- 06 Aug '11 06:11 / 4 editsSuppose I give you a white, blank card. Suppose it is square with side length of 5 (arbitrary) units. Suppose I have a bunch of crayons of all different colors. Suppose I tell you that you need to color one side of the card (either partially colored or fully colored, however you want to do it) such that when you finish there will be no set of two points on that side that satisfies both the following: (1) the two points are 1 unit apart and (2) the two points have the same color.

What is the minimum number of crayons you need? - 06 Aug '11 18:40

Exactly one unit? Not "one unit or more"? Then certainly no more than four. Perhaps, with a different pattern, three. In fact, I suspect that if fractal patterns are allowed you could get it down to two, but fractal patterns don't work very well with crayons.*Originally posted by LemonJello***Suppose I give you a white, blank card. Suppose it is square with side length of 5 (arbitrary) units. Suppose I have a bunch of crayons of all different colors. Suppose I tell you that you need to color one side of the card (either partially colored or fully colored, however you want to do it) such that when you finish there will be no set of two point ...[text shortened]... and (2) the two points have the same color.**

What is the minimum number of crayons you need?

Richard - 07 Aug '11 00:56

I presume the blank white card counts as a colour?*Originally posted by LemonJello***Suppose I give you a white, blank card. Suppose it is square with side length of 5 (arbitrary) units. Suppose I have a bunch of crayons of all different colors. Suppose I tell you that you need to color one side of the card (either partially colored or fully colored, however you want to do it) such that when you finish there will be no set of two point ...[text shortened]... and (2) the two points have the same color.**

What is the minimum number of crayons you need? - 07 Aug '11 14:22Minimum 8 colors

Divide the square in adjacent circles each of diameter 1 unit. Each circle must have 4 surrounding circles touching its circumference.

1st row of circles-> color circles in alternate colors (A and B)

2nd row of circle-> color circles in alternate colors (C and D)

3rd row of circle-> color circles in alternate colors (A and B)

4th row of circle-> color circles in alternate colors (C and D)

and so on

Also there are gaps between circles.

1st row of gaps-> color gaps in alternate colors (E and F)

2nd row of gaps-> color gaps in alternate colors (G and H)

3rd row of gaps-> color gaps in alternate colors (E and F)

4th row of gaps-> color gaps in alternate colors (G and H)

and so on

Therefore, distinct colors used are, A B C D E F G H (total 8)

Since the color of the paper is white, we need 7 non-white colors. - 07 Aug '11 18:25

Test of private -- ignore.*Originally posted by LemonJello***Suppose I give you a white, blank card. Suppose it is square with side length of 5 (arbitrary) units. Suppose I have a bunch of crayons of all different colors. Suppose I tell you that you need to color one side of the card (either partially colored or fully colored, however you want to do it) such that when you finish there will be no set of two point ...[text shortened]... and (2) the two points have the same color.**

What is the minimum number of crayons you need?

[private] mmm [/private] - 07 Aug '11 18:29
*Originally posted by LemonJello*

What is the minimum number of crayons you need?

This is my guess. Don't reply & quote or you will reveal it.

Five layers -- ABCDE, CDEBA, ABCDE, CDEBA, ABCDE where each letter is a color but one is white so four crayons. - 08 Aug '11 18:14
*Originally posted by LemonJello*

What is the minimum number of crayons you need?

5 layers or ranks, alternating ABCDA with CDABC with A = white, so 3 colors (B, C, and D. - 09 Aug '11 01:55 / 1 editThis is an interesting problem! I can't think of anything better than smartarin's solution, though I see no proof that his is the best (and if we can use fractals, I doubt that it is). How do you see to do four Shallow Blue?

2 colors (1 crayon) is impossible even with fractals because there exist triples of points that define 1 unit sided equilateral triangles on the card, but it wouldn't surprise me if it was possible to get down to 3 or at least fewer than smartarin's 8 with fractals.

JS357 -Are you dividing the paper up into a 1x1 square grid and then coloring? If so, then the solutions dont work, because within each square there are two points that are within 1 unit of one another (along the diagonal for instance) - 09 Aug '11 16:08 / 1 edit

It's fairly easy to prove that you can't do it with two colors:*Originally posted by Shallow Blue***In fact, I suspect that if fractal patterns are allowed you could get it down to two, but fractal patterns don't work very well with crayons.**

Richard

Assume an equilateral triangle with side = 1 unit.

The color at each vertex must be different since each vertex is one unit away from both other vertices. - 09 Aug '11 17:42

Got it. I oversimplified it to a special case of the map coloring problem.*Originally posted by Anthem***This is an interesting problem! I can't think of anything better than smartarin's solution, though I see no proof that his is the best (and if we can use fractals, I doubt that it is). How do you see to do four Shallow Blue?**

2 colors (1 crayon) is impossible even with fractals because there exist triples of points that define 1 unit sided equilateral trian ...[text shortened]... two points that are within 1 unit of one another (along the diagonal for instance)[/hidden] - 10 Aug '11 22:27 / 1 edit

Yes, exactly 1 unit.*Originally posted by Shallow Blue***Exactly one unit? Not "one unit or more"? Then certainly no more than four. Perhaps, with a different pattern, three. In fact, I suspect that if fractal patterns are allowed you could get it down to two, but fractal patterns don't work very well with crayons.**

Richard

The subject of fractal patterns is an interesting line of thought here. Can you develop that more fully? Also, can you offer support for your claim of "certainly no more than four"? - 10 Aug '11 22:33 / 3 edits

If you have some circle of diameter 1 and of some solid color, then wouldn't opposite terminal points on any diameter violate the rule? That is, such points would be 1 unit apart and of the same color.*Originally posted by smartarin***Minimum 8 colors**

Divide the square in adjacent circles each of diameter 1 unit. Each circle must have 4 surrounding circles touching its circumference.

1st row of circles-> color circles in alternate colors (A and B)

2nd row of circle-> color circles in alternate colors (C and D)

3rd row of circle-> color circles in alternate colors (A and B)

4th ...[text shortened]... e, A B C D E F G H (total 8)

Since the color of the paper is white, we need 7 non-white colors.

Also, how are you treating, for example, those points that lie at the meeting points between adjacent circles? The circles you propose will all have specific points that are common to more than one circle, correct? How do we treat those?