3) Look at 2 in Be8 : whether e9=2, in which i3=2%row, or d8=2 : (pentagonal situation [(i8=2)--(i3=2)==(e3=2)--(e9=2)==(d8=2)--(i8=2)], what's that?) .

We also have : 6 in Ce are in e13, which forbids {f1=6, f2=6}, and we arrive below :

4) Look at 3 in Bh8 : whether it's in R7, in which b7=2%cell, or it's in Ci, in which i4=6%cell,i8=5%cell : d8=2 (%row or %cell) (11-agonal situation !).

Follow then easy eliminations, (if needed :e9=5%block, i8=5%block, e6=1%cell, d1=1%block, a4=1%row, f4=2%row, f5=6%block, i4=6%block, d4=3%row, a6=3%row, d6=5%row, f6=8%block, d9=8%block, f9=9%block, d2=9%block, d5=4%block, g6=4%block, c5=8%cell, b3=8%block, c3=1%block) up to :

5) Look now at i9. Whether i9=2, in which g7=7%cell or i9=3, in which i5=7%cell : g5=1%cell and i7<>7(%row or %col) (heptagonal situation). Follow then easy eliminations (if needed : h8=1%block, g9=6%block, b8=6%block, b9=1%block), up to :

Here the chain (a2=4)==( f2=4)--( f2=3)==( f1=3)-- (c1=3)==( c1=7) forbids {a2=7} that would close it in a heptagon. If you better have it in words , look 3 in Cf ; whether f1=3 (in which case c1=7%cell) or f2=3 (in which case a2=4%row) : 7 is not in a2.
Now the chain (h1=8)==( i1=8)--( i1=4)==( i3=4)--( i3=2)==(e3=2)-- (e3=6)==(e1=6) forbids {h1=6} that would close it in a nonagon. If you better have it in words, look 4 in Ci ; whether it's in i1 (in which case h1=8%row) or it's in i3 (in which case e3=2%row,e1=6%cell ), in both cases 6 can't be in h1 .

From here it's straight to unique solution (details if needed : h2=6%col, g2=7%row, i5=7%col, h7=7%col, h1=8%col, i7=8%col, b7=3%row, i9=3%row, h5=3%row, f2=3%row, a2=4%row, i3=4%row, f1=4%row, c1=3%row, a9=2%row, g7=2%row, b5=2%row, e3=2%row, i1=2%row, c9=7%row, a1=7%row, a3=6%row, e1=6%row, a5=5%row, b2=5%row, g1=5%row).

By far the toughest one I've ever seen here (needed a pentagon, a heptagon, a nonagon and a 11-agon !).

Note : Equivalent puzzles are : 09/13, 09/16, 09/22, 10/03.

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