Related to Open Problem Number 1: In what spots CAN'T you place a king or knight?

Assuming an infinite chess board with tiles in { 1 , 2 , 3 , . . . } 2 \left\{1,2,3,...\right \}^2 , and assuming that every piece (king or knight) is attacking a total of two kings and two knights, how many of the following tiles can a king or a knight be placed in?

( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 1 ) , ( 2 , 2 ) , ( 3 , 2 ) (1,1),\quad (1,2),\quad (1,3),\quad (2,1),\quad (2,2),\quad (3,2)


Bonus: Can you find any other limitations on how pieces can be placed?

All Some None

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Marcus Luebke
Nov 15, 2017

(Via Stefan Van der Waal) If we assume some smallest, finite board exists, we know the following about that board: 1. The board will have an edge on the left, on the top, on the right, and on the bottom 2. To guarantee the board can’t get shrunken, the board needs at least one piece on each edge

With those two ideas I started looking at a corner. I assumed a piece is put in the corner and kept extrapolating until I arrived at a contradiction. All of my work is over here: https://docs.google.com/document/d/1fEALESsS72xeT1e8YamxNqrrOEg6qAn-3mTHk9Rgxnk/edit?usp=sharing

Assuming I made no mistakes, the conclusion I reached was: There are no pieces on the squares A1, A2, A3, B1, and C1.

From there, I was able to prove that A4 and D1 cannot have knights on them, by focusing on A4, D1, and B2

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...