In The Logic of Sudoku, Andrew Stuart introduces a UR method in which corners on one side are known values, i.e. clues, and remaining candidates are prevented from forming a deadly rectangle with these clues. This is his Avoidable Rectangle.
Andrew shares his puzzle maker’s experience with this UR formation, and shows several examples. The simplest case is shown in this slink marked version of Andrew’s first example.
Let’s say that 8 was just added at r3c7, confining 6 candidates to r12, proving N6, implying the naked pair 89 in N. If you are the least bit aware of the avoidable UR, you can now fill the N box.
Additional UR applications apply to avoidable rectangles. Logic provides two examples similar to these:
In Avoidable 1, the eliminations do not depend on which 7 will be in the solution. Additional 7 -candidates in c23 would not be eliminated as demonstrated in Avoidable 2
Logic follows up with a puzzle in which an avoidable rectangle provides the decisive removal. I have a similar challenge for you, as you continue on with Tom Sheldon’s Apprentice Level # 3.
Here is the box marking checkpoint, before clean up.
Below is the box marking trace.
For next time, let’s do line marking as a Sheldon apprentice. Watch out for a UR with bv on opposite corners!
Tom tells his apprentices that #3 contains a unique rectangle, but he does not hint what kind it is. And he doesn’t tell, in Sudoku Master Class, how to handle such a UR. That’s what happens to apprentices, isn’t it? They are assigned ambiguous tasks without adequate training, and then get all the blame when they mess up.
Even though you may never have seen such a UR, you have some experience and won’t mess up. I will tell you that the target does not emerge until line marking is almost completed. Come back next week with a removal, or at least, knowing where the UR is located.
Systematic Sudoku 
So considering UR(56) = r6c14=56, r5c1=256, r5c4=567, the floor(56) is r6c14 which leaves r5c14 as the roof. So either r5c1=2 or r5c3=7. Assuming r5c1=2 leads to a conflict, so r5c3=7 and from there it is naked singles to the solution:
871369524246578193935214876197483265623795481584621937458937612312856749769142358
Richard, you are a step away from T & E, because your choice of the trial candidate is far from arbitrary. I think the choice of
any
bv for a trial is pure T & E, by contrast. Now if you demonstrate the logic of that conflict by a forcing chain, you have escaped T & E altogether.
The statement “Assuming r5c1=2 leads to a conflict” is reporting a fact, but it is not logical justification. The demonstrated forcing chain is logical justification.
(I wanted to leave some grid shots of another solution, but NOT sure how to do that without WordPress messing it up! I have tried HTML, PDF, JPEG and can’t seem to get there!)
Anyway I noticed if you solved all singletons and then find the following X-Wing:
X-Wing(5) (n5r26c14) => (r4c15, r3c45, r5c45)
A BV map would show you an X-Wing(6) composed entirely of BV cells. And r6c1=56 produced very quickly and clearly on the BV map that r6c1=6 caused a conflict on the BV map. Then setting r6c1=5 lead to the solution using Only naked singles.
Richard, the order of battle starting with solving all singletons belongs on your blog, not mine. What I use for grid shots is to use Paint to capture grids from MS PowerPoint presentations, and insert them as pictures. Its very easy.
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