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Tough Sudoku for 13/July/2019

                 
                 
                 
                 
                 
                 
                 
                 
                 

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Sudoku Solving Techniques

Submitted by: sudoku lover

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Basics to 41
1.W-wing 45 [g4,h7] =>h6<>4 if h6=9 =>g12=29 g2=7
=> box9 <>5 =>h6=7 basics to 44
2.W-wing 24 [d7,g1]=> d1<>2 =>g1=2 solve
13/Jul/19 1:41 AM
Did what? Cute photo .
13/Jul/19 2:00 AM
12:32.
13/Jul/19 5:16 AM
1. Unique possibilities to 41.
2. Whether g4=4=f6,f2=1,c2=9=g6;OR g4=5=h2,h6=7,i5=1=f6,f2=2;gh1=29.UP81.
13/Jul/19 7:23 AM
Basics to 41
1. (1)d4 = a4 - a5 = (1-5)i5 = i7 - h7 = HT[(5*)h2|(7)h6|(9^)h1] - [(5*|9^)=2]g2 - (2=1)f2 => -1f6 ; UP = 81
13/Jul/19 11:13 AM
Hi rwm. The following logic is similar to yours, but seems to need fewer cells. I am not so good at Eureka notation, so don't know how it would look in Eureka.
1. Whether i5=5,i6=1;OR g4=5,h261=579,g2=2;f2=1.UP81.

Best regards, Alfred.
13/Jul/19 1:50 PM
The sudoku Jackie.
13/Jul/19 8:36 PM
14:52
14/Jul/19 5:33 AM
Alfred,
Thanks for commenting. You are correct. I would incorporate your idea by replacing the first 3 candidates in my chain with the single candidate (1)i6.
1. (1)i6 = (1-5)i5 = i7 - h7 = HT[(5*)h2|(7)h6|(9^)h1] - [(5*|9^)=2]g2 - (2=1)f2 => -1f6 ; UP = 81
I confess I'm not sure how More...
15/Jul/19 12:11 AM
Does this make sense?

1. (1)i6=(1-5)i5=i7-h7=h2-(7)h2=h6-(9)h6=h1-(59)g2=(2)g2-(2=1)f2.
15/Jul/19 8:05 AM
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