Sudoku Puzzles

I first project each no. into the 3x3 boxes & fill some places where the no. can't go elsewhere. Then I look at each col., row & box & start on the ones with the least spaces to fill. I put the possibilities into the empty places & fill more spaces by deduction. Either a space has only one number that can go there, or I narrow it down to doubles or triples. This puzzle was solved this way.

Got stuck and shamefully had to use one guess. Otherwise easy. But took a long time to do.

19:08

was it too long a weekend for that lad? good priest joke
all those Jesuits would like it

7:19 minutes total.

I think you aren't supposed to stand that close to the camera, are you?

Buenas MAEN, amigo/as.

easy to 29
look at c5 if either 3 or 8 h7=9 then easy eliminations to end

Hi crazy solvers.
Hints for a proof (very close from Valy's I guess).

1) easy fillings to 28.
2) eliminate h78=1 (1set), c12=3 (1set)
3) looking at g8h8 (2 sets), eliminate some possibles. Easy to 29.
4) looking at 6 in Ch, eliminate a1,a2,c8=1 (3 sets). easy to 47.
5) More...

gb check out c5=3 or 8 then h7=9. let me know if i missed something since this was a simple solution

Note : the proof sketched above is pure fc.
Note on the issue 'advanced techniques/guessing'.

Some argued that techniques like fcs require blind guess+luck.

I claim that trying to rely only on basic techniques with the hope they'll do the job is more blindful than a careful More...

Alot of different methods here - mine follows:
I6=7, b6=6, b8=9,
349 triple in col 9, g4=4
18 double in Bf8, 23 double in Bf8
17 double in Bb5, 49 double in row 6
2369 quad in col g, G2=5
24 in Bb2, 3’s in Bb5, c9=2
6’s in col c, remove elsewhere in Bb2
6’s in row 3, remove elsewhere in Be2
29 filled

If c2=1 or 6, c8=5
To 47 filled
If a8 or a9 =7, then f8=4
Triple 239 in col f, f5=7, d5=2
Triple 169 in row 7 yields 23 double in col f, f1=9
149 triple in col d, d2=7
1246 quad in row 2, i2=9, 53 filled
If either c1 or c2=1, then a7=6. Easy to the end.

finally get stuck... at 29 filled ! In my proof I get to the same point in two steps, looking inside 4 sets. My whole proof carefully examines 14 sets in 6 steps, BUT the RIGHT ONES !

So imho (blind search+luck) is no more typical in advanced techniques than in basic techniques... In both More...

Reminds me an anecdote on chess worldchampion Capablanca who once lost a game against a New York expert. Amateurs asked the players how far ahead a grandmaster could think over the chessboard. Capa said 'Ten moves ! '. His partner replied 'One move ! but always the best one...'

19:05 great fun.

Hi mark from ny.
Could you detail how you get h7=9 at 29 filled ? I couldn't figure it.

12:42, again one of the easiest I've done. I didn't do anything but eliminations (box/line) and was able to complete without guessing. See you all on the morrow.

gb from france....also(imho)can't get past fill 29...I've tried all my extreme tricks but still dead...Approached from 2 points but still stuck at 29...Not sure how Taz Box from North Carolina eliminated...it's probably very obvious but 'me stuck' nevertheless....

INNOCENCE IS PRICELESS

One Sunday morning, the pastor noticed little Alex standing in the foyer of the church staring up at a large plaque. It was covered with names with small American flags mounted on either side of it.

The seven year old had been staring at the plaque for some More...

The pastor said,' Well, son, it's a memorial to all the young men and women who died in the service.'

Soberly, they just stood together, staring at the large plaque. Finally, little Alex's voice, barely audible and trembling with fear,
asked, 'Which service, the 9:45 or the 11:15?'

Hi Jack From Colorado, complete solution graphically detailed in sandbox.

fc solution:
at 28 filled
(i8=1)==(i9=1) forbids {h8=1, h7=1}
(c6=3)==(c5=3) forbids {c1=3, c2=3}
(g8=2)==(g8=3)--(h8=3)==(h8=2) forbids {h7=2, f8=2, c8=2, g9=2, e8=2}
(h8=3)==(h8=2)--(g8=2)==(g8=3) forbids {i8=3, h7=3, e8=3, f8=3}
at 29 filled
(a7=1)==(a7=6)--(h7=6)==(h2=6)-- More...

gb from france....gracias !!

Always about 'fcs=guessing'...

To have a look at what is a 'trial and error' proof, juste paste today's puzzle :
_3_____4_+___6_____+_84___7_5+2____18 __+4___5___6+__93____2+9_7___18_+_____8_ __+_5_____7_
inside the solver at
http://www.sudokusolver.co.uk/index.ht ml
and see More...

Late start today. No snow, just rain overnight in Utah. Lots at the slopes though. Who is that guy? Does he post here? I'm going to have nightmares. 14 + minutes. Still learning

Simple eliminations yiels b6=6, b8=9, c9=2, g2=5, g4=4, and i6=7
Forbidding Chain looking at 1's. (c2=1)==(c2=6)--(h2=6)==(h7=6)--(a7=6)== (a7=1)--(a2/1=1) Since a2/1 are weak linked at start and end of chain, a2 and a1 do not equal 1, a2=3, a1=8. Since 1 mst be in c1 or c2, c8=5. Simple More...

BTW, who's goat is that?

Simple to the end.

Noticed typo. ab should be a8 I believe.

Since I am doing multiple posts, I might as well add that what in my mind I call Box 4, most would call Box b5.

hello! Mr. Funny Guy

Great one, Dixie! Sending it on to my dad, a retired Presbyterian minister, and the three Methodist pastors I used to work for.

To bed I must go - I'll try again tomorrow - too many 7's and not enough 9's. Bonne nuit tout le monde, à demain

23:18 with one guess, so it doesn't really count. I think it was tough. After seeing today's pic I think I prefer the cats and dogs.

8:14

why would anyone submit a picture of themself that is so...not flattering?

I was able to solve the puzzle using conventional methods...no guesses...just simple eliminatin...I am not skilled in many of the solving methods that many of the players describe in this blog, but I am able to struggle through most puzzles.

An apology to Amelie from VA...A couple of days More...

the abbreviation for Virginia.

This Sudoku does require advance techniques to solve. There are two points in todays puzzle that simple eliminations(conventional methods) will not be able to advance.

Looked at this at lunch - noticed that at 29 filled, looking at just the 6's, that a7 and b2 both =16 and are remote pairs. This fills to 47. Now I see, still remmembering the remote pairs, lookng at 1's now, a7=1 leaves impossible 7's on grid, therefor a7=6. Easy to end. Could probably write this in a different way, but the remote pairs make this puzzle trivial.