in a 4x4 kmap what happens if i get a full horizon...

If you have a 4×4 K-map and you get a full horizontal row of 1’s encircled, that group simplifies to an expression that depends only on the row variables – the column variables drop out.

Why that happens

In a standard 4-variable K-map:

• Rows are usually labeled with two variables, say A B
• Columns are labeled with the other two variables, say C D

So each cell is a minterm like $A'B' C D$, $A'B C' D$, etc.

If you encircle an entire row of 4 ones, that means:

• That row has a fixed value of A and B (e.g., $A'B'$, or $AB$, etc.)
• Across the row, C and D take on all four possible combinations (00, 01, 11, 10 in Gray code)

Because C and D appear in all possible combinations, they get eliminated in simplification. Only the part that is constant over the entire group (the row’s A/B combination) remains.

Example

Say the top row (AB = 00) is all ones and you encircle it:

• Cells are:
  $A'B' C'D'$, $A'B' C'D$, $A'B' CD$, $A'B' CD'$

Factoring that:

$$A'B'(C'D' + C'D + CD + CD') = A'B' \cdot 1 = A'B'$$

So that group simplifies to just:

$A'B'$

General rule

• Full horizontal row of 1s (4 cells): result is a 2-variable term (just the row variables).
• Full vertical column of 1s (4 cells): result is a 2-variable term (just the column variables).
• If you had two adjacent full rows of 1s (8 cells), then only one variable would remain, etc.

If you tell me how your rows/columns are labeled (which variables you put where), I can write your exact simplified term.