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', | + | 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 03:26, 14 December 2023
Problem
On a 
board the squares have coordinates 
with 
integers, 
and 
. A ship on the board moves as follows: Before each movement, the ship is in a position and has a speed 
where 
and 
are integers. The ship chooses a new speed 
such that 
be equal to 
or 
and 
be equal to or . The new position of the ship will be 
where 
is the reminder of the division of 
by 2000, and 
is the reminder of the division of 
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 
. 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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