Sudoku Puzzles

Wishing you and your families a very,

" HAPPY EASTERN " .

With all my best wishes to all of you.

Another lovely photo from your visit to southern CA, Wagdy! "Happy Eastern" to you too!

Fill to 28, then try i2=5. So i3=8=c2,b7=4=a2, f6=1=e9, and the e column has no 4's. Thus i2=8 and the puzzle can be completed.

22:48

all of you!!!

Thanks for another one of your California palm 'gifts', Wagdy!

Similar to Mathman's:
1. Note pair 36 at b45.Unique possibilities to 28.
2. When i3=8,c2=8=b6,b7=4=e9,f9=1,g6=9,e6=2,a6=4, and row 2 is devoid of 4.So i2=8 and UPs to 81.

11:17 An alternate path to Mathman's: I had b7 solved down to either a 1 or 4. I tried 4 and that didn't work, which left 1 and that solved the board.

1.Start at 22,basic techniques to UP=28
2.AUR(89)gh69,
=>(7)h6=(5)gh9-(5)i7=(5-8)i2=(8)i3-(8=1)b3-(1=4)b7-(4)e7=(4-1)e9=(1)f9-(1) f6=(1)d6,
=>d6<>7.UP=29
3.(1)d6=(1)f6-(1)f9=(1-4)e9=(4)e7-(4)b7=(4)b6
=>b6<>4.U P=81

3.=>d6<>4

Appologise for the extra n in Easster.
Happy Easter to all of you.

Oh Wagdy, thanks! I just love those palm trees! These are really tall ones!

28:09, I need coffee, that was too long for a reasonably easy puzzle! See you all tomorrow.

Hi Sotir.Your step 3 is a very quick and efficient way to solve the puzzle.I don't think your step 2 is necessary, because step 3 still works ,whether you prove h6=7 or not.What do you think? Regards, Alfred.

Afternoon all. Just dropped in to say it was wonderful to meet you Alfred...

Hi All,

This is my first attempt at a formal proof, so if anybody spots any mistakes, please tell me.

1) - start at 22, UPs to 28. Note locked 8 in a789
2) (2=7]c1-(7)e3=(7)e5-(7=4)d5-(4)c5=(4-8)b6=(8-1)b3=(28)c23,=>c4<>2, UP = 81.

Best regards, Neil

Thanks Alfred,step 2 is not necessary.
Regards,Sotir

Hi Neil!

I took a look and here is what I found:

2) (2=7]c1-(7)e3=(7)e5-(7=4)d5-(4)c5=(4-8)b6=(8-1)b3=(28)c23,=>c4<>2, UP = 81.

The beginning of your chain is invalid because 7 is not forbidden at e3, in fact it is forbidden at c3 which enables 7 at e3: (2=7)c1 - More...

16:42

Hi Dave

Very many thanks for your detailed notes on my first 'proof' - I really appreciate it. You are absolutely correct about the first bit of the chain, as I forgot to enter my 'real step 2' - the elimination of 7 in d6.

I have taken carefuls 'notes of your notes' as you have More...