Tough Sudoku 2005-09-18.
Come from http://www.sudoku.com.au/default.aspx?ID=4&Go=V18-9-2005
Starts with
We expose proof
of uniqueness using abbreviated Andrei notations, see
comments on archives 2005/09/16 and 5005/09/17.
Following
simple eliminations :
b2=4%row, c5=4%box, a2=2%row, b3=7%box, a9=7%box, b9=8%box, g2=7%box,
h6=7%box, c3=5%row, e3=9%row, f1=8%box, d1=5%box, g7=9%row, e8=3%cell,
d8=7%cell, f5=7%box, i5=2%row, h1=2%box, i1=4%box, g1=3%box, d5=3%row,
f2=3%box, d2=6%box, e2=1%box, a5=9%row, c1=9%box, a1=6%box, b1=1%box, g5=8%row,
f8=5%cell, a7=5%cell, b6=5%row,
give
us :
On the 2d
diagram we indicated remaining possibilities in some cells. Looking cells h3
and h8, we see that they must contain 6 and 8, no matter the order. So h7≠6%col. It follows c7=6%row and c8=1%. We’re now to :
Here, what
if b4=3? Then b7=2%, f9=2%box and remaining values in f4,g4,h4,i4
are 569 which won’t fill these four cells. Hence b4=2%.
From here,
simple eliminations lead to unique solution. (Details if needed : c6=3%box,
b7=3%box, c9=2%box, e7=2%box, d9=4%box, f9=9%box, h7=4%box, e4=4%box, e6=8%box,
a4=8%box, f6=2%box, f4=6%box, i6=6%box, i4=9%box, d6=9%box, h8=6%box, i8=8%box,
h3=8%box, g3=6%box, h4=3%box, g4=5%box, h9=5%box, i9=3%box, g9=1%box, d4=1%box,
a6=1%box, i3=1%box).
Thanks to
Leo for rereading and finding a typing flaw (hope no other remains...)