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 ...
- Code: Select all
. 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.