A more secure steganography based on adaptive pixel-value ...

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Multimed Tools Applcase 4.2:case 4.3:if |d 1 | > T & |d 2 |≤Trange g ′i+1= [ max ( g i+2 − T, g i + T + 1 ) ],...,g i+2if |d 1 |≤T & |d 2 | > Trange g ′i+1= [ g i ,...,min ( g i+2 − T − 1, g i + T )]Please note that the range range g ′i+1is independent of the original pixel valueg i+1 in all cases. And what’s <strong>more</strong>, it can be proven that the sort order and therelationships as mentioned previously will be preserved for all gi+1 ′ ∈ range g i+1 ′ ,which is very important to guarantees the validity of data extraction. Please referto Appendix for <strong>more</strong> details.– Step 4 Calculaten = min (⌊ ∣ ∣⌋ )log ∣rangeg 2 ′ i+1 , kwhere |range g ′i+1| denotes the number of the elements in the set range ′ g i+1, kdenotes the maximal number of embedded bits into a single pixel g i+1 , we limitk ≤ 4. In this work, we set k = 4. We can change it to adjust the embeddingcapacity and visual quality.If n > 0, covert n secret bits into a decimal value b, and calculate{∣ ∣∣ ∣ ∣g i+1 ′ (= arg min ∣e − gi+1 ∣e − gi ≡ b mod 2n ) }, e ∈ range g ′ei+1– Step 5 Select the next embedding unit <strong>based</strong> on the key key 2 , and repeat aboveoperations from Step 3, until all the message are embedded.If all the available embedding units have been traveled while the secret messagehas not been embedded completely, decrease the threshold T to T − 1, andrestart the operations from Step 2. If T = 1 and the message can not beembedded completely, the cover image has not enough space to embed the givensecret message.– Step 6 Re-rotate the resulting image <strong>based</strong> on the block size Bz × Bz and thesecret key key 1 . Finally, we obtain the stego image.For instance, we are dealing with an embedding unit [g i , g i+1 , g i+2 ]=[40, 70, 50].Assuming that the threshold T = 10 and the secret bit stream is 0011001 ... (2) .Wefirstly calculate the differences d 1 = g i+1 − g i = 30, d 2 = g i+2 − g i+1 =−20, thenwehave g i+1 > g i and g i+1 > g i+2 ,soCase 2 is considered. And since |d 1 | > T & |d 2 | >T, the range of the centre pixel values can be determined by (case 2.1)range g ′i+1= [ max ( g i + T + 1, g i+2 + T + 1 ) ,...,255 ] =[61,...,255]Then calculate n = min(⌊log 2 195⌋, 4) = 4 and extract 4 bits from the secret messagestream, i.e. 0011 (2) , and then converted it into decimal value b = 3. Finally thevalue of g i+1 after data embedding becomesg i+1 ′ { (= arg min |e − 70|| |e − 40| ≡3 mod 24 ) , e ∈[61,...,255] } = 75eIt can be checked that the sort order and the relationships of the values g i =40, gi+1 ′ = 75, g i+2 = 50 and T = 10 remain the same, namelyg i+1 ′ > g i, g i+1 ′ > g i+2, ∣ ∣ ∣ ∣ ∣ ∣ ∣ d′ 1 = ∣g ′i+1 − g i > T, ∣d ′ 2 = ∣gi+2 − g i+1′ ∣ > T
Multimed Tools Appl3.2 Data extractionWe first divide the stego image into Bz × Bz blocks which are then rotated by randomdegrees <strong>based</strong> on the secret key key 1 , and obtain non-overlapping embeddingunits with three consecutive pixels just as Steps 1 and 2 in data embedding. We thentravel all the embedding units in a pseudo-random order <strong>based</strong> on the secret keykey 2 .For a given unit, say [g i , g ′ i+1 , g i+2], 2 we execute the following operations:First calculate the differences of the adjacent pixel values byd ′ 1 = g′ i+1 − g i, d ′ 2 = g i+2 − g ′ i+1If |d ′ 1 |≤T and |d′ 2 |≤T,3 then skip to next embedding unit. Otherwise determinethe range of gi+1 ′ according to the four cases just as shown in the process in dataembedding, also denoted as range g ′i+1. 4Then determine the number of the embedded bits in pixel gi+1 ′ byn = min (⌊ ∣ ∣⌋ )log ∣rangeg 2 ′ i+1 , kwhere k = 4And then the embedded value b in decimal representation isb ≡ ∣ ∣( g′i+1 − g i mod 2n )Finally convert the decimal value b into the binary representation and obtain thehidden message bits.For instance, we want to extract secret bits from an embedding unit[g i , gi+1 ′ , g i+2] =[40, 75, 50] with a threshold T = 10. First, calculate the differencesd ′ 1 = 75 − 40 = 35, d′ 2 = 50 − 75 =−25, then we get g′ i+1 > g i, gi+1 ′ > g i+2, henceCase 2 is considered. Since |d ′ 1 |=35 > T, |d′ 2|=25 > T, the range of the centre pixelis determined by (case 2.1)g ′ i+1range g ′i+1= [ max ( g i + T + 1, g i+2 + T + 1 ) ,...,255 ] =[61, 255]So the number of embedded bits isn = min (⌊ log 2∣ ∣rangeg ′i+1∣ ∣⌋, k)= min(⌊log2 195 ⌋ , 4 ) = 4Finally we obtain the secret bits in the decimal representation byb = 3 ≡ 74 − 40 ( mod 2 4)namely 0011 (2)2 Note that only the centre one has been changed after data hiding.3 The threshold T can be extracted in a preset region in the stego.4 It can be proven that such region is the same before and after data embedding using our embeddingscheme. Please refer to Appendix for <strong>more</strong> details.
Page 1 and 2: Multimed Tools ApplDOI 10.1007/s110
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