Suppose the information content of a packet is the bit pattern 1110 0110 1001 1101 and an even parity scheme is being used. What would the value of the field containing the parity bits be for the case of a two-dimensional parity scheme? Your answer should be such that a minimum-length EDC (error dectection and correction) field is used.
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Key Concepts
- Even parity
- Row parity
- Column parity
- Corner bit
- Minimum-length EDC (4 × 4 layout)
Answer Approach
- Arrange the 16 bits in the layout that minimises parity bits (4 rows of 4).
- Compute an even-parity bit for each row.
- Compute an even-parity bit for each column.
- Add the corner bit covering the parity row and column.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
Arrange as a 4 × 4 matrix, which needs the fewest parity bits for 16 data bits (4 + 4 + 1 = 9, versus 2 + 8 + 1 = 11 for 2 × 8): 1110 / 0110 / 1001 / 1101. Row parities (even): 1, 0, 0, 1. Column parities: 1, 1, 0, 0. Corner (parity of the parity bits): 0. The EDC field is therefore 9 bits: row parity 1001, column parity 1100, and corner bit 0 (e.g. 1001 1100 0).
EDC bits = (i + 1)(j + 1) − ij = i + j + 1 → 9 for a 4 × 4 layoutShortcuts: K concepts · A approach · F answer · R reviewed · B bookmark · ← / → previous / next