Reversible Steganographic Scheme for AMBTC-Compressed Image Based on (7,4) Hamming Code
Abstract
:1. Introduction
2. Related Works
2.1. Absolute Moment Block Truncation Coding (AMBTC)
2.2. (7,4) Hamming Code
3. Proposed Scheme
3.1. Prediction Method and Selections of Embeddable Blocks
Example of the Selection of Embeddable Blocks
3.2. Data Embedding Phase
- Step 1. Partition the original image into pixel blocks and conduct AMBTC encoding; a set of compressed trios, i.e., , is obtained, where H is the high mean table, L is the low mean table, is the bitmap, n = 4, and is the coordinate of the pixel block where and .
- Step 2. Based on their positions, use the pixels located in the odd rows to predict the pixels located in the even rows with Equations (9)–(11), respectively. If all of the pixels in a 1 × 7-sized block satisfy Equation (13), determine the block to be embeddable and denote it as “1” in the location map, LP. Otherwise, denote it as “0” in the location map, LP.
- Step 3. After all blocks have been evaluated, concatenate LP and secret data SD as the final secret message S, where S = LP||SD, and “||” denotes the concatenation of LP and SD.
- Step 4. Scan all AMBTC-compressed blocks in a zig-zag direction to embed the final secret message, S, into bitmap Bm. If H = L, 4 × 7 bits of S are selected and used to replace the original Bm. If not, H is not equal to L, select one bit, s, of S and embed it into the H and L pair by swapping the order when s = 1. Note that, if H ≠ L and s = 0, then the order of H and L is not changed.
- Step 5. After all H and L pairs have been checked, take three bits of the remaining S as m and embed them into block x, which is marked with “1” in the location map LP by using Equation (14). The detailed description can be referred to Section 2.2.
Example of Data Embedding
3.3. Data Extraction and Recovery Phase
- Step 1. Scan stego bitstream , and if H = L, extract 1 × 7 secret bits from the Bm. If H > L, extract secret bit “0”; otherwise, extract secret bit “1”, and swap H and L.
- Step 2. Decide which 1 × 7-sized blocks located in even rows are embeddable or not according to the extracted location map, LP. If the current block is an embeddable block, 3 bits are extracted from the corresponding 1 × 7 bits of Bm according to Equation (15) [34].
- Step 3. Use Equation (12) to check each pixel of the embeddable block to see it is satisfied. If one pixel is not satisfied, modify its bit value to its complementary bit value, i.e., if the current bit value is “0”, change it to “1” and vice versa.
- Step 4. Check the next embeddable block until all blocks are preceded. Output all extracted secret data and replace the corrected bitmaps with the corresponding Hs and Ls to obtain the reconstructed AMBTC-image.
Example of the Data Extraction and Recovery Phase
4. Experimental Results
5. Discussion and Conclusions
Author Contributions
Funding
Conflicts of Interest
References
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Image | Lana | Airplane | Barbara | Goldhill | Wine | Bird | Zelda | Boat |
---|---|---|---|---|---|---|---|---|
2 × 7 | 0 | 1 | 0 | 6 | 398 | 8 | 0 | 0 |
4 × 7 | 0 | 0 | 0 | 1 | 89 | 0 | 0 | 0 |
Image | Lana | Airplane | Barbara | Goldhill | Wine | Bird | Zelda | Boat |
---|---|---|---|---|---|---|---|---|
Number of embeddable blocks | 3980 | 3482 | 3924 | 3037 | 5314 | 3676 | 4386 | 3155 |
Number of blocks with one bit changed | 3614 | 3134 | 3556 | 2686 | 5060 | 3332 | 4006 | 2821 |
Number of blocks without bits changed | 366 | 348 | 368 | 351 | 254 | 344 | 380 | 334 |
Ratio 1 | 0.21 | 0.19 | 0.21 | 0.16 | 0.29 | 0.20 | 0.24 | 0.17 |
2 × 7 (18,688) Criteria | AMBTC PSNR (dB) | PSNR (dB) | H ≠ L HC (bits) | Proposed Scheme H ≠ L IC (bits) | H ≠ L EHC (bits) | H = L HLC (bits) | Total of HC 1 (bits) |
---|---|---|---|---|---|---|---|
Lena | 32.05 | 29.88 | 11,940 | 8612 | 3328 | 0 | 22,016 |
Airplane | 31.9 | 29.17 | 10,446 | 7953 | 2493 | 14 | 21,194 |
Barbara | 28.9 | 27.66 | 11,772 | 8266 | 3506 | 0 | 22,194 |
Goldhill | 32.52 | 31.07 | 9111 | 7020 | 2091 | 84 | 20,857 |
Wine | 32.02 | 29.9 | 15,942 | 10,456 | 5486 | 5572 | 29,348 |
Bird | 29.57 | 28.3 | 11,028 | 8821 | 2207 | 112 | 20,999 |
Zelda | 35.67 | 33.96 | 13,158 | 9794 | 3364 | 0 | 22,052 |
Boat | 30.72 | 28.73 | 9465 | 7096 | 2369 | 0 | 21,057 |
Average | 31.66 | 29.83 | 11,607 | 8502 | 3105 | 720 | 22,464 |
2 × 7 (9344) Criteria | AMBTC PSNR (dB) | PSNR (dB) | H ≠ L HC (bits) | Proposed Scheme H ≠ L IC (bits) | H ≠ L EHC (bits) | H = L HLC (bits) | Total of HC 1 (bits) |
---|---|---|---|---|---|---|---|
Lena | 31.29 | 29.45 | 11,907 | 8902 | 3005 | 0 | 12,349 |
Airplane | 30.49 | 28.21 | 12,159 | 8873 | 3286 | 0 | 12,630 |
Barbara | 28.25 | 27.17 | 12,102 | 8553 | 3549 | 0 | 12,893 |
Goldhill | 31.35 | 30.03 | 11,286 | 8063 | 3223 | 28 | 12,594 |
Wine | 30.84 | 29.02 | 17,511 | 11,280 | 6231 | 2492 | 17,978 |
Bird | 28.46 | 27.43 | 11,685 | 9352 | 2333 | 0 | 11,677 |
Zelda | 34.66 | 33.25 | 13,440 | 10,223 | 3217 | 0 | 12,561 |
Boat | 29.60 | 27.99 | 10,983 | 8037 | 2946 | 0 | 12,290 |
Average | 30.61 | 29.06 | 12,634 | 9160 | 3474 | 315 | 13,121 |
Performance | Scheme | Lena | Airplane | Barbara | Goldhill | Wine | Bird | Zelda | Boat |
---|---|---|---|---|---|---|---|---|---|
PSNR | AMBTC | 32.05 | 31.9 | 28.9 | 32.52 | 32.02 | 29.57 | 35.67 | 30.72 |
Chen [23] | 32.05 | 31.9 | 28.9 | 32.52 | 32.02 | 29.57 | 35.67 | 30.72 | |
Proposed | 29.88 | 29.17 | 27.66 | 31.07 | 29.9 | 28.3 | 33.96 | 28.73 | |
HC | Chen [23] | 18,688 | 18,701 | 18,688 | 18,766 | 23,864 | 18,688 | 18,688 | 18,792 |
Proposed | 22,016 | 21,194 | 22,194 | 20,857 | 29,348 | 20,999 | 22,052 | 21,057 | |
CS1 | Chen [23] | 564,736 | 564,736 | 564,736 | 564,736 | 564,736 | 564,736 | 564,736 | 564,736 |
Proposed | 564,736 | 564,736 | 564,736 | 564,736 | 564,736 | 564,736 | 564,736 | 564,736 | |
ER | Chen [23] | 0.033 | 0.033 | 0.033 | 0.033 | 0.042 | 0.033 | 0.033 | 0.033 |
Proposed | 0.039 | 0.038 | 0.039 | 0.037 | 0.052 | 0.037 | 0.039 | 0.037 |
Schemes | Types | Average PSNRs | max PSNR/ min PSNR | Average HC | max HC/ min HC | Average ER | max ER/ min ER |
---|---|---|---|---|---|---|---|
Proposed | I(code) | 29.83 | 33.96/27.66 | 22,464 | 29,348/20,999 | 0.04 | 0.052/0.037 |
Chen et al. [23] | I(code) | 32.28 | 35.67/28.90 | 19,359 | 23,864/18,688 | 0.037 | 0.042/0.033 |
Lo et al. [24] | I(code) | 33.1 | 33.23/32.97 | 3615 | 4570/2660 | 0.006 | 0.008/0.005 |
Chang et al. [25] | I(code) | 31.74 | 32.89/30.59 | 16,381 | 12,683/20,080 | 0.031 | 0.024/0.038 |
Sun et al. [26] | II(code) | 29.7 | 33.40/26.0 | 64,008 | 64,008/64,008 | 0.122 | 0.122/0.122 |
Hong et al. [27] | II(code) | 30.19 | 33.39/26.91 | 64,516 | 64,516/64,516 | 0.12 | 0.116/0.125 |
Lin et al. [28] | II(code) | 33.36 | 37.23/30.91 | 90489 | 114,533/70,889 | 0.240 | 0.241/0.217 |
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Lin, J.; Lin, C.-C.; Chang, C.-C. Reversible Steganographic Scheme for AMBTC-Compressed Image Based on (7,4) Hamming Code. Symmetry 2019, 11, 1236. https://doi.org/10.3390/sym11101236
Lin J, Lin C-C, Chang C-C. Reversible Steganographic Scheme for AMBTC-Compressed Image Based on (7,4) Hamming Code. Symmetry. 2019; 11(10):1236. https://doi.org/10.3390/sym11101236
Chicago/Turabian StyleLin, Juan, Chia-Chen Lin, and Chin-Chen Chang. 2019. "Reversible Steganographic Scheme for AMBTC-Compressed Image Based on (7,4) Hamming Code" Symmetry 11, no. 10: 1236. https://doi.org/10.3390/sym11101236
APA StyleLin, J., Lin, C. -C., & Chang, C. -C. (2019). Reversible Steganographic Scheme for AMBTC-Compressed Image Based on (7,4) Hamming Code. Symmetry, 11(10), 1236. https://doi.org/10.3390/sym11101236