Timeline for Steganography - How many bytes of information can you hide in a $1024 \times 768$ photo?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2019 at 19:18 | comment | added | AgapwIesu | This is not really accurate. Undetectability is not so much on the amount of data you add, and since OP specified last bit of each color, this does not really answer the question. If using lsb, undetectability is entirely on whether the data gets randomized (encrypted with a good enough encryption algorithm) before injecting it into the image. Without the original image to compare, random noise is meaningless. Capacity and mathematical deniability makes the all-noise image you have here to be no better than any other image. That image is only more likely to raise suspicions than the rest. | |
| Jun 4, 2017 at 21:19 | comment | added | bashbin | Thank you for all these tips! I read quite a lot of steganography materials and didn't see these included (compression, the photo examples). Cheers! | |
| Jun 4, 2017 at 21:18 | vote | accept | bashbin | ||
| Jun 4, 2017 at 21:18 | vote | accept | bashbin | ||
| Jun 4, 2017 at 21:18 | |||||
| Jun 4, 2017 at 16:50 | comment | added | Paul Uszak | I'm upping the feasible compression ratio for English to 4. | |
| Jun 4, 2017 at 16:49 | history | edited | Paul Uszak | CC BY-SA 3.0 |
Some algebra added
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| Jun 4, 2017 at 11:48 | comment | added | Paul Uszak | @daniel 1024*768*3*2 =~ 5 Mbits with 2 being an approximate compression ratio for English prose. | |
| Jun 4, 2017 at 9:00 | comment | added | daniel | Nice answer but " 5 Mbits of capacity" is that out of a 5 Mbit image? Plus the picture that helped me see that the LSB usually looks random is this one: intechopen.com/source/html/49098/media/fig3.png | |
| Jun 4, 2017 at 3:01 | history | answered | Paul Uszak | CC BY-SA 3.0 |
