Sudoku Puzzles

If everyone would light just one candle...

First? Easy one today

This picture looks familiar, but I love to burn candles in the winter. It puts heat into the air and warms the soul! Have a great day and week-end everyone.

That is a really pretty photo!

Rather easy today, though I lost count of how many guesses I needed.

Hey, are we getting duplicate pictures here, also! I know I've seen this one here before.

Good Maen!

Good morning to all! Debby and Missielely: You are correct, and I think it might actuallt be a triplicate!

Actually ... Spell checker again Gath ... or a typing course!

Very pretty picture...

Nice pic, see the ligh and think about the life ...!

In a hurry to get to games, so... likely a typo-laden proof, certain to be improved upon:
Possible proof of tough sudoku of 09 24 06:
1) Start at 22 filled - the given puzzle. Unique Possibilities to 22 filled. (UP 22).
2) Triple 235 at h789 forbids gi9=2, h15=2, i9h56=3,h1256=5 UP More...

The depth 5 matrix step:
d8=6 d8=4
0000 f9=4 f5=4 f2=4
0000 c8=4 c5=4 0000
0000 0000 0000 f2=1 f1=1 (closure point for elims)
d1=6 0000 0000 0000 f1=6
forbids d7=6,f2=57, f1=578

I think this is 'actually' at least the 4th time for this one! We need to send in some new photos, everyone!

1) Start 22, unique possibilities to (UP) 22.
2) Naked triplet h7=235, h8=25, h9=23 forbids g9=2, i9=23, g1=25, g2=5, g5=235, g6=35. UP 32.
3) Hidden pair 14 at b49 forbids b4=7, b9=37.
4) c8=4 == c5=4 -- b4=4 == e4=4 forbids e8=4.
5) b9=1 == b9=4 -- c8=4 == d8=4 -- d8=6 == a8=6 More...

typo corrections on my proof:
Possible proof of tough sudoku of 09 24 06:
1) Start at 22 filled - the given puzzle. Unique Possibilities to 22 filled. (UP 22).
2) Triple 235 at h789 forbids gi9=2, h15=2, i9h56=3,h1256=5 UP 32
3a) Hidden pair 14 at b49 forbids b4=7, b9=37
3b) ALS: More...

If one uses Clark's beggining, which is better then mine plus parts of my ending, can make a better combined proof.

AFter Clark's step 10, use depth 5 wrap around chain similar to my step 7. Then one can finish with my steps - or probably even Clark's.

I would suggest the following: after Clark 10:
11) f2=4 == f5=4 -- c5=4 == c8=4 -- d8=4 == d8=6 -- d1=6 == f1=6 -- f1=1 == f2=1
forbids f2=5, and forbids f1=58 and forbids de5=4=4 and forbids d7=6 UP 41
12) f3=5 == f7=5 -- f7=6 == d8=6 -- d1=6 == d1=7 -- b1=7 == b1=5 forbids b3e1=5
13) using 12) f3=5 == d1=7 -- e1=7 == e1=8 forbids f3=8 UP 81

Keeps max depth at 5... keeps pretty wrap around chains, which I am partial to... hehe!

This picture is so old!!! I can't believe they are running it again. It must be at least the 10 time this year.

Steve - you seem to be a wizz at this, and I appreciate your solutions. I often refer to your comments when i get stuck. At times, i can finish a tough puzzle on my own and other times i can't. You seem to be using some techniques I'm just not familiar with, but i want to learn more. For example, More...

Hi Joey!
I would suggest the site: http://www.scanraid.com/BasicStrategies.htm#PP

Note that you will have to probably delete some spaces if you cut and paste this link. This site then has some links to some sudoku forums that explain other techniques in some detal.

Also, the More...

BTW - matrices are not a real solving tool, but rather more of a presentation tool. Understanding how the matrices make the presentation, though, can open one's eyes towards some creative steps to take to find a solution.

The linking of ideas together - combining coloring, for example, More...

Hi Steve. I've often wondered if Scanraid's 'Alternate Inference Chains' are the same as the 'Forbidding Chains' used here. What do you think?

Kaz - they are the same, in my opinion. The terminology is very different though. The defintions even are different. But, logically, they are equivalent, except that I think the forbidding chains are more transparent and have simpler terminology - and more easily generalized.

If however, one writes out what one is actually doing:

A -- B == C means A implies C. Using this idea repeatedly, I can find no logical difference between the two - other opinions, though are most strongly encouraged.

Steve. BTW Scanraid does mention in his notes that
---' 'Guardians' will never solve anything while 'Multi-colouring' is switch on since they both attack similar configurations.'

Thanks Kaz - I did not see that note - albeit I suspect that I may have motivated that note...
The idea of guardians, expanded beyond just one type of candidated, though - could perhaps occassionally yield something - although it is rare - I have only been able to use it rarely - and in none of More...

proof of today's tough - max chain depth 3
max classical depth Steve?

1) Start with 22, UP 23
2) Naked Triple: b321=357 => ~(b954, ac1)=357
3) Naked Triple: h987=235 => ~h6521=235, UP 33
4) Naked Pair: (h5, i4)=19 => ~i5=19
5) Naked Pair: ac1=29 => ~deh1=29
6) More...

note: the AUR isn't required (just fluff)

or using hidden pair at step 7:
c8=2 == c8=4 -- b9=4 == b9=1 -- a8=1 == e8=1 -- HP: eh8=25 == ac8=2

step 11 could be replaced with:
f7=6 == f1=6 -- f1=1 == i1=1 -- i1=8 == i3=8 -- f3=8 == f3=5
=> ~f17=5, UP 81.

somewhat simplified proof:

1) Start with 22, UP 23
2) Naked Triple: b321=357 => ~(b954, ac1)=357
3) Naked Triple: h987=235 => ~h6521=235, UP 33
4) Naked Pair: (h5, i4)=19 => ~i5=19
5) Naked Pair: ac1=29 => ~deh1=29
6) c8=2 == c8=4 -- b9=4 == b9=1 -- a8=1 == e8=1 More...

Some hints for the non mathemeticians amongst us - with apologies to those it irritates.

At 31 filled: If h8=2, c8=4, d8=6 therefore a8&b9=1 Impossible therefore h8=5
In the same way eliminate e8=2
Therefore a8 or c8= 2 . Eliminate other possibilities for 2 in Bb8 up to 34
More...

17:31

This was one of the original pics from when this site started showing pics over a year ago and have seen it quite a number of times!! Still lovely and peaceful

Hi Pat!
In your last proofs - 'classical depth' of the steps:
7.1) 4 - strong sets being the four cells considered: acd8,b9
10b.1)4- strong sets being the four cells considerd again
11.1) 7 - again, just the cells considered

6.2) - 4 - but the language you use has me a bit More...

Hi Pat again!
6.2) sorry about my density! strong sets are 5 the way you present it... 2's and 5's and 1's in row 8 plus the two cells.

BTW - I missed the depth 3 elimination of 7 at f5 - otherwise my first proof would have essentially looked like yours, as the ALS on 7's was among the More...

steve why tell every1 wat da solution iz? i 4 1 like 2 figer them out 4 myself.

u 2 pat & clark & fi, stop ruining it 4 me!!!

What the?? You are funny!

20:11