Jeff from Maryland

Hi Jeff,

Very many thanks for showing your solution - I would never, ever, have found that one.

Best Regards,

Neil
~

Your welcome, but I was merely the pilot for Andrew's sudoku solver, I use it frequently as a learning tool. I thought that chain was interesting and so I posted.

Hi Jeff,

Have you considered using a vector drawing package instead of MS Paint. I use CorelDraw, but I'm sure that there are a lot of others.

Best Regards, Neil

Thank you for your thoughts Neil. I'm going to check out the free software that someone suggested on the the tough puzzle chat Of May 15th (I think thats the date).

Jeff, I left a comment on today's comments for the tough.When d1=4,a1 to i1=625489371 and a3 to i3 =813752496,contradicting the 3 on c8.How does this help?

Jeff, regarding the Tough from July 3rd.,after UP 29,you are right that when d1 is not 4 , the rest are UPs to 81. The following forced sequence proves that d1 is not 4, but may be a bit long for you:When d1=4,e3=5,c1=5,g1=3,b1=2,f3=2,a3=3,g3=4; and now there is no room for 6 and 8 on the third rank.I think I got the gist of your proof of today's puzzle, but I think there are some typos.See you, Alfred.

Hi Jeff! I just saw your "secret code" comment on today's tough. I've always suspected something of the sort
I was wondering where in Maryland you call home.
I'm in Bowie, about halfway between Annapolis and Washington. Should I be saying, "Howdy, neighbor"?

Jeff, I checked the July 4th Tough again.It was OK this time.Last time there were some brackets and /s in some places. I went thru your proof again and after you prove g3 is not 2 (in step 4),a3=2 ,c2=8 etc and I got UPs to 81, without needing step 5. See you, Alfred.

Jeff, great proof today(6th.July).Took me a while, but finally saw the logic in each of your steps.BTW,there is a definite typo today.In the note to step 1, the locked 6s are at a2 and c2, not a3 and c3. See you ,Alfred.

Hi Jeff, I think Steve meant no 9s at ef4.The reasoning; 9 at d4 or d7. If d7=9,e9=5 and g6 becomes 1, forcing a 9 at e6 or f6.A 9 at d4 obviously disallows 9s at ef4 too.See you, Alfred.

Sorry,should read f9=5 (not e9=5).As a chess player,I am used to an 8x8 board and am prone to mistakes on a 9x9.

Hi Jeff,I left a comment on todays tough page for you.I think you have set a record today.One would wish for a shorter solution,but it may be like Fermat's Last Theorem; there don't seem to be any short cuts.See you,Alfred.

Jeff,regarding today,s solution:After your step 1, I found that by testing for 4 and 8 at g7,another pair 48 is created at i2,which in turn creates another pair 48 at g1,which helps to fill three more cells.I think it is similar to your steps 4 and 5. At this stage,however, I got bogged down and had to resort to trial and error.Thanks for posting your solution,Alfred.