CN108648133A - A kind of rotation of combined block and mosaic without embedded camouflage method - Google Patents
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Abstract
Description
技术领域technical field
本发明属于图像信息安全和数字图像信号处理交叉领域,涉及一种伪装方法,具体涉及一种结合块旋转和马赛克的密图伪装与恢复方法。The invention belongs to the cross field of image information security and digital image signal processing, relates to a camouflage method, in particular to a dense image camouflage and restoration method combined with block rotation and mosaic.
背景技术Background technique
当前随着深度学习的不断深入和人工智能的继续发展以及量子计算机的初现端倪,使得传统以图像音频为主要传输介质的多媒体信息安全形势变得更为严峻。而与此同时,一方面,压缩技术的持续发展,也使得传统基于修改式嵌入的信息隐藏可供利用的冗余空间越来越小,而另一方面,基于机器学习的隐写分类器维数不断提高,甚至出现了34761维空域富特征模型,导致传统信息隐藏的隐匿信息不被发现的可能性也越来越小,而所有这些也使得传统基于修改式嵌入的信息隐藏技术发展陷入了瓶颈。At present, with the continuous deepening of deep learning, the continued development of artificial intelligence and the emergence of quantum computers, the traditional multimedia information security situation with image and audio as the main transmission medium has become more severe. At the same time, on the one hand, the continuous development of compression technology also makes the redundant space available for traditional modified embedding-based information hiding smaller and smaller; on the other hand, the steganographic classifier dimension based on machine learning The number of data continues to increase, and even a 34,761-dimensional spatial domain-rich feature model appears, which makes it less and less likely that the hidden information of traditional information hiding will not be discovered, and all these also make the development of traditional information hiding technology based on modified embedding fall into a trap bottleneck.
如何有效地进行下一代信息隐藏技术研究,来自北京和上海的专家于2014年5月召开的全国信息隐藏与多媒体安全专家研讨会首次提出了“无载体信息隐藏”。2015年3月在武汉召开的第12届全国信息隐藏大会上,北京电子技术应用研究所所长郭云彪研究员在大会特邀报告:信息隐藏之我见,将无载体信息隐藏技术列入了未来信息隐藏的前沿阵地。2016年10月在合肥召开的第13届全国信息隐藏大会将无载体信息隐藏正式定位为第2代信息隐藏技术,2个会议主题报告都与无载体信息隐藏直接相关。传统基于无载体信息隐藏技术主要有两种典型的方法,分别是基于搜索式无载体信息隐藏方法和基于纹理合成的生成式无载体信息隐藏。How to effectively research the next generation of information hiding technology, experts from Beijing and Shanghai held the National Symposium on Information Hiding and Multimedia Security Experts held in May 2014 for the first time to put forward "carrier-free information hiding". At the 12th National Conference on Information Hiding held in Wuhan in March 2015, researcher Guo Yunbiao, director of the Beijing Electronics Technology Application Research Institute, was invited to report at the conference: My opinion on information hiding, including carrierless information hiding technology in future information Hidden frontier. The 13th National Information Hiding Conference held in Hefei in October 2016 officially positioned carrierless information hiding as the second generation of information hiding technology, and the two conference theme reports are directly related to carrierless information hiding. There are two typical methods based on traditional blanking information hiding technology, which are search-based blanking information hiding method and generative blanking information hiding method based on texture synthesis.
传统基于搜索的无载体信息隐藏方法,主要通过检索数据库中包含指定秘密矢量的载体文本或图像来传递秘密信息,由于自然图像文本对不相关秘密信息的表达能力十分有限,导致这类方法的嵌密容量极低。例如,Zhou Z L,2015.(Zhou Z L,Sun H Y,Harit RH,Chen X Y,Sun X M.Coverless image steganography without embedding[C]//International Conference on Cloud Computing and Security.SpringerInternational Publishing,2015:123-132.)基于块均值比较,将图像分为9个块,并进一步利用块均值比较的方式将图像映射为对应的8比特哈希值,最后检索数据库中哈希值与秘密比特小段相等的图像作为含密载体,这类方法每幅图像仅能隐藏8比特。Zhou Z,2017(Zhou Z,Wu Q M J,Yang C N,et al.Coverless image steganography usinghistograms of oriented gradients-based hashing algorithm[J].Journal ofInternet Technology,2017,18(5):1177-1184)在Zhou Z L,2015.的基础上引入用户标识来确定秘密信息所嵌入的图像块位置,并进一步将图像块的像素梯度幅值与梯度方向映射为20比特,从而嵌密容量仅为20比特每图像。为进一步提升单图像的嵌密容量并缩减数据库规模,吴建斌,2018(吴建斌,贾炎柯,刘逸雯.基于图像编码及拼接的无载体信息隐藏[C]//第14届全国信息隐藏暨多媒体信息安全学术会议(CIHW2018),广州,2018:45-52)通过哈希函数将图像的信息熵编码为7比特哈希值,并通过排列组合将4张图像拼接在一起,利用4张图像所表达的哈希值以及4张图像的16种拼接顺序隐藏32比特信息。由于基于搜索的无载体信息隐藏方法嵌密容量极低,需在信道中传输海量的载体文本或图像来完整表达秘密信息,从而易引起攻击者的怀疑,导致传输的秘密信息容易遭受破坏。Traditional search-based carrierless information hiding methods mainly transmit secret information by retrieving carrier text or images containing specified secret vectors in the database. Due to the limited ability of natural image text to express irrelevant secret information, the embedding of such methods Very low density. For example, Zhou Z L, 2015. (Zhou Z L, Sun H Y, Harit RH, Chen X Y, Sun X M. Coverless image steganography without embedding[C]//International Conference on Cloud Computing and Security. Springer International Publishing, 2015:123-132 .) Based on the comparison of the block mean value, the image is divided into 9 blocks, and the image is further mapped to the corresponding 8-bit hash value by means of the block mean value comparison, and finally the image in the database whose hash value is equal to the secret bit segment is retrieved as With a dense carrier, this method can only hide 8 bits per image. Zhou Z,2017(Zhou Z,Wu Q M J,Yang C N,et al.Coverless image steganography using histograms of oriented gradients-based hashing algorithm[J].Journal of Internet Technology,2017,18(5):1177-1184) in Zhou Z L , 2015. Based on the introduction of user identification to determine the position of the image block where the secret information is embedded, and further map the pixel gradient magnitude and gradient direction of the image block to 20 bits, so the embedding capacity is only 20 bits per image. In order to further increase the embedding capacity of a single image and reduce the size of the database, Wu Jianbin, 2018 (Wu Jianbin, Jia Yanke, Liu Yiwen. Carrier-free information hiding based on image coding and splicing [C]//The 14th National Information Hiding and Multimedia Information Security Academic Conference (CIHW2018, Guangzhou, 2018: 45-52) encodes the information entropy of the image into a 7-bit hash value through the hash function, and stitches the 4 images together by permutation and combination, using the hash value expressed by the 4 images Hidden values and 16 splicing sequences of 4 images hide 32 bits of information. Due to the extremely low embedding capacity of the search-based carrierless information hiding method, a large amount of carrier text or images need to be transmitted in the channel to fully express the secret information, which easily arouses the suspicion of the attacker, and the transmitted secret information is easily damaged.
传统基于纹理合成的生成式无载体信息隐藏方法,它的主要思想是通过纹理生成的方式来表达秘密信息。例如,Wu K C,2015(Wu K C,Wang C M.Steganography usingreversible texture synthesis[J].IEEE Transactions on Image Processing.2015,24(1):130-139)将简单质地的纹理图像分成不重叠等大图像小块,利用MSE将图像小块划分为不同等级。在嵌密时,将秘密信息和图像小块的MSE等级相对应,从中选择合适的候选小块来行成嵌密掩体。又比如,Qian Z X,2017(Qian Z X,Zhou H,Zhang W M,et al.Robuststeganography using texture synthesis.Advances in Intelligent InformationHiding and Multimedia Signal Processing.Smart Innovation,Systems andTechnologies,2017,63:25-33)根据小块核心区域的复杂度将小块分类,将每一分类代表一种秘密信息,并将所选的小块随机放置在掩体图像中,利用其它小块对其进行拼接来对秘密信息掩盖。但这类方法只能生成简单质地的纹理图像,而难以生成复杂有意义的图像。The main idea of the traditional generative carrierless information hiding method based on texture synthesis is to express secret information through texture generation. For example, Wu K C, 2015 (Wu K C, Wang C M. Steganography using reversible texture synthesis[J]. IEEE Transactions on Image Processing. 2015, 24(1): 130-139) divides simple texture texture images into non-overlapping equal-sized Image small blocks, using MSE to divide image small blocks into different levels. When embedding, the secret information is corresponding to the MSE level of the small image block, and the appropriate candidate small block is selected to form the embedding mask. Another example, Qian Z X, 2017 (Qian Z X, Zhou H, Zhang W M, et al. Robuststeganography using texture synthesis. Advances in Intelligent Information Hiding and Multimedia Signal Processing. Smart Innovation, Systems and Technologies, 2017, 63:25-33) according to small The complexity of the core area of the block classifies the small blocks, and each classification represents a kind of secret information, and the selected small blocks are randomly placed in the cover image, and the secret information is concealed by splicing them with other small blocks. However, such methods can only generate texture images with simple textures, and it is difficult to generate complex and meaningful images.
传统基于马赛克拼图的信息隐藏方法可产生有意义的图像,但需要借助修改载体的方式进行秘密信息嵌入。例如,Lai I J,2011(Lai I J,Tsai W H.Secret-fragment-visible mosaic image–a new computer art and its application to informationhiding[J].IEEE Transactions on Information Forensics&Security,2011,6(3):936-945)寻找与密图等大且最为相似的掩体图像,利用密图划分小块来拼接公开图像并通过可逆LSB嵌入方法来嵌入映射参数,从而借助拼接后的公开图像来传递密图,但该方法仅限于将密图伪装成与之视觉质量相接近的公开图像,导致其适用面较窄。为提高伪装后公开图像的视觉质量和密图恢复质量,并且避免公开图像不能自由选择,Zhai S Y,2015(Zhai SY,Li F,Chang C C,et al.A meaningful scheme for sharing secret images usingmosaic images[J].International Journal of Network Security,2015,17(5):643-649)通过密图小块对公开图像小块替换的方法,将密图隐藏在4张任意选定的掩体图像中,并利用LSB法来嵌入密图恢复参数。张梦,2016(张梦,翟圣云,苏栋骐.基于马赛克技术的秘密图像共享改进算法[J].计算机应用研究,2016,33(11):3480-3484)结合差值扩展可逆信息隐藏方法,基于马赛克拼接的方法给出了密图到公开图像的伪装策略,使得密图小块在参数提取后可完整恢复。Lee Y L,2014(Lee Y L,Tsai W H.A new secure imagetransmission technique via secret-fragment-visible mosaic images by nearlyreversible color transformations[J].IEEE Transactions on Circuits&Systems forVideo Technology,2014,24(4):695-703)将秘密图像小块和掩体图像小块根据均值和标准差分别排序,然后按照排序顺序建立一一映射关系,并引入角度变换来提高密图小块对公开图像小块的视觉匹配质量。Hou D D,2016(Hou D,Zhang W,Yu N.Image camouflageby reversible image transformation[J].Journal of Visual Communication&ImageRepresentation,2016,40:225-236)在Lee Y L,2014基础上引入了K均值聚类对密图小块和掩体小块进行分类和匹配,并对嵌入参数进行了优化。刘小凯,2018(刘小凯,姚恒,秦川.一种基于图像块分类阈值优化的改进可逆图像伪装[C]//第14届全国信息隐藏暨多媒体信息安全学术会议(CIHW2018),广州,2018:37-44)对Hou D D,2016做了改进,引入了分类阈值优化算法使匹配的密图小块与掩体小块之间的均方差更小,并在角度变换中进一步添加了水平翻转,使得生成的公开图像视觉质量较好。Traditional mosaic-based information hiding methods can produce meaningful images, but they need to modify the carrier to embed secret information. For example, Lai I J,2011(Lai I J,Tsai W H.Secret-fragment-visible mosaic image–a new computer art and its application to informationhiding[J].IEEE Transactions on Information Forensics&Security,2011,6(3):936- 945) to find the cover image that is as large and most similar to the secret map, use the secret map to divide small blocks to splice the public image, and use the reversible LSB embedding method to embed the mapping parameters, so as to transfer the secret map with the public image after splicing, but this The method is limited to disguising the dense image as a public image with a visual quality close to it, resulting in a narrow range of application. In order to improve the visual quality of public images after masquerading and the restoration quality of secret images, and to avoid free selection of public images, Zhai S Y, 2015 (Zhai SY, Li F, Chang C C, et al. A meaningful scheme for sharing secret images using mosaic images[ J].International Journal of Network Security,2015,17(5):643-649) through the method of replacing the small block of the public image with the small block of the dense image, the dense image is hidden in 4 randomly selected bunker images, and The LSB method is used to embed the recovery parameters of the dense graph. Zhang Meng, 2016 (Zhang Meng, Zhai Shengyun, Su Dongqi. Improved secret image sharing algorithm based on mosaic technology [J]. Computer Application Research, 2016, 33(11): 3480-3484) Combined with the difference expansion reversible information hiding method, based on The method of mosaic mosaic provides a camouflage strategy from the dense image to the public image, so that the small blocks of the dense image can be completely restored after parameter extraction. Lee Y L,2014(Lee Y L,Tsai W H.A new secure imagetransmission technique via secret-fragment-visible mosaic images by nearly reversible color transformations[J].IEEE Transactions on Circuits&Systems for Video Technology,2014,24(4):695-703) The secret image patch and the cover image patch are sorted according to the mean and standard deviation respectively, and then a one-to-one mapping relationship is established according to the sorting order, and an angle transformation is introduced to improve the visual matching quality of the secret image patch to the public image patch. Hou D D, 2016 (Hou D, Zhang W, Yu N. Image camouflageby reversible image transformation [J]. Journal of Visual Communication & Image Representation, 2016, 40: 225-236) introduced K-means clustering pair based on Lee Y L, 2014 The dense image patches and cover patches are classified and matched, and the embedding parameters are optimized. Liu Xiaokai, 2018 (Liu Xiaokai, Yao Heng, Qin Chuan. An improved reversible image camouflage based on image block classification threshold optimization [C]//The 14th National Conference on Information Hiding and Multimedia Information Security (CIHW2018), Guangzhou, 2018: 37-44) made improvements to Hou D D, 2016, introduced the classification threshold optimization algorithm to make the mean square error between the matched dense map small block and the mask small block smaller, and further added horizontal flip in the angle transformation, so that The resulting public images have better visual quality.
除了将密图和公开图像划分为小块进行拼接的马赛克拼图以外,还有通过小图像拼接为大图像的,图像马赛克拼图方法。这种拼图方法局部细节都是小图像,而将所有的小图像组织在一起,则可构成完整的具有任意表现力的其他图像。基于图像马赛克拼图方法,Lin W L,2004(Lin W L,Tsai W H.Data hiding in image mosaics by visibleboundary regions and its copyright protection application against print-and-scan attacks[J].2004)将掩体图像分块,然后将每个分块对应为矩形图像,通过加噪调整矩形图像左右上下四条边的方差来决定所表示的2位秘密信息并按大于阈值的最小边方差来提取秘密信息。In addition to the mosaic puzzle that divides the secret image and the public image into small pieces for splicing, there is also an image mosaic puzzle method that stitches small images into large images. The local details of this puzzle method are small images, and all the small images can be organized together to form a complete other image with any expressive force. Based on the image mosaic puzzle method, Lin W L, 2004 (Lin W L, Tsai W H. Data hiding in image mosaics by visible boundary regions and its copyright protection application against print-and-scan attacks[J].2004) divided the mask image into blocks, Then each block is corresponding to a rectangular image, and the variance of the left, right, upper and lower sides of the rectangular image is adjusted by adding noise to determine the 2-bit secret information represented, and the secret information is extracted according to the minimum side variance greater than the threshold.
以上给出的基于马赛克拼图的信息隐藏方法均涉及对掩体图像的修改,从而会在载体中留下修改的痕迹,同时所给出的参数嵌入方法,鲁棒性较差,在遭受攻击时,容易导致嵌入的参数丢失,从而不能对重构密图的真实性和可靠性进行准确鉴别。The mosaic-based information hiding methods given above all involve the modification of the cover image, which will leave traces of modification in the carrier. At the same time, the parameter embedding method given is less robust, and when attacked, It is easy to cause the loss of embedded parameters, so that the authenticity and reliability of the reconstructed dense map cannot be accurately identified.
发明内容Contents of the invention
本发明的目的在于克服现有技术缺陷,提出一种结合块旋转和马赛克的无嵌入伪装重构方法,采用图像马赛克,基于含密载体直接生成的方法,利用任意选取的图像来生成公开有意义含密掩体,在生成过程中,将任意选取的图像转换为圆形图像,利用添加随机转角的圆形图像在不同的位置表达不同的秘密信息,所提方法严格依赖于用户密钥,只有提供正确的密钥才能获取秘密信息并对所提取的秘密信息的准确性进行验证,而其他情况下将无法获取。由于圆形图像仅涉及随机转角,而不涉及任何参数的修改嵌入,从而具备较强的鲁棒性。The purpose of the present invention is to overcome the defects of the prior art, and propose a non-embedded camouflage reconstruction method combining block rotation and mosaic, using image mosaic, based on the method of direct generation of encrypted carrier, using arbitrarily selected images to generate open and meaningful In the generation process, the randomly selected image is converted into a circular image, and the circular image with random corners is used to express different secret information at different positions. The proposed method is strictly dependent on the user key. Only the provided The correct key is required to obtain the secret information and verify the accuracy of the extracted secret information, which cannot be obtained otherwise. Since the circular image only involves random corners and does not involve any parameter modification embedding, it has strong robustness.
为实现上述目的,本发明采用以下技术方案:To achieve the above object, the present invention adopts the following technical solutions:
一种结合块旋转和马赛克的无嵌入伪装方法,其特征在于,包括以下步骤:A non-embedding camouflage method combining block rotation and mosaic, characterized in that it comprises the following steps:
第1步:输入分辨率为m0×n0的P0阶灰度掩体图像分辨率为m1×n1的P1阶灰度秘密图像以及l个分辨率为(2r-1)×(2r-1)两两不等的P2阶灰度图像k=0,1,…,l-1,其中,P0,P1为大于等于0的整数,r为大于0的正整数,l为2的整数幂次且满足 Step 1: Input a P 0 -level grayscale mask image with a resolution of m 0 ×n 0 P 1 -level grayscale secret image with resolution m 1 ×n 1 And l P 2 -level grayscale images with a resolution of (2r-1)×(2r-1) in pairs k=0,1,...,l-1, among them, P 0 , P 1 are integers greater than or equal to 0, r is a positive integer greater than 0, l is an integer power of 2 and satisfies
第2步:由密钥k0产生随机整数indk∈[0,n-1],k=0,1,…,l-1,且n为大于0的整数;Step 2: Generate a random integer ind k ∈ [0,n-1] from the key k 0 , k=0,1,...,l-1, and n is an integer greater than 0;
第3步:将indk映射为[0,2π)之间的随机转角αind,然后将所有的Hk转换为具有随机转角αind且半径为r的圆形图像 Step 3: Map ind k to a random rotation angle α ind between [0,2π), then convert all H k into circular images with random rotation angle α ind and radius r
第4步:将S扫描为2值比特序列由密钥k1产生l0个两两不等的m0×n0范围内的随机整数坐标序列其中,为第i个随机整数坐标,即 Step 4: Scan S as a sequence of 2-valued bits Generate l 0 random integer coordinate sequences in the range of m 0 ×n 0 from the key k 1 in, is the ith random integer coordinate, namely
第5步:初始化分辨率为m2×n2的P0阶灰度掩体图像且满足m2=m0(2r-1),n2=n0(2r-1),由密钥k2产生m0n0个[0,l-1]范围内的随机整数矩阵且ri,j∈[0,l-1];Step 5: Initialize the P 0 -level grayscale mask image with a resolution of m 2 ×n 2 And satisfy m 2 =m 0 (2r-1), n 2 =n 0 (2r-1), generate m 0 n 0 random integer matrices in the range of [0,l-1] from key k 2 And r i, j ∈ [0, l-1];
第6步:对于从B中截取2值比特Bcur,将其转换为整数XCur,并进一步将XCur映射为[0,l-1]范围内的整数然后从H′k,k=0,1,…,l-1中选取第张图像并计算的随机转角αind,将逆时针旋转为将其放置在M上选取的以(x·(2r-1),y·(2r-1))为左上角起点、大小为(2r-1)×(2r-1)的矩阵小块上;Step 6: For Intercept the 2-value bit B cur from B, convert it to an integer X Cur , and further map X Cur to an integer in the range [0,l-1] Then select the first from H′ k , k=0,1,…,l-1 images and calculate The random rotation angle α ind will be Rotate counterclockwise to Place it on the matrix block selected on M with (x·(2r-1), y·(2r-1)) as the starting point of the upper left corner and a size of (2r-1)×(2r-1);
第7步:计算与灰度掩体图像T上的像素tx,y之间的差异Δx,y,然后将Δx,y传递给tx,y周围未处理的像素 Step 7: Calculate The difference Δx ,y between the pixel t x,y and the grayscale mask image T, and then pass Δx, y to the unprocessed pixels around t x,y
第8步:反复执行第6步到第7步,直至处理完T0中的所有坐标位置;Step 8: Repeat steps 6 to 7 until all coordinate positions in T 0 are processed;
第9步:对于从H′k,k=0,1,…,l-1中寻找与T上像素tx,y最接近的小块并计算的随机转角αind,将逆时针旋转为将其放置在M上选取的以(x·(2r-1),y·(2r-1))为左上角起点、大小为(2r-1)×(2r-1)的矩阵小块上;Step 9: For From H′ k , k=0,1,…,l-1, find the closest small block to pixel t x, y on T and calculate The random rotation angle α ind will be Rotate counterclockwise to Place it on the matrix block selected on M with (x·(2r-1), y·(2r-1)) as the starting point of the upper left corner and a size of (2r-1)×(2r-1);
第10步:计算与tx,y之间的差异Δx,y,然后将Δx,y传递给tx,y周围未处理的像素 Step 10: Calculate The difference between t x,y and Δ x,y , then pass Δ x,y to the unprocessed pixels around t x,y
第11步:反复执行第9步到第10步,直至处理完不属于T0的所有坐标位置;Step 11: Repeat steps 9 to 10 until all coordinate positions that do not belong to T 0 are processed;
第12步:将最终M输出作为嵌密后的马赛克图像。Step 12: Use the final M output as an embedding mosaic image.
优选地,第3步中,将indk映射为[0,2π)之间的随机转角αind,然后将所有的Hk转换为具有随机转角αind且半径为r的圆形图像的具体方法是:Preferably, in the third step, ind k is mapped to a random rotation angle α ind between [0,2π), and then all H k are converted into a circular image with a random rotation angle α ind and a radius of r The specific method is:
按式(1)将所有的Hk转换为具有随机转角αind且半径为r的圆形图像其中,αind=indk·2π/n;Transform all H k into circular images with random rotation angle α ind and radius r according to formula (1) Among them, α ind = ind k · 2π/n;
H′k=Rot(Hk,r,αind) (1);H′ k =Rot(H k ,r,α ind ) (1);
式(1)中,Rot()是逆时针旋转函数,Hk对应为输入原始图像,r为输出圆环图像半径,αind是逆时针旋转角度。In formula (1), Rot() is a counterclockwise rotation function, H k corresponds to the input original image, r is the radius of the output ring image, and α ind is the counterclockwise rotation angle.
优选地,式(1)中Rot()函数的具体实现步骤如第3.1步~第3.4步所示:Preferably, the specific implementation steps of the Rot() function in formula (1) are shown in steps 3.1 to 3.4:
第3.1步:按式(2)对H′k进行初始化:Step 3.1: Initialize H′ k according to formula (2):
第3.2步:将Hk半径r范围内的像素按式(3)转换为极坐标将逆时针旋转αind角度并按式(4)转换为直角坐标(i1,j1),然后将赋值给H′k中的元素其中,当i=r-1,j=r-1时,则直接将作为 Step 3.2: Pixels within H k radius r Convert to polar coordinates according to formula (3) Will Rotate the angle α ind counterclockwise and transform it into Cartesian coordinates (i 1 , j 1 ) according to formula (4), then Assign to elements in H′ k Among them, when i=r-1, j=r-1, then directly as
式(4)中,[]为4舍5入操作符;In formula (4), [] is a rounding operator;
第3.3步:在每个以为中心的5×5领域内寻找的像素点,将这些不为-1的像素点以此记为P0,P1,…,Pcount-1,与之对应的与的2次距离依次记为d0,d1,…,dcount-1,然后利用式(5)对进行插值拟合;Step 3.3: After each Find within the centered 5x5 field The pixels of which are not -1 are recorded as P 0 , P 1 ,…,P count-1 , and the corresponding and The secondary distances of are recorded as d 0 ,d 1 ,…,d count-1 in turn, and then use formula (5) to Perform interpolation fitting;
第3.4步:将H′k输出。Step 3.4: Output H′ k .
优选地,第6步中,从B中截取2值比特Bcur的具体方法是:Preferably, in step 6, the specific method of intercepting the 2-value bit B cur from B is:
按式(6)从B中截取log2l个2值比特Bcur,若截取总数不足log2l个,则截取剩余全部比特;Intercept log 2 l binary bits B cur from B according to formula (6), if the total number of intercepts is less than log 2 l, then intercept all remaining bits;
Bcur=Trim(B,Cur·log2l,log2l) (6);B cur = Trim(B, Cur·log 2 l, log 2 l) (6);
式(6)中,Trim()是2值序列截断函数,其中Trim()函数的第1个输入参数为输入2值比特序列,第2个参数为截取起始索引位置,第3个参数对应为截取长度,Cur是当前截取的2值比特索引且Cur≥0;In formula (6), Trim() is a binary sequence truncation function, where the first input parameter of the Trim() function is the input binary bit sequence, the second parameter is the starting index position of the interception, and the third parameter corresponds to is the truncation length, Cur is the binary bit index of the current truncation and Cur≥0;
第6步中,将XCur映射为[0,l-1]范围内的整数的具体方法是式(7):In step 6, map X Cur to an integer in the range [0,l-1] The specific method is formula (7):
第6步和第9步中,计算随机转角αind,将逆时针旋转为的具体方法是:In steps 6 and 9, the random rotation angle α ind is calculated, and the Rotate counterclockwise to The specific method is:
按式(8)计算[0,n-1]范围内的整数ind,然后按式(1)将逆时针旋转αind得到其中,αind=ind·2π/n:Calculate the integer ind in the range of [0, n-1] according to formula (8), and then use formula (1) to Rotate α ind counterclockwise to get where α ind =ind·2π/n:
ind=(XCur+x·y+y·rx,y+x·rx,y)mod n (8);ind = (X Cur + x y + y r x, y + x r x, y ) mod n (8);
H′k=Rot(Hk,r,αind) (1);H′ k =Rot(H k ,r,α ind ) (1);
优选地,第9步中,对于从H′k,k=0,1,…,l-1中寻找与T上像素tx,y最接近的小块的具体方法是:Preferably, in step 9, for From H′ k , k=0,1,…,l-1, find the closest small block to pixel t x, y on T The specific method is:
按式(9)从H′k,k=0,1,…,l-1中寻找与T上像素tx,y均值最接近的小块 Find the small block closest to the average value of pixel t x, y on T from H′ k ,k=0,1,…,l-1 according to formula (9)
式(9)中,mean()为圆形图像块均值计算函数,计算的是以输入图像的中心为圆心,半径为r的圆形图像范围内的像素均值;In formula (9), mean() is a circular image block mean value calculation function, which calculates the mean value of pixels within a circular image range whose radius is r with the center of the input image as the center;
第10步中,计算与tx,y之间的差异Δx,y的具体方法是式(10):In step 10, calculate The specific method of the difference Δ x, y between t x, y is formula (10):
第10步中,将Δx,y传递给tx,y周围未处理的像素的具体方法是式(11):In step 10, pass Δ x,y to the unprocessed pixels around t x,y The specific method is formula (11):
与权利要求1相对应的一种结合块旋转和马赛克的无嵌入恢复方法,包括如下步骤:A non-embedded restoration method combining block rotation and mosaic corresponding to claim 1, comprising the following steps:
第1步:输入分辨率为m2×n2的P0阶灰度马赛克图像l个分辨率为(2r-1)×(2r-1)两两不等的P2阶灰度图像秘密图像分辨率m1×n1、秘密图像的灰度阶P1、初始化二值比特序列B=Φ和B′=Φ、输入密钥k0,k1,k2并设定阈值参数T,T>0;Step 1: Input a P 0 -level grayscale mosaic image with a resolution of m 2 ×n 2 l P 2 -level grayscale images with a resolution of (2r-1)×(2r-1) in pairs Secret image resolution m 1 ×n 1 , gray scale P 1 of secret image, initialize binary bit sequence B=Φ and B′=Φ, input key k 0 , k 1 , k 2 and set threshold parameter T ,T>0;
第2步:由密钥k0产生随机整数indk∈[0,n-1],k=0,1,…,l-1且n为大于0的整数,将indk映射为[0,2π)之间的随机转角αind,然后将所有的Hk转换为具有随机转角αind且半径为r的圆形图像 Step 2: Generate a random integer ind k ∈ [0,n-1] from the key k 0 , k=0,1,...,l-1 and n is an integer greater than 0, and map ind k to [0, 2π) with a random rotation angle α ind , and then convert all H k into a circular image with a random rotation angle α ind and radius r
第3步:对所有H′k旋转校正为H″k且满足H″k质心在H″k圆心约定的方向上;Step 3: Correct all H′ k rotations to H″ k and satisfy the H″ k centroid in the direction agreed by the H″ k circle center;
第4步:初始化分辨率为m1×n1的P1阶灰度秘密图像由密钥k1产生l0个两两不等的m0×n0范围内的随机整数坐标序列T0,其中由密钥k2产生m0×n0个[0,l-1]范围内的随机整数矩阵其中m0=m2/(2r-1),n0=n2/(2r-1);Step 4: Initialize the P 1 -level grayscale secret image with resolution m 1 ×n 1 Generate l 0 pairs of random integer coordinate sequences T 0 in the range of m 0 ×n 0 from the key k 1 , where Generate m 0 ×n 0 random integer matrices in the range [0,l-1] from the key k 2 Where m 0 =m 2 /(2r-1), n 0 =n 2 /(2r-1);
第5步:依次读取T0中整数坐标(x,y),计算(x,y)在M中的坐标位置,从M中截取(2r-1)×(2r-1)的马赛克图像Mt;Step 5: Read the integer coordinates (x, y) in T 0 sequentially, calculate the coordinate position of (x, y) in M, and intercept the mosaic image M of (2r-1)×(2r-1) from M t ;
第6步:将Mt旋转校正为M′t且满足Mt质心在M′t圆心约定的方向上,然后从所有的H″k,k=0,1,…,l-1中寻找和Mt′距离最接近的对应的编号将转换为秘密值XCur;Step 6: Correct the rotation of M t to M′ t and satisfy that the center of mass of M t is in the direction agreed by the center of M′ t , and then find the sum from all H″ k , k=0,1,…,l-1 M t ′ is the closest corresponding number Will Convert to secret value X Cur ;
第7步:按权利要求1给出的第6步同样的方法计算随机转角αind∈[0,2π),将Mt逆时针旋转-αind得到M″t;The 7th step: calculate random angle of rotation α ind ∈ [0,2π) by the 6th step same method that claim 1 provides, M t is rotated counterclockwise-α ind to obtain M "t;
第8步:将XCur转换为log2l位2进制数添加至2值序列B中,计算M″t和的像素点的差值d,若d<T,则秘密信息XCur没有被破坏,则将log2l位1构成的2进制数添加至B′中,反之则将log2l位0构成的2进制数添加至B′中;Step 8: Convert X Cur to log 2 l-digit binary numbers and add them to the binary sequence B, and calculate M″ t and The pixel point difference d, if d<T, the secret information X Cur is not destroyed, then the binary number composed of log 2 l bits 1 is added to B′, otherwise, log 2 l bits 0 are formed The binary number of is added to B';
第9步:反复执行第5步~第8步,直至T0中所有整数坐标(x,y)读取完毕;Step 9: Repeat steps 5 to 8 until all integer coordinates (x, y) in T 0 are read;
第10步:将B和B′以P1个2进制位为1组转换为10进制数,然后将其组织为m1×n1的P1阶灰度秘密图像和认证图像并输出。Step 10: Convert B and B′ to a decimal number with P 1 binary digits as a group, and then organize it into a P 1 -level grayscale secret image of m 1 ×n 1 and the authentication image and output.
优选地,第2步将indk映射为[0,2π)之间的随机转角αind,然后将所有的Hk转换为具有随机转角αind且半径为r的圆形图像的具体方法是式(1),其中αind=indk·2π/n:Preferably, step 2 maps ind k to a random rotation angle α ind between [0,2π), and then converts all H k into circular images with random rotation angle α ind and radius r The specific method is formula (1), where α ind = ind k · 2π/n:
H′k=Rot(Hk,r,αind) (1);H′ k =Rot(H k ,r,α ind ) (1);
第3步对所有H′k旋转校正为H″k且满足H″k质心在H″k圆心约定的方向上的具体方法是对所有H′k按式(12)旋转校正为H″k且满足H″k质心在H″k圆心右侧的水平方向上:The third step is to correct all H′ k rotations to H″ k and satisfy the H″ k centroid in the direction agreed by the center of H″ k . The specific method is to correct all H′ k rotations according to formula (12) to H″ k and Satisfied that the H″ k centroid is in the horizontal direction on the right side of the H″ k circle center:
H″k=Rotmass(H′k) (12);H″ k = Rot mass (H′ k ) (12);
第6步将Mt旋转校正为M′t且满足M′t质心在M′t圆心约定的方向上的具体方法是按式(13)旋转校正为M′t且满足M′t质心在M′t圆心右侧的水平方向上:Step 6. The specific method to correct the rotation of M t to M′ t and satisfy the centroid of M′ t in the direction agreed by the center of M′ t is to correct the rotation to M′ t according to formula (13) and satisfy the centroid of M′ t to be in the direction of M′ t. ′ t in the horizontal direction on the right side of the center of the circle:
M′t=Rotmass(Mt) (13);M′ t = Rot mass (M t ) (13);
式(12)和式(13)中,函数Rotmass()为质心旋转函数,其特征在于包括如下步骤:In formula (12) and formula (13), function Rot mass () is centroid rotation function, it is characterized in that comprising the following steps:
第3.1步:记计算H′k的质心(xmass,ymass):Step 3.1: Remember Calculate the centroid of H′ k (x mass , y mass ):
第3.2步:计算H′k质心相对于H′k圆心的几何倾角;Step 3.2: Calculate the geometric inclination angle of the center of mass of H′ k relative to the center of circle of H′ k ;
第3.3步:将H′k逆时针旋转得到H″k,然后将H″k输出;Step 3.3: Rotate H′ k counterclockwise Get H″ k , and then output H″ k ;
优选地,第3.1步计算H′k的质心(xmass,ymass)的具体方法是式(14):Preferably, the specific method for calculating the centroid (x mass , y mass ) of H′ k in step 3.1 is formula (14):
第3.2步计算H′k质心相对于H′k圆心的几何倾角的具体方法是按式(15)计算H′k质心相对于H′k圆心的几何倾角 The specific method for calculating the geometric inclination of the H′ k centroid relative to the H′ k circle center in step 3.2 is to calculate the geometric inclination angle of the H′ k centroid relative to the H′ k circle center according to formula (15)
第3.3步将H′k逆时针旋转得到H″k的具体方法是式(16):Step 3.3 Rotate H′ k counterclockwise The concrete method that obtains H " k is formula (16):
优选地,第5步计算(x,y)在M中的坐标位置从M中截取(2r-1)×(2r-1)的马赛克图像Mt的具体方法是式(17):Preferably, the step 5 calculates the coordinate position of (x, y) in M. The specific method of intercepting the mosaic image M t of (2r-1)×(2r-1) from M is formula (17):
第6步从所有的H″k,k=0,1,…,l-1中寻找和Mt′距离最接近的对应的编号的具体方法是式(18):Step 6 Find the closest distance to M t ′ from all H″ k , k=0,1,…,l-1 corresponding number The specific method is formula (18):
式(18)中,为M′和H″k的2次距离。In formula (18), is the secondary distance between M′ and H″ k .
优选地,第6步将转换为秘密值XCur的具体方法为式(19):Preferably, step 6 will The specific method of converting to the secret value X Cur is formula (19):
第7步将Mt逆时针旋转-αind得到M″t的具体方法是:In the 7th step, M t is rotated counterclockwise-α ind to obtain M″ t . The specific method is:
M″t=Rot(Mk,r,-αind) (20);M″ t = Rot(M k ,r,-α ind ) (20);
第8步计算M″t和的像素点的差值d的具体方法是式(21):Step 8 Calculate M″ t and The specific method of the difference d of the pixel point is formula (21):
与现有技术相比,本发明有益效果是:Compared with prior art, the beneficial effect of the present invention is:
①传统基于搜索式无载体信息隐藏方法信息隐藏容量极低,涉及大量的图像和文本在信道中密集传输,尽管单个图像和文本都是未修改的图像和文本,而大量的图像和文本在信道中密集传输也必将引起攻击者的怀疑,从而所提策略无法抵抗密写分析;基于纹理合成的生成式无载体信息隐藏仅管能产生简单质地的纹理图像,但无法生成有意义的图像;而传统基于马赛克拼图的信息隐藏方法均涉及对掩体图像的修改,从而会在载体中留下修改的痕迹,从而难以抵抗密写分析攻击。①The information hiding capacity of the traditional search-based carrierless information hiding method is extremely low, involving a large number of images and texts intensively transmitted in the channel, although a single image and text are unmodified images and texts, and a large number of images and texts in the channel Medium-intensive transmission will also arouse the suspicion of attackers, so the proposed strategy cannot resist steganalysis; generative carrierless information hiding based on texture synthesis can only produce texture images with simple textures, but cannot generate meaningful images; The traditional information hiding methods based on mosaic puzzles all involve the modification of the cover image, which will leave traces of modification in the carrier, making it difficult to resist steganalysis attacks.
同传统方法不同,本发明将多个分辨率为(2r-1)×(2r-1)的灰度图像转换为具有伪随机转角且半径为r的圆形图像,通过生成随机整数坐标序列来决定密图划分的比特位串在掩体上的隐藏位置,并将放置圆形图像所产生的均值误差分散至周围位嵌密的圆形图像上。在嵌密过程中,不涉及变换参数的任何修改嵌入,而仅利用圆形图像的旋转角度和放置位置来对秘密信息比特进行充分有效的表达。掩体嵌密和未嵌密像素都是添加了几何转角的圆形图像,并且与初始转角没有任何关系,因此不会在嵌密载体上留下与秘密信息有关的任何修改痕迹,从而隐藏的秘密难以发现,因此可以抵抗密写分析。Different from the traditional method, the present invention converts a plurality of grayscale images with a resolution of (2r-1)×(2r-1) into a circular image with a pseudo-random corner and a radius of r, by generating a sequence of random integer coordinates to Determine the hidden position of the bit string divided by the dense image on the cover, and disperse the mean error generated by placing the circular image to the surrounding circular image with dense bit embedding. In the process of embedding, it does not involve any modified embedding of transformation parameters, but only uses the rotation angle and placement position of the circular image to fully and effectively express the secret information bits. The embedding and non-embedding pixels of the bunker are circular images with geometric corners added, and have nothing to do with the initial corners, so no modification traces related to secret information will be left on the embedding carrier, thus hiding the secret Difficult to discover and thus resistant to steganalysis.
②传统基于马赛克拼图的信息隐藏方法需要将变换参数直接嵌入到含密掩体中,由于隐藏的容量问题,通常采用基于LSB或与之相关的空域可逆信息隐藏方法。这些嵌入方法不仅会在嵌密载体上留下修改痕迹,同时由于非显著比特位的修改,在遭受攻击时会导致嵌入的参数无法提取,从而无法对秘密信息进行准确重构。②The traditional mosaic-based information hiding method needs to embed the transformation parameters directly into the dense bunker. Due to the hidden capacity problem, the space-domain reversible information hiding method based on LSB or related to it is usually used. These embedding methods will not only leave modification traces on the embedding carrier, but at the same time, due to the modification of non-significant bits, the embedded parameters cannot be extracted when attacked, so that the secret information cannot be accurately reconstructed.
同传统基于马赛克拼图的信息隐藏方法不同,本发明是将秘密信息划分的2值比特位串通过圆形图像进行表达,采用质心旋转图像匹配恢复秘密信息,利用中值滤波器修复JPEG压缩或噪声攻击后的异常像素点,每个圆形图像仅通过均值逼近掩体图像上的1个像素,而JPEG压缩和随机噪声攻击都不会对圆形图像的均值产生较大影响,从而圆形图像所表示的秘密信息可以在很大程度上能进行精确恢复,因而相对于传统方法具有较强的鲁棒性。Different from the traditional information hiding method based on mosaic puzzles, the present invention expresses the binary bit string divided by the secret information through a circular image, uses centroid rotation image matching to restore the secret information, and uses the median filter to repair JPEG compression or noise For the abnormal pixels after the attack, each circular image only approximates 1 pixel on the cover image through the mean value, and neither JPEG compression nor random noise attack will have a large impact on the mean value of the circular image, so the circular image The represented secret information can be recovered accurately to a large extent, so it is more robust than traditional methods.
③传统基于马赛克拼图的密图伪装方法由于嵌入容量小,未考虑对提取秘密信息的正确性进行检验,同时所考虑的安全性也十分有限。接收方根据信道接收到的含密掩体对秘密信息进行重构,但无法对重构秘密信息的真实性进行有效检验。③The traditional mosaic-based secret image camouflage method does not consider the correctness of the extracted secret information due to the small embedding capacity, and the security considered is also very limited. The receiver reconstructs the secret information according to the secret cover received by the channel, but cannot effectively verify the authenticity of the reconstructed secret information.
而本发明所述方法利用添加随机转角的圆形图像在不同的位置表达不同的秘密信息,并严格依赖于用户密钥,只有提供正确的密钥才能获取秘密信息并对所提取的秘密信息的准确性进行验证,而其他情况下将无法获取。However, the method of the present invention uses circular images with random corners to express different secret information at different positions, and is strictly dependent on the user key. Only by providing the correct key can the secret information be obtained and the extracted secret information Verified for accuracy, otherwise unavailable.
附图说明Description of drawings
图1是秘密信息嵌入流程图;Figure 1 is a flowchart of secret information embedding;
图2是秘密信息提取流程图;Fig. 2 is a flowchart of secret information extraction;
图3是实施例,分辨率为64×64的8位灰度秘密图像;Fig. 3 is an embodiment, and the resolution is an 8-bit grayscale secret image of 64×64;
图4是实施例,分辨率为128×128的8位灰度掩体图像;Fig. 4 is an embodiment, and the resolution is an 8-bit grayscale mask image of 128 * 128;
图5a是实施例,32个分辨率为65×65的8位灰度图像;Fig. 5 a is an embodiment, 32 resolutions are 8-bit grayscale images of 65 * 65;
图5b是实施例,图5a经圆形初始化和添加随机转角后对应的32个圆形图像;Fig. 5b is an embodiment, and Fig. 5a corresponds to 32 circular images after circular initialization and adding random corners;
图5c是实施例,图5b经计算质心后,将质心坐标点旋转到X正半轴上对应的32个圆形图像;Fig. 5c is an embodiment. After calculating the centroid in Fig. 5b, rotate the coordinate point of the centroid to the corresponding 32 circular images on the positive semi-axis of X;
图6a是实施例,利用图5b圆形图像,形成的分辨率为8320×8320的8位含密灰度马赛克图像的局部图像(图6b分辨率较高,这里通过图6a给出图6b局部细节);Fig. 6 a is an embodiment, utilizes Fig. 5 b circular image, the partial image (Fig. 6 b resolution is higher, provides Fig. 6 b part by Fig. 6 a here by Fig. detail);
图6b是实施例,利用图5b圆形图像,形成的分辨率为8320×8320的8位含密灰度马赛克图像;Figure 6b is an embodiment, using the circular image in Figure 5b to form an 8-bit dense grayscale mosaic image with a resolution of 8320×8320;
图7是实施例,由图6b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 7 is an embodiment, the secret image extracted by Fig. 6b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图8是实施例,由图6b按图2流程提出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证;Fig. 8 is an embodiment, by Fig. 6 b according to Fig. 2 flow process, proposes the authentication map that secret image is generated, is the blank image of all white, represents that all pixels of secret image all pass authentication;
图9a是实施例,对图6b进行质量因子为50的JEPG压缩,形成的JPEG灰度图像的局部图像(图9b分辨率较高,这里通过图9a给出图9b局部细节);Fig. 9a is an embodiment, carrying out the JPEG compression that quality factor is 50 to Fig. 6b, the partial image of the JPEG grayscale image that forms (Fig. 9b resolution is higher, provides Fig. 9b local details by Fig. 9a here);
图9b是实施例,对图6b进行质量因子为50的JEPG压缩,形成的JPEG灰度图像;Figure 9b is an embodiment, the JPEG grayscale image formed by performing JPEG compression with a quality factor of 50 on Figure 6b;
图10是实施例,由图9b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 10 is an embodiment, the secret image extracted by Fig. 9b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图11是实施例,由图9b按图2流程提取出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证;Fig. 11 is an embodiment, the authentication map generated by extracting the secret image according to the process of Fig. 2 from Fig. 9b is a completely white blank image, indicating that all pixels of the secret image have passed the authentication;
图12a是实施例,对图6b进行质量因子为80的JPEG压缩,形成的JPEG灰度图像的局部图像(图12b分辨率较高,这里通过图12a给出图12b局部细节);Fig. 12a is an embodiment, carrying out the JPEG compression that quality factor is 80 to Fig. 6b, the partial image of the JPEG gray scale image that forms (Fig. 12b resolution is higher, provides Fig. 12b local details by Fig. 12a here);
图12b是实施例,对图6b进行质量因子为80的JPEG压缩,形成的JPEG灰度图像;Figure 12b is an embodiment, the JPEG grayscale image formed by performing JPEG compression with a quality factor of 80 on Figure 6b;
图13是实施例,由图12b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 13 is an embodiment, the secret image extracted by Fig. 12b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图14是实施例,由图12b按图2流程提取出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证;Fig. 14 is an embodiment, the authentication map generated by extracting the secret image according to the process of Fig. 2 from Fig. 12b is a completely white blank image, indicating that all pixels of the secret image have passed the authentication;
图15a是实施例,对图6b进行强度为8%的噪声攻击后形成的含噪声马赛克图像的局部图像(图15b分辨率较高,这里通过图15a给出图15b局部细节);Fig. 15a is an embodiment, the partial image of the noise-containing mosaic image formed after the strength of 8% noise attack is carried out on Fig. 6b (Fig. 15b has a higher resolution, here the local details of Fig. 15b are provided by Fig. 15a);
图15b是实施例,对图6b进行强度为8%的噪声攻击后形成的含噪声马赛克图像;Fig. 15b is an embodiment, a noise-containing mosaic image formed after performing a noise attack with an intensity of 8% on Fig. 6b;
图16是实施例,由图15b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 16 is an embodiment, the secret image extracted by Fig. 15b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图17是实施例,由图15b按图2流程提取出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证;Fig. 17 is an embodiment. The authentication map generated by extracting the secret image according to the process of Fig. 2 from Fig. 15b is a completely white blank image, indicating that all pixels of the secret image have passed the authentication;
图18a是实施例,对图6b进行强度为20%的噪声攻击后形成的含噪声马赛克图像的局部图像(图18b分辨率较高,这里通过图18a给出图18b局部细节);Fig. 18a is an embodiment, a partial image of a mosaic image containing noise formed after a 20% noise attack is carried out on Fig. 6b (Fig. 18b has a higher resolution, here the local details of Fig. 18b are provided by Fig. 18a);
图18b是实施例,对图6b进行强度为20%的噪声攻击后形成的含噪声马赛克图像;Fig. 18b is an embodiment, a noise-containing mosaic image formed after performing a noise attack with a strength of 20% on Fig. 6b;
图19是实施例,由图18b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 19 is an embodiment, the secret image extracted by Fig. 18b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图20是实施例,由图18b按图2流程提取出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证。Fig. 20 is an embodiment. The authentication map generated by extracting the secret image according to the process of Fig. 2 from Fig. 18b is a completely white blank image, indicating that all pixels of the secret image have passed the authentication.
具体实施方式Detailed ways
以JAVA jdk1.8.0_20作为案例实施环境,结合附图对本发明实施方式进行详细说明,但不局限于本实施案例,其中图1是秘密信息伪装流程图,图2是秘密信息恢复流程图。其中,结合块旋转和马赛克的无嵌入伪装方法,包括以下步骤:Taking JAVA jdk1.8.0_20 as the case implementation environment, the embodiment of the present invention will be described in detail in conjunction with the accompanying drawings, but not limited to this implementation case, wherein Fig. 1 is a flow chart of secret information camouflage, and Fig. 2 is a flow chart of secret information recovery. Among them, the embedding-free camouflage method combining block rotation and mosaic includes the following steps:
第1步:输入分辨率为m0×n0的P0阶灰度掩体图像分辨率为m1×n1的P1阶灰度秘密图像以及l个分辨率为(2r-1)×(2r-1)两两不等的P2阶灰度图像其中l为2的整数幂次且满足 Step 1: Input a P 0 -level grayscale mask image with a resolution of m 0 ×n 0 P 1 -level grayscale secret image with resolution m 1 ×n 1 And l P 2 -level grayscale images with a resolution of (2r-1)×(2r-1) in pairs where l is an integer power of 2 and satisfies
例如:取分辨率为m0×n0=4×4的灰度掩体图像T=(ti,j)4×4,ti,j∈{0,1,…,255},像素值为{(10,20,30,40),(50,60,70,80),(90,100,110,120),(130,140,150,160)},其中(10,20,30,40),(50,60,70,80),(90,100,110,120),(130,140,150,160)分别对应为灰度掩体图像第0行、第1行、第2行和第3行;取分辨率为m1×n1=2×2的灰度图像作为灰度秘密图像S=(si,j)2×2,si,j∈{0,1,…,255},取l=32个分辨率为65×65的两两不等的灰度图像其中r=33,P0,P1,P2=8, For example: take a grayscale mask image T=(t i,j ) 4×4 with a resolution of m 0 ×n 0 =4×4, t i,j ∈{0,1,…,255}, the pixel value is {(10,20,30,40),(50,60,70,80),(90,100,110,120),(130,140,150,160)}, where (10,20,30,40),(50,60,70,80) , (90, 100, 110, 120), (130, 140, 150, 160) respectively correspond to the 0th, 1st, 2nd and 3rd lines of the grayscale mask image; the grayscale image with resolution m 1 ×n 1 = 2×2 is taken as gray Degree secret image S=(s i,j ) 2×2 , s i,j ∈{0,1,…,255}, take l=32 pairwise grayscale images with a resolution of 65×65 where r=33, P 0 , P 1 , P 2 =8,
第2步:由密钥k0产生随机整数indk∈[0,n-1],k=0,1,…,l-1且n为大于0的整数;Step 2: Generate a random integer ind k ∈ [0,n-1] from the key k 0 , k=0,1,...,l-1 and n is an integer greater than 0;
例如:取密钥k0=9,取n=8可产生32个[0,7]范围内的随机整数(5,2,6,6,6,6,4,7,7,1,3,2,4,4,4,0,3,0,7,7,2,5,0,3,3,5,2,6,5,6,1,5);For example: taking the key k 0 =9, taking n=8 can generate 32 random integers in the range of [0,7] (5,2,6,6,6,6,4,7,7,1,3 ,2,4,4,4,0,3,0,7,7,2,5,0,3,3,5,2,6,5,6,1,5);
第3步:按式(1)将indk映射为[0,2π)之间的随机转角αind,然后将所有的Hk转换为具有随机转角αind且半径为r的圆形图像 Step 3: Map ind k to a random rotation angle α ind between [0,2π) according to formula (1), and then convert all H k into circular images with random rotation angle α ind and radius r
H′k=Rot(Hk,r,αind),αind=indk·2π/n (1);H' k = Rot(H k , r, α ind ), α ind = ind k · 2π/n (1);
式(1)中,Rot()是逆时针旋转函数,Hk对应为输入原始图像,r为输出圆环图像半径,αind是逆时针旋转角度,Rot()函数的具体实现步骤如第3.1步~第3.4步所示:In formula (1), Rot() is a counterclockwise rotation function, H k corresponds to the input original image, r is the radius of the output ring image, and α ind is the counterclockwise rotation angle. The specific implementation steps of the Rot() function are shown in Section 3.1 Steps to 3.4 show:
第3.1步:按式(2)对H′k进行初始化Step 3.1: Initialize H′ k according to formula (2)
第3.2步:将Hk半径r范围内的像素按式(3)转换为极坐标将逆时针旋转αind角度并按式(4)转换为直角坐标(i1,j1),然后将赋值给H′k中元素其中当i=r-1,j=r-1时,则直接将作为 Step 3.2: Pixels within H k radius r Convert to polar coordinates according to formula (3) Will Rotate the angle α ind counterclockwise and transform it into Cartesian coordinates (i 1 , j 1 ) according to formula (4), then Assign to the elements in H′ k Wherein when i=r-1, j=r-1, then directly as
式(4)中,“[]”为4舍5入操作符;In formula (4), “[]” is a rounding operator;
第3.3步:在每个以为中心的5×5领域内寻找的像素点,将这些不为-1的像素点以此记为P0,P1,…,Pcount-1,与之对应的与的2次距离依次记为d0,d1,…,dcount-1,然后利用式(5)对进行插值拟合;Step 3.3: After each Find within the centered 5x5 field The pixels of which are not -1 are recorded as P 0 , P 1 ,…,P count-1 , and the corresponding and The secondary distances of are recorded as d 0 ,d 1 ,…,d count-1 in turn, and then use formula (5) to Perform interpolation fitting;
第3.4步:将H′k输出;Step 3.4: output H′ k ;
例如:取k=0,假设H0对应的图像为图5a(0),取ind0=5,n=8,按式(1)可计算对应的随机转角αind=ind0×2π/n=5·2π/8≈3.92699For example: take k=0, assume that the image corresponding to H 0 is Figure 5a(0), take ind 0 =5, n=8, and calculate the corresponding random rotation angle α ind =ind 0 ×2π/n according to formula (1) =5·2π/8≈3.92699
①按第3.1步,对5a(0)初始化,将以图像中心点坐标为圆心,半径r=33范围内的像素点初始化为-1,其余像素点初始化为0;① According to step 3.1, initialize 5a(0), and the coordinates of the center point of the image will be is the center of the circle, the pixels within the range of radius r=33 are initialized to -1, and the rest of the pixels are initialized to 0;
②按第3.2步,将半径r=33范围内的像素按式(3)转换为极坐标(ρ,θ),其中θ=arctan((33-i-1)/(j-33+1)),比如像素点(0,32)对应的极坐标为旋转后的新坐标将图5a(0)图像中坐标位置(0,32)的像素值复制到图5b(0)图像中(32,3)坐标点对应的位置上;②According to step 3.2, the pixels within the range of radius r=33 Convert to polar coordinates (ρ, θ) according to formula (3), where θ=arctan((33-i-1)/(j-33+1)), for example, the polar coordinates corresponding to the pixel point (0,32) are new coordinates after rotation Copy the pixel value of the coordinate position (0,32) in the image of Figure 5a (0) to the position corresponding to the coordinate point (32,3) in the image of Figure 5b (0);
③按第3.3步在图5b(0)中寻找像素点为-1的值,假设坐标点(32,32)的像素值为-1,搜索(32,32)为中心的5×5领域内寻找像素值不等于-1的像素点,利用count进行计数,并计算与中心点的距离,利用加权平均法对中心像素点(32,32)进行插值拟合。③According to step 3.3, find the value of the pixel point of -1 in Figure 5b(0), assuming that the pixel value of the coordinate point (32,32) is -1, and search for the 5×5 area centered on (32,32) Find the pixel points whose pixel value is not equal to -1, use count to count, and calculate the distance from the center point, and use the weighted average method to interpolate and fit the center pixel point (32,32).
第4步:将S扫描为2值比特序列由密钥k1产生l0个两两不等的m0×n0范围内的随机整数坐标序列其中,为第i个随机整数坐标,即 Step 4: Scan S as a sequence of 2-valued bits Generate l 0 random integer coordinate sequences in the range of m 0 ×n 0 from the key k 1 in, is the ith random integer coordinate, namely
例如,分辨率为2×2的灰度秘密图像S的4个像素点为{1,2,3,4},将S扫描为2值比特序列B=(1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,),其中B共含有32个元素,由密钥k1=5可产生l0=7个两两不等的4×4范围内的随机整数坐标序列T0=((2,3),(0,2),(0,1),(2,1),(1,3),(3,3),(3,1));For example, the 4 pixels of the grayscale secret image S with a resolution of 2×2 are {1,2,3,4}, and S is scanned as a binary bit sequence B=(1,0,0,0,0 ,0,0,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,0,0,0 ,0,0,), where B contains a total of 32 elements, and the key k 1 =5 can generate l 0 =7 random integer coordinate sequences T 0 =((2 ,3),(0,2),(0,1),(2,1),(1,3),(3,3),(3,1));
第5步:初始化分辨率为m2×n2的P0阶灰度掩体图像且满足m2=m0(2r-1),n2=n0(2r-1),由密钥k2产生m0n0个[0,l-1]范围内的随机整数矩阵且ri,j∈[0,l-1];Step 5: Initialize the P 0 -level grayscale mask image with a resolution of m 2 ×n 2 And satisfy m 2 =m 0 (2r-1), n 2 =n 0 (2r-1), generate m 0 n 0 random integer matrices in the range of [0,l-1] from key k 2 And r i, j ∈ [0, l-1];
例如:取m2=m0(2r-1)=4·65=260,n2=n0(2r-1)=4·65=260,可初始化分辨率为260×260的灰度掩体图像,由密钥k2=5产生4×4个[0,l-1=31]范围内的随机整数矩阵R={(23,5,2,19),(15,30,14,26),(14,13,22,24),(8,11,24,26)},其中(23,5,2,19),…,(8,11,24,26)分别对应为R矩阵的第0行到第3行;For example: take m 2 =m 0 (2r-1)=4·65=260, n 2 =n 0 (2r-1)=4·65=260, the grayscale mask image with a resolution of 260×260 can be initialized , generate 4×4 random integer matrices R={(23,5,2,19),(15,30,14,26) in the range of [0,l-1=31] from key k 2 =5 ,(14,13,22,24),(8,11,24,26)}, where (23,5,2,19),…,(8,11,24,26) correspond to the R matrix Line 0 to line 3;
第6步:对于按式(6)从B中截取2值比特Bcur,将其转换为整数XCur,并进一步按式(7)将XCur映射为[0,l-1]范围内的整数然后从H′k,k=0,1,…,l-1中选取第张图像并按式(8)计算[0,n-1]范围内的整数ind,然后按式(1)将逆时针旋转αind得到其中αind=ind·2π/n,将逆时针旋转为将其放置在M上以(x·(2r-1),y·(2r-1))为左上角起点,(2r-1)×(2r-1)的矩阵小块上;Step 6: For Intercept 2-value bit B cur from B according to formula (6), convert it to an integer X Cur , and further map X Cur to an integer in the range of [0,l-1] according to formula (7) Then select the first from H′ k , k=0,1,…,l-1 images And calculate the integer ind in the range of [0,n-1] according to formula (8), then press formula (1) to Rotate α ind counterclockwise to get where α ind =ind·2π/n, the Rotate counterclockwise to Place it on M with (x · (2r-1), y · (2r-1)) as the starting point of the upper left corner, (2r-1) × (2r-1) matrix small block;
Bcur=Trim(B,Cur·log2l,log2l) (6);B cur = Trim(B, Cur·log 2 l, log 2 l) (6);
式(6)中,Trim()是2值序列截断函数,其中B为输入2值比特序列,第2个参数为截取起始索引位置,第3个参数对应为截取长度;In formula (6), Trim() is a binary sequence truncation function, where B is the input binary bit sequence, the second parameter is the starting index position of the interception, and the third parameter corresponds to the interception length;
ind=(XCur+x·y+y·rx,y+x·rx,y)mod n (8);ind = (X Cur + x y + y r x, y + x r x, y ) mod n (8);
H′k=Rot(Hk,r,αind) (1);H′ k =Rot(H k ,r,α ind ) (1);
例如:以(x,y)=(2,3)为例,由于(2,3)∈T0,按式(6)可从B中截取5个2值比特BCur=(1,0,0,0,0)这里取Cur=0,将BCur转化为十进制数XCur,计算得XCur=1·20+0·21+0·22+0·23=1,并进一步由式(7)将XCur映射为即其中r2,3=24;取n=8,由式(8)可得到ind=(1+2×3+2×24+3×24)mod8=7,由此可将即图5b(25)按式(1)逆时针旋转α7角度得到然后放置在M中以(2,3)为左上角起点的矩阵小块上,其中α7=7×π×0.25=5.49779;For example: taking (x, y)=(2,3) as an example, since (2,3)∈T 0 , according to formula (6), five binary bits B Cur =(1,0, 0,0,0) where Cur=0 is taken, and B Cur is converted into a decimal number X Cur , and X Cur =1·2 0 +0·2 1 +0·2 2 +0·2 3 =1 is calculated, and Further map X Cur by formula (7) as which is Where r 2,3 = 24; take n = 8, from formula (8) can get ind = (1+2×3+2×24+3×24) mod8=7, thus can be That is, Fig. 5b(25) is rotated counterclockwise by an angle of α 7 according to formula (1) to obtain Then place it on the matrix block starting from (2,3) in M, where α 7 =7×π×0.25=5.49779;
第7步:按式(10)计算与灰度掩体图像T上的tx,y之间的差异Δx,y,然后按式(11)将Δx,y传递给tx,y周围未处理的像素 Step 7: Calculate according to formula (10) and the difference Δ x, y between t x,y on the gray-scale mask image T, and then transfer Δ x,y to the unprocessed pixels around t x,y according to formula (11)
式(10)中,mean()为圆形图像块均值计算函数,计算的是以输入图像的中心为圆心,半径为r的圆形图像范围内的像素均值In formula (10), mean() is the mean value calculation function of the circular image block, which calculates the mean value of the pixels within the circular image range with the center of the input image as the center and the radius r
例如:假设M上坐标位置(0,1)点要放置取即对应图5b(5),假设以图5b(5)中心为圆心,半径r=33范围内的所有像素点的均值为27,则按式(10)计算与T中(0,1)上的像素点t0,1=20的差值Δ0,1=27-20=7,按式(11)将差值7传递给周围未处理的像素t1,0,t1,1,t1,2,其中 这里之所以不扩散到t0,2,原因是(0,2)∈T0=((2,3),(0,2),(0,1),(2,1),(1,3),(3,3),(3,1))。For example: Assume that the coordinate position (0,1) on M is to be placed Pick That is, corresponding to Figure 5b(5), assuming that the center of Figure 5b(5) is the center of the circle, and the mean value of all pixels within the radius r=33 is 27, then it is calculated according to formula (10) and (0,1) in T The difference Δ 0,1 =27-20=7 of the pixel point t 0,1 =20, according to formula (11), transfer the difference 7 to the surrounding unprocessed pixels t 1,0 ,t 1,1 ,t 1,2 where The reason why it does not spread to t 0,2 here is that (0,2)∈T 0 =((2,3),(0,2),(0,1),(2,1),(1, 3), (3,3), (3,1)).
第8步:反复执行第6步到第7步,直至处理完T0中的所有坐标位置;Step 8: Repeat steps 6 to 7 until all coordinate positions in T 0 are processed;
第9步:对于按式(9)从H′k,k=0,1,…,l-1中寻找与T上像素tx,y最接近的小块按式(8)计算[0,n-1]范围内的整数ind,然后按式(1)将逆时针旋转αind得到其中αind=ind·2π/n,XCur=0,将其放置在M上以(x·(2r-1),y·(2r-1))为左上角起点,(2r-1)×(2r-1)的矩阵小块上;Step 9: For Find the closest small block to pixel t x,y on T from H′ k ,k=0,1,…,l-1 according to formula (9) Calculate the integer ind in the range of [0, n-1] according to formula (8), and then use formula (1) to Rotate α ind counterclockwise to get where α ind =ind·2π/n, X Cur =0, place it on M with (x·(2r-1), y·(2r-1)) as the starting point of the upper left corner, (2r-1)× On the matrix block of (2r-1);
式(9)中,mean()为圆形图像块均值计算函数,计算的是以输入图像的中心为圆心,半径为r的圆形图像范围内的像素均值;In formula (9), mean() is a circular image block mean value calculation function, which calculates the mean value of pixels within a circular image range whose radius is r with the center of the input image as the center;
例如:以(x,y)=(0,0)为例,由于假设按式(9)从H′k,k=0,1,…,l-1中寻找与T上像素t0,0最接近的小块是H′1,H′1对应的图像如图5b(1)所示,取XCur=0,由R={(23,5,2,19),(15,30,14,26),(14,13,22,24),(8,11,24,26)}知r0,0=23,则按式(8)可计算ind=(0+0×0+0×23+0×23)mod8=0,从而αind=ind·2π/n=0,然后按式(1)可将图5b(1)逆时针旋转0,并其放置在M中以(0·(2r-1),0·(2r-1))=(0,0)为左上角起点的矩阵小块上。For example: Take (x,y)=(0,0) as an example, because Assuming that according to the formula (9), the small block closest to the pixel t 0,0 on T is found from H′ k ,k=0,1,…,l-1 is H′ 1 , and the image corresponding to H′ 1 is shown in the figure As shown in 5b(1), take X Cur =0, by R={(23,5,2,19),(15,30,14,26),(14,13,22,24),(8, 11,24,26)} Knowing that r 0,0 = 23, then according to formula (8), it can be calculated that ind = (0+0×0+0×23+0×23) mod8=0, thus α ind =ind· 2π/n=0, then according to formula (1), Figure 5b(1) can be rotated counterclockwise by 0, and placed in M with (0·(2r-1),0·(2r-1))=( 0,0) is the matrix block starting from the upper left corner.
第10步:计算与tx,y之间的差异Δx,y,然后将Δx,y传递给tx,y周围未处理的像素 Step 10: Calculate The difference between t x,y and Δ x,y , then pass Δ x,y to the unprocessed pixels around t x,y
例如:假设M上坐标位置(0,0)点要放置取即对应图5b(1),假设以图5b(1)中心为圆心,半径r=33范围内的所有像素点的均值为36,则按式(10)计算与T中(0,0)上的像素点t0,0=10的差值Δ0,0=36-10=26,按式(11)将差值26传递给周围未处理的像素t1,0,t1,1,其中 这里之所以不扩散到t0,1和t0,-1,原因是(0,1)∈T0=((2,3),(0,2),(0,1),(2,1),(1,3),(3,3),(3,1))且像素点t0,-1不存在。For example: Assume that the coordinate position (0,0) on M is to be placed Pick That is, corresponding to Figure 5b(1), assuming that the center of Figure 5b(1) is the center of the circle, and the mean value of all pixels within the range of radius r=33 is 36, then it is calculated according to formula (10) and on (0,0) in T The difference Δ 0,0 =36-10=26 of the pixel point t 0,0 =10, according to formula (11), transfer the difference 26 to the surrounding unprocessed pixels t 1,0 ,t 1,1 , where The reason why it does not spread to t 0,1 and t 0 ,- 1 here is that (0,1)∈T 0 =((2,3),(0,2),(0,1),(2, 1), (1,3), (3,3), (3,1)) and pixel t 0, -1 does not exist.
第11步:反复执行第9步到第10步,直至处理完不属于T0的所有坐标位置;Step 11: Repeat steps 9 to 10 until all coordinate positions that do not belong to T 0 are processed;
执行以上的步骤,最终生成一副嵌密的马赛克图像。Perform the above steps to finally generate an embedded mosaic image.
与之对应的密图恢复方法的具体实施步骤如下:The specific implementation steps of the corresponding secret image recovery method are as follows:
第1步:输入分辨率为m2×n2的P0阶灰度马赛克图像l个分辨率为(2r-1)×(2r-1)两两不等的P2阶灰度图像秘密图像分辨率m1×n1,秘密图像的灰度阶P1,初始化二值比特序列B=Φ和B′=Φ,输入密钥k0,k1,k2并设定阈值参数T,T>0;Step 1: Input a P 0 -level grayscale mosaic image with a resolution of m 2 ×n 2 l P 2 -level grayscale images with a resolution of (2r-1)×(2r-1) in pairs Secret image resolution m 1 ×n 1 , secret image gray scale P 1 , initialize binary bit sequence B=Φ and B′=Φ, input key k 0 , k 1 , k 2 and set threshold parameter T ,T>0;
例如:输入分辨率为m2×n2=260×260的8阶灰度马赛克图像M=(mi,j)260×260,32个分辨率为(2r-1)×(2r-1)=(2×33-1)×(2×33-1)=65×65两两不等的8阶灰度图像如图5a(0)~图5a(31)所示,这里取r=33秘密图像分辨率为2×2,秘密图像灰度阶为8阶,初始化二值比特序列B=Φ和B′=Φ,设定密钥k0=9,k1=5,k2=5并设定阈值参数T=70;For example: the input resolution is m 2 ×n 2 =260×260 8-level gray mosaic image M=(m i,j ) 260×260 , 32 resolutions are (2r-1)×(2r-1) =(2×33-1)×(2×33-1)=65×65 8-level grayscale images of different pairs As shown in Fig. 5a(0) ~ Fig. 5a(31), here r = 33, the resolution of the secret image is 2×2, the gray scale of the secret image is 8, and the binary bit sequence B=Φ and B′= Φ, set key k 0 =9, k 1 =5, k 2 =5 and set threshold parameter T=70;
第2步:由密钥k0产生随机整数indk∈[0,n-1],k=0,1,…,l-1且n为大于0的整数,按式(1)将indk映射为[0,2π)之间的随机转角αind,然后将所有的Hk转换为具有随机转角αind且半径为r的圆形图像 Step 2: Generate a random integer ind k ∈ [0,n-1] from the key k 0 , k=0,1,...,l-1 and n is an integer greater than 0, according to formula (1) will ind k is mapped to a random rotation angle α ind between [0,2π), and then all H k are transformed into a circular image with a random rotation angle α ind and radius r
例如:取密钥k0=9,产生32个[0,7]范围内的随机整数(5,2,6,6,6,6,4,7,7,1,3,2,4,4,4,0,3,0,7,7,2,5,0,3,3,5,2,6,5,6,1,5),假设取n=8,k=0,则由随机整数(5,2,6,6,6,6,4,7,7,1,3,2,4,4,4,0,3,0,7,7,2,5,0,3,3,5,2,6,5,6,1,5)可知:ind0=5,按式(1)将ind0映射为[0,2π)之间的随机转角α5=5·2π/8=3.9269909,然后将即图5a(0),转换为具有随机转角α5=3.9269909且半径为r=33的圆形图像即图5b(0);For example: take the key k 0 =9, generate 32 random integers in the range of [0,7] (5,2,6,6,6,6,4,7,7,1,3,2,4, 4,4,0,3,0,7,7,2,5,0,3,3,5,2,6,5,6,1,5), assuming n=8, k=0, then consisting of random integers (5,2,6,6,6,6,4,7,7,1,3,2,4,4,4,0,3,0,7,7,2,5,0, 3, 3, 5, 2, 6, 5, 6, 1, 5) we know: ind 0 = 5, according to formula (1), ind 0 is mapped to a random rotation angle between [0, 2π) α 5 = 5· 2π/8=3.9269909, then the That is, Figure 5a(0), transformed into a circular image with random rotation angle α 5 =3.9269909 and radius r=33 That is, Figure 5b(0);
第3步:对所有H′k按式(12)旋转校正为H″k且满足H″k质心在H″k圆心约定的方向上;Step 3: For all H′ k , the rotation is corrected to H″ k according to formula (12) and the center of mass of H″ k is in the direction agreed by the center of H″ k circle;
H″k=Rotmass(H′k) (12);H″ k = Rot mass (H′ k ) (12);
式(12)中,函数Rotmass()为质心旋转函数,具体的执行功能如下:In formula (12), the function Rot mass () is the centroid rotation function, and the specific execution function is as follows:
第3.1步:记按式(14)计算H′k的质心(xmass,ymass):Step 3.1: Remember Calculate the centroid (x mass , y mass ) of H′ k according to formula (14):
第3.2步:按式(15)计算H′k质心相对于H′k圆心的几何倾角 Step 3.2: Calculate the geometric inclination angle of H′ k centroid relative to H′ k circle center according to formula (15)
第3.3步:按式(16)将H′k逆时针旋转得到H″k,然后将H″k输出;Step 3.3: Rotate H′ k counterclockwise according to formula (16) Get H″ k , and then output H″ k ;
例如:取k=0,可将即图5b(0)按式(12)旋转校正为图5c(0)且满足图5c(0)质心在图5c(0)圆心右侧的水平方向上,具体实现步骤如下:For example: take k=0, can be That is, Fig. 5b(0) is rotated and corrected to Fig. 5c(0) according to formula (12) and satisfies that the centroid of Fig. 5c(0) is in the horizontal direction on the right side of the center of Fig. 5c(0). The specific implementation steps are as follows:
首先按式(14)计算H′k=0的质心(xmass.ymass),即图5b(0)的质心(xmass.ymass),First calculate the centroid (x mass .y mass ) of H′ k=0 according to formula (14), that is, the centroid (x mass .y mass ) of Figure 5b(0),
接着按式(15)计算图5b(0)质心相对于图5b(0)圆心的几何倾角最后按式(16)将图5b(0)逆时针旋转得到图5c(0),然后将图5c(0)输出; Then calculate the geometric inclination angle of the center of mass in Figure 5b(0) relative to the center of circle in Figure 5b(0) according to formula (15) Finally, rotate Figure 5b(0) counterclockwise according to formula (16) Get Figure 5c(0), and then output Figure 5c(0);
第4步:初始化分辨率为m1×n1的P1阶灰度秘密图像由密钥k1产生l0个两两不等的m0×n0范围内的随机整数坐标序列T0,其中由密钥k2产生m0n0个[0,l-1]范围内的随机整数矩阵其中m0=m2/(2r-1),n0=n2/(2r-1);Step 4: Initialize the P 1 -level grayscale secret image with resolution m 1 ×n 1 Generate l 0 pairs of random integer coordinate sequences T 0 in the range of m 0 ×n 0 from the key k 1 , where Generate m 0 n 0 random integer matrices in the range [0,l-1] from the key k 2 Where m 0 =m 2 /(2r-1), n 0 =n 2 /(2r-1);
例如:初始化分辨率为m1×n1=2×2的P1=8阶灰度秘密图像S=(si,j=0)2×2,由密钥k1=5产生l0=7个两两不等的m0×n0=4×4范围内的随机整数坐标序列T0=((2,3),(0,2),(0,1),(2,1),(1,3),(3,3),(3,1)),由密钥k2=5产生m0n0=16个[0,31]范围内的随机整数矩阵R=((23,5,2,19),(15,30,14,26),(14,13,22,24),(8,11,24,26)),其中(23,5,2,19),…,(8,11,24,26)分别对应为R矩阵的第0行到第3行;For example: P 1 =8-level grayscale secret image S=(s i,j =0) 2×2 with initial resolution of m1×n1=2×2, l 0 =7 generated by key k 1 =5 Random integer coordinate sequence T 0 in the range of m 0 ×n 0 = 4 × 4 unequal in pairs T 0 =((2,3),(0,2),(0,1),(2,1),( 1,3), (3,3), (3,1)), m 0 n 0 = 16 random integer matrix R=((23, 5,2,19),(15,30,14,26),(14,13,22,24),(8,11,24,26)), where (23,5,2,19),… , (8,11,24,26) correspond to the 0th row to the 3rd row of the R matrix;
第5步:依次读取T0中整数坐标(x,y),按式(17)计算(x,y)在M中的坐标位置(u,v),然后以(u,v)为起点,从M中截取(2r-1)×(2r-1)的马赛克图像Mt;Step 5: Read the integer coordinates (x, y) in T 0 sequentially, calculate the coordinate position (u, v) of (x, y) in M according to formula (17), and then use (u, v) as the starting point , intercept (2r-1)×(2r-1) mosaic image M t from M;
例如:首先读取T0=((2,3),(0,2),(0,1),(2,1),(1,3),(3,3),(3,1))中的整数坐标(2,3),按式(17)计算(2,3)在M中的坐标位置(u=130,v=195),其中u=x·(2r-1)=2×(2×33-1)=130,v=y·(2r-1)=3×(2×33-1)=195,然后以(130,195)为起点,从M中截取(2r-1)×(2r-1)=(2×33-1)×(2×33-1)=65×65的马赛克图像M0;For example: first read T 0 = ((2,3),(0,2),(0,1),(2,1),(1,3),(3,3),(3,1) ), calculate the coordinate position (u=130, v=195) of (2,3) in M according to formula (17), where u=x·(2r-1)=2 ×(2×33-1)=130, v=y·(2r-1)=3×(2×33-1)=195, then take (130,195) as the starting point, intercept (2r-1) from M ×(2r-1)=(2×33-1)×(2×33-1)=65×65 mosaic image M 0 ;
第6步:将Mt代入式(12),按式(12)几何校正为M′t且满足Mt质心在M′t圆心约定的方向上,然后按式(18)从所有的H″k,k=0,1,…,l-1中寻找距离最接近的对应的编号按式(19)将转换为秘密值XCur;Step 6: Substituting M t into formula (12), geometrically correcting to M′ t according to formula (12) and satisfying that the center of mass of M t is in the direction agreed by the center of M′ t , and then according to formula (18) from all H″ k ,k=0,1,…,l-1 to find the closest distance corresponding number According to formula (19) will Convert to secret value X Cur ;
式(18)中,为M′和H″k的2次距离;In formula (18), is the 2nd distance between M′ and H″ k ;
例如将M0代入式(12),按式(12)几何校正为M′0,然后按式(18)从所有的H″k,k=0,1,…,l-1,然后从图5c中寻找距离最接近的对应的编号,假设找到的编号由于第5步中的坐标(2,3)属于T0=((2,3),(0,2),(0,1),(2,1),(1,3),(3,3),(3,1))中的第一个坐标,因此r2,3对应随机整数矩阵R=((23,5,2,19),(15,30,14,26),(14,13,22,24),(8,11,24,26))中的第一个值23,所以r2,3=23;将r2,3=23,l=32,x=2,y=3代入式(19)中,可计算得到 For example, M 0 is substituted into formula (12), geometrically corrected to M′ 0 according to formula (12), and then according to formula (18) from all H″ k , k=0,1,...,l-1, and then from Fig. Find the closest distance in 5c The corresponding number, assuming the found number Since the coordinates (2,3) in step 5 belong to T 0 =((2,3),(0,2),(0,1),(2,1),(1,3),(3, 3), the first coordinate in (3,1)), so r 2,3 corresponds to the random integer matrix R=((23,5,2,19),(15,30,14,26),(14 ,13,22,24),(8,11,24,26)), the first value 23 in (8,11,24,26)), so r 2 , 3 =23; set r 2,3 =23, l=32, x=2, y=3 are substituted into formula (19), and can be calculated as
第7步:按权利要求1给出的第6步同样的方法按式(8)计算[0,n-1]范围内的整数ind,将Mt按式(20)逆时针旋转-αind从而得到M″t,其中αind=ind·2π/nThe 7th step: by the 6th step same method that claim 1 provides, calculate the integer ind in [0,n-1] scope by formula (8), M t is rotated counterclockwise by formula (20)-α ind Thus M″ t is obtained, where α ind =ind·2π/n
M″t=Rot(Mk,r,-αind) (20);M″ t = Rot(M k ,r,-α ind ) (20);
例如:按式(8)计算[0,7]范围内的整数ind=(XCur+x·y+y·rx,y+x·rx,y)mod n=(1+2×3+3×23+2×23)mod8=7,将M0按式(20)逆时针旋转-αind,其中αind=7×π×0.25=5.49779,从而得到M″0;For example: calculate the integer ind=(X Cur +x·y+y·r x,y +x·r x,y ) mod n=(1+2×3) in the range of [0,7] according to formula (8) +3×23+2×23) mod8=7, rotate M 0 counterclockwise according to formula (20)-α ind , where α ind =7×π×0.25=5.49779, thereby obtaining M″ 0 ;
第8步:将XCur转换为log2l位2进制数添加至2值序列B中,然后按式(21)计算M″t和的像素点的差值d,若d<T,则秘密信息XCur没有被破坏,则将log2l位1构成的2进制数添加至B′中,反之则将log2l位0构成的2进制数添加至B′中;Step 8: Convert X Cur to log 2 l-digit binary numbers and add them to the binary sequence B, then calculate M″ t and The pixel point difference d, if d<T, the secret information X Cur is not destroyed, then the binary number composed of log 2 l bits 1 is added to B′, otherwise, log 2 l bits 0 are formed The binary number of is added to B';
例如:将XCur=1以二进制的形式(10000)2顺序存放在B中,按式(21)计算M″0和图5b(25)的像素点的差值d=29<T=70,证明该坐标点的秘密信息XCur没有被破坏,将5位1构成的2进制数(11111)2添加至B′中,反之则将5位0构成的2进制数(00000)2添加至B′中;For example: X Cur =1 is stored in B in binary form (10000) 2 order, calculate M " 0 and the difference d=29<T=70 of the pixel point of Fig. 5b (25) by formula (21), To prove that the secret information X Cur of this coordinate point has not been destroyed, add the binary number (11111) 2 composed of 5 digits 1 to B′, otherwise, add the binary number (00000) 2 composed of 5 digits 0 into B';
第9步:反复执行第5步~第8步,直至T0中所有的整数坐标(x,y)被读取完毕;Step 9: Repeat steps 5 to 8 until all integer coordinates (x, y) in T 0 are read;
第10步:将B和B′以P1个2进制位为1组转换为10进制数,然后将其组织为m1×n1的P1阶灰度秘密图像和认证图像并输出。Step 10: Convert B and B′ to a decimal number with P 1 binary digits as a group, and then organize it into a P 1 -level grayscale secret image of m 1 ×n 1 and the authentication image and output.
例如:假设最终得到的B=(10000000010000001100000000100000000)2,以8个2进制位为1组转换为10进制数(1,2,3,4),B′=(11111111111111111111111111111111111)2,以8个2进制位为1组转换为10进制数(255,255,255,255),然后将其组织为2×2的8阶灰度秘密图像S=(si,j)2×2和认证图像A=(ai,j)2×2并输出。For example: Assuming that the final B=(10000000010000001100000000100000000) 2 is converted into a decimal number (1,2,3,4) with 8 binary digits as a group, B'=(111111111111111111111111111111111) 2 , with 8 digits One group of binary digits is converted into a decimal number (255,255,255,255), and then organized into a 2×2 8-level grayscale secret image S=( si,j ) 2×2 and an authentication image A=(a i, j ) 2×2 and output.
图3是实施例,分辨率为64×64的8位灰度秘密图像;Fig. 3 is an embodiment, and the resolution is an 8-bit grayscale secret image of 64×64;
图4是实施例,分辨率为128×128的8位灰度掩体图像;Fig. 4 is an embodiment, and the resolution is an 8-bit grayscale mask image of 128 * 128;
图5a是实施例,32个分辨率为65×65的8位灰度图像;Fig. 5 a is an embodiment, 32 resolutions are 8-bit grayscale images of 65 * 65;
图5b是实施例,图5a经圆形初始化和添加随机转角后对应的32个圆形图像;Fig. 5b is an embodiment, and Fig. 5a corresponds to 32 circular images after circular initialization and adding random corners;
图5c是实施例,图5b经计算质心后,将质心坐标点旋转到X正半轴上对应的32个圆形图像;Fig. 5c is an embodiment. After calculating the centroid in Fig. 5b, rotate the coordinate point of the centroid to the corresponding 32 circular images on the positive semi-axis of X;
图6a是实施例,利用图5b圆形图像,形成的分辨率为8320×8320的8位含密灰度马赛克图像的局部图像(图6b分辨率较高,这里通过图6a给出图6b局部细节);Fig. 6 a is an embodiment, utilizes Fig. 5 b circular image, the partial image (Fig. 6 b resolution is higher, provides Fig. 6 b part by Fig. 6 a here by Fig. detail);
图6b是实施例,利用图5b圆形图像,形成的分辨率为8320×8320的8位含密灰度马赛克图像;Figure 6b is an embodiment, using the circular image in Figure 5b to form an 8-bit dense grayscale mosaic image with a resolution of 8320×8320;
图7是实施例,由图6b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 7 is an embodiment, the secret image extracted by Fig. 6b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图8是实施例,由图6b按图2流程提出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证;Fig. 8 is an embodiment, by Fig. 6 b according to Fig. 2 flow process, proposes the authentication map that secret image is generated, is the blank image of all white, represents that all pixels of secret image all pass authentication;
图9a是实施例,对图6b进行质量因子为50的JEPG压缩,形成的JPEG灰度图像的局部图像(图9b分辨率较高,这里通过图9a给出图9b局部细节);Fig. 9a is an embodiment, carrying out the JPEG compression that quality factor is 50 to Fig. 6b, the partial image of the JPEG grayscale image that forms (Fig. 9b resolution is higher, provides Fig. 9b local details by Fig. 9a here);
图9b是实施例,对图6b进行质量因子为50的JEPG压缩,形成的JPEG灰度图像;Figure 9b is an embodiment, the JPEG grayscale image formed by performing JPEG compression with a quality factor of 50 on Figure 6b;
图10是实施例,由图9b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 10 is an embodiment, the secret image extracted by Fig. 9b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图11是实施例,由图9b按图2流程提取出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证;Fig. 11 is an embodiment, the authentication map generated by extracting the secret image according to the process of Fig. 2 from Fig. 9b is a completely white blank image, indicating that all pixels of the secret image have passed the authentication;
图12a是实施例,对图6b进行质量因子为80的JPEG压缩,形成的JPEG灰度图像的局部图像(图12b分辨率较高,这里通过图12a给出图12b局部细节);Fig. 12a is an embodiment, carrying out the JPEG compression that quality factor is 80 to Fig. 6b, the partial image of the JPEG gray scale image that forms (Fig. 12b resolution is higher, provides Fig. 12b local details by Fig. 12a here);
图12b是实施例,对图6b进行质量因子为80的JPEG压缩,形成的JPEG灰度图像;Figure 12b is an embodiment, the JPEG grayscale image formed by performing JPEG compression with a quality factor of 80 on Figure 6b;
图13是实施例,由图12b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 13 is an embodiment, the secret image extracted by Fig. 12b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图14是实施例,由图12b按图2流程提取出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证;Fig. 14 is an embodiment, the authentication map generated by extracting the secret image according to the process of Fig. 2 from Fig. 12b is a completely white blank image, indicating that all pixels of the secret image have passed the authentication;
图15a是实施例,对图6b进行强度为8%的噪声攻击后形成的含噪声马赛克图像的局部图像(图15b分辨率较高,这里通过图15a给出图15b局部细节);Fig. 15a is an embodiment, the partial image of the noise-containing mosaic image formed after the strength of 8% noise attack is carried out on Fig. 6b (Fig. 15b has a higher resolution, here the local details of Fig. 15b are provided by Fig. 15a);
图15b是实施例,对图6b进行强度为8%的噪声攻击后形成的含噪声马赛克图像;Fig. 15b is an embodiment, a noise-containing mosaic image formed after performing a noise attack with an intensity of 8% on Fig. 6b;
图16是实施例,由图15b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 16 is an embodiment, the secret image extracted by Fig. 15b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图17是实施例,由图15b按图2流程提取出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证;Fig. 17 is an embodiment. The authentication map generated by extracting the secret image according to the process of Fig. 2 from Fig. 15b is a completely white blank image, indicating that all pixels of the secret image have passed the authentication;
图18a是实施例,对图6b进行强度为20%的噪声攻击后形成的含噪声马赛克图像的局部图像(图18b分辨率较高,这里通过图18a给出图18b局部细节);Fig. 18a is an embodiment, a partial image of a mosaic image containing noise formed after a 20% noise attack is carried out on Fig. 6b (Fig. 18b has a higher resolution, here the local details of Fig. 18b are provided by Fig. 18a);
图18b是实施例,对图6b进行强度为20%的噪声攻击后形成的含噪声马赛克图像;Fig. 18b is an embodiment, a noise-containing mosaic image formed after performing a noise attack with a strength of 20% on Fig. 6b;
图19是实施例,由图18b按图2流程提取出的秘密图像,相对于图3的PSNR=∞;Fig. 19 is an embodiment, the secret image extracted by Fig. 18b according to the process of Fig. 2, relative to Fig. 3 PSNR = ∞;
图20是实施例,由图18b按图2流程提取出秘密图像生成的认证图,为全白的空白图像,表示秘密图像的所有像素均通过认证。Fig. 20 is an embodiment. The authentication map generated by extracting the secret image according to the process of Fig. 2 from Fig. 18b is a completely white blank image, indicating that all pixels of the secret image have passed the authentication.
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