Difference between revisions of "2001 OIM Problems/Problem 5"

 
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== Problem ==
 
== Problem ==
  
On a <math>2000 \times 2001</math> board the squares have coordinates <math>(x, y)</math> with <math>x, y</math> integers, <math>0 \le x \le 1999</math> and <math>0 \le y \le 2000</math>. A ship on the board moves as follows:  Before each movement, the ship is in a position <math>(x, y)</math> and has a speed <math>(h, v)</math> where <math>h</math> and <math>v</math> are integers. The ship chooses a new speed <math>(h',v')</math> such that <math>h'-h</math> be equal to <math>-1, 0</math> or <math>1</math> and <math>v'-v</math> be equal to <math>-1, 0</math> or <math>1</math>. The new position of the ship will be <math>(x',y')</math> where <math>x'</math> is the reminder of the division of <math>x+h'</math> by 2000, and <math>y'</math> is the reminder of the division of <math>y+v'</math> by 2001.  There are two ships on the board: the Martian one and the Earth one that wants to catch the Martian one. Initially each ship is on a square on the board and has speed <math>(0, 0)</math>. First moves Earth ship and continue moving alternately.  Is there a strategy that always allows the Earth ship to catch the Martian ship, whatever the initial positions are?
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On a <math>2000 \times 2001</math> board the squares have coordinates <math>(x, y)</math> with <math>x, y</math> integers, <math>0 \le x \le 1999</math> and <math>0 \le y \le 2000</math>. A ship on the board moves as follows:  Before each movement, the ship is in a position <math>(x, y)</math> and has a speed <math>(h, v)</math> where <math>h</math> and <math>v</math> are integers. The ship chooses a new speed <math>(h',v')</math> such that <math>h'-h</math> be equal to <math>-1, 0</math> or <math>1</math> and <math>v'-v</math> be equal to <math>-1, 0</math> or <math>1</math>. The new position of the ship will be <math>(x',y')</math> where <math>x'</math> is the reminder of the division of <math>x+h'</math> by 2000, and <math>y'</math> is the reminder of the division of <math>y+v'</math> by 2001.  There are two ships on the board: the Martian one and the Earth one that wants to catch the Martian one. Initially each ship is on a square on the board and has speed <math>(0, 0)</math>. Earth ship moves first and continue moving alternately.  Is there a strategy that always allows the Earth ship to catch the Martian ship, whatever the initial positions are?
  
 
'''Note:''' the Earth ship, which always sees the Martian, catches the Martian if after one of its movement falls at the same position as the Martian.
 
'''Note:''' the Earth ship, which always sees the Martian, catches the Martian if after one of its movement falls at the same position as the Martian.

Latest revision as of 04:26, 14 December 2023

Problem

On a $2000 \times 2001$ board the squares have coordinates $(x, y)$ with $x, y$ integers, $0 \le x \le 1999$ and $0 \le y \le 2000$. A ship on the board moves as follows: Before each movement, the ship is in a position $(x, y)$ and has a speed $(h, v)$ where $h$ and $v$ are integers. The ship chooses a new speed $(h',v')$ such that $h'-h$ be equal to $-1, 0$ or $1$ and $v'-v$ be equal to $-1, 0$ or $1$. The new position of the ship will be $(x',y')$ where $x'$ is the reminder of the division of $x+h'$ by 2000, and $y'$ is the reminder of the division of $y+v'$ by 2001. There are two ships on the board: the Martian one and the Earth one that wants to catch the Martian one. Initially each ship is on a square on the board and has speed $(0, 0)$. Earth ship moves first and continue moving alternately. Is there a strategy that always allows the Earth ship to catch the Martian ship, whatever the initial positions are?

Note: the Earth ship, which always sees the Martian, catches the Martian if after one of its movement falls at the same position as the Martian.


~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also