{"id":1335,"date":"2020-04-11T12:57:09","date_gmt":"2020-04-11T07:27:09","guid":{"rendered":"https:\/\/binaryterms.com\/?p=1335"},"modified":"2022-12-20T10:51:52","modified_gmt":"2022-12-20T05:21:52","slug":"substitution-technique-in-cryptography","status":"publish","type":"post","link":"https:\/\/binaryterms.com\/substitution-technique-in-cryptography.html","title":{"rendered":"Substitution Technique in Cryptography"},"content":{"rendered":"<p><strong>Substitution technique<\/strong> is a classical encryption technique where the characters present in the <strong>original message<\/strong> are <strong>replaced<\/strong> by the <strong>other characters or numbers or by symbols.<\/strong> If the plain text (original message) is considered as the string of bits, then the substitution technique would replace bit pattern of plain text with the bit pattern of cipher text.<\/p>\n<p>We will discuss some of the substitution techniques which will help us to understand the procedure of converting plain text o cipher text.\u00a0 In this section, we will study the following substitution techniques:<\/p>\n<h2>Substitution Technique:<\/h2>\n<ol>\n<li><a href=\"#CaesarCipher\">Caesar Cipher<\/a><\/li>\n<li><a href=\"#MonoalphabeticCipher\">Monoalphabetic Cipher<\/a><\/li>\n<li><a href=\"#PlayfairCipher\">Playfair Cipher<\/a><\/li>\n<li><a href=\"#HillCipher\">Hill Cipher<\/a><\/li>\n<li><a href=\"#PolyalphabeticCipher\">Polyalphabetic Cipher<\/a><\/li>\n<li><a href=\"#One-TimePad\">One-Time Pad<\/a><\/li>\n<\/ol>\n<p><a name=\"CaesarCipher\"><\/a><\/p>\n<h3>Caesar Cipher<\/h3>\n<p>This the simplest substitution cipher by Julius Caesar. In this substitution technique, to encrypt the plain text, each alphabet of the plain text is replaced by the alphabet three places further it. And to decrypt the cipher text each alphabet of cipher text is replaced by the alphabet three places before it.<\/p>\n<p>Let us take a simple example:<\/p>\n<p><strong>Plain Text:<\/strong> meet me tomorrow<\/p>\n<p><strong>Cipher Text:<\/strong> phhw ph wrpruurz<\/p>\n<p>Look at the example above, we have replaced, \u2018m\u2019 with \u2018p\u2019 which occur three places after, \u2018m\u2019. Similarly, \u2018e\u2019 is replaced with \u2018h\u2019 which occurs in three places after \u2018e\u2019.<\/p>\n<p><strong>Note:<\/strong> If we have to replace the letter \u2018z\u2019 then the next three alphabets counted after \u2018z\u2019 will be \u2018a\u2019 \u2018b\u2019 \u2018c\u2019. So, while counting further three alphabets if &#8216;z&#8217; occurs it circularly follows &#8216;a&#8217;.<\/p>\n<p>There are also some drawbacks of this simple substitution technique. If the hacker knows that the Caesar cipher is used then to perform brute force cryptanalysis, he has only to try 25 possible keys to decrypt the plain text.<br \/>\nThe hacker is also aware of the encryption and decryption algorithm.<br \/>\n<a name=\"MonoalphabeticCipher\"><\/a><\/p>\n<h3>Monoalphabetic Cipher<\/h3>\n<p>Monoalphabetic cipher is a substitution cipher, where the cipher alphabet for each plain text alphabet is fixed, for the entire encryption.<\/p>\n<p>In simple words, if the alphabet \u2018p\u2019 in the plain text is replaced by the cipher alphabet \u2018d\u2019. Then in the entire plain text wherever alphabet \u2018p\u2019 is used, it will be replaced by the alphabet \u2018d\u2019 to form the ciphertext.<br \/>\n<a name=\"PlayfairCipher\"><\/a><\/p>\n<h3>Playfair Cipher<\/h3>\n<p>Playfair cipher is a substitution cipher which involves a 5X5 matrix. Let us discuss the technique of this Playfair cipher with the help of an example:<\/p>\n<p><strong>Plain Text:<\/strong> meet me tomorrow<\/p>\n<p><strong>Key:<\/strong> KEYWORD<\/p>\n<p>Now, we have to convert this plain text to ciphertext using the given key. We will discuss the further process in steps.<\/p>\n<p><strong>Step 1:<\/strong> Create a 5X5 matrix and place the key in that matrix row-wise from left to right. Then put the remaining alphabet in the blank space.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3741\" src=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/Play-fair-Cypher-matrix-1.jpg\" alt=\"Play fair Cypher matrix 1\" width=\"464\" height=\"356\" srcset=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/Play-fair-Cypher-matrix-1.jpg 464w, https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/Play-fair-Cypher-matrix-1-300x230.jpg 300w\" sizes=\"auto, (max-width: 464px) 100vw, 464px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><strong>Note:<\/strong> If a key has duplicate alphabets, then fill those alphabets only once in the matrix, and I &amp; J should be kept together in the matrix even though they occur in the given key.<\/p>\n<p><strong>Step 2:<\/strong> Now, you have to break the plain text into a pair of alphabets.<\/p>\n<p><strong>Plain Text:<\/strong> meet me tomorrow<\/p>\n<p><strong>Pair:<\/strong> <u>me<\/u> <u>et<\/u> <u>me<\/u> <u>to<\/u> <u>mo<\/u> <u>rx<\/u> <u>ro<\/u> <u>wz<\/u><\/p>\n<p><strong>Note<\/strong><\/p>\n<ul>\n<li>Pair of alphabets must not contain the same letter. In case, pair has the same letter then break it and add \u2018x\u2019 to the previous letter. Like in our example letter \u2018rr\u2019 occurs in pair so, we have broken that pair and added \u2018x\u2019 to the first \u2018r\u2019.<\/li>\n<li>In case while making pair, the last pair has only one alphabet left then we add \u2018z\u2019 to that alphabet to form a pair as in our above example, we have added \u2018z\u2019 to \u2018w\u2019 because \u2018w\u2019 was left alone at last.<\/li>\n<li>If a pair has \u2018xx\u2019 then we break it and add \u2018z\u2019 to the first \u2018x\u2019, i.e. \u2018xz\u2019 and \u2018x_\u2019.<\/li>\n<\/ul>\n<p><strong>Step 3:<\/strong> In this step, we will convert plain text into ciphertext. For that, take the first pair of plain text and check for cipher alphabets for the corresponding in the matrix. To find cipher alphabets follow the rules below.<\/p>\n<p><strong>Note<\/strong><\/p>\n<ul>\n<li>If both the alphabets of the pair occur in the <strong>same row<\/strong> replace them with the alphabet to their <strong>immediate right<\/strong>. If an alphabet of the pair occurs at extreme right then replace it with the first element of that row, i.e. the last element of the row in the matrix circularly follows the first element of the same row.<\/li>\n<li>If the alphabets in the pair occur in the <strong>same column<\/strong>, then replace them with the alphabet<strong> immediate below<\/strong> them. Here also, the last element of the column circularly follows the first element of the same column.<\/li>\n<li>If the alphabets in the pair are <strong>neither in the same column and nor in the same row,<\/strong> then the alphabet is replaced by the element in its own row and the corresponding column of the other alphabet of the pair.<\/li>\n<\/ul>\n<p><strong>Pair:<\/strong> <u>me<\/u> <u>et<\/u> <u>me<\/u> <u>to<\/u> <u>mo<\/u> <u>rx<\/u> <u>ro<\/u> <u>wz<\/u><\/p>\n<p><strong>Cipher Text:<\/strong> kn ku kn kz ks ta kc yo<\/p>\n<p>So, this is how we can convert a plain text to ciphertext using Playfair cipher. When compared with monoalphabetic cipher Playfair cipher is much more advanced. But still, it is easy to break.<\/p>\n<p><a name=\"HillCipher\"><\/a><\/p>\n<h3>Hill Cipher<\/h3>\n<p>Hill cipher is a polyalphabetic cipher introduced by Lester Hill in 1929. Let us discuss the technique of hill cipher.<\/p>\n<p><strong>Plain text:<\/strong> Binary<\/p>\n<p><strong>Key:<\/strong> HILL<\/p>\n<p><strong>Choose the key<\/strong> in such a way that it always forms a <strong>square matrix<\/strong>. With HILL as the key, we can form a 2&#215;2 matrix.<\/p>\n<p>Now, of plain text, you have to form a column vector of length similar to the key matrix. In our case, the key matrix is 2&#215;2 then the column vectors of plain text would be 2&#215;1.<\/p>\n<p>The general equation to find cipher text using hill cipher is as follow:<\/p>\n<p style=\"text-align: center\"><strong><em>C = KP mod 26<\/em><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1338\" src=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-1.jpg\" alt=\"hill cipher 1\" width=\"344\" height=\"62\" srcset=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-1.jpg 344w, https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-1-300x54.jpg 300w\" sizes=\"auto, (max-width: 344px) 100vw, 344px\" \/><\/p>\n<p>For our example, our key matrix would be:<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1339\" src=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-2.jpg\" alt=\"hill cipher 2\" width=\"90\" height=\"65\" \/><\/p>\n<p>And our plain text matrices of 2&#215;1 will be as follow:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1340\" src=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-3.jpg\" alt=\"hill cipher 3\" width=\"130\" height=\"55\" \/><\/p>\n<p>Now, we have to convert the key matrix and plain text matrices into numeric matrices. For that number the alphabets such as A=0, B=1, C=2, \u2026\u2026\u2026\u2026, Z=25. So, considering the alphabet numbering:<\/p>\n<p>Key matrix will be:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1341\" src=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-4.jpg\" alt=\"hill cipher 4\" width=\"177\" height=\"62\" \/><\/p>\n<p>Plain text matrices would be:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1342\" src=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-5.jpg\" alt=\"hill cipher 5\" width=\"142\" height=\"58\" \/><\/p>\n<p>In the first calculation, we would get two cipher alphabets for plain text alphabet \u2018B\u2019 &amp; \u2018I\u2019.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1343\" src=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-6.jpg\" alt=\"hill cipher 6\" width=\"333\" height=\"322\" srcset=\"https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-6.jpg 333w, https:\/\/binaryterms.com\/wp-content\/uploads\/2020\/04\/hill-cipher-6-300x290.jpg 300w\" sizes=\"auto, (max-width: 333px) 100vw, 333px\" \/><\/p>\n<p>So, the cipher alphabet for plain text alphabet \u2018B\u2019 &amp; \u2018I\u2019 is \u2018T\u2019 &amp; \u2018V\u2019. Similarly, we have to calculate ciphertext for remaining plain text. And then accumulate them to form the ciphertext.<\/p>\n<p>The calculated <strong>ciphertext<\/strong> for \u2018<strong>Binary<\/strong>\u2019 using hill cipher is \u2018<strong>TVNNZJ<\/strong>\u2019.<br \/>\n<a name=\"PolyalphabeticCipher\"><\/a><\/p>\n<h3>Polyalphabetic Cipher<\/h3>\n<p>Polyalphabetic cipher is far more secure than a monoalphabetic cipher. As monoalphabetic cipher maps a plain text symbol or alphabet to a ciphertext symbol and uses the same ciphertext symbol wherever that plain text occurs in the message.<br \/>\nBut polyalphabetic cipher, each time replaces the plain text with the different ciphertext.<br \/>\n<a name=\"One-TimePad\"><\/a><\/p>\n<h3>One-Time Pad<\/h3>\n<p>The one-time pad cipher suggests that the <strong>key length<\/strong> should be <strong>as long as the plain text<\/strong> to prevent the repetition of key. Along with that, the <strong>key<\/strong> should be <strong>used<\/strong> only <strong>once<\/strong> to encrypt and decrypt the single message after that the key should be discarded.<\/p>\n<p>Onetime pad suggests a new key for each new message and of the same length as a new message. Now, let us see the one-time pad technique to convert plain text into ciphertext. Assume our plain text and key be:<\/p>\n<p><strong>Plain text:<\/strong> Binary<\/p>\n<p><strong>Key:<\/strong> Cipher<\/p>\n<p>Now again convert the plain text and key into the numeric form. For that number the alphabets such as A=0, B=1, C=2, \u2026\u2026\u2026\u2026, Z=25. So, our plain text and key in numeric form would be:<\/p>\n<p><strong>Plain text:<\/strong> 1 8 13 0 17 24<\/p>\n<p><strong>Key:<\/strong> 2 8 15 7 4 17<\/p>\n<p>Now, you have to add the number of the plain text alphabet, to the number of its corresponding key alphabet. That means, for this example, we will add:<\/p>\n<p style=\"text-align: center\"><strong>B+C = 1+2 = 3<\/strong><\/p>\n<p style=\"text-align: center\"><strong>I+I = 8+8 = 16<\/strong><\/p>\n<p style=\"text-align: center\"><strong>N+P = 13+15 = 28<\/strong><\/p>\n<p style=\"text-align: center\"><strong>A+H = 0+7 = 7<\/strong><\/p>\n<p style=\"text-align: center\"><strong>R+E = 17+4 = 21<\/strong><\/p>\n<p style=\"text-align: center\"><strong>Y+R = 24+17 = 41<\/strong><\/p>\n<p>The resultant ciphertext numbers we get are (3, 16, 28, 7, 21, 41)<\/p>\n<p>If the addition of any plain text number and the key number is &gt;26, then subtract only that particular number from 26. We have the addition of two pair of plain text number and a key number, greater than 26, i.e. N+P=28 &amp; Y+R=41.<\/p>\n<p>Subtract them by 26.<\/p>\n<p>N+P = 28 \u2013 26 = 2<\/p>\n<p>Y+R = 41 \u2013 26 = 15<\/p>\n<p>So, the final <strong>ciphertext numbers are (3, 16, 2, 7, 21, 15)<\/strong>. Now convert this number to alphabets assuming A to be numbered 0 and B to be 1\u2026..Z to 25.<\/p>\n<p><strong>Ciphertext:<\/strong> dqchvp.<\/p>\n<p>In this way, we can convert plain text to cipher text using a one-time pad.<\/p>\n<p>So, this is all about the substitution cipher techniques. It has a monoalphabetic cipher and polyalphabetic cipher technique. Substitution technique is also called classical substitution technique.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Substitution technique is a classical encryption technique where the characters present in the original message are replaced by the other characters or numbers or by symbols. If the plain text (original message) is considered as the string of bits, then the substitution technique would replace bit pattern of plain text with the bit pattern of [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[9],"tags":[],"class_list":{"0":"post-1335","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-security","7":"entry"},"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>What is Substitution Technique in Cryptography? - Binary Terms<\/title>\n<meta name=\"description\" content=\"Substitution technique is a classical encryption technique where the characters present in the original message are replaced by the other characters.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/binaryterms.com\/substitution-technique-in-cryptography.html\" \/>\n<meta property=\"og:locale\" content=\"en_GB\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"What is Substitution Technique in Cryptography? 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