Unsolvable 23 Page 1

It has been quite some time since I have added to the blog. My time has not been as free as of late, thus the blog pages will be generally rare for a while.

The following illustrated proof for the diabolical & extreme Unsolvable #23 builds heavily upon the previously blogged eight page proof of the tough sudoku of 11/10/06. Ideas that one may need from that solution are: Super cells and Appending Chains.

If this is your first visit to this blog, WELCOME!!

Certainly, the terminology used here is not the standard terminolgy used in many other Sudoku sites. Therefor, Other previous blog pages may be helpful. Links to these pages are found to the right, under Sudoku Techniques. The list is long, so specifically, one may want to refer to the following pages:

• Locked Candidates
• Pairs, Triples, Quads
• Coloring
• Finding Coloring
• Y Wing Styles
• Finding Y Wing Styles
• Forbidding Chains 101 The Theory
• Forbidding Chains 102 The Practice How to Find 'em
• Advanced Forbidding Chains The theory of using techniques within Chains
• Definitions: Most terms used on this page have a defintion somewhere on the defintions page, albeit I have yet to update the definitions page to reflect the last few blog pages.

The illustrations of forbidding chains, also called Alternating Inference Chains (AIC), shown in this proof will share this key:

• black line = strong inference performed upon a set (strong link)
• red line = weak inference performed upon a set (weak link)
• black containers define a partioning of a strong set(s)
• candidates crossed out in red = candidates proven false

Please be aware that, for me, strong and weak need not be mutually exclusive properties.

This puzzle is meant to be very difficult, thus the proof is often complex.

Unsolvable #23

Finned X Wing with candidate 5

Above, the fives conspire to eliminate 5 from i2:

• i4 == e4 -- e1 == gi1 => i2≠5

In order to bring the symbols used herein more in line with those used at other sudoku sites for Alternating Inference Chains (AIC), I will for the rest of this proof attempt to use Eureka notation. However, I shall keep the grid conventions. Thus, one could write the elimination above as:

• (5)i4=e4-e1=gi1 => i2≠5

A short chain using candidate 8 and an ALS with candidates 26

The step above is not very complicated. In my previous style:

• h1=8 == {pair 26 at h14} -- h6=2 == b6=2 -- b6=8 == a6=8 => a1≠8

conforming to a more accepted style:

• (8)h1=(26)h14-(2)h6=(2-8)b6=(8)a6 => a1≠8

Eliminations now seem a bit harder to find.

A complex, or advanced, chain using candidates 57 - an appended Almost X wing

Above, but for the 7's at cd2 and the 7 at c9, one would have a column XWing at cd35. Much like one can append an Almost Hidden Pair, one can also append this configuration. In other words, one can see this elmination by understanding that considering only the 7s, we have:

• {Xwing at cd35} == {cd2 == c9}

One can of course distribute the =='s anyway that is convenient. Thus one also has:

cd2 == {{Xwing at cd35} == c9}

Now, a few links are added:

• i1=7 == i2=7 -- cd2=7 ==₁₂{fc on 7's: {{Xwing at cd35₁₂} ==₁ c9 -- e9 == f8}} -- f5=7 == f5=5 -- e4=5 == i4=5
  • => i1≠5

One might alternatively write:(I think this symtax is correct!)

• (7)i1=i2-cd2=₁₂{Xwing₁₂cd35=₁c9-e9=f8}-(7=5)f5-(5)e4=(5)i4 => i1≠5

In cases such as this, one can almost always (probably no need for saying almost), represent the idea in more than one way. For example, one could think:

• i1=7 == i2=7 -- {Xwing on 7's at cd23₁₂} == {fc on 7s: cd5 ==₁₂ c9 -- e9 == f8} -- f5=7 .....

Chain pieces such as this are powerful tools when faced with truly difficult puzzles.

After making this elimination, an easy elimination is available:

• Locked 5's at i45 => g5≠5

The following two steps were seen as one. I will later attempt to graph what I saw.

Overlapping Super Cells - also overlapping Hidden Pairs 25

Above, in row 1, the 5s form a Super Cell that provide the equivalent of a bilocation strong link with the 2s. Also, in box e2, the 2s form a Super Cell that provide the equivalent of a bilocation strong link with the 5s. The 8s form a bridge between the two Super Cells, and the intersection of these two Super Cells can contain only 25.

One way to write the chain:

• {Hidden pair 25 at eg1} == h1=2 -- h1=8 == h3=8 -- f3=8 == f2=8 -- f2=5 == {Hidden Pair 25 at d2e1}
  • => e1=25 only => e1≠679

An alternative way to write the chain:

• (2&5)eg1=(2-8)h1=(8)h3-(8)f3=(8-5)f2=(2&5)d2e1 => (2=5)e1

With such an almost hidden pair configuration, there generally exists an almost locked naked tuple idea that essentially accomplishes the same thing. In this case:

• {Triple 679 at def3} == f3=8 -- h3=8 == h1=8 -- b1=8 == {Triple 679 at abi1}
  • => e1≠679 & c3≠9

I much prefer the Hidden Pair usage. Perhaps, though, the reason that I prefer the Hidden Pair usage is primarily rooted in the concept of Super Cells. Furthermore, to me the usage of Hidden Pairs is more efficient and easier to find. However, others may well find that the use of Almost Locked Sets is easier.

After e1 is limited to only 25, Locked 9s at ef3 => c3≠9.

Wrap Around Y wing style - Or - Hub, Rim and Spoke

Illustrated above is a typical wrap around, or continuous, Y Wing Style. It can also be understood as the Hub, Rim and Spokes concept as noted. One can write the chain:

• e1=2 == e1=5 -- g1=5 == g2=5 -- g2=2 == d2=2 =>
  1. e1=5 == g1=5, but we already had this
  2. g2=5 == g2=2 => g2≠46
  3. e1=2 == d2=2, but again we already had this

Alternatively, one might write:

• (2=5)e1-(5)g1=(5-2)g2=(2)d2 => (5=2)g2 => g2≠46

Again, if one wishes, one can achieve the same result using Naked Almost Locked Sets:

• {Quad 3467 at i12g3h3} == h3=8 -- h1=8 == b1=8 --ac2=8 == {Triple 467 at aci2}
  • => g2≠46

However, one may find it simpler to combine the two steps as illustrated below.

Simulataneous overlapping Hidden Pairs 25 - also simul. overlapping Super Cells

Above, one has hidden pair 25 at eg1 and g12 (yellow) OR h1=2. One also has hidden pair 25 at e1d2 and dg2 (blue) OR g2=5. Combining these with the strong 8s as before, the (Yellow + blue = green) intersections are pared down to only 25. One could write:

• (2&5)e1g1g2=(2-8)h1=(8)h3-(8)f3=(8-5)f2=(2&5)e1d2g2 => (2=5)e1 & (2=5)g2

As strange as it may seem, I find uncovering such eliminations to be things of beauty.

This concludes the first page of this proof. There remain some very interesting, and also non-trivial, steps. To find some of these, please visit the next page.