{"id":215,"date":"2018-07-12T07:33:51","date_gmt":"2018-07-12T07:33:51","guid":{"rendered":"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/?post_type=chapter&#038;p=215"},"modified":"2018-07-31T09:07:44","modified_gmt":"2018-07-31T09:07:44","slug":"elliptic-curve","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/chapter\/elliptic-curve\/","title":{"rendered":"Elliptic Curve"},"content":{"raw":"<strong>Elliptic Curve<\/strong>\r\n\r\n<img class=\"size-full wp-image-216 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-125.png\" alt=\"\" width=\"480\" height=\"360\" \/>\r\n\r\nWhich steps are involved?\r\n<ul>\r\n \t<li>1) choose an elliptic curve Ep(a, b).<\/li>\r\n \t<li>2) Chooses another prime q the private key d.<\/li>\r\n \t<li>3) Choose e1(\u2026, \u2026), a point on the curve.<\/li>\r\n \t<li>4) calculates e2(\u2026, \u2026) = d \u00d7 e1(\u2026, \u2026).<\/li>\r\n \t<li>5) Public key is (a, b, p, q, e1, e2); private key is d.<\/li>\r\n<\/ul>\r\nECDSA signing and verifying:\r\n\r\n&nbsp;\r\n\r\n<img class=\"size-full wp-image-217 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-126.png\" alt=\"\" width=\"898\" height=\"468\" \/>\r\n\r\n<strong>Elliptic curves in Cryptography<\/strong>\r\n<ul>\r\n \t<li style=\"text-align: justify\">Elliptic Curve (EC) systems as applied to cryptography were first proposed in 1985 independently by Neal Koblitz and Victor Miller.<\/li>\r\n \t<li style=\"text-align: justify\">The discrete logarithm problem on elliptic curve groups is believed to be more difficult than the corresponding problem in (the multiplicative group of nonzero elements of) the underlying finite field.<\/li>\r\n<\/ul>\r\n<strong>Elliptic Curve on a finite set of Integers<\/strong>\r\n<ul>\r\n \t<li>Consider y2 = x3 + 2x + 3 (<strong>mod 5<\/strong>)<\/li>\r\n \t<li>x = 0 \u00de y2 = 3 \u00de no solution (mod 5)<\/li>\r\n \t<li>x = 1 \u00de y2 = 6 = 1 \u00de y = 1,4 (mod 5)<\/li>\r\n \t<li>x = 2 \u00de y2 = 15 = 0 \u00de y = 0 (mod 5)x = 3 \u00de y2 = 36 = 1 \u00de y = 1,4 (mod 5)x = 4 \u00de y2 = 75 = 0 \u00de y = 0 (mod 5)<\/li>\r\n \t<li>Then points on the elliptic curve are (1,1) (1,4) (2,0) (3,1) (3,4) (4,0) and the point at infinity: \u00a5<\/li>\r\n<\/ul>\r\n<strong>ECDSA involves four elements<\/strong>\r\n<ol>\r\n \t<li>All participants use the same global domain parameters .<\/li>\r\n \t<li style=\"text-align: justify\">First public, private key pair is generated by sender. Signer selects a random or pseudorandom number for the private key. Signer computes another point on the elliptic curve using the random number and the point of origin and that is signer\u2019s public key.<\/li>\r\n \t<li style=\"text-align: justify\">A hash value is generated for the message to be signed. Signature is generated using the private key, the domain parameters and hash value. The signature consists of two integers r and s.<\/li>\r\n \t<li style=\"text-align: justify\">To verify the signature, the verifier inputs signer\u2019s public key, the domain parameters and the integer s. The output is a value v and is compared with r. The signature is verified if v=r.<\/li>\r\n<\/ol>\r\n<strong>Global Domain Parameters<\/strong>\r\n<ul>\r\n \t<li>Prime curves over Zp are used.<\/li>\r\n \t<li>q - a prime number<\/li>\r\n \t<li style=\"text-align: justify\">a,b \u2013 integers that specify the elliptic curve equation defined over Zq with the equation y2=x3+ax+b<\/li>\r\n \t<li style=\"text-align: justify\">G \u2013 a base point represented by G=(xg,yg) on elliptic curve equation<\/li>\r\n \t<li style=\"text-align: justify\">n- order of point G. n is smallest positive integer such that nG=O. These are points on curve.<\/li>\r\n<\/ul>\r\n<strong>Simple elliptic curve<\/strong>\r\n\r\n&nbsp;\r\n\r\n<img class=\"size-full wp-image-218 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-127.png\" alt=\"\" width=\"528\" height=\"252\" \/>\r\n\r\n&nbsp;\r\n\r\nEncryption\r\n<ul>\r\n \t<li>Consider <em>\u2018m\u2019<\/em> has the point <em>\u2018M\u2019<\/em> on the curve <em>\u2018E\u2019.<\/em> Randomly select \u2018k\u2019 from [1 \u2013 (n-1)].<\/li>\r\n \t<li>Two cipher texts will be generated let it be C1 and C2. C1 = k*PC2 = M + k*Q<\/li>\r\n \t<li>C1 and C2 will be sent.<\/li>\r\n<\/ul>\r\n<strong>Decryption<\/strong>\r\n<ul>\r\n \t<li>M = C2 \u2013 d * C1\u00a0 M is the original message that has been sent.<\/li>\r\n<\/ul>\r\n<strong>Proof<\/strong>\r\n<ul>\r\n \t<li>M = C2 \u2013 d * C1<\/li>\r\n \t<li>\u2018M\u2019 can be represented as \u2018C2 \u2013 d * C1\u2019<\/li>\r\n \t<li>C2 \u2013 d * C1 = (M + k * Q) \u2013 d * ( k * P )( C2 = M + k * Q and C1 = k * P )<\/li>\r\n \t<li>= M + k * d * P \u2013 d * k *P ( canceling out k * d * P ) = M ( Original Message )<\/li>\r\n<\/ul>\r\nKey generation\r\n<ul>\r\n \t<li>Signer generates private and public key in following way.<\/li>\r\n<\/ul>\r\n<ol>\r\n \t<li>Select a random integer d, d\u03b5[1,n-1]<\/li>\r\n \t<li>Compute Q=dG. This is a point in Eq(a,b)<\/li>\r\n \t<li>Bob\u2019s public key is Q and private key is d.<\/li>\r\n<\/ol>\r\nDigital signature generation and Authentication\r\n<ul>\r\n \t<li>Digital signature of 320 bytes for message m is generated using following steps:<\/li>\r\n<\/ul>\r\n<ol>\r\n \t<li>Select a random or pseudorandom integer k, k\u03b5[1,n-1]<\/li>\r\n \t<li>Compute point P=(x,y) =kG and r=x mod n. if r=0 then goto step 1<\/li>\r\n \t<li>Compute t=k-1 mod n<\/li>\r\n \t<li>Compute e=H(m) where H is the SHA-1 hash function, which produces a 160 bit hash function.<\/li>\r\n \t<li>Compute s=k-1 (e+dr) mod n. If s=O then goto step 1.<\/li>\r\n \t<li>The signature of message m is the pair (r,s)<\/li>\r\n<\/ol>\r\nThe receiver having message m, digital signature \u2013 verifies the signature using following steps.\r\n<ol>\r\n \t<li>Verify that r and s are integers in the range 1 through n-1<\/li>\r\n \t<li>Using SHA-1 compute the 160 bit hash value e=H(m)<\/li>\r\n \t<li>Compute w=s-1 mod n<\/li>\r\n \t<li>Compute u1=ew and u2=rw<\/li>\r\n \t<li>Compute the point X=(x1,y1)=u1G+u2Q<\/li>\r\n \t<li>If X=O, reject the signature else compute v=x1 mod n<\/li>\r\n \t<li>Accept Sender\u2019s signature if and only if v=r. consider an elliptic curve over the field F23. Ref:https:\/\/www.certicom.com\/index.php\/31-example-of-an-elliptic-curve-group-over-fp<\/li>\r\n<\/ol>\r\n<ul>\r\n \t<li>With a = 1 and b = 0, the elliptic curve equation is y2= x3 + x.<\/li>\r\n \t<li>The point (9,5) satisfies this equation since: y2 mod p = x3 + x mod p<\/li>\r\n<\/ul>\r\n25 mod 23 = 729 + 9 mod 23\r\n\r\n25 mod 23 = 738 mod 232 = 2\r\n\r\nThe 23 points which satisfy this equation are: (0,0) (1,5) (1,18) (9,5) (9,18) (11,10) (11,13) (13,5)(13,18) (15,3) (15,20) (16,8) (16,15) (17,10) (17,13) (18,10) (18,13) (19,1) (19,22) (20,4) (20,19) (21,6) (21,17)\r\n\r\nAll points on graph\r\n\r\n<img class=\"size-full wp-image-219 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-128.png\" alt=\"\" width=\"415\" height=\"427\" \/>\r\n\r\n&nbsp;\r\n<div>\r\n\r\n1.\u00a0 Point Addition, L = J + K\r\n\r\n2.\u00a0 Point Doubling, L = 2J Example:\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">If k = 23;<\/span>\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">then, kP = 23*P<\/span><span style=\"text-align: initial;font-size: 1em\">= 2(2(2(2P) + P) + P) + P<\/span>\r\n\r\n<\/div>\r\n<strong>Point Addition :( Geometrically)<\/strong>\r\n\r\n&nbsp;\r\n\r\n<img class=\"size-full wp-image-221 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-130.png\" alt=\"\" width=\"565\" height=\"345\" \/>\r\n<ul>\r\n \t<li>Consider two distinct points J and K such that J = (xJ, yJ) and K = (xK, yK) Let L = J + K where L = (xL, yL), then<\/li>\r\n \t<li>xL = s2 - xJ \u2013 xK<\/li>\r\n \t<li>yL = -yJ + s (xJ \u2013 xL)<\/li>\r\n<\/ul>\r\ns = (yJ \u2013 yK)\/(xJ \u2013 xK), s is slope of the line through J and K\r\n\r\n<img class=\"size-full wp-image-222 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-131.png\" alt=\"\" width=\"603\" height=\"364\" \/>\r\n\r\n&nbsp;\r\n\r\nConsider a point J such that J = (xJ, yJ), where yJ \u2260 0\u00a0 Let L = 2J where L = (xL, yL), Then\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>\u2022<\/td>\r\n<td>xL = s2 \u2013 2xJ<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\u2022<\/td>\r\n<td>yL = -yJ + s(xJ - xL)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\ns = (3xJ2 + a) \/ (2yJ), s is the tangent at point J and a is one of the parameters chosen with the elliptic curve\r\n\r\n&nbsp;\r\n\r\n<strong>Suggested Reading:<\/strong>\r\n<ol>\r\n \t<li>Cryptography and Network Security Principles and Practice by William Stallings, sixth Edition, PEARSON.<\/li>\r\n \t<li>Security in Computing by Charles Pfleeger &amp; Shari Lawrence Pfleeger, fourth Edition, PEARSON.<\/li>\r\n \t<li>Network Security by Charlie Kaufman, Radia Perlman, Mike Speciner, second Edition, PHI.<\/li>\r\n \t<li>The Complete Reference \u2013 Network Security by Roberta Bragg, Mark Rhodes-Ousley &amp; Keith Strassberg, Tata McGraw Hill<\/li>\r\n \t<li>Network Security Bible by Eric Cole, Ronald Krutz, James Conley, Wiley<\/li>\r\n \t<li>Hacking 6 Exposed by Stuart McClure, Joel Scambray &amp; George Kurtz , Tata McGraw Hill .<\/li>\r\n \t<li><a href=\"http:\/\/www.snort.org\/\">www.snort.org<\/a><\/li>\r\n \t<li><a href=\"https:\/\/nmap.org\/\">https:\/\/nmap.org<\/a><\/li>\r\n<\/ol>\r\n&nbsp;","rendered":"<p><strong>Elliptic Curve<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-216 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-125.png\" alt=\"\" width=\"480\" height=\"360\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-125.png 480w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-125-300x225.png 300w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-125-65x49.png 65w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-125-225x169.png 225w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-125-350x263.png 350w\" sizes=\"auto, (max-width: 480px) 100vw, 480px\" \/><\/p>\n<p>Which steps are involved?<\/p>\n<ul>\n<li>1) choose an elliptic curve Ep(a, b).<\/li>\n<li>2) Chooses another prime q the private key d.<\/li>\n<li>3) Choose e1(\u2026, \u2026), a point on the curve.<\/li>\n<li>4) calculates e2(\u2026, \u2026) = d \u00d7 e1(\u2026, \u2026).<\/li>\n<li>5) Public key is (a, b, p, q, e1, e2); private key is d.<\/li>\n<\/ul>\n<p>ECDSA signing and verifying:<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-217 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-126.png\" alt=\"\" width=\"898\" height=\"468\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-126.png 898w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-126-300x156.png 300w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-126-768x400.png 768w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-126-65x34.png 65w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-126-225x117.png 225w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-126-350x182.png 350w\" sizes=\"auto, (max-width: 898px) 100vw, 898px\" \/><\/p>\n<p><strong>Elliptic curves in Cryptography<\/strong><\/p>\n<ul>\n<li style=\"text-align: justify\">Elliptic Curve (EC) systems as applied to cryptography were first proposed in 1985 independently by Neal Koblitz and Victor Miller.<\/li>\n<li style=\"text-align: justify\">The discrete logarithm problem on elliptic curve groups is believed to be more difficult than the corresponding problem in (the multiplicative group of nonzero elements of) the underlying finite field.<\/li>\n<\/ul>\n<p><strong>Elliptic Curve on a finite set of Integers<\/strong><\/p>\n<ul>\n<li>Consider y2 = x3 + 2x + 3 (<strong>mod 5<\/strong>)<\/li>\n<li>x = 0 \u00de y2 = 3 \u00de no solution (mod 5)<\/li>\n<li>x = 1 \u00de y2 = 6 = 1 \u00de y = 1,4 (mod 5)<\/li>\n<li>x = 2 \u00de y2 = 15 = 0 \u00de y = 0 (mod 5)x = 3 \u00de y2 = 36 = 1 \u00de y = 1,4 (mod 5)x = 4 \u00de y2 = 75 = 0 \u00de y = 0 (mod 5)<\/li>\n<li>Then points on the elliptic curve are (1,1) (1,4) (2,0) (3,1) (3,4) (4,0) and the point at infinity: \u00a5<\/li>\n<\/ul>\n<p><strong>ECDSA involves four elements<\/strong><\/p>\n<ol>\n<li>All participants use the same global domain parameters .<\/li>\n<li style=\"text-align: justify\">First public, private key pair is generated by sender. Signer selects a random or pseudorandom number for the private key. Signer computes another point on the elliptic curve using the random number and the point of origin and that is signer\u2019s public key.<\/li>\n<li style=\"text-align: justify\">A hash value is generated for the message to be signed. Signature is generated using the private key, the domain parameters and hash value. The signature consists of two integers r and s.<\/li>\n<li style=\"text-align: justify\">To verify the signature, the verifier inputs signer\u2019s public key, the domain parameters and the integer s. The output is a value v and is compared with r. The signature is verified if v=r.<\/li>\n<\/ol>\n<p><strong>Global Domain Parameters<\/strong><\/p>\n<ul>\n<li>Prime curves over Zp are used.<\/li>\n<li>q &#8211; a prime number<\/li>\n<li style=\"text-align: justify\">a,b \u2013 integers that specify the elliptic curve equation defined over Zq with the equation y2=x3+ax+b<\/li>\n<li style=\"text-align: justify\">G \u2013 a base point represented by G=(xg,yg) on elliptic curve equation<\/li>\n<li style=\"text-align: justify\">n- order of point G. n is smallest positive integer such that nG=O. These are points on curve.<\/li>\n<\/ul>\n<p><strong>Simple elliptic curve<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-218 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-127.png\" alt=\"\" width=\"528\" height=\"252\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-127.png 528w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-127-300x143.png 300w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-127-65x31.png 65w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-127-225x107.png 225w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-127-350x167.png 350w\" sizes=\"auto, (max-width: 528px) 100vw, 528px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Encryption<\/p>\n<ul>\n<li>Consider <em>\u2018m\u2019<\/em> has the point <em>\u2018M\u2019<\/em> on the curve <em>\u2018E\u2019.<\/em> Randomly select \u2018k\u2019 from [1 \u2013 (n-1)].<\/li>\n<li>Two cipher texts will be generated let it be C1 and C2. C1 = k*PC2 = M + k*Q<\/li>\n<li>C1 and C2 will be sent.<\/li>\n<\/ul>\n<p><strong>Decryption<\/strong><\/p>\n<ul>\n<li>M = C2 \u2013 d * C1\u00a0 M is the original message that has been sent.<\/li>\n<\/ul>\n<p><strong>Proof<\/strong><\/p>\n<ul>\n<li>M = C2 \u2013 d * C1<\/li>\n<li>\u2018M\u2019 can be represented as \u2018C2 \u2013 d * C1\u2019<\/li>\n<li>C2 \u2013 d * C1 = (M + k * Q) \u2013 d * ( k * P )( C2 = M + k * Q and C1 = k * P )<\/li>\n<li>= M + k * d * P \u2013 d * k *P ( canceling out k * d * P ) = M ( Original Message )<\/li>\n<\/ul>\n<p>Key generation<\/p>\n<ul>\n<li>Signer generates private and public key in following way.<\/li>\n<\/ul>\n<ol>\n<li>Select a random integer d, d\u03b5[1,n-1]<\/li>\n<li>Compute Q=dG. This is a point in Eq(a,b)<\/li>\n<li>Bob\u2019s public key is Q and private key is d.<\/li>\n<\/ol>\n<p>Digital signature generation and Authentication<\/p>\n<ul>\n<li>Digital signature of 320 bytes for message m is generated using following steps:<\/li>\n<\/ul>\n<ol>\n<li>Select a random or pseudorandom integer k, k\u03b5[1,n-1]<\/li>\n<li>Compute point P=(x,y) =kG and r=x mod n. if r=0 then goto step 1<\/li>\n<li>Compute t=k-1 mod n<\/li>\n<li>Compute e=H(m) where H is the SHA-1 hash function, which produces a 160 bit hash function.<\/li>\n<li>Compute s=k-1 (e+dr) mod n. If s=O then goto step 1.<\/li>\n<li>The signature of message m is the pair (r,s)<\/li>\n<\/ol>\n<p>The receiver having message m, digital signature \u2013 verifies the signature using following steps.<\/p>\n<ol>\n<li>Verify that r and s are integers in the range 1 through n-1<\/li>\n<li>Using SHA-1 compute the 160 bit hash value e=H(m)<\/li>\n<li>Compute w=s-1 mod n<\/li>\n<li>Compute u1=ew and u2=rw<\/li>\n<li>Compute the point X=(x1,y1)=u1G+u2Q<\/li>\n<li>If X=O, reject the signature else compute v=x1 mod n<\/li>\n<li>Accept Sender\u2019s signature if and only if v=r. consider an elliptic curve over the field F23. Ref:https:\/\/www.certicom.com\/index.php\/31-example-of-an-elliptic-curve-group-over-fp<\/li>\n<\/ol>\n<ul>\n<li>With a = 1 and b = 0, the elliptic curve equation is y2= x3 + x.<\/li>\n<li>The point (9,5) satisfies this equation since: y2 mod p = x3 + x mod p<\/li>\n<\/ul>\n<p>25 mod 23 = 729 + 9 mod 23<\/p>\n<p>25 mod 23 = 738 mod 232 = 2<\/p>\n<p>The 23 points which satisfy this equation are: (0,0) (1,5) (1,18) (9,5) (9,18) (11,10) (11,13) (13,5)(13,18) (15,3) (15,20) (16,8) (16,15) (17,10) (17,13) (18,10) (18,13) (19,1) (19,22) (20,4) (20,19) (21,6) (21,17)<\/p>\n<p>All points on graph<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-219 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-128.png\" alt=\"\" width=\"415\" height=\"427\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-128.png 415w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-128-292x300.png 292w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-128-65x67.png 65w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-128-225x232.png 225w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-128-350x360.png 350w\" sizes=\"auto, (max-width: 415px) 100vw, 415px\" \/><\/p>\n<p>&nbsp;<\/p>\n<div>\n<p>1.\u00a0 Point Addition, L = J + K<\/p>\n<p>2.\u00a0 Point Doubling, L = 2J Example:<\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">If k = 23;<\/span><\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">then, kP = 23*P<\/span><span style=\"text-align: initial;font-size: 1em\">= 2(2(2(2P) + P) + P) + P<\/span><\/p>\n<\/div>\n<p><strong>Point Addition \ud83d\ude41 Geometrically)<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-221 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-130.png\" alt=\"\" width=\"565\" height=\"345\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-130.png 565w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-130-300x183.png 300w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-130-65x40.png 65w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-130-225x137.png 225w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-130-350x214.png 350w\" sizes=\"auto, (max-width: 565px) 100vw, 565px\" \/><\/p>\n<ul>\n<li>Consider two distinct points J and K such that J = (xJ, yJ) and K = (xK, yK) Let L = J + K where L = (xL, yL), then<\/li>\n<li>xL = s2 &#8211; xJ \u2013 xK<\/li>\n<li>yL = -yJ + s (xJ \u2013 xL)<\/li>\n<\/ul>\n<p>s = (yJ \u2013 yK)\/(xJ \u2013 xK), s is slope of the line through J and K<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-222 aligncenter\" src=\"http:\/\/itp4.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-131.png\" alt=\"\" width=\"603\" height=\"364\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-131.png 603w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-131-300x181.png 300w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-131-65x39.png 65w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-131-225x136.png 225w, https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-content\/uploads\/sites\/25\/2018\/07\/Untitled-131-350x211.png 350w\" sizes=\"auto, (max-width: 603px) 100vw, 603px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Consider a point J such that J = (xJ, yJ), where yJ \u2260 0\u00a0 Let L = 2J where L = (xL, yL), Then<\/p>\n<table>\n<tbody>\n<tr>\n<td>\u2022<\/td>\n<td>xL = s2 \u2013 2xJ<\/td>\n<\/tr>\n<tr>\n<td>\u2022<\/td>\n<td>yL = -yJ + s(xJ &#8211; xL)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>s = (3xJ2 + a) \/ (2yJ), s is the tangent at point J and a is one of the parameters chosen with the elliptic curve<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Suggested Reading:<\/strong><\/p>\n<ol>\n<li>Cryptography and Network Security Principles and Practice by William Stallings, sixth Edition, PEARSON.<\/li>\n<li>Security in Computing by Charles Pfleeger &amp; Shari Lawrence Pfleeger, fourth Edition, PEARSON.<\/li>\n<li>Network Security by Charlie Kaufman, Radia Perlman, Mike Speciner, second Edition, PHI.<\/li>\n<li>The Complete Reference \u2013 Network Security by Roberta Bragg, Mark Rhodes-Ousley &amp; Keith Strassberg, Tata McGraw Hill<\/li>\n<li>Network Security Bible by Eric Cole, Ronald Krutz, James Conley, Wiley<\/li>\n<li>Hacking 6 Exposed by Stuart McClure, Joel Scambray &amp; George Kurtz , Tata McGraw Hill .<\/li>\n<li><a href=\"http:\/\/www.snort.org\/\">www.snort.org<\/a><\/li>\n<li><a href=\"https:\/\/nmap.org\/\">https:\/\/nmap.org<\/a><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n","protected":false},"author":4,"menu_order":24,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["miss-hiteishi-diwanji"],"pb_section_license":""},"chapter-type":[],"contributor":[58],"license":[],"class_list":["post-215","chapter","type-chapter","status-publish","hentry","contributor-miss-hiteishi-diwanji"],"part":3,"_links":{"self":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/pressbooks\/v2\/chapters\/215","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/wp\/v2\/users\/4"}],"version-history":[{"count":5,"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/pressbooks\/v2\/chapters\/215\/revisions"}],"predecessor-version":[{"id":406,"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/pressbooks\/v2\/chapters\/215\/revisions\/406"}],"part":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/pressbooks\/v2\/chapters\/215\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/wp\/v2\/media?parent=215"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/pressbooks\/v2\/chapter-type?post=215"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/wp\/v2\/contributor?post=215"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/itp4\/wp-json\/wp\/v2\/license?post=215"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}