Nick70 wrote:I didn't say it was a corollary to your technique.

I see it as the same technique. Starting with your illustration ... and adding a few candidates for completeness ... we have ...

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`. a .|e . .|* . . `

. b .|f . .|. * .

c w d|x g h|+ + .

-----------------

m y n|z r s|. + +

. i .|p . .|. * .

. j .|q . .|* * .

-----------------

* + *|. . *|. . .

. . *|+ * .|. . .

. + .|. . .|. . .

From an equations perspective (equalities and inequalities) ...

Constraining equalities(K):

(a+b)+(c+d)+w=1 (box 1)

(e+f)+(g+h)+x=1 (box 2)

(i+j)+(m+n)+y=1 (box 4)

(p+q)+(r+s)+z=1 (box 5)

"Constrained" inequalities(U):

(a+b)+(i+j)+w+y<=1 (col 2)

(c+d)+(g+h)+w+x<=1 (row 3)

(e+f)+(p+q)+x+z<=1 (col 4)

(m+n)+(r+s)+y+z<=1 (row 4)

In order to eliminate the other candidates in the rows and cols, one must "convert" the inequalities to equalities. Specifically, one must prove equalities(V):

(a+b)+(i+j)+w+y=1 (col 2)

(c+d)+(g+h)+w+x=1 (row 3)

(e+f)+(p+q)+x+z=1 (col 4)

(m+n)+(r+s)+y+z=1 (row 4)

Without proof, I hypothesize that proving equalities(V) is only possible if w=x=y=z=0. If this hypothesis is true, then the technique that eliminates other candidates from the rows and columns is the *same* technique that eliminates the candidates at the intersections of the rows and columns.