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edit: This is, i think, one of the cooler puzzles so do give it a shot!
Last day's puzzle was first solved by Origami, good job! + Show Spoiler [solution] + First do 5 races of 5 horses, call these 5 races group A B C D E. Then race the winner of each race, say a b c d e, call this race F. The winner of F will be the top one horse. Suppose the first/second/third place for race F is b/a/c in that order.
Suppose b wins race F, then in group B, there can potentially be the 2nd and 3rd place, beaten by B. So take the 2nd and 3rd horse from group B. (say they're b_2, b_3)
Suppose a is 2nd place in race F, then in group A, a could've beaten a real 3rd place. So take the 2nd place from group A. (say it's a_2)
So now we race a, a_2, b_2, b_3, c to find out which is the real 2nd and 3rd place.
I'm reading this combinatorics book, so it has some great problems, I'll share one now.
Suppose we take all the integer coordinates of R^2, (i.e. (0,0), (1,1), (4,5), (-1,-4), and so on), and color each of those coordinates with one of SIX colors, for instance, say we color (4, -2) with RED, and (-4, 5) with BLUE, and so on...
Prove that for any coloring scheme, there exists a rectangle, which has four verticies that has the same color. (For instance, maybe (0,0) (1,0) (0,1) (1,1) all has GREEN, then we'd have a rectangle who's verticies are all green).
Again, answers in spoilers and if you cannot solve it before you leave this blog, post your partially formed thoughts so others can draw some inspirations (And for me personally I want to see lots of replies since it makes me think these blogs are worthwhile <3).
GL HF(this one's tricky, I didn't solve it till I read the solutions)!
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+ Show Spoiler +e.g. take the color C at (0,0), arbitrarily move along the positive x-axis until you reach the same color C at some coordinate (X,0). move in the positive y direction for both aforementioned coords until (0,Y) and (X,Y) share the same color? if you take all of R2, this is bound to happen --> infinity.
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On May 29 2009 07:04 paper wrote:+ Show Spoiler +e.g. take the color C at (0,0), arbitrarily move along the positive x-axis until you reach the same color C at some coordinate (X,0). move in the positive y direction for both aforementioned coords until (0,Y) and (X,Y) share the same color? if you take all of R2, this is bound to happen --> infinity
This is why we play dota.
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On May 29 2009 07:04 paper wrote:+ Show Spoiler +e.g. take the color C at (0,0), arbitrarily move along the positive x-axis until you reach the same color C at some coordinate (X,0). move in the positive y direction for both aforementioned coords until (0,Y) and (X,Y) share the same color? if you take all of R2, this is bound to happen --> infinity
+ Show Spoiler + what if the color at (0,0) is green, and there's only 1 green on our entire coordinate system?
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On May 29 2009 07:06 evanthebouncy! wrote:Show nested quote +On May 29 2009 07:04 paper wrote:+ Show Spoiler +e.g. take the color C at (0,0), arbitrarily move along the positive x-axis until you reach the same color C at some coordinate (X,0). move in the positive y direction for both aforementioned coords until (0,Y) and (X,Y) share the same color? if you take all of R2, this is bound to happen --> infinity + Show Spoiler + what if the color at (0,0) is green, and there's only 1 green on our entire coordinate system?
+ Show Spoiler +since there is a finite number of colors, you are bound to end up with a rectangle somewhere (and the origin choice was arbitrary).
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On May 29 2009 07:09 paper wrote:Show nested quote +On May 29 2009 07:06 evanthebouncy! wrote:On May 29 2009 07:04 paper wrote:+ Show Spoiler +e.g. take the color C at (0,0), arbitrarily move along the positive x-axis until you reach the same color C at some coordinate (X,0). move in the positive y direction for both aforementioned coords until (0,Y) and (X,Y) share the same color? if you take all of R2, this is bound to happen --> infinity + Show Spoiler + what if the color at (0,0) is green, and there's only 1 green on our entire coordinate system?
+ Show Spoiler +since there is a finite number of colors, you are bound to end up with a rectangle somewhere
This is why we play dota.
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Been lurking TL, never posted.
Edit added Spoiler tag.
+ Show Spoiler +Consider the X-axis, (X,0). Since there is a finite number of colours, and an infinite number of points, at least 1 of the colours is repeated infinitely many times. Call this colour ONE. Remove the columns where the colours are not ONE.
Now repeat for (X,1), at least one of the colours is infinite, call this colour TWO. Remove all columns not colour TWO in the (X,1) row.
So now we have a map with all ONE in the (X,0) and all TWO in the (X,1).
Keep going for (X,2) ... (X,6). We have 7 rows, and 6 colours. One of the colours is repeated. So there are infinitely many rectangles of the repeated colour.
Edit: Actually, you can keep going along the Y direction, one of the colours will appear infinitely many times. So you can have the existence of a infinite two-d rectangular grid of 1 colour.
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Konfustikator, I don't think your argument is correct, since I can tactically set the colours in a way such that the repeated colour do not have anything that forms a rectangle. For example, when you get the 7th column you argued that there will be two columns of the same colour, but out of these two columns, one may have the coloured ones only at odd numbers and the other one only at even, then you cannot form a rectangle.
Here's my solution. + Show Spoiler + If it's too hard to understand I will draw some pictures later. First choose a horizontal line that contains 6 points of the same colour (this must be possible). Through these 6 points we draw 6 vertical lines, and consider each 6-tuple of points that are on the 6 vertical lines that are on a horizontal line. Although there are infinitely many of these 6-tuples we will only consider 7*6^6 + 1 many of them (7*6^6 + 1 is a very conservative estimate, but it works). Note that each 6-tuple has 6^6 different ways of colouring, so for 7*6^6 + 1 many of 6-tuples, at least 7 6-tuples have the same colours. Furthermore, for each of these 3 6-tuples, each of the 6 points must have difference colours (otherwise we have a rectangle of the same colour). Now out of these 7 6-tuples, draw horizontal lines across them, and consider the intersection of these 7 horizontal lines with another vertical line (there are 7 intersections). Thus out of these 7 points at least two have the same colour, which will form a rectangle with two other points on some of the 7 6-tuples.
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Reserved for a picture in case it is needed.
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I think you misread my solution. I remove any columns that do not have the infinite colour.
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On May 29 2009 09:08 Konfustikator wrote: I think you misread my solution. I remove any columns that do not have the infinite colour.
I don't see how my example contradicts with this statement.
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I'm pretty sure I got it...
+ Show Spoiler +I didn't read anything in spoilers, so here's my reasoning. Take a column centered on the x-axis with length 2y. Since you have 6 colors to use on those 2y+1 points, there are P(2y+1,6) ways to set up this column. This is a finite number, so if you take exactly that number of columns (all centered on the x-axis but with different x-values), you'll either have two identical ones or every single possible coloring scheme. If you take one more, you have two identical ones for sure. This will necessarily produce a rectangle as the colors line up (say, 2 blues in both columns, which have respective common y-values) as long as 2y+1 > 6, so as long as y>=3.
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When I make the (X,0) row all the colour ONE, I remove all columns which do not have ONE in their (X,0) position. So the new map which I work off, has all colour ONE on the x-axis. eg If the x-axis is 1 2 1 3 1 4 1 5 ...., then the 2,4,6,8th colums are all removed, so the map has 1 1 1 1 1 on the x-axis.
The new map is infinite, same as the old map. But now it has all 1s in the (x,0) row.
Then repeat for the 2nd (x,1) row.
The new map is infinite, with all 1s in the 1st row, and all 2s in the second row.
etc.
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Pseudo_Utopia's solution seems to be right, but it applies to any number of colours, not just 6.
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On May 29 2009 10:49 Konfustikator wrote: When I make the (X,0) row all the colour ONE, I remove all columns which do not have ONE in their (X,0) position. So the new map which I work off, has all colour ONE on the x-axis. eg If the x-axis is 1 2 1 3 1 4 1 5 ...., then the 2,4,6,8th colums are all removed, so the map has 1 1 1 1 1 on the x-axis.
The new map is infinite, same as the old map. But now it has all 1s in the (x,0) row.
Then repeat for the 2nd (x,1) row.
The new map is infinite, with all 1s in the 1st row, and all 2s in the second row.
etc.
yes urs works, his works too.
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Wow that's cool stuff. Is this stuff graduate level?
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On May 29 2009 18:39 Monoxide wrote: Wow that's cool stuff. Is this stuff graduate level? no i'm just undergrad reading combinatorics book haha :D
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On May 29 2009 19:34 evanthebouncy! wrote:Show nested quote +On May 29 2009 18:39 Monoxide wrote: Wow that's cool stuff. Is this stuff graduate level? no i'm just undergrad reading combinatorics book haha :D
Oh ya?? I just finished first year, soo my math is only at a first year level.
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On May 29 2009 07:04 paper wrote:+ Show Spoiler +e.g. take the color C at (0,0), arbitrarily move along the positive x-axis until you reach the same color C at some coordinate (X,0). move in the positive y direction for both aforementioned coords until (0,Y) and (X,Y) share the same color? if you take all of R2, this is bound to happen --> infinity.
This is trial and error, this is not proof. I am pretty sure you'd fail with this answer at any graduate course. But I suppose its acceptable here. You are thinking from a programmers perspective and not from a mathmaticians point of view.
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