SPOILERS for User Puzzle #251,#253 using Color Rules |
Naivoj Kwon-Tom Addict Puzzles: 314 Best Total: 33m 50s | Posted - 2007.12.05 10:07:02 This is as far as I get for #251 without using FP, Highlander rules or Color rules: After that I color the inside paths red, and the outside paths yellow. - I like it that way because the outside path merge with the frame which is yellow in Kwon-Tom.
Then I only need the following simple color rule: a clue 2 square is always surrounded by 2 red "adjacent cross squares" and 2 yellow cross squares: - (adjacent) cross squares: the 4 squares which are part of the "5 squares Cross" centered by the referred clue(clue 2 in this example). - I call the other 4 surrounding squares the "(adjacent) diagonal squares"
I apply this rule on clue 2 @c10r8: - top and bottom squares are red=inside - right square is outbound=outside=yellow = so the left square must be yellow=outside Next a very basic color rule gives a line at the left of clue 2 @c9r8
Afterward the rest of the puzzle can be solved with regular rules.
Last edited by Naivoj - 2008.01.08 09:03:01 |
bear Kwon-Tom Obsessive Puzzles: 1066 Best Total: 34m 35s | Posted - 2007.12.05 14:26:34 Thanks. That was interesting. I realized that I'd missed a common pattern.
Presented with:
I now see you can always get:
But I'd only seen the top line, not the two side x's. |
Para Kwon-Tom Obsessive Puzzles: 1923 Best Total: 19m 28s | Posted - 2007.12.06 03:14:02 This is the reasoning i used in the bottom right corner.
Of the 3 marked questionmarks one of the left and right has to be a line and the other an x. The same goes for the top and right questionmark. The left and top questionmark can't both be lines. From this we can conclude that the only option is that the right questionmark is a line and the other 2 are x's.
Para
Last edited by Para - 2007.12.06 21:56:50 |
procrastinator Kwon-Tom Obsessive Puzzles: 1083 Best Total: 12m 56s | Posted - 2007.12.08 08:45:40 Thanks Naivoj, that's exactly what I was hoping for. I don't see a pattern-based way to work that line out as quickly - it certainly wouldn't occur to me quickly to use Para's logic - so I think really might be a case where colours can make something significantly easier and maybe quicker once you get used to visualizing them and don't need to spend the time actually colouring. I'm quite hopeful I can learn some new techniques from this, maybe quite a range. Thanks a bunch.
Now let me tell you how I would solve this one: there are two things fairly apparent to me here. A vertical line in the third-last slot on the fifth line would force seven lines through that row. A line in the third-last slot of the ninth row would force three lines into the corner. |
Brian Kwon-Tom Obsessive Puzzles: 4900 Best Total: 9m 6s | Posted - 2007.12.09 05:28:14 This can be done by counting arguments, which Jankonyex has explained. If you look at square c9r9, the bottom left corner has one edge coming in, the bottom right corner has two or zero edges coming in, and the top right corner has one edge coming in. Therefore the top left corner must have two or zero edges coming in--and since it can't have two, it must have zero. So we get the line on the left edge of square c9r8.
(As procrastinator said, I would first see that the left edge of square c9r9 can't be used, since three lines would enter that corner.) |
Naivoj Kwon-Tom Addict Puzzles: 314 Best Total: 33m 50s | Posted - 2007.12.09 12:18:16 Of all these explanations, the easiest to understand for me is the procrastinator 2nd one: left of square c9r9 can't be a line. Note that after that the rest of the puzzle solves with regular patterns.
Quote: Originally Posted by brian Therefore the top left corner must have two or zero edges coming in--and since it can't have two, it must have zero. So we get the line on the left edge of square c9r8. |
Brian, shouldn't the conclusion be: since it can't have zero edge(there is already one), it must have two. |
Brian Kwon-Tom Obsessive Puzzles: 4900 Best Total: 9m 6s | Posted - 2007.12.09 14:45:48
Quote: Originally Posted by naivoj Brian, shouldn't the conclusion be: since it can't have zero edge(there is already one), it must have two. |
I don't think I explained it well. By "coming in" I mean that we consider the two edges of the square at that corner. So it couldn't be two edges because that would make it at least three lines at that vertex. In this case though, it's probably clearer to just say that c9r9 is a virtual 2.
Last edited by Brian - 2007.12.09 14:46:05 |
Naivoj Kwon-Tom Addict Puzzles: 314 Best Total: 33m 50s | Posted - 2007.12.10 00:44:16
Quote: Originally Posted by brian I don't think I explained it well. |
Actually you did explain it well the first time. My mistake: when I wrote this I was incorrectly thinking "coming out" instead of "coming in". |
Naivoj Kwon-Tom Addict Puzzles: 314 Best Total: 33m 50s | Posted - 2008.01.08 10:16:25 I get this far for #253 without using FP, Highlander rules or Color rules:
Basic path coloring (inside paths red, outside paths yellow) give a line at the top of c3r4 (which lead to 4 more lines: left and right of c1r4 and left and top of c2r3) and a line at the top of c3r6.
Furthermore as the left & right squares of clue 2 @c5c7 are both yellow (outside) therefore the bottom & top squares must be red(inside)leading to a line at the right of c4c8 permitting to complete the first 3 columns and giving 2 crosses top and right of clue 1 @c6r6, ..., resulting in:
Again since the left and right squares of clue 2 @c10r6 are yellow(outside) the top and right squares must be red: after that the puzzle solve easily.
I realize now that I did not introduce new color rules describing #253, next time I will choose one that will. |
Jankonyex Kwon-Tom Obsessive Puzzles: 5680 Best Total: 9m 35s | Posted - 2008.01.09 17:09:33
nice highlander pattern
Last edited by Jankonyex - 2008.01.09 17:11:56 |
procrastinator Kwon-Tom Obsessive Puzzles: 1083 Best Total: 12m 56s | Posted - 2008.01.10 15:32:41
Quote: Originally Posted by jankonyex |
Just beautiful. |