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Tough Sudoku for 2/September/2013


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Submitted by: Gath

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02/Sep/13 12:32 AM
#1. UP27; d8=8->C(S) :=> e7=8; UP81
02/Sep/13 1:27 AM
Hello everyone!
Nice one, Kate!!
02/Sep/13 7:56 AM
Site puzzle with AICs (and some of ALS/AHS)
Standard techniques UP 27
1. (3)b7=3c9-(3=126)def9-(2)e7=(2-356)b127=(6-3)f2=(3)bc2 =>b1<>3
2. (1)c1=b1-(356)b127=(6-3)f2=(3)bc2 c1<>3
3. (8)d1=(8-76)dg8=(7-24)h89=(4)g89-g6=(4-3)e6=(3)d6 =>d1<>3
4. More...
02/Sep/13 8:43 AM
Posted a path to P708.
02/Sep/13 8:45 AM
1. Unique possibilities to 27.
2. Whether b7=2;OR e7=2,d8=8,f7=7;b7=23.
3(A). If g6=4,h89=24.Whether g8=7,d8=6;OR d8=7;d1=8.Also:
3(B). If e6=4,d6=3.Now d1=5=b2,de3=67,and Box B2 is devoid of 6.So d1=8.UP81.
02/Sep/13 10:11 AM
(28) 6?@d8=>e7=8=>b7=2=>c9=3=>df9=1=>d8<>6
(29) 3?7?@e7=>[d8=8&b7=2=>b8=1]=>ac8=4=>e7<>37
1?@d8=>[e7=8=>e9=2&h8=7=& GT;h9=2]=>d8<>1
1?@f7=>d8=7=>e7=8=>a7=5=&G More...
02/Sep/13 10:38 AM
Thanks Kate!
02/Sep/13 1:09 PM
Another way of stating:
3(B). If e6=4,d6=3.Whether b1=5;OR b2=5,a3=6,e3=7,d3=5;d1=8.UP81.
02/Sep/13 1:58 PM
SSTS to UP=27
1)(59=5)acg9-(15=7)gi7-gh8=(7-8)d8=(8-2)e7=(2-3)b7=(3)c9,=>gh89<>1,d8=78,e7=28,b7=23 ,a9=59,c9=359
3) (4)f1=(4-1)c1=b1-(1=2)b8-h8=(2-4)h9=g9-g6=(4)e6,=>e3<>4
4)(5)b1=(5-6)b2=f2-(67=5)ed3,=> df1<>5.UP=81
03/Sep/13 1:23 AM
Hi sotir,
Your step 1) is tricky, but I don't see how you eliminate 1gh89 with that loop. Indeed, your first SIS are 59ac9=5g9-5gi7=7gi7...
Here 5gi7=7gi7, infered from ALS gi7, means that gi7 contains 5 OR 7, but not 5 XOR 7, i.e. the final solution might be gi7=57
I don't see another More...
04/Sep/13 7:36 AM
Hi Cenoman,

Sotir's step 1) is a continuous loop.
So either g9=5->gi7=17, or gh8=7->gi7=15.
Therefore, 1 is locked in gi7 :=> -1gh89 or(=>and) -1abf7 depending on the choice of the cover set for 1gi7 (1Box h8 or 1Row7)!

As 'Eureka notation' can sometimes be More...
04/Sep/13 8:33 PM
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