This post encounters a very odd situation. All the givens in your homework Dave Green Saturday 4-star of 7/18/21 are on corner cells of the boxes. No double line exclusion (dublex) or cross hatch clues in the bypass. The past few posts on 4-fills suggest what you might do with a grid full of 5-fills. Here we’ll solve the 4-star in two ways. After we examine how to use the 5-fills to wring out line slinks, you get to find them all. Then we’ll ignore that anything is different, and carry out normal box and line marking, while exploiting every 4-fill, as in the previous post. Then we’ll see what happens when we start box marking with your line slinks.
Here’s another look at the Saturday 4-star. Values 2, 7 and 9 are missing from r1. Given 3r3c3 bars 3 from r1c12 and 3r6c7 bars a 3rd cell, leaving two cells, and a 3-slink in r1. Similarly, given 9r3c7 and 9r7c1 produce a 9-slink in r1. Going to r3, value 5 is missing and 5r1c4 bars 5 from three cells, for a 5-slink in r3.
Now follow up with 2 missing from r6, 8 and 3 from r4, 6 and 3 from r6, 9, 7, 8 and 5 from r9. Then do the same for columns. The slinks may produce naked pairs, 4-fills and 3-fills.
Normally you would follow up the naked pairs and their resulting 3-fils immediately. Naked pairs in lines r7, r9, c1, and c3 produce 3-fills, and their resolutions can bring clues to add 4-fills and 3 -fills
This traces the follow up of the naked pairs, bringing . . .
this grid, with several clues, to the bypass
The bypass now closes out the 4-star like a 4-star
Here’s where we leave off tracing.
Next we dip into Friday 4-stars again to see an extreme subset I ran across, an octuplet. That’s 8 cells containing 8 values. Of course there is a complementary subset with one cell containing one value, but you can judge which is easier to spot. This Dave Green hit the Akron Beacon Journal 11/05/2021.