A checker is placed on a square of an infinite checkerboard, where each square is 1cm by 1 cm. It moves according to the following rules:
1/ In the first move, the checker moves 1 square North
2/ All odd numbered moves are North or South and all even numbered moves are East or West
3/ In the n-th move, the checker makes a move of n squares in the same direction.
The checker makes 12 moves so that the distance between the centers of its initial and final squares is as small as possible. What is this minimum distance?
For the question above, I reasoned, that the shortest distance, will be worked out, if the checker, moves around, in a spiral fashion. Unforunately, that does not produce the right answer.
Can you guys, please inform me, what kind of shape, I should have thought of using, in order to achieve the minimum distance, and with that shape, what the distance works out to be?
Thanking you in advance
Best Answer
The horizontal displacement of the checker after the 12 moves is of the form $$\pm2\pm4\pm6\pm8\pm10\pm12$$ Similarly, the vertical displacement is of the form $$1\pm3\pm5\pm7\pm9\pm11$$ and we want to minimise the absolute value of both expressions by some choice of signs.
The best displacement we can achieve for the odd steps is $$1+3+5-7+9-11=0$$ But for a horizontal displacement of 0 some subset of the even numbers would have to sum to the impossibly odd 21, so we make do with a 22/20 split and $$2+4+6+8-10-12=-2$$ Translate a positive term to a movement north or east, and a negative term to a movement south or west. The displacement of the checker after these moves is then the smallest, 2 cm horizontally.