Hi coloin,

I am travelling with irregular links to the net. Some quick answers

champagne wrote:... all possibilities to speed up the process limiting stacks to 6 clues are used.

so this would mean you are already optimizing the band 3 additions...

although maybe a gain in not searching

222

221

xxx

since

321

122

222

puzzle would still be found [?]

In your example, one box has more than 2 clues. This is assumed filtered in my answer

incidentally - I am not sure how you make the band1and2 {6,5} reductions - {50000,4000} is a big product !

surely it cant be quick enough to just - solve and discard if sol >1

I answer as if you were here in the general search problem. My rough estimate is an average number of XY over 1 billion as written in the comments. The first reduction is done using UAs located in bands 1+2 and more precisely bit field vectors already settled for each band. This is explained with many details in the comments and a simple “and” on 2 64 bits fields already clear more than 90% of the {valid band1} x {valid band2} XY

choosing band 3 to have 6 clues

Pros

- the number of {6,6} is too large ?

Cons

- only one pass is needed at searching a {6,6,5}

- band 3 search is quicker with 5 clues

In the 6 6 5 (5 clues in band3) the threat is not only the bigger number of XY but more the intrinsic bigger number of finally valid Bands 1+2 (toward the double gangster). As you wrote, the brute force remains a costly process, especially when the solution is unique. I am hesitating to-day because the filter on GUAs is now so efficient that, applied just after the previous one, it can save most of the calls to the uniqueness check for bands 1+2. This can make the 665 search better than a 656 search with another entry file.

and just out of interest ...

for a single band1and2 approx how many {5,6} and {6,5} completions are there per band 3 gangster representative ? and how many valid {6,6}

I keep this for later but just a remark. Having a given ED band 1+2, the gangster in band 3 is unique. The number of attached bands 3 vary.

how do you make sure that you consider every ED band in band 3 gangster - adding a GUA to band 3 excludes some ED bands. Is this your worry ?

Again, I have some problems with the wording. The entry is an ED band 1+2. The gangster in band 3 is just the consequence of missing digits in each column, and the attached bands 3 are all possible fills of the band 3 gangster filtered to avoid redundancy (the total number of bands 3 is here equal to the number of ED solution grids).

In this context, a band 3 is considered if the band 1+2 has been checked valid and if known GUAs did not pass the limits (band 3 number of clues or stack limit).

Then, a brute force for the entire grid can produce a new UA with a corresponding GUA.

And yes, such GUAs added to known ones can filter other ED bands 1+2.

Is this the answer to you point ?? If yes, it’s not a worry, it’s a chance to find so many such GUAs