dailysudoku.com

[Earl]

A multi-task puzzle. I could not find a single step solution.

Earl

[Asellus]

I found a 3-step path. Anyone come up with something shorter?

[storm_norm]

[Marty R.]

Thanks Earl, I had fun with that one.

XY (seemed useless)

Two ERs

XY with pincer coloring

XY

At this point there was a BUG+1 pattern, but I finished it off with another XY.

[Asellus]

One of my steps is a not often used technique present in boxes 4 and 7. See if anyone can spot it.

The other is a Skyscraper which is present from the beginning and independent of the first step.

These two result in a BUG+1.

[storm_norm]

[tlanglet]

The best I have achieved is to use multi-coloring on <8> which eliminates several cells and after more basics, an xy-wing on <348> with pivot at r3c8 deleting the <3> at r1c4 finished it off.

Ted

[tlanglet]

Norm,

I believe that we did very similar operations on <8>. You used extensions to a kite and I used multi-coloring. The <348> xy-wing I used to complete the puzzle is evident in your code after the kite eliminations.

Ted

[storm_norm]

[Asellus]

[storm_norm]

[cathyv]

Help please - I got this far and can trace the 89 pair marked with * in the post above, but you guys lost me with how you extended it from there. I'm learning...but these APEs and ALSs and long chains are still a struggle. Thanks!

[storm_norm]

cathyv, hello.
in the above image is my chain, its a long, long, long chain to try and get a grip on. its like jumping into the deep end of the W-wing pool.

the green line is the Kite on the number 8 (as you said you figured out).
When you have this kite that "sees" both {8,9}'s in r4c1 and r5c5, all this means is that either 9 is true ( in r4c1 and r5c5 )

with this knowlegde, you can "extend" A.K.A. "color" those 9's.

first is the "blue" lines
the 9 in r4c1, if that 9 is true, that means,
that the 9 in r1c1 is not true
this means that the 8 in r1c1 is true... remember that...

next is the red lines
xx over to the 9 in r3c5, if that 9 is true
leads to r5c5 is not 9
" r5c5 is 2
" r9c5 is not 2, r9c5 is 8
" r9c3 is not 8, r9c3 is 2
" r7c2 is not 2, r7c2 is 8, so this 8 is true.

now we have the same situation as we did with the original 9's. both the 8 in r7c2 and r1c1 can be true, thus we can eliminate the 8 in r1c2.

note: you don't need this "overkill" move to continue solving this puzzle from your posted position. if you understand xy-chains and xy-wings, then I think 2 or 3 xy-chains should do the trick.
also, the challenge was made to solve it in less than 3 moves which prompted this type of chain. I was trying to "nuke" it.

[Asellus]

Hi cathyv,

Norm's chain is an interesting creation. But, if it is bewildering you, then perhaps you'd prefer to look at coloring of <8>s. Let's start with basic coloring. Perhaps you are familiar with the technique already. But, just in case, here is what I've done: I've marked a sequence of end-to-end conjugate pairs of <8>s in the grid using "A" and "a" to label the opposite pairs, or "polarity". A conjugate link or conjugate pair (also commonly called a strong link) is an "either/or" relationship: one must be true and one must be false. This is the case when there are only two instances of a particular candidate digit within a house (row, column or box).

[Asellus]

Norm,

Your notation is very close to Eureka notation. In fact, the only difference that matters is that you aren't alternating the inference symbols. We know that the link between those two <8>s in r3 must be weak, so it must be notated using "-". Then, we could take what you've written and just alternate the symbols:

(8)r1c1=(9)r1c1-(9)r4c1=(8)r4c1-(8)r5c3=(8)r5c9-(8)r4c8=(8)r3c8-(8)r3c5=(9)r3c5-
(9)r5c5=(2)r5c5-(2)r9c5=(8)r9c5-(8)r9c3=(8)r7c2 => r1c2<>8

We now know for sure that there is a strong inference between the <8>s on the ends of the chain without counting links. It is important to realize that conjugate links are both weak and strong. When they are joined up end-to-end in a chain, they still have alternating weak and strong inferences. You can shorten the notation by combining links contained within a single cell into a single set of parentheses:

(8=9)r1c1-(9=8)r4c1-(8)r5c3=(8)r5c9-(8)r4c8=(8)r3c8-(8=9)r3c5-
(9=2)r5c5-(2=8)r9c5-(8)r9c3=(8)r7c2 => r1c2<>8

This would be acceptable Eureka to me. Some prefer to write these in "discontinous loop" form, that is with the victim(s) weakly linked at both the beginning and end of the chain:

(8)r1c2-(8=9)r1c1-(9=8)r4c1-(8)r5c3=(8)r5c9-(8)r4c8=(8)r3c8-(8=9)r3c5-
(9=2)r5c5-(2=8)r9c5-(8)r9c3=(8)r7c2-(8)r1c2; r1c2<>8

I've changed your "implies" symbol, the "=>", to a semicolon. This is what is usually used in Eureka. The victims are at a discontinuity in the loop because the links on both sides of them are weak rather than alternating. This means they must be false. (You can also have a discontinuity where the links on both sides are strong. This causes the entities at the discontinuity to be true. Chains of this sort aren't seen as often.)

From what I have seen, there is some variety in how people notate chains within this general Eureka approach. And, that is fine with me. I understand them even if they aren't notated exactly as I would do it. For example, I use braces to group different candidates together since that's how I have seen it done in several places. Recently, Myth Jellies posted a chain notated with ampersands to accomplish the same thing. (In that case, the groups were only two candidates. I write "{37}" and he writes "3&7". I don't know if he would write {3479} as 3&4&7&9. If so, I prefer the braces!)

So, keeping with that variety... some people shorten chain notation further by combining cell references for links between cells, especially if they are strong links. Here's what I mean using your chain from above:

(8)r1c2-(8=9)r1c1-(9=8)r4c1-(8=8)r5c39-(8=8)r34c8-(8=9)r3c5-
(9=2)r5c5-(2=8)r9c5-(8)r9c3=(8)r7c2-(8)r1c2; r1c2<>8

For an example using groups in braces, I'll write the chain notation for the ALS elimination in this puzzle:

(2)r6c3|r7c2-(2=1)r6c2-({178}=2)r459c3-(2)r6c3|r7c2; r6c3|r7c2<>2

There's another new thing here: the vertical line symbol, "|". It means "and". Here, it joins the cell references for the two victim <2>s. The 1278 ALS is expressed as a {178} Locked Set strongly linked with the "extra candidate" <2>: They can't both be false. (It is important to note that the {178} Locked Set is false as a Locked Set if any one of its digits is false.) Notice then that the <1> from the other ALS (the 12 bivalue) is weakly linked with the {178} Locked Set: They can't both be true.

You don't really need much more than this to do Eureka notation.

[cgordon]

I found two ERs on <8> and <1> and an xy wing <289>. But I can't see the two wings mentioned above xy<348> and xyz <134> because I have too many 9's in Box 3. Is a chain or ALS solution the only way to get rid of these 9's?

[tlanglet]

[cgordon]

Ted said:

[tlanglet]

Hi Again Craig,

I could not find my notes, so I solved it again in what I believe is the same manner.

First, I used multi-coloring on <8>. One set, say Blue and Green, has 8 cells; the second set, say Red and Orange, contains 6 cells. One cell in the second set "sees" both colors in the first set and therefore that color must be "false". Given my colors, row 1 contains both a Blue and a Red, and box 2 contains a Green and a Red. I deleted the <8>s in r1c2, r3c5, r7c4 and r9c3.

After cleaning up all the resultant singles, I used normal coloring on <8>, forming a set of 8 cells. Cell r3c3, <58>, sees both colors and the cell is assigned the value <5> after deleting the <8>. This, in turn, deletes the <5> in r1c3 thereby providing the pair of naked <89>s.

Hope this helps ..........

Ted

[cgordon]

Thanks Ted: I’m not into colouring – tried it - all those conjugate pairs, multiple peers, parities and synchronization – too much like hard work.

But at least I spell it properly – the Colonial way.