Opened 12 years ago

Closed 12 years ago

# edgeIdx=4 for SaoTypeIdx=2

Reported by: Owned by: sorin bbross minor Text D8 (K0030) v3 SAO edgeIdx bbross, wjhan, jct-vc@…

### Description

In equation (8-322), edgeIdx=4 if

recPicture[ xC + i ][ yC + j ] > recPicture[ xC + i + hPos[ 0 ] ][ yC + j + vPos[ 0 ] ] &&
recPicture[ xC + i ][ yC + j ] > recPicture[ xC + i + hPos[ 1 ] ][ yC + j + vPos[ 1 ] ]

edgeIdx should be in the [0, 3] range.

### comment:1 Changed 12 years ago by DefaultCC Plugin

• Cc bbross wjhan jct-vc@… added

### comment:2 Changed 12 years ago by bbross

• Resolution set to invalid
• Status changed from new to closed

Thanks Elena for the following explanation:

" 1. The edgeIdx is index of SaoOffsetVal array:
The array SaoOffsetVal is derived as follows.
SaoOffsetVal[ cIdx ][ rx ][ ry ][ 0 ] = 0 (7 49)
SaoOffsetVal[ cIdx ][ rx ][ ry ][ i + 1 ] =

offsetSign*sao_offset_abs[ cIdx ][ rx ][ ry ][ i ] << ( bitDepth – Min( bitDepth, 10 ) ) (7 50)

If edgeIdx ==0 then there is no correction by SAO. If 1<= edgeIdx <=4 then SAO is applied.

1. The equation (8-322) is followed by another one:

edgeIdx = 2 + ∑k( Sign3( recPicture[ xC + i ][ yC + j ] –

recPicture[ xC + i + hPos[ k ] ][ yC + j + vPos[ k ] ] ) ) with k = 0..1 (8‑322)

When edgeIdx is equal to 0, 1, or 2, it is modified as follows.
edgeIdx = (edgeIdx = = 2) ? 0 : (edgeIdx + 1) (8‑323)
The result of (8-322) and (8-323) is as follows:
edgeIdx = 1 for local 1D minimim
edgeIdx = 2 for local 1D concave corner
edgeIdx = 3 for local 1D convex corner
edgeIdx = 4 for local 1D maximum
edgeIdx = 0 for non of listed above categories.
I guess current spec text is correct.
Kind regards,
Elena"

Note: See TracTickets for help on using tickets.

This list contains all users that will be notified about changes made to this ticket.

These roles will be notified: Reporter, Owner, Subscriber, Participant

• Benjamin Bross(Owner, Subscriber, Participant)
• jct-vc@…(Subscriber)
• karl.sharman@…(Always)
• Karsten Suehring(Always)
• Sorin Cismas(Reporter)
• Woo-Jin Han(Subscriber)