rugram from India

Gday and rugram. I remember your name from a while ago. It is great to have you back with us.
I look forward to seeing you around the pages.
Have fun.

rugram, nice to have you with us here in Sudokuland. Hope you are enjoying the puzzles and the nice people here.

Hi Rugram nice to know u are a sudoku fan ......hope to be in touch.

Hi Rugram, I also prefer not to use the more technical language.I can usually follow what they mean,but am not happy using the = sign to mean "or" instead of equal.I left a comment on today's tough.See you, Alfred.

Thanks for the reply,Rugram.If you look at my site, you will see others have been happy with my "plain english" solutions too.There is a demand for this. I only seem to solve the puzzles at the easier end of the spectrum, but with experience, I have been getting more of them out.Today,I got three pairs leading to all the 1's coming out, but that is as far as I got so far.Have other things to do now, will try again later in the day. See you, Fred.

Thanks for the email, Rugram.The other day I googled "Sudoku Proof Notation" and about 10 answers came back (not using the quotation marks).It seems the technical notation used is Boolean algebra,which is "a mathematical means of representing statements in logic" (dictionary definition).Although mathematics was always one of my favourite subjects , I had never come across Boolean algebra when I was young,and find it a bit hard going now.(Maybe I am just a bit lazy too.)Anyway, my approach is more experimental rather than mathematical. I try out different possibilities, and see if they lead to contradictions or confirmations.I haven't been able to crack the harder Toughs with my method, so maybe Boolean algebra is required there.

Hi Rugram,I tried to send a message on the other system, but I got an error message that there were more than a 100 characters in the message. I followed your tough solution today. I think once you proved d8=3,there was no need to go further, as the rest comes out with basic technigues.At first I didn't understand the 168 triple, until found the 34 pair at d58. See you, Alfred.

Hi again Rugram, still can't get the other site to accept a message. Anyway, I.ve posted a solution on the comments page.It has more steps than yours, but each step is fairly short.See you, Alfred.

Very neat Rugram.You nearly stumped the experts,but now ttt has seemingly translated your idea into technical jargon. By the way, you are also prone to chess player type typos! : 8 is at d3 or f2.See you, Alfred.

Rugram, I am saddened by what is happening in your country at the moment and all are in my thoughts.
Please stay safe.

rugram! Thank you for all the solutions you post on the Tough - I find them useful! I have no trouble with basic techniques (the 'Hards' just fall into place) but I am having difficulty making the leap to Tough without prompting! Getting there though!

This is my detailed solution to the Tough of 25th Dec., 2011. A condensed version of this solution is given on the puzzle page of 25th Dec., 2011. I am sure that there would be many shorter solutions compared to mine.
I acknowledge with thanks the help I have received from my friend Alfred from Sydney, without which I wouldnt have been able to post this solution.
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Tough Sudoku puzzle of 25.12.2011 – A possible solution (long)
1.i8=7=g5, i7=1, (locked 5 at fg3 and so) i6=5. UP=26.
2.If d4=7, pairs 24 at d89, d2=3, d6=6, f6=3, e9=1, which leaves the middle block without 1. So e4=7. UP=27.
3.a) Assume e5=2. Then pairs 24 at d89, e9=1, pairs 13 at df6, d4=6, i5=9, i4=2, pairs 34 at i12, g1=2=a2. So h2 can be 8 or 9.
b) If h2=8, g6=8, g3=5, f4=8, g9=4, d9=2=c7, leaving both 4 & 8 to satisfy b7.
So h2<>8.
c) If h2=9, b2=8, pairs 69 at ef1, d3=1=b1=f6, b5=3, c5=8=a7=f4, leaving the top middle block without 8. So h2<>9.
d) From (b) and (c) above, it follows that the assumption in 3(a) that e5=2, is not correct. So d4=2. UP=28.
4. d2 can be 3 or 7. If d2=3, d3=7=b2, f6=3=i1, pairs 59 at ef2, h3=9, g3=5, h2=8=g6, g7=6;
(g9 cannot be 2 as then c7=2=a2, leaving bottom right block without 2, and so) g9=4, g1=2=a2, d8=4=i2=h6, h5=6, i4=9, i5=2, d6=6, d9=1=f1, e1=6=f8, f2=9, e2=5=f7, h7=2=c9. This leaves both 2 and 8 to satisfy e8. Hence d2<>3, but d2=7, b3=7. UP=30.
5.a) If d8=4, d9=1, d6=6, d3=3=f6, f1=1=a3=b5, e1=6=h5=g7=f8=a4, i5=2, e6=1. So c3 has to be 8 or 9.
b) If c3=9, a9=9=b4, i4=4, f4=8=e8, e9=2=g1, g9=4, c7=2. This leaves both 4 and 8 to satisfy b7. Hence c3<>9.
c) If c3=8, g3=5, h3=9, h2=8, g6=8, g1=2, g9=4, h6=4,i5=2, i4=9=e5, a2=4. This leaves the bottom left block without 2. So c3 <>8.
d) From (b) and (c) above, it is clear that d8 cannot be 4. Hence d9=4. Follows that d8=6, e9=1, e8=2, e2=5. UP=35.
6. If g6=8, g7=6, g3=5, g1=4, g9=2=c7=a2=i1, i2=3, i4=4, leaving column c without 4. So g6<>8, but g3=8. Follows h3=5. UP=37.
7. g1 can be 2 or 4. If g1=2, g9=5. If g1=4, i4=4=a6=c7; locked 2 at ac 9 and so g9=5. In both cases, therefore, g9=5. UP=38.
8. If h8=3, h9=2=c7=a2, b2=8=a8=c4, pairs 45 at b78, b4=9. This leaves the left middle block without 5. So h8<>3, but h9=3, h8=4. UP=40.
9. If g7=2, h7=6, g6=6, g1=4=i4=c7, c4=5, c5=8=h6. This leaves the central block without
8. So g7<>2, but g7=6. This takes us to UP=81.