Paint the tablecloth

It is a CRT pixel array.

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Next.

Laying out an array of dots does not constitute "painting the surface".

*Every* point needs to be painted.

-- Mark Brader "Great things are not done by those Toronto who sit down and count the cost snipped-for-privacy@vex.net of every thought and act." --Daniel Gooch

Expand each "center point" into a hex.

Expand the scale such that the hexes are compliant with the rule.

I think the condition is

'no two points "exactly" 1cm apart have the same color'

instead of

'less or equal to 1cm apart'.

It is trivial as long as you can leave some parts of the surface/tablecloth uncoloured. Just divide it into columns of squares with sides x cm where

0.5 < x < sqrt(0.5) then in each even column colour squares alternately in two different colours and in each odd column colour alternate squares in the third colour leaving the rest uncoloured.

--
Ian Malcolm.   London, ENGLAND.  (NEWSGROUP REPLY PREFERRED)
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk
[at]=@, [dash]=- & [dot]=. *Warning* HTML & >32K emails --> NUL:

If you have to paint _all_ the points, it's impossible.

Consider a unit square (with distances in cm).

Two of the vertices P,Q say must have the same color, red say, hence to avoid having a distance of 1, P,Q must be diagonally opposite. The midpoint, M say, of the line segment PQ, can't be red, so assume M is blue. But then,other than P,Q,M, all other points on the line segment PQ must be green. But any two green points on the line segment PM will be have a distance to each other less than 1, contradiction.

quasi

I take it back -- I misunderstood the restriction.

The restriction applies only to pairs of points which are _exactly_ 1 unit apart. I mistakenly thought the restriction applied to pairs of distinct points with distance less than or equal to 1.

Ignore the above argument.

It's not that simple.

If it can be done, it may require the axiom of choice, but I'm not sure that it can be done.

quasi

Yes, I just realized that.

Which I think makes the problem non-trivial.

quasi

I don't choose to accept it: it is impossible. E.g. search for "Moser spindle".

--
Tim

The answer will then depend on whether two pen 'pixels' or 'droplets' occupying the same space, causing a different color is regarded as a different color, or "some" of the set of three, causing a fail.

If mixing is allowed, then it is easy.

If the hex wall segment is more than a .414cm, the cell itself fails to comply. If the hex wall segment is less than 1 cm, the group fails to comply.

So hex won't do it.

RL

Whoops. That *doesn't* work. The constraint has to be the diagonal distance across a square is less than one while the distance between squares is greater than one. NOT POSSIBLE using squares.

--
Ian Malcolm.   London, ENGLAND.  (NEWSGROUP REPLY PREFERRED)
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk
[at]=@, [dash]=- & [dot]=. *Warning* HTML & >32K emails --> NUL:

h

You can't even do that if you assume that you can only paint Lebesgue- measurable sets.

Riddles obfustigate life. I hate riddles.

--
Failure does not prove something is impossible, failure simply
indicates you are not using the right tools...
nico@nctdevpuntnl (punt=.)
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Can you mix / blend the red, green and blue together to make other colours?

Which might have meaning if it was actually a real word.

It's a perfectly cromulent word.

-- Richard

Can't be done - here is my proof which may be more complicated than necessary...

Assuming it is possible:

Applying the usual x- and y-axes to the plain, look at the points (n,0) where n is an integer. By constructing a lattice of equilateral triangles incorporating these points, we see that they must be coloured in succession ...R,B,G,R,B,G,R,B,G,R,B,G,..., or the reverse. In any case, the points (0,0) and (3,0) must be the same colour.

(Here is a diagram to help make this clear:

. . . . G B R . . .

. . . B R G B . . .

. . . . . . . . . .

having fixed 2 corners of a triangle, the 3rd is forced, and this extends in a ladder pattern. The pattern repeats with period 3cm)

But this applies whatever the orientation and origin chosen for the axes, so in fact all points on any circle of radius 3cm will be the same colour! Hence there will be points on the circle 1cm apart which are the same colour (a contradiction).

Mike.