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#1: Tom King (sgusa) on Jan 13, 2013 [HINT]

I couldn't solve without internal edge logic. Thought the solve was fairly advanced. The Supreme Commander will simply order his minions to GIVE him the solution.

By the way, Aldege, I attempted to make this in "Aldege's style." ;) Strayed a bit to make it solvable...

Didn't finish, I'm afraid. I know some people enjoy the advanced "guessing games" but I'm afraid I'm not one of them.

There is no guessing, Norma. It is tough, but logically solvable. Look at the vertical 7 and 8 as well as the bottom 5 rows...

Wow Tom,this could be a remake of "Planet Of The Apes". Maybe my nickname should be monkeyboy. Glad you joined in on the fun,thx sgusa.:)

Thanks for a Royal entry! I enjoyed solving very much! I first placed most of the bottom rows with edge logic and then the eight on C18.

thx Tom

Comment Suppressed:Click below to view spoilers

lol it does doesn't it?

The face kept getting wider as I attempted to make it solvable...

Double post...

Hmm, I placed the ears using edge logic, though I suppose the ear on the left side was using visual logic, since I matched the ear on the right. It was a pretty easy solve after that :)

Deep lookahead is BS. Dot r5, c8. 10. 12
EL 13 r1
EL 5 r20, dot c1-5
EL 4 c20, dot r1, 12-20
LL
EL 7 r19, dot c15-19
EL 10 r17, dot c17-19
LL
EL 6 r15, dot c7
LL
EL 2 r13, dot c2
EL 8 c18, dot r1-5, 15, 16
LL to finish.

Very difficult solve, and I created it and proved it twice. It does not require deep lookahead. It is logical, and deserves a very difficult rating (not the crap under 3 that it has received). It requires a lot of internal edge logic. How come Gator gets 5s on difficulty and this gets a 2?

It is now trending north of 3. I solve every puzzle that I create. I don't guess. Many forget the first step in this solve, back-to-back ones or the possibility which allows you to do the first step. If it is a 1,2, and there is a mark in the 3rd column, you can always dot behind it. It is either a one (negating row 2), or a 2 with a one behind it (again negating row 2). Same goes for anything with a 1, number starting it. If row or column 3 is marked, then row or column two will be dotted.

Tom, you need to remember that there are people like Karl out there who deliberately mess with the ratings, sometimes high, sometimes low, so you can't pay attention to the ratings.

It is marked deep lookahead.

I solved it again. Don't know if I solved it the same way. I hate the question mark.

EL r1, c1,20
Dot r5, c8,10,12
Black r3, c9,11
EL 5 r20, dot c2-5,15-20
LL
EL 8 c18 dot r16.17
LL
EL c3, dot r16,17
EL 8 c18, Dot r14,15
EL 8 c18 not r1-4
EL 7 c3, nor r2-4
EL 5 C17, not r18
EL 2 r15, not c2-4
LL to finish

Two different logical solutions without deep lookahead.

Started with a logical guess that the top 13 ran between the two 5's (because this married up well with the checkerboard of the next 4 rows). After that, it all fell into place. It helps to know that Aldege has a beard.

Thanks, Gaynor. Sorry that it looks so "monkey-like"

aldege.

Found to be solvable with moderate lookahead by infrapinklizzard.

Tom's second solve falls apart on the third line when he makes r3c9 &c11 black for no discernible reason.

Tom's first solve fails at the sixth line where he says c15-19 dotted (white). The EL on the 7 can only give c17-19 white. In order to get c16-19 white you must use deep lookahead to show that the 5 cannot fit under it after filling the vertical clues. You cannot logically get c15 white at this time. This makes line 9 etc fail.


Here's my streamlined moderate-lookahead solve:

After Line Logic
Edge Logic 13 r1 = c1-3 White
LL

Deep EL 5 r20 (vs r17) = c15-17 w
LL
Internal EL 10 r17 = c17-19 w {conflicts 2 r16}
LL

IEL 8 c18: notice the 1|2|2|3|1 in r4-8. There is no way the 8 can cross all those clues without causing a conflict in either c17 or c19. Therefore c18r1-4 must be white.
LL (don't forget the whites around the 1s in r9)

two-clue EL:
If the 1,2 in c19 were all the way down (in r13 and and r15-16 respectively) then
> the 2 would trigger the 3 in r15 which would go into c17
> the rest of c19 would be white, causing
> r10&11 to force the 3s in those rows left into c17
= causing a block of six in c17. Therefore, our initial assumption must be wrong and the 1 in c19 cannot be in r13. (And so c19r13 must be white)

Then LL to finish.


****

Here's my unedited solve in case you want to see how I got to that solution:
After Line Logic
Edge Logic 13 r1 = c1-3 White
LL
EL 4 c1 = r12-16 w
EL 4 c20 = r1, r12-16 w
LL
Deep EL 5 r20 (vs r17) = c2-5, w {conflict with the 10}
LL
Internal EL 8 c18 (vs 5 c17) = r20 w
LL
DEL 5 r20 (vs r17) = c15-17 w
LL
IEL 10 r17 = c2 w {conflicts 3 r16}
LL
IEL 10 r17 = c17-19 w {conflicts 2 r16}
LL
Here I'm stuck for moderate lookahead that goes anywhere.
EL 3 r16 = c2-3 w
No LL
Here I'm stuck for moderate lookahead, period.

Deep lookahead:
if the 7 in r19 were in c10-16, then:
> the 4s in c15&16 would trigger the 10 in r17 thus
> would force the 3 in c14 up into r17 thus
> would make r20 c14 w
meanwhile,
> the 7's placement would also make r19c9 white thus
> would force r20c9 white
= leaving no room for the 5 in r20. Therefore the initial assumption is wrong and so r19c16 must be white.

LL (one pixel)

More deep lookahead:
if the 5 in r20 were to go right into c14, then:
> the 10 in r17 would be forced all the way left into c4-13, thus
> crossing clues would immediately make r16r4-6 black.
> However, note that the 7 in r19 would gain a black, thus making r19c7 white,
> which would force the 3 in c7 up into r16
= that would make the fourth pixel in a block [r16c4-7] where the biggest clue is a 3. Therefore our initial assumption must be wrong and so r20c14 must be white.

LL (two pixels)

Two-clue edge logic:
If the 3,2 in r16 were all the way right {in c14-16 and c18-19 respectively} then columns 16,18,&19 would all be forced up into r15 causing a conflict. So our inital assumption was wrong and r16's 3 must be in the left section [c4-8] instead, making r16c6 black.
(No LL)

Hmm, here's some moderate lookahead that doesn't depend on the deep lookahead. I'll have to see later if it can be done before the deep lookahead. (NB: It could.)
IEL 8 c18: notice the 1|2|2|3|1 in r4-8. There is no way the 8 can cross all those clues without causing a conflict in either c17 or c19. Therefore c18r1-4 must be white.
LL

Similar to the last, but harder to see is IEL on the 7 in c3. If it were to go in r3 or above,
> the 2 r3 would be forced left into c2 fulfilling the 1 in that column and
> even with r6 crossing the 11 instead of the 1, there would be no way of arranging rows 3-7 without causing a conflict in either c2 or c4
= Therefore c3r2-3 must be white
LL

EL 2 c2 = r13 w

two-clue EL:
If the 1,2 in c19 were all the way down (in r13 and and r15-16 respectively) then
> the 2 would trigger the 3 in r15 which would go into c17
> the rest of c19 would be white, causing
> r10&11 to force the 3s in those rows left into c17
= causing a block of six in c17. Therefore, our initial assumption must be wrong and the 1 in c19 cannot be in r13. (And so c19r13 must be white)

LL to finish.

I couldn't figure it out without a DEEP lookahead...but then, I've never been a moderate type person lol. I loved chewing on this one. Thanks Tom, for creating it for us...and thank you Jota, for the detailed explanations of how one can solve it. I love this place.

No problem, JoDeen, but while my name starts with a jota, it then continues with an o and an e. ;)

LOL! BTW We haven't seen sgusa for a long time.

Lol...sorry Joe. Sheesh...trying to keep everyone straight and not having a mind picture on which to hang your names, plus having your "handle" names thrown in there too, plays havoc with my brain. I'll try to be more careful in the future. ;)

Just think of me as a giant infrapink lizard, and you'll be sure to recognize me the next time you see me.

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