After the hidden unique rectangle hinted last week, a pair of Single Alternate Sue de Coq trials gets the review solver Beeby moving again. ALS-wings and a finned X-wing get us to a colorful finish.
The hidden unique rectangle left on grid is signaled by the bv 6 and 7 repeated in three more corners. The resolution method is marked by the three side slinks in the UR value 6. If true, 7r8c9 generates a reversable rectangle in a double solution.
Both solvers now quiet, my choice for a trial is the Single Alternate Sue de Coq Cr5, with contents described by 7(5+9)(1+8) +817.
That’s because it can’t contain 5 and 9. It must be 5 or 9 or neither. We test neither, i.e. 817.
The test reaches a contradiction on the 9th breadth first level, leaving the Sue de Coq
Cr5 = 7(5+9)(1+8).
Now the (5+9) and the bv 59 sweep other 5’s and 9’s from r5, and the 9r5c9 removal brings the boxline removing 9r3c7.
This leaves Sue tugging on our sleaves with two Single Alternate SdC’s
Er4 =1(2+8)(7+9)+971 and
Wr4 = 5(7+9)(2+8)+582.
Both Er4 =971 and Wr4 =582 produce the same contradiction as before, and both remove 7r4c8, leaving the pair E28 to remove 8r6c8.
After a 2-chain ANL, Beeby does a pair of ALS-wing, the first, depending on singles values,
The second, sharing an ALS with the first. The 7r4c2 removal allows
a finned 7-wing. The 7r7c9 fin sees the victims, so if it’s true, they’re toast and if it’s false, the 7-wing toasts them.
Next, after a 7-chain ANL,
and NE3 => N3,
Beeby’s next overlapped ALS-wing.
We sit back and watch as r5 naked triple generates W1 and the West box naked quad.
Now what is that thing?
It’s a classic boomerang with an ALS providing the closing slink.
It’s also a 1-way. If 8r7c4 is true, 2r7c4 is not, and if 8r7c4 is false, the ALS AIC promotes another 2 in the South box.
The removal adds a critical slink for the hidden unique rectangle.
Seeing an XY chain,
And an easy AIC, I think that closing this down with coloring may be overdue.
On first coloring, the bridge not(green and orange) becomes mute as green is wrapped by a trap induced boxline.
In the follow up and red/orange expansion,
1, 2 and 4 squeeze out 7 and 9 and the Sc5 boxline and trap removing 7r8c4 confirms orange.
Here’s the colored solution.
One more ultrahardcore to do, UHC 487. The site and the portable versions of the Guide need attention, and the only reason for doing another collection is to confirm any reader opinion that it is likely to reveal more about human solving. Stefan Heine’s ultrahardcores has certainly done that.