Just a hint...

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Just a hint...

Postby Cysis » Tue Nov 20, 2012 3:28 pm

Hi there. I got this from a friend and doesnt understand it.
After a week now, I registred here and hope someone can help me to solve it.
Thanks in advance :)
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Re: Just a hint...

Postby civiliza » Wed Nov 21, 2012 4:37 am

A set of instructions for the puzzle would have helped.

Without that, my best guess is that
1) It will result in a standard Sudoku
2) The clues are based on the 7 bars that an old style LED would use to depict a number
3) Each of the colours has an implication on the presence or absence of that bar
4) Grey bars are just placeholders and imply nothing

In traffic lights, Green is go and Red is stop, so I would guess that Green means "this bar must be here", while Red means "this bar must not be here"

Using this logic, the Red 3's in the fifth and sixth columns would actually be 1's (the only possible digit after the five red bars making up the 3 are eliminated).
Likewise, the green 3 in column eight would be either a 3, 8 or 9 (the only digits containing all five of the green bars).

But where does that leave Blue?
There is a digit in the third column that contains one Red, one Blue and one Green, So the interpretation of Blue as "this bar might be there" would be useless.
One (admitedly wild) guess is that Blue is a puzzle wide "All Or None" ie the puzzle can only work if all the Blue bars must be there, or all the Blue bars must not be there.

That said, all this (and especiallly the Blue interpretation) is pure conjecture, as I said at the start, a set of instructions would have helped.
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Re: Just a hint...

Postby civiliza » Wed Nov 21, 2012 5:37 am

Drat - Just tried using a graphics package to Selectively Filter out different combinations of colours, and my previous interpretation won't work with either all blues on, or all blues off - the 1's always fail on the bottom three rows.

Sorry - It seems that this interpretation if the colour scheme is a bust.
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Re: Just a hint...

Postby civiliza » Wed Nov 21, 2012 6:59 am

The Colour Filtering Method was getting confusing, finally had to trace the picture into Vectors and group all Reds together, all Blues together etc.

Anyway, having done this I tried all eight permutations of "Colour A must be there", "Colour B must not be there" etc, and they all fail due to either producing impossible digit shapes, or houses that must contain the same digit more than once.

I think that any solution based on certain colours indicating that the bar is mandatory while others indicate that the bar must not be there is doomed to failure.
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Re: Just a hint...

Postby Smythe Dakota » Tue Nov 27, 2012 10:31 am

In each cell, each of the 7 LED segments can be in any of 4 states, so for each cell there are 4^7 possibilities, i.e. 16384 different possible values.

Actually, there is also a 16385th possibility, namely, that the cell is completely blank. (This alternative could, however, simply be an abbreviation for "all LED segments are grey" or something.)

So it seems highly likely this could be some kind of N-puzzles-in-one thing, where N is perhaps 3 or 4.

For example, with N=3, the upper left and upper right LED segments could be assigned a value as follows:

red-red = blank
red-green = 1
red-blue =2
red-grey = 3
green-red = 4
green-green = 5
green-blue = 6
green-grey = 7
blue-red = 8
blue-green = 9

-- and we could continue this pattern up to 15, if we needed to, but 1-9 (plus blank) is enough for ordinary sudoku, thank you.

Similarly, we could pair the lower left and lower right, which would generate a second puzzle completely unrelated to the first.

And the top and bottom could be paired, generating yet a third puzzle.

The middle segment could just be there for decoration.

For that matter, we could make N=4 (instead of N=3) by using base-10 instead of base-16. For example, the four possibilities for each LED segment could be assigned values as follows:

upper left = 0,1,2,3
upper right = 0,4,8,12
lower left = 0,16,32,48
lower right = 0,64,128,192
top = 0,256,512,768
bottom = 0,1024,2048,3072
middle = 0,4096,8192,12288

By adding these values together, we can generate any number from 0 through 16383 for each cell.

The rightmost digits in the resulting numbers in each cell would then constitute Puzzle A, the next (tens) digits Puzzle B, the third (hundreds) Puzzle C, and the fourth (thousands) Puzzle D.

My only other observation is that, since all the completely blank cells in the original puzzle are on two intersecting diagonals (one NE-SW, the other NW-SE), that Tarek probably has something to do with this, with his wrap-around toroidal and projectively planar ideas.

Bill Smythe
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Re: Just a hint...

Postby tarek » Tue Nov 27, 2012 5:46 pm

Smythe Dakota wrote:My only other observation is that, since all the completely blank cells in the original puzzle are on two intersecting diagonals (one NE-SW, the other NW-SE), that Tarek probably has something to do with this, with his wrap-around toroidal and projectively planar ideas.

To me, This has a Set-do-ku feel to it :D . With the Setdoku, different combinations could create Sudoku Subpuzzles. which might be the case here.

3 different colours in 4 LEDs (vertical) you get 81 different symbols. 3 different colours in 3 horizontal LEDs gives 27 different symbols unless there is some constraint rule that would limit this ( I can't see any example having 3 different horizontal LED colours, if the top 2 are always similar then you will have your 9 symbols for each line or box)
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Re: Just a hint...

Postby Cysis » Tue Dec 04, 2012 10:06 pm

Sorry for my late reply, I was in a hospital.
I will try to understand all ur helping answers. After all, may I will crack that hard puzzle :)
P.S: Think that the "RGB color model" will may help. Dunno why, but its red, green and blue.
Any Idea?
P.P.S: That puzzle is the "easy" version.
The hard version is in the attachment.
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