too many equality cases
Source: 2023 ISL C6
July 17, 2024
combinatoricssussypartition
Problem Statement
Let be a positive integer, and consider an grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence.Prove that the cells of the grid cannot be partitioned into less than right-down or right-up paths. For example, the following partition of the grid uses paths.
[asy]
size(4cm);
draw((5,-1)--(0,-1)--(0,-2)--(5,-2)--(5,-3)--(0,-3)--(0,-4)--(5,-4),gray+linewidth(0.5)+miterjoin);
draw((1,-5)--(1,0)--(2,0)--(2,-5)--(3,-5)--(3,0)--(4,0)--(4,-5),gray+linewidth(0.5)+miterjoin);
draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,black+linewidth(2.5)+miterjoin);
draw((0,-1)--(3,-1)--(3,-2)--(1,-2)--(1,-4)--(4,-4)--(4,-3)--(2,-3)--(2,-2),black+linewidth(2.5)+miterjoin);
draw((3,0)--(3,-1),black+linewidth(2.5)+miterjoin);
draw((1,-4)--(1,-5),black+linewidth(2.5)+miterjoin);
draw((4,-3)--(4,-1)--(5,-1),black+linewidth(2.5)+miterjoin);
[/asy]
Proposed by Zixiang Zhou, Canada