CAT 2020 Slot 2 DILR Set 1: Grids (4 Questions) with Solutions

Identify Valid Color Pairs:

Determine Arrangement Pattern:

Construct Possible Grids:

Two Reds in a line need a Blue and a Green between them, so their positions differ by at least 3. In a line of 5 cells the only possible Red pairs are positions {1,4}, {1,5} and {2,5}. Three Reds would need at least 7 cells. So every row and every column holds at most 2 Reds.

Position 3 appears in none of these pairs. A Red in the middle column cannot share its row with another Red, because the other Red would sit at distance 1 or 2 from it. The same holds for a Red in the middle row and its column.

Suppose the middle column holds no Red. Then Reds lie only in columns 1, 2, 4 and 5, with at most 2 each, giving at most 8.

Suppose the middle column holds 2 Reds. Those two rows each have only 1 Red, so the total is at most $2\times1+3\times2=8$.

Suppose the middle column holds exactly 1 Red. That row has 1 Red and the other four rows have at most 2 each, so the total is at most $1+4\times2=9$.

Nine is reachable. Place Reds at (1,1), (1,4), (4,1), (4,4), (2,2), (2,5), (5,2), (5,5) and (3,3). Every row and column pair is one of the allowed pairs, and the remaining cells can be filled with Blue and Green to satisfy the rules.

A line with no Blue uses only Red and Green. Adjacent beads must differ, so it alternates, giving R G R G R or G R G R G. In both, two Reds have no Blue between them, which breaks Rule 3. So every row and column has at least 1 Blue, and the total is at least 5.

Exactly 5 Blues would mean exactly one Blue in every row and every column.

List the valid one-Blue rows. With one Blue, a row cannot have 4 Greens, because Greens cannot be adjacent. So it has 1 or 2 Reds.

In all these rows the middle cell is Green or Blue, never Red.

If every row has one Blue, the middle column has no Red. It is then built only from Green and Blue with no two adjacent equal, which forces at least 2 Blues (G B G B G).

That contradicts one Blue per column, so 5 Blues is impossible.

A valid configuration with 6 Blues exists, where one line carries a second Blue.

A Red in the middle column of a row cannot have another Red in that row. Likewise a Red in the middle row of a column cannot have another Red in that column.

The Red at (2,3) is in the middle column, so row 2 has no other Red. Its column 3 can take one more Red, at row 5, since {2,5} is allowed.

The Red at (3,2) is in the middle row, so column 2 has no other Red. Its row 3 can take one more Red, at column 5.

The Red at (3,5) is in the middle row of column 5, so column 5 has no other Red. The Red at (5,3) is in the middle column of row 5, so row 5 has no other Red.

Column 2 is closed, column 3 is full, and column 5 is closed. Rows 1 and 4 can therefore use only columns 1 and 4, which is the allowed pair {1,4}. Add (1,1), (1,4), (4,1) and (4,4). Each of columns 1 and 4 then has Reds in rows 1 and 4, which is allowed.

The total is 8 Reds: the two given plus (5,3), (3,5), (1,1), (1,4), (4,1) and (4,4).

Nine cannot be reached. It would need row 3 to hold 2 Reds and rows 1, 4 and 5 to hold 2 each, drawn from columns 1 and 4 only. Column 1 would then hold 3 Reds, which is impossible.

So 6 more Reds can be added.