#Combinatorics problem from All-Russian Olympiad

7 messages · Page 1 of 1 (latest)

hearty lotus
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Can I please have a hint for this problem? Here is what I got so far: ||I am suspecting that k=1. Define a cell to be “connected” if it is possible to paint this cell without using a move. Call a cell “disconnected” if otherwise. Notice that there can only be a maximum of 1 “disconnected” cells (since it needs to be the largest of a column and the smallest of a row). Indeed, suppose that there are two cells x,y that are both the largest of their columns and smallest of their rows. Without loss of generality, let x>y. Notice that x,y needs to be in distinct rows and columns, however, x<the cell in the same column as y. Which contradicts the assumption that y is the largest of its columns. ||
||I need to show that in the absence of a “disconnected” cell, painting a number will be sufficient and that when there is a “disconnected” cell, painting the disconnected cell will be sufficient||

errant oracleBOT
hearty lotus
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Combinatorics problem from All-Russian Olympiad

tough jungle
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Yeah i think you are along the right track, for the disconnected cell, paint it, then you can paint everything else in the same row and column.

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If there are no disconnected cells, you can consider alternatingly

  • go to smallest in row
  • go to biggest in column
    and eventually you must loop.

Maybe consider this graph

hearty lotus
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Would this work? || Now, if we have a "disconnected" cell, then we paint it. So now we are left with a board without "disconnected" cells. Suppose that after Petya has painted all of the cells that he could paint, there still remains cells that are unpainted. Consider cell $x$ in row $i$ and column $j$, it is left unpainted because the smallest cell in row $i$ is left unpainted and cell $x$ is larger than all of the painted cells in column $j$. In this column $j$, there must be a biggest cell in row $k$ ($k$ can be equal to $i$). Now consider row $k$, it has an unpainted cell, which means that the smallest cell in row $k$ is left unpainted, which means that it must be the largest cell in its column. This is a contradiction. ||

fossil saffronBOT
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One more