Critical mode in the band 3 search of 17 clues.

Entering the band 3 critical mode, the situation is the following


All missing clues are in the remaining list of active GUA2s GUA3s

And the “in field” is made of the corresponding cells.


The stack constraint has normally been verified before. (not more than 6 clues in a stack)

Some clues can have been assigned before the entry 


The B sub list of UAs is empty


Usually, the A sub list of UAs is not empty. Each of these UAs must have one cell in the remaining “in field”. This was by construction when A and B sub lists have been produced. A new assignment should  not create “out field “ UAs  (dead branches)


In critical mode, in mini rows having 2 GUA2s, the common cell is a compulsory clue and must be assigned (one source of possible new UA “out field”.)


Compulsory UA


The A sub list of UAs is done of band 3 UAs plus GUA4 plus GUA6 having one or more “in field “ cells.

During the process, 3 things can happen


Cells assigned can hit a UA, this UA is discarded

A UA can have only one cell “in field” then this cell must be assigned (and the UA discarded)

A UA can have no left cell “in field”, this is a dead branch.


The A table is reduced step by step and the new version loaded behind the previous one in a buffer.

Once a new assignment is done, remaining UAs  have to be checked again, new dead branches or compulsory UAs can come.

As this process is applied after assignment of mini rows having 2 cells, a 1 cell UA will do one of the followings


Assign a mini row with one pair or one triplet, the mini row is solved

Assign a mini row with 3 pairs, the pair not hit remains


In most cases, at the end of this process the 6 clues in band 3 will be assigned.


What must be done at the end will vary


Some UAs are still there after compulsory uas


If the 6 clues are known, we have no solution with 17 clues.


If not we have to choose to solve a GUA2 GUA3 or to solve a UA

If 5 of the 6 clues are assigned, the last clue must hit all remaining UAs. This will be the first choice.

If this is not true, we can not find a smaller UA than a GUA2 (a UA with one cell “in field”  would have been assigned). With no GUA2, we still have a chance to find a UA with 2 cells “in field” and we have also the possibility to assign one clue in a triplet. So the worst case will be a triplet.


No more UA are there after compulsory uas.


This can happen with any number of given in band 3, although usually only one or 2 clues are missing.

The process is here quite simple, a mini row is solved and the process is called in recursive mode to solve the next mini row if other clues are missing. .


Final check


When the band 3 is filled (6 clues), the brute force is called to check whether the solution is unique. Collecting new UAs at this stage has no clear advantage, so this is a simple brute force check to see if the solution is unique.