Posers and Puzzles

Posers and Puzzles

  1. Joined
    24 Apr '05
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    06 Aug '11 06:114 edits
    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 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?
  2. Joined
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    06 Aug '11 18:40
    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?
    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
  3. Standard memberwolfgang59
    Mr. Wolf
    at home
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    07 Aug '11 00:56
    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?
    I presume the blank white card counts as a colour?
  4. Joined
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    07 Aug '11 14:22
    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 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.
  5. Joined
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    07 Aug '11 18:25
    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?
    Test of private -- ignore.

    [private] mmm [/private]
  6. Joined
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    07 Aug '11 18:29
    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?
    Oops, it's "hidden".

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

    Reveal Hidden Content
    Five layers -- ABCDE, CDEBA, ABCDE, CDEBA, ABCDE where each letter is a color but one is white so four crayons.
  7. DonationAnthem
    The Ferocious Camel
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    08 Aug '11 16:333 edits
    ignore this post
  8. Joined
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    08 Aug '11 18:14
    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?
    How about this one:

    Reveal Hidden Content
    5 layers or ranks, alternating ABCDA with CDABC with A = white, so 3 colors (B, C, and D.
  9. DonationAnthem
    The Ferocious Camel
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    09 Aug '11 01:551 edit
    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 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 - Reveal Hidden Content
    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)
  10. San Diego, CA
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    09 Aug '11 06:06
    Sounds like the answer you're looking for is six.
  11. Standard memberforkedknight
    Defend the Universe
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    09 Aug '11 16:081 edit
    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
    It's fairly easy to prove that you can't do it with two colors:
    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.
  12. Joined
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    09 Aug '11 17:42
    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]
    Got it. I oversimplified it to a special case of the map coloring problem.
  13. Joined
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    10 Aug '11 22:271 edit
    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
    Yes, exactly 1 unit.

    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"?
  14. Joined
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    10 Aug '11 22:28
    Originally posted by wolfgang59
    I presume the blank white card counts as a colour?
    Yes, for this problem, let's agree that the white counts as a color.
  15. Joined
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    10 Aug '11 22:333 edits
    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.
    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.

    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?
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