Approach to Daily 1271

Here is an approach to Daily 1271.

To get started on this puzzle the extended rule of 45 needs to be applied. Lets start in the lower right nonet. The red cell-blue cell = 3.

Also, if we apply rule-of-45 to the middle nonet we get that see that the red cell must 9,8,or 7. (To sum to 11).

The rule of 45 on the rightmost nonets indicates that the two red cells must sum to 15.    
Again, the rule of 45 in the top middle nonet indicates the two red cells must sum to 13.    
The rule of 45 on the top-left nonet shows that the cells marked in red and blue must have the same value.    
Lets look at the top-right nonet. First, the red cell cannot contain 7 (due to the 24-cage below it).

So, go ahead and eliminate the invalid combinations.

We can now fill in some cells due to the fact that 87 cannot be in any of the other cells in the 6th column.    
The red + blue cells must sum to 20. So, the red cells are either 15 or 14.

If they are 15 (blue cell is 5), then they must 87.

If they are 16 (blue cell is 6), then they must be 95.

Note, however, that if the red cells sum to 14 (9+5), then the blue cell must be a 7; which is illegal. So, the red cells must sum to 15.    
Some clean up can now be performed.    
The blue cell determines the remaining possibilities in the top-middle nonet.

Like wise the red cells only have one possibility

The 4 in the blue cell means that in the right-middle nonet the value 4 can only be in either the 11-cage (in red) or as part of the 12-cage. If it were part of the 11-cage, then the 12-cage would be 5+7 and the 24-cage would have a 7&9 in the right-middle nonet. This would be illegal. So, the 11-cage (in red) must be 6+5.    
We can do some clean up.    
The 20-cage in the lower-right nonet is now completely specified. The red cell is the only place for a 7 in that nonet. The remaining sum of 13 can only be 8+5.    
Only legal sum for the 11-cage (in red) is 9+2.

The open cells in the lower-left nonet (in blue) must sum to 9 and only 6+3 is legal.

This means the 9-cage in the bottom-middle nonet must be 8+1.

And, the red cell in the bottom-middle nonet must be a 2.

Blue cell is only legal place for a 7 in the right-middle nonet.

The red cell in the lower-right nonet is the only legal place for a 2.

The blue cell in the lower-left nonet is the only legal place for a 7.

We can clean up the 3rd row. By putting a 9 in the red cell.

We know the 8-cage in the middle nonet must be 7+1.

We can fill in the remaining cells in the bottom-nonet and the middle-nonet.

We can also finish the bottom-right nonet.

The blue cell must be 3. Which also determines the red cells.

Lets fill in the remaining possibilies in the lower-left nonet.

This limits the choices for the two 11-cages in the middle-left nonet.

The two red cells determine the blue cell.

The blue cell in the 4th row must be a 1.

This makes the red cell a 2.

We can now do some elimination on the 4th and 5th rows.

Once the 4 is in the red cell, we know that the blue cell cannot be a 6 (since the remaining parts of the 16-cage sum to 12.

A lot falls out after that.


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