2013 Supporter

Posts: 475
Joined: 13 January 2008

### Re: August 5, 2015

Don,

it is not clear to me, what you are out for. First of all, this is hardly a puzzle for newcomers.

I looked at your solution here and cannot find a step, which is not easy to write with "normal" logic symbols.
E.g. step 5:
Code: Select all
`+----------------------+----------------------+----------------------+| 456    356    9      |h457   h45     8      | 1      2      3457   || 7      8      1      | 26-45  26-45  239-45 |a359   a3459  a345    || 245   d35     24     |h4579   1     c34579  | 8     b34579  6      |+----------------------+----------------------+----------------------+| 259   e579    6      | 8      2457   24579  | 357    1      3457   || 1258  e57     278    | 12457  3      2457   | 567    4567   9      ||f19     4      3      |g19     57     6      | 2      57     8      |+----------------------+----------------------+----------------------+| 3      1      5      | 467    9      47     | 67     8      2      || 4689   679    478    | 2567   25678  257    | 35679  35679  1      || 689    2      78     | 3      5678   1      | 4      5679   57     |+----------------------+----------------------+----------------------+`

5. als(459=3)r2c789-r2c6=r3c6-(3=5)r3c2-als(5=79)r45c2-(9)r6c1=r6c4-als(9=457)r3c6,r1c45 => -45r2c456

ALS 3459r2c789: triple 459r2c789 or 3 in r2c789
3r2c78 (-> r3c8<>3) -> 3r3c6 -> 5r3c2 -> 79r45c2 -> 1r6c1 -> 9r6c4 -> triple 457r1c45,r3c4
=> r2c456<>45

Step 10 is wrong, you missed the 2 in r2c4 (which is hidden in your "irresistible" picture).
Code: Select all
`+----------------------+-----------------------+----------------------+| 456    356    9      | *57     45     8      | 1      2      37     || 7      8      1      | #26     26     39     | 59     49     345    || 245    35     24     | *4579   1      34579  | 8      79     6      |+----------------------+-----------------------+----------------------+| 259    579    6      |  8      2457   24579  | 3      1      457    || 1258   57     278    |*#12457  3      2457   | 67     4567   9      || 19     4      3      |  19     57     6      | 2      57     8      |+----------------------+-----------------------+----------------------+| 3      1      5      |  467    9      47     | 67     8      2      || 4689   679    478    |*#2567   25678  257    | 59     3      1      || 689    2      78     |  3      5678   1      | 4      5679   57     |+----------------------+-----------------------+----------------------+`

10. grp(5)r13c4=aHP(25-1)r58c4=(1-9)r6c4=(9-4)r3c4=(4)r1c5 => r1c5<>5=4 -> r4c9=4

Also note, that if there was no 2 in r2c4 (and the chain correct from left to right), it would not be correct from right to left in this form:
1r6c4=(1-hp25)r58c4 then must be replaced by 1r6c4=(hp12-5)r58c4
So maybe better is (hp25-hp12)r58c4, which is correct, but unnecessarily hard to read.
eleven

Posts: 2075
Joined: 10 February 2008

### Re: August 5, 2015

eleven wrote:Don,
it is not clear to me, what you are out for.

Apparently, though I don't think I could have made it more clear! While I appreciate your error-checking my solution and finding what is, by any measure, a careless error (thankfully in one of the final steps rather than early steps), the challenge was not to error-check my solution. I will address corrections to my solution of Gurth's puzzle in a separate thread.

Let's refresh. This was your post to sultan vinegar:
'Simply because you have learned that stuff on the Eureka forum. So you are no real newcomer, but biased.
You can notate your things as you want, so far i understood each notation. But i insist, that knowing AIC's is no requirement (and only a very limited help) for being a good solver.
Letting others know, that they would need to learn it, is all but newcomers friendly
.'

The first statement is a dismissive statement implying that anyone using Eureka notation is biased as if Eureka notation is of minor value and there are realistic alternatives, presumably your 'freehand' method. This general dismissal of Eureka notation (and therefore promotion of your own as of equal value) is a continuation of several of your posts going back several years.

Ordinarily, I would have no interest in once again addressing this tiresome agenda of yours, but since 'newcomers' are mentioned, I don't want newcomers misled into thinking that learning Eureka notation is not necessary or is a waste of time when it comes to taking part in a sudoku-solving forum.

So, I'll both restate and further clarify my point(s):

1. Even though it is true that it is possible for a good solver to not know AICs, it is categorically false that knowing the present standard notation is (to quote 'eleven') 'only a very limited help'. Learning and understanding a standard notation means that one can grow from a 'newcomer' solving status to an excellent solver by following the solutions of other more experienced solvers who are using the standard notation.

2. Even with one-steppers, the more complex the step particularly nets, the more difficult it will be for others to understand it if you are using your own, non-standard (ie. not Eureka-AIC) notation.

3. When it comes to puzzles beyond one-steppers, it will be increasingly difficult, as the more steps are required or the more complex the solution is (even if a few steps) for anyone to know if you are a good solver if you are using your own non-standard notation (such as your freehand method), because those using a standard notation will be less likely to even read it or, particularly for newcomers, it will just be too hard to follow. Not to mention that the less standard the notation, the less others are able to find mistakes in that solution.

So, eleven, my challenge to you was not to error-check my solution (though I do appreciate it) nor was it to in any way use my solution. It was to come up with your own entire solution of Gurth's puzzle using your freehand method and let us see how easy it is to follow (for newcomers or allcomers) vs. Eureka-AIC notation.

Finally, my guess is that you were able to quickly error-check my solution because the notation was so clear. Also, it did not surprise me that you could take one of my steps and express it in your freehand method because each of my steps were relatively simple non-network chains. But again, doing that was not the challenge.

EDIT: Oh yes: a fair comparison would exclude the use of nets.
DonM
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Posts: 475
Joined: 13 January 2008

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