Ming-Shi Wang *, Wei-Che Chen 11 November 2006 present: 沈雅楓

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Presentation transcript:

Ming-Shi Wang *, Wei-Che Chen 11 November 2006 present: 沈雅楓 A majority-voting based watermarking scheme for color image tamper detection and recovery Ming-Shi Wang *, Wei-Che Chen 11 November 2006 present: 沈雅楓

Outline Introduction Proposed scheme Block-based watermark embedding Watermark verification Tampered image recovery Experimental results /17

Introduction Two categories of conveying the authentication data: labeling-based schemes and watermarking based schemes. The watermarking-based authentication scheme can be classified as either fragile watermarking, which is designed to detect any possible modification of the pixel values, or semi-fragile watermarking, which just distinguish content-preserving operations form malicious alteration. /17

Proposed scheme /17

Block division M*M Zij 8-bit R 8-bit B 8-bit G /17

Example (Block mapping) block Z1,1 k1=2, k2=3, M=10 So the recovery information of Z1,1 will hide in block Z2,1 /17

Block-based watermark embedding In the proposed scheme, a 36-bit watermark consists of the authentication data( 16 bits) and the recovery data( 20 bits) is embedded into the LSBs of each image block. The authentication data of length 16 bits is obtained by exploiting both the local features of each image block and the global features of the image. /17

Example (local feature) the total number of 1s in the five MSBs of all the color components of pixel l The parity-check bit for pixel l: =01010101 =10110110 1 =11110001 so = 0101010110111101 The local feature of block Z1,1 : 0000000000000001 0000000000000001 /17

Example (global feature) Hash function, MD5 0110110110110011………0 converted from 128 to 16 bits: d0=d1||d2||…||d16= 0110110110110011 the global features of image Z : /17

Authentication data Authentication =local (xor) global /17

Example (recovery data) =01010101 =10110110 1 =11110001 (u,v) is mapped to (i,j) That is because we just want 20 Bits for recovery data. /17

Example (recovery data) Assume Yu,v=01010011, Cbu,v=00001110 Cru,v=11011101 ,(u,v) mapped to(1,1) So the recovery data in block Zi,j is: C1,1=01010011000011110111 /17

Example(LSB embedding) authentication data(16): 0011010101010101 recovery data(20): 00101101001101010011 R=……001 B=……101 G=……010 R=……101 B=……010 G=……100 R=……101 B=……101 G=……001 R=……101 B=……010 G=……011 /17

Watermark verification and Tampered image recovery /17

Experimental results(1/3) PSNR=37.96 PSNR=37.83 PSNR=37.96 /17

Experimental results(2/3) Rtd=100% Rtd=100% Rtd=100% /17

Experimental results(3/3) /17