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Apr 16, 2017 at 0:45 history edited Jamal CC BY-SA 3.0
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Mar 15, 2017 at 1:23 vote accept tbadams45
Mar 5, 2017 at 22:09 history edited tbadams45 CC BY-SA 3.0
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Mar 5, 2017 at 21:02 answer added JS1 timeline score: 3
Mar 5, 2017 at 13:26 comment added tbadams45 In terms of (x, y) position then, if I understand you correctly: bit 0 of first (or "left") byte is (7, 0). Bit 7 (MSB) of first byte is (0, 0). bit 0 of second (or "current") byte is (8, 0), whereas bit 7 is (15, 0). Bit 0 of third byte is (16, 0), and bit 7 is (23, 0).
Mar 5, 2017 at 13:23 history edited tbadams45 CC BY-SA 3.0
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Mar 5, 2017 at 13:13 comment added tbadams45 I'm running this on an XMOS processor (XUF216-512-TQ128). The code actually is xC, a variant of C, but in this section I'm just using straight C. As for the bit: good catch. The original input data is exactly like I've described it. However, because each byte is overwritten by the result of DenoiseAndFlipByte(), once DenoiseAndFlipByte() gets to a new byte, the orientation is 0 1 2 3 4 5 6 7 | 15 14 13 12 11 10 9 8 | 23 22 21 20 19 18 17 16. So rightBit = left & 0x1 obtains 7, while leftBit = right & 0x1 obtains 16.
Mar 5, 2017 at 9:39 comment added JS1 What type of cpu are you running this on? Also, something isn't right with your explanation of the bit orientation. It can't be that both rightBit = left & 0x1 and leftBit = right & 0x1. One of those should be 0x80 instead of 0x1. If you think your code is correct, could you clarify which bits appear in which order in terms of (x,y) position? For example, is bit 0 of the first byte at (0,0) or (7,0)? Is bit 0 of the second byte at (8,0) or (15,0)?
Mar 5, 2017 at 8:06 history edited 200_success
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Mar 5, 2017 at 5:09 answer added Stephen Rauch timeline score: 3
Mar 5, 2017 at 4:56 review First posts
Mar 5, 2017 at 4:59
Mar 5, 2017 at 4:52 history asked tbadams45 CC BY-SA 3.0