The New Sudoku Players' Forum

[Fredrik]

I have finally been able to develop chains through coloring. However, with the chain completed, I have to resort to trial-and-error to find which color is true and which is false.

What is the logical way to contine without trial-and-error?

:?: [size=18][/size][size=9][/size]

[Karyobin]

Presumably at the beginning of your chain you write a little note for yourself somewhere saying, for example, 'right, for this chain, red means it is x and blue means it isn't'. Then when you get to the end of your chain you find the little note and remind yourself of what you've just discovered.

That's what I'd do anyway. Probably because the colour doesn't actually mean anything by itself.

[simes]

But if you know that red means it is X, then why both with the chain? Why not just put X into the cell?

The reason is that you don't know which links in the chain are true and which are false. You construct the chain hoping you can find exclusions without knowing which are the true links, or to find contradictions and so determining which are the true links.

Simes

[Fredrik]

Thanks for feedback. Some insight gained.

Assume I have generated a chain with four pairs of true/false. No color doubles in any row or column. At this poin, as you said, I don't know which of the colors represents true. At the same time, I see no "violations".

When attempting trial-and-error, I arbitrarily try one color out (on Angus's Simple Sudoku program). 50/50 I get a correct answer. If I don't, I try the other color, and will of course get it right.

What tells me, without this trial, which color is "right"?

[Jeff]

With 'Angus solving with colours', the idea is not to determine which colour is true and which is false, but to find exclusions in cells that are inside the intersection of 2 different colours. Sometimes, an exclusion will in turn confirm that one of the colours is true.

[angusj]

With 'Angus solving with colours', the idea is not to determine which colour is true and which is false, but to find exclusions in cells that are inside the intersection of 2 different colours.

That's certainly one thing to look for with colors. Another thing to look for is two cells of the same color in the same group. When that happens, that color must be the 'false' color.

Note: Just to clarify - it's not my colours - I've just documented (and perhaps promoted the technique) described by others elsewhere.

[Karyobin]

Hanging head in shame. Suffered from catastrophic jargon failure. Hoping to resume normal service soon.

[simes]

Fredrik,

My explanation can be found here, perhaps it'll help.

Simes

[emm]

Karyobin - And I thought you were one of the geniuses (?genii)

[Karyobin]

Me? Certainly not one of the genii loci. I merely try to embody the zeitgeist, and get told off for it.

[emm]

Ah, the burdens we bear!

[simes]

By PM:

Thanks,

Your web site DID help.

My blockage was in believing that the chain (e.g. of 8's) was to "solve its own true/false" rather than that the members of the chain would serve to knock off the number 8's NOT part of the chain.

I have crawled through sveral examples and DO get them right, eventually. However, I have difficulties picking the proper chain members, so as to zero in on those I want to eliminate.

Most of the time, I find that the chain "members" I am using to eliminate non-chain numbers are of the same color (to be expected, if I linked them properly). I.e., if I am intersecting a target using two numbers I have assigned to be TRUE, what should be my conclusion as to the T/F of the target, and how would that go if I had assigned them a FALSE status.

Thanks for your help,

Fred

Fred,

(I hope you don't mind me posting your PM, and replying in the public forum, but the information could also be useful to other readers of this thread.)

If I understand you correctly, and you are eliminating non-chain candidate because they are at the intersection of two same-colour chain members, then you're doing it wrong!

Any non-chain candidate to be eliminated must be at the intersection of different colour chain members. Then we know that whichever colour members in the chain are the true ones, one of them will exclude the candidate at the intersection. This is an "exclusion".

The other useful information that can sometimes be found through colouring is a contradiction. If, when colouring the chain, you come across a unit (row, column or block) that contains two nodes of the same colour, then you know that colour must be false. You can then eliminate the candidate from all cells of that colour. This is because as the unit cannot contain two cells with the same value, both nodes cannot be true, so they must both be false. The other colour must be the true one.

S

[Fredrik]

Simes,

No problem with the public reply. Thanks for the quick turnaround.

My message was ambiguous in that it conveyed that I eliminated a non-chain number with two chain numbers of the same color. What I ment to convey, was that each time I constructed a chain, I seemed to end up with two chain members of the same color when attempting to eliminate the non-chain number.

Where I now need help, if the problem lends itself to that, is in setting up a chain so that I can eliminate a non-chain number with two different chain members of different colors. Is there a trick to setting up the chain properly (so as to get the different colors aiming at the elimination candidate), or is it just brute force?

Again, thanks for taking time.

fred

[simes]

No, there's no trick, it just takes a suitable puzzle. You can't set up the chain to do something specific - the chain is either there or not, and if it's there, it's either useful or not. You just have to take it a you find it.

S