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# Tough Sudoku for 2/September/2013

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Submitted by: Gath

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 everybody! 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 271. (3)b7=3c9-(3=126)def9-(2)e7=(2-356)b127=(6-3)f2=(3)bc2 =>b1<>32. (1)c1=b1-(356)b127=(6-3)f2=(3)bc2 c1<>33. (8)d1=(8-76)dg8=(7-24)h89=(4)g89-g6=(4-3)e6=(3)d6 =>d1<>34. 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<>371?@d8=>[e7=8=>e9=2&h8=7=& GT;h9=2]=>d8<>12@g56=>g7<>21?@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 SE=8.3SSTS to UP=271)(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=3592)(3)d6=(3-4)e6=g6-(4=5)g9-ac9=(5-8)a7=e7-e1=d1-(8=7)d8-(7=3)f7,=>d19<>33) (4)f1=(4-1)c1=b1-(1=2)b8-h8=(2-4)h9=g9-g6=(4)e6,=>e3<>44)(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=57I 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