witzar
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Posts posted by witzar
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No matter how 9 coins are placed on 6x6 board, according to pigeonhole principle, there have to be three columns containing 6 or more coins total. After removing such three columns (a greedy approach works here for Player B, he chooses best columns one by one), there will be no more than three coins left on the board. Obviously Player B can choose three rows to collect all remaining coins. There is nothing Player A can do about it.
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How about making just one big loop that splits the playing field into two equal parts?
After this just repeat copying your opponent moves (what he does in one of the parts, copy in another),
until he runs out of moves.
Good idea but does that guarantee a win?
The playing field is separated into two equivalent areas before my opponent's move.
The above statement is invariant in my strategy (it's true after every move I make).
Therefore after every my move either there is no move for my opponent (I win) or there is one.
In later case, no matter what he does, the invariant guarantees, that I can duplicate his move in the other area.
This way (1) I always have a move after my opponent, (2) the invariant is conserved after my move.
Number of moves in the game is obviously finite, therefore this strategy guarantees a win.
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How about making just one big loop that splits the playing field into two equal parts?
After this just repeat copying your opponent moves (what he does in one of the parts, copy in another),
until he runs out of moves.
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If I understand correctly, the following pattern is not considered "crossing":
__Ω_Ω_Ω_Ω_Ω__
The loops are closed, but it is "touching" not "crossing".
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You obviously have the general formula. Care to share it?
(2/3) * (4/5) * (6/7) * ... * ((n-2)*(n-1))
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So you can win instantly in first move connecting two dots and cutting off each of the remaining dots with a loop around it?
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Obviously I am missing something.
Me too.
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the sequence must start at 1 and go to some n (consecutively) where n does not equal 1.
Now it's a different story. BTW "best answer" sequence starts with 3 not with 1.
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1. Player 2 has the advantage. He wins 8 times out of 15 on average.
2. 40 lines. -
Clearly empty sequence (length 0) wins the contest..
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There is no way to paint vertices of regular 13-gon with three colors without forming monochromatic isosceles triangle.
And I don't have an elegant proof. (I was hoping that someone here will find it.)
But it can be proved as follows.
As it was said before (and according to pigeonhole principle) there will be 5 points of the same color.
It is enough to prove that some three of the five form isosceles triangle.
Five vertices of regular 13-gon inscribed in circle divide the circle into 5 arcs.
We can assign to each arc number of sides of 13-gon that this arc covers.
This way each selection of 5 points gives some partition of number 13 with 5 parts.
Here is the list of all partitions of 13 with 5 parts:
[9, 1, 1, 1, 1]
[8, 2, 1, 1, 1]
[7, 3, 1, 1, 1]
[7, 2, 2, 1, 1]
[6, 4, 1, 1, 1]
[6, 3, 2, 1, 1]
[6, 2, 2, 2, 1]
[5, 5, 1, 1, 1]
[5, 4, 2, 1, 1]
[5, 3, 3, 1, 1]
[5, 3, 2, 2, 1]
[5, 2, 2, 2, 2]
[4, 4, 3, 1, 1]
[4, 4, 2, 2, 1]
[4, 3, 3, 2, 1]
[4, 3, 2, 2, 2]
[3, 3, 3, 3, 1]
[3, 3, 3, 2, 2]
If partition contains three or more equal parts, than two arc of the same length have to be next to each other.
Such two arcs give us isosceles triangle, so we can remove all such partitions from the list.
The list of partitions to investigate becomes shorter:
[7, 2, 2, 1, 1]
[6, 3, 2, 1, 1]
[5, 4, 2, 1, 1]
[5, 3, 3, 1, 1]
[5, 3, 2, 2, 1]
[4, 4, 3, 1, 1]
[4, 4, 2, 2, 1]
[4, 3, 3, 2, 1]
Now observe that each partition of type ababc also gives isosceles triangle,
because the only order of parts that avoids two a's next to each other and two b's next to each other
contains abab pattern, but in such case there are two arcs of length (a+b) next to each other.
Removing such partitions leaves us with four partitions to investigate:
[6, 3, 2, 1, 1]
[5, 4, 2, 1, 1]
[5, 3, 2, 2, 1]
[4, 3, 3, 2, 1]
These partitions can be investigated one by one.Let's investigate [6, 3, 2, 1, 1].
The only way to avoid two 1's next to each other is to go with the sequence (6, 1, x, 1, y).
(Two sequences are equivalent if one can be produced from the other using any combination of circular shifts and/or reversals.)
Second 1 is surrounded by 2 and 3, so there are two arcs of lenght 3 (2+1=3) next to each other.
Let's investigate [5, 4, 2, 1, 1].
The only way to avoid two 1's next to each other is to go with the sequence (5, 1, x, 1, y).
No matter where we choose to put 4, there will be two arcs of length 5 (4+1=5) next to each other.
Let's investigate [5, 3, 2, 2, 1].
The only way to avoid two 2's next to each other is to go with the sequence (5, 2, x, 2, y).
Second 2 is surrounded by 1 and 3, so there are two arcs of lenght 3 (1+2=3) next to each other.
Let's investigate [4, 3, 3, 2, 1].
The only way to avoid two 3's next to each other is to go with the sequence (4, 3, x, 3, y).
No matter where we choose to put 1, there will be two arcs of length 4 (3+1=4) next to each other.
This concludes the proof. -
If we divide the circumference into 10 equal portions with 10 equally spaced points, then at least 4 of the points must be the same color, and there is no way to pick 4 of the points without 3 of them forming an isosceles triangle.
Label those ten points with digits 0-9 clockwise starting with some point. Suppose points 0, 1, 4, 5 are of the same color. Which three of them form isosceles triangle?
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reg pentagon vertex makes it, but i'm thinking of arms longer than base?
Equilateral triangle is an isosceles triangle.
I guess my last answer makes no sense.
What I meant to say is this: arms can be longer, shorter or equal to the base.
As long as arms are of equal length, a triangle is isosceles.
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Each point of the circle is painted either red, blue or green.
Prove, that there exists isosceles triangle inscribed in the circle with all its vertices of the same color.
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reg pentagon vertex makes it, but i'm thinking of arms longer than base?
Equilateral triangle is an isosceles triangle.
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Each point of the circle is painted either red or blue.
Prove, that there exists isosceles triangle inscribed in the circle with all its vertices of the same color.
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This is not true.If a 1 unit length circle with red center point is made of green and blue points,
there are 2 points with same color on the circle that is 1 unit away.
If you pick a point on the green-blue circle..there is a chance that there are 2 or 1 that is 1 length away from it or 25% that there is none..with that probability..applying to all the points in the circle, you get almost 100% for the whole circle to have at least one. Suppose at first there is no same color,then next,then next...so on, until you are back to the first point - the probability approaches zero. Therefore the statement above is not 100% True but not False..Unless statistics is out of the question it proves it
Pick a random real number from interval [0,100]. What is the probability that picked number is not 7? Clearly 100%. Does it follow, that picking 7 is impossible or that number 7 does not exists? Clearly not.
Divide unit circle into 6 equal arcs. Paint those arc blue and green alternately (so that six end points of the arcs are painted alternately too).
You have an "impossible" blue-green circle.
In fact you can take just one arc (1/6 of circle) and paint it any way you want. Then for each point P of the arc take a regular hexagon H inscribed in the unit circle such that P is a vertex of H, and paint the other 5 vertices of H alternately (regarding color of P). Again you have an impossible blue-green circle.
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Each point of the plane is painted either red or blue.
Prove, that there exists a rectangle with all vertices of the same color.
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..if you pick the set of points, and the painter picks the color...
5
4 2 6
1 3 7
for example:
o o o
o o o
o o o
No matter how you paint this set of 9 points, one-colored equilateral triangle will appear -
as Yoruichi-san proved using only 7 of these points.
But your set is not minimal and your answer lacks the reasoning Yoruichi-san provided.
Therefore I consider your former answer a sort of a partial proof.
Still I would give you the "best answer", if no real proof (as provided by Yoruichi-san) would appear.
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This is not true.If a 1 unit length circle with red center point is made of green and blue points,
there are 2 points with same color on the circle that is 1 unit away.
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Each point of the plane is painted either red, green or blue.
Prove, that there exists segment of length 1 with both ends of the same color.
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Each point of the plane is painted either red or blue.
Prove, that there exists equilateral triangle with all vertices of the same color.
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Nine points are given on the plane, no three of which are collinear.
Each pair of points is connected with segment, that is painted either white or black.
Prove that there exists either quadrilateral with white sides and diagonals or triangle with black sides.
A Necktie Paradox
in New Logic/Math Puzzles
Posted · Edited by witzar