GATE CS 2007 Q26 — Partial Orders and Lattices

A partition $\pi_i$ refines $\pi_j$ (written as $\pi_i \prec \pi_j$) if every block of $\pi_i$ is a subset of some block of $\pi_j$. Let us analyze the refinement relations among the given partitions on $S = \{a, b, c, d\}$:

• Bottom Element: $\pi_4 = \{\bar{a}, \bar{b}, \bar{c}, \bar{d}\}$ consists of only singleton blocks. Since singletons are subsets of any block, $\pi_4$ refines all other partitions. Thus, $\pi_4 \prec \pi_2$, $\pi_4 \prec \pi_3$, and $\pi_4 \prec \pi_1$.
• Top Element: $\pi_1 = \{\overline{abcd}\}$ consists of a single block containing all elements. Thus, every other partition refines $\pi_1$. So, $\pi_2 \prec \pi_1$ and $\pi_3 \prec \pi_1$.
• Incomparability: Compare $\pi_2 = \{\overline{ab}, \overline{cd}\}$ and $\pi_3 = \{\overline{abc}, \overline{d}\}$:
  • The block $\{c, d\} \in \pi_2$ is not a subset of any block in $\pi_3$. Hence, $\pi_2 \not\prec \pi_3$.
  • The block $\{a, b, c\} \in \pi_3$ is not a subset of any block in $\pi_2$. Hence, $\pi_3 \not\prec \pi_2$.
  Therefore, $\pi_2$ and $\pi_3$ are incomparable.

This structure forms a diamond-shaped Hasse diagram with $\pi_4$ at the bottom, $\pi_1$ at the top, and the incomparable elements $\pi_2$ and $\pi_3$ in the middle. This corresponds to the diagram in Option C.