(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 5126362, 173360] NotebookOptionsPosition[ 4640687, 161106] NotebookOutlinePosition[ 4834501, 166065] CellTagsIndexPosition[ 4793122, 165058] WindowFrame->Normal ContainsDynamic->False*) (* Beginning of Notebook Content *) Notebook[{ Cell["\<\ FUNDAMENTALS OF CRYPTOLOGY A Professional Reference and Interactive Tutorial\ \>", "Title", TextAlignment->Center, TextJustification->0], Cell["\<\ Henk van Tilborg Eindhoven University of Technology\ \>", "Subtitle", TextAlignment->Center, TextJustification->0], Cell[CellGroupData[{ Cell["Contents", "Chapter", TabSpacings->3], Cell[TextData[{ "\t", ButtonBox["Preface", BaseStyle->"Hyperlink", ButtonData:>"Chap Preface"] }], "Text", GeneratedCell->True, 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FontFamily->"Helvetica", FontSize->16, FontWeight->"Bold", CellTags->"Chap Preface"], Cell[TextData[{ "The protection of sensitive information against unauthorized access or \ fraudulent changes has been of prime concern throughout the centuries. Modern \ communication techniques, using computers connected through networks, make \ all data even more vulnerable for these threats. Also, new issues have come \ up that were not relevant before, e.g. how to add a (digital) signature to an \ electronic document in such a way that the signer can not deny later on that \ the document was signed by him/her. \n", StyleBox["\n", FontSize->11], "Cryptology addresses the above issues. It is at the foundation of all \ information security. The techniques employed to this end have become \ increasingly mathematical of nature. This book serves as an introduction to \ modern cryptographic methods. After a brief survey of classical \ cryptosystems, it concentrates on three main areas. First of all, stream \ ciphers and block ciphers are discussed. These systems have extremely fast \ implementations, but sender and receiver have to share a secret key. Public \ key cryptosystems (the second main area) make it possible to protect data \ without a prearranged key. Their security is based on intractable \ mathematical problems, like the factorization of large numbers. The remaining \ chapters cover a variety of topics, such as zero-knowledge proofs, secret \ sharing schemes and authentication codes. Two appendices explain all \ mathematical prerequisites in great detail. One is on elementary number \ theory (Euclid's Algorithm, the Chinese Remainder Theorem, quadratic \ residues, inversion formulas, and continued fractions). The other appendix \ gives a thorough introduction to finite fields and their algebraic structure.\ \n", StyleBox["\n", FontSize->11], "This book differs from its 1988 version in two ways. That a lot of new \ material has been added is to be expected in a field that is developing so \ fast. Apart from a revision of the existing material, there are many new or \ greatly expanded sections, an entirely new chapter on elliptic curves and \ also one on authentication codes. The second difference is even more \ significant. The whole manuscript is electronically available as an \ interactive ", ButtonBox["Mathematica", BaseStyle->"Hyperlink", ButtonData:>{ URL["http://www.wolfram.com/"], None}], " manuscript. So, there are hyperlinks to other places in the text, but more \ importantly, it is now possible to work out non-trivial examples. Even a \ non-expert can easily alter the parameters in the examples and try out new \ ones. It is our experience, based on teaching at the California Institute of \ Technology and the Eindhoven University of Technology, that most students \ truly enjoy the enormous possibilities of a computer algebra notebook. \ Throughout the book, it has been our intention to make all ", StyleBox["Mathematica", FontSlant->"Italic"], " statements as transparent as possible, sometimes sacrificing elegant or \ smart alternatives that are too dependent on this particular computer algebra \ package. " }], "Text", FontFamily->"Times New Roman"], Cell[TextData[{ StyleBox["There are several people that have played a crucial role in the \ preparation of this manuscript. In alphabetical order of first name, I would \ like to thank Fred Simons for showing me the full potential of ", FontFamily->"Times New Roman"], StyleBox["Mathematica", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" for educational purposes and for enhancing many the ", FontFamily->"Times New Roman"], StyleBox["Mathematica ", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["commands, Gavin Horn for the many typo's that he has found as well \ as his compilation of solutions, Lilian Porter for her feedback on my use of \ English, and Wil Kortsmit for his help in getting the manuscript camera-ready \ and for solving many of my Mathematica questions. I also owe great debt to \ the following people who helped me with their feedback on various chapters: \ Berry Schoenmakers, Bram van Asch, Eric Verheul, Frans Willems, Mariska Sas, \ and Martin van Dijk.", FontFamily->"Times New Roman"] }], "Text", FontFamily->"Times New Roman"], Cell[TextData[{ StyleBox["Henk van Tilborg\nDept. of Mathematics and Computing Science\n\ Eindhoven University of Technology\nP.O. Box 513\n5600 MB Eindhoven\nthe \ Netherlands\nemail: ", FontFamily->"Times New Roman"], StyleBox[ButtonBox["henkvt@win.tue.nl", BaseStyle->"Hyperlink", ButtonData:>{ URL["mailto:henkvt@win.tue.nl"], None}], FontFamily->"Times New Roman"], StyleBox[".", FontFamily->"Times New Roman"] }], "Text", ParagraphSpacing->{0, 0}, FontFamily->"Times New Roman"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], "\t", "Introduction" }], "Chapter", CounterAssignments->{{"Chapter", 0}}, CellTags->"Chap Intro"], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], "\t", "Introduction and Terminology" }], "Section", CellTags->"SectIntro Intro"], Cell[TextData[{ StyleBox["Cryptology", FontSlant->"Italic"], ", the study of cryptosystems, can be subdivided into two disciplines. ", StyleBox["Cryptography", FontSlant->"Italic"], " concerns itself with the design of cryptosystems, while ", StyleBox["cryptanalysis", FontSlant->"Italic"], " studies the breaking of cryptosystems. These two aspects are closely \ related; when setting up a cryptosystem the analysis of its security plays an \ important role. At this time we will not give a formal definition of a \ cryptosystem, as that will come later in this chapter. We assume that the \ reader has the right intuitive idea of what a cryptosystem is." }], "Text", TextAlignment->Left, TextJustification->1, CellTags->"DefIntro Cryptology"], Cell["\<\ Why would anybody use a cryptosystem? There are several possibilities:\ \>", "Text"], Cell[TextData[{ StyleBox["Confidentiality", FontSlant->"Italic"], ": When transmitting data, one does not want an eavesdropper to understand \ the contents of the transmitted messages. The same is true for stored data \ that should be protected against unauthorized access, for instance by \ hackers." }], "Text", CellTags->"DefIntro Privacy"], Cell[TextData[{ StyleBox["Authentication", FontSlant->"Italic"], ": This property is the equivalent of a signature. The receiver of a message \ wants proof that a message comes from a certain party and not from somebody \ else (even if the original party later wants to deny it)." }], "Text", CellTags->"DefIntro Authen"], Cell[TextData[{ StyleBox["Integrity", FontSlant->"Italic"], ": This means that the receiver of certain data has evidence that no changes \ have been made by a third party." }], "Text", CellTags->"DefIntro Integrity"], Cell[TextData[{ "Throughout the centuries (see ", ButtonBox["[Kahn67]", BaseStyle->"Hyperlink", ButtonData:>"RefKahn67"], ") cryptosystems have been used by the military and by the diplomatic \ services. The nowadays widespread use of computer controlled communication \ systems in industry or by civil services, often asks for special protection \ of the data by means of cryptographic techniques." }], "Text", GeneratedCell->True], Cell["\<\ Since the storage, and later recovery, of data can be viewed as transmission \ of this data in the time domain, we shall always use the term transmission \ when discussing a situation when data is stored and/or transmitted.\ \>", "Text", PageBreakBelow->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], "\tShannon's Description of a Conventional Cryptosystem" }], "Section", CellTags->"SectIntro Shannon"], Cell[TextData[{ "Chapters ", ButtonBox["2", BaseStyle->"Hyperlink", ButtonData:>"Chap Class"], ", ", ButtonBox["3", BaseStyle->"Hyperlink", ButtonData:>"Chap Shift"], ", and ", ButtonBox["4", BaseStyle->"Hyperlink", ButtonData:>"Chap Block"], " discuss several so-called conventional cryptosystems. The formal \ definition of a conventional cryptosystem as well as the mathematical \ foundation of the underlying theory is due to C.E. Shannon ", ButtonBox["[Shan49]", BaseStyle->"Hyperlink", ButtonData:>"RefShan49"], ". 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With |\[ScriptCapitalA]| we denote the cardinality of \[ScriptCapitalA]. \ We shall often use ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["\[DoubleStruckCapitalZ]", "q"], "=", RowBox[{"{", RowBox[{"0", ",", "1", ",", "\[Ellipsis]", ",", " ", RowBox[{"q", "-", "1"}]}], "}"}]}], TraditionalForm]]], " as alphabet, where we work with its elements modulo ", Cell[BoxData[ FormBox["q", TraditionalForm]]], " (see the beginning of ", ButtonBox["Subsection A.3.1", BaseStyle->"Hyperlink", ButtonData:>"SubSecA Congruences"], " and ", ButtonBox["Section B.2", BaseStyle->"Hyperlink", ButtonData:>"SecAppB Field Constr"], ". The alphabet ", Cell[BoxData[ FormBox[ SubscriptBox["\[DoubleStruckCapitalZ]", "26"], TraditionalForm]]], " can be identified with the set ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"a", ",", " ", "b", ",", " ", "\[Ellipsis]", ",", " ", "z"}], "}"}], TraditionalForm]]], ". In most modern applications ", Cell[BoxData[ FormBox["q", TraditionalForm]]], " will often be 2 or a power of 2." }], "Text", GeneratedCell->True, CellTags->"DefIntro Alphabet"], Cell[TextData[{ "A concatenation of ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " letters from \[ScriptCapitalA] will be called an ", Cell[BoxData[ FormBox["n", TraditionalForm]], FontSlant->"Italic"], StyleBox["-gram", FontSlant->"Italic"], " and denoted by ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["a", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "=", RowBox[{"(", RowBox[{ SubscriptBox["a", "0"], ",", " ", RowBox[{ SubscriptBox["a", RowBox[{"1", ","}]], "\[Ellipsis]"}], ",", " ", SubscriptBox["a", RowBox[{"n", "-", "1"}]]}], ")"}]}], TraditionalForm]]], ". Special cases are ", StyleBox["bi-grams", FontSlant->"Italic"], " (", Cell[BoxData[ FormBox[ RowBox[{"n", "=", "2"}], TraditionalForm]]], ") and ", StyleBox["tri-grams", FontSlant->"Italic"], " (", Cell[BoxData[ FormBox[ RowBox[{"n", "=", "3"}], TraditionalForm]]], "). The set of all ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "-grams from \[ScriptCapitalA] will be denoted by ", Cell[BoxData[ FormBox[ SuperscriptBox["\[ScriptCapitalA]", "n"], TraditionalForm]]], ". " }], "Text", CellTags->"DefIntro bigram"], Cell[TextData[{ "A ", StyleBox["text", FontSlant->"Italic"], " is an element from ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["\[ScriptCapitalA]", "*"], "=", RowBox[{ SubscriptBox["\[Union]", RowBox[{"n", "\[GreaterEqual]", "0"}]], SuperscriptBox["\[ScriptCapitalA]", "n"]}]}], TraditionalForm]]], ". A ", StyleBox["language", FontSlant->"Italic"], " is a subset of ", Cell[BoxData[ FormBox[ SuperscriptBox["\[ScriptCapitalA]", "*"], TraditionalForm]]], ". In the case of programming languages this subset is precisely defined by \ means of recursion rules. In the case of spoken languages these rules are \ very loose." }], "Text", CellTags->"DefIntro Language"], Cell[TextData[{ "Let \[ScriptCapitalA] and \[ScriptCapitalB] be two finite alphabets. Any \ one-to-one mapping ", Cell[BoxData[ FormBox["E", TraditionalForm]]], " of ", Cell[BoxData[ FormBox[ SuperscriptBox["\[ScriptCapitalA]", "*"], TraditionalForm]]], " to ", Cell[BoxData[ FormBox[ SuperscriptBox["\[ScriptCapitalB]", "*"], TraditionalForm]]], " is called a ", StyleBox["cryptographic transformation", FontSlant->"Italic"], ". In most practical situations |\[ScriptCapitalA]| will be equal to |\ \[ScriptCapitalB]|. Also often the cryptographic transformation ", Cell[BoxData[ FormBox["E", TraditionalForm]]], " will map ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "-grams into ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "-grams (to avoid data expansion during the encryption process)." }], "Text", CellTags->"DefIntro Crypto Transf"], Cell[TextData[{ "Let ", Cell[BoxData[ FormBox[ StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], TraditionalForm]]], " be the message (a text from ", Cell[BoxData[ FormBox[ SuperscriptBox["\[ScriptCapitalA]", "*"], TraditionalForm]]], ") that Alice in ", ButtonBox["Figure 1.1", BaseStyle->"Hyperlink", ButtonData:>"FigIntro Classic CS"], " wants to transmit in secrecy to Bob. It is usually called the ", StyleBox["plaintext", FontSlant->"Italic"], ". Alice will first transform the plaintext into ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "=", RowBox[{"E", "(", StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], ")"}]}], TraditionalForm]]], ", the so-called ", StyleBox["ciphertext", FontSlant->"Italic"], ". It will be the ciphertext that she will transmit to Bob. " }], "Text", CellTags->"DefIntro plaintext"], Cell[TextData[{ StyleBox["Definition ", FontWeight->"Bold"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold"], StyleBox[ CounterBox["Definition"], FontWeight->"Bold"], StyleBox["\n", FontWeight->"Bold"], "A ", StyleBox["symmetric ", FontSlant->"Italic"], "(or", StyleBox[" conventional", FontSlant->"Italic"], ")", StyleBox[" cryptosystem", FontSlant->"Italic"], " \[GothicCapitalE] is a set of cryptographic transformations \ \[GothicCapitalE] = ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ SubscriptBox["E", "k"], "|", " ", RowBox[{"k", "\[Element]", "\[ScriptCapitalK]"}]}], "}"}], TraditionalForm]]], ". \nThe index set \[ScriptCapitalK] is called the ", StyleBox["key space", FontSlant->"Italic"], ", and its elements ", Cell[BoxData[ FormBox["k", TraditionalForm]]], " ", StyleBox["keys", FontSlant->"Italic"], "." }], "Definition", CellTags->"DefIntro Crypto Syst"], Cell[TextData[{ "Since ", Cell[BoxData[ FormBox[ SubscriptBox["E", "k"], TraditionalForm]]], " is a one-to-one mapping, its inverse must exist. We shall denote it with \ ", Cell[BoxData[ FormBox[ SubscriptBox["D", "k"], TraditionalForm]]], ". Of course, the ", Cell[BoxData[ FormBox["E", TraditionalForm]]], " stands for ", StyleBox["encryption", FontSlant->"Italic"], " (or enciphering) and the ", Cell[BoxData[ FormBox["D", TraditionalForm]]], " for ", StyleBox["decryption", FontSlant->"Italic"], " (or deciphering). One has" }], "Text", CellTags->"DefIntro encryption"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["D", "k"], RowBox[{"(", RowBox[{ SubscriptBox["E", "k"], "(", StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], ")"}], ")"}]}], "=", StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]]}], TraditionalForm]]], ", \tfor all plaintexts ", Cell[BoxData[ FormBox[ RowBox[{"m", "\[Element]", SuperscriptBox["\[ScriptCapitalA]", "*"]}], TraditionalForm]]], "and keys ", Cell[BoxData[ FormBox[ RowBox[{"k", "\[Element]", "\[ScriptCapitalK]"}], TraditionalForm]]], "." }], "DisplayFormula"], Cell[TextData[{ "If Alice wants to send the plaintext ", Cell[BoxData[ FormBox[ StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], TraditionalForm]]], " to Bob by means of the cryptographic transformation ", Cell[BoxData[ FormBox[ SubscriptBox["E", "k"], TraditionalForm]]], ", both Alice and Bob must know the particular choice of the key ", Cell[BoxData[ FormBox["k", TraditionalForm]]], ". They will have agreed on the value of ", Cell[BoxData[ FormBox["k", TraditionalForm]]], " by means of a so-called ", StyleBox["secure channel", FontSlant->"Italic"], ". This channel could be a courier, but it could also be that Alice and Bob \ have, beforehand, agreed on the choice of ", Cell[BoxData[ FormBox["k", TraditionalForm]]], ".\nBob can decipher ", Cell[BoxData[ FormBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], TraditionalForm]]], " by computing" }], "Text", CellTags->"DefIntro secure channel"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["D", "k"], "(", StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], ")"}], "=", RowBox[{ RowBox[{ SubscriptBox["D", "k"], RowBox[{"(", RowBox[{ SubscriptBox["E", "k"], "(", StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], ")"}], ")"}]}], "=", StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]]}]}], TraditionalForm]]], "." }], "DisplayFormula"], Cell[TextData[{ "Normally, the same cryptosystem \[GothicCapitalE] will be used for a long \ time and by many people, so it is reasonable to assume that this set of \ cryptographic transformations \[GothicCapitalE] is also known to the \ cryptanalist. It is the frequent changing of the key that has to provide the \ security of the data. This principle was already clearly stated by the \ Dutchman Auguste Kerckhoff (see ", ButtonBox["[Kahn67]", BaseStyle->"Hyperlink", ButtonData:>"RefKahn67"], ") in the 19-th century." }], "Text", GeneratedCell->True], Cell[TextData[{ "The ", StyleBox["cryptanalist", FontSlant->"Italic"], " (Eve) who is connected to the transmission line can be:" }], "Text", CellTags->"DefIntro Cryptanalist"], Cell[TextData[{ "\[FilledVerySmallSquare] ", StyleBox["passive", FontSlant->"Italic"], " (eavesdropping): The cryptanalist tries to find ", Cell[BoxData[ FormBox[ StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], TraditionalForm]]], " (or even better ", Cell[BoxData[ FormBox["k", TraditionalForm]]], ") from ", Cell[BoxData[ FormBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], TraditionalForm]]], " (and whatever further knowledge he has). By determining ", Cell[BoxData[ FormBox["k", TraditionalForm]]], " more ciphertexts may be broken." }], "Text", CellTags->"DefIntro Cryptanalist passive"], Cell[TextData[{ "\[FilledVerySmallSquare] ", StyleBox["active", FontSlant->"Italic"], " (tampering): The cryptanalist tries to actively manipulate the data that \ are being transmitted. For instance, he transmits his own ciphertext, \ retransmits old ciphertext, substitutes his own texts for transmitted \ ciphertexts, etc.." }], "Text", CellTags->"DefIntro Cryptanalist active"], Cell["In general, one discerns three levels of cryptanalysis:", "Text"], Cell[TextData[{ "\[FilledVerySmallSquare] ", StyleBox["Ciphertext only attack", FontSlant->"Italic"], ": Only a piece of ciphertext is known to the cryptanalist (and often the \ context of the message)." }], "Text", CellTags->"DefIntro Cipher Only Att."], Cell[TextData[{ "\[FilledVerySmallSquare] ", StyleBox["Known plaintext attack", FontSlant->"Italic"], ": A piece of ciphertext with corresponding plaintext is known. If a system \ is secure against this kind of attack the legitimate receiver does not have \ to destroy deciphered messages." }], "Text", CellTags->"DefIntro Known Pl. Att."], Cell[TextData[{ "\[FilledVerySmallSquare] ", StyleBox["Chosen plaintext attack", FontSlant->"Italic"], ": The cryptanalist can choose any piece of plaintext and generate the \ corresponding ciphertext. The public-key cryptosystems that we shall discuss \ in Chapters ", ButtonBox["7", BaseStyle->"Hyperlink", ButtonData:>"Chap Public"], "-", ButtonBox["12", BaseStyle->"Hyperlink", ButtonData:>"Chap Knapsack"], " have to be secure against this kind of attack." }], "Text", CellTags->"DefIntro Chosen Pl. Att."], Cell[TextData[{ "This concludes our general description of the conventional cryptosystem as \ depicted in ", ButtonBox["Figure 1.1", BaseStyle->"Hyperlink", ButtonData:>"FigIntro Classic CS"], ". " }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], "\tStatistical Description of a Plaintext Source" }], "Section", CellTags->"SectIntro Statistical"], Cell[TextData[{ "In cryptology, especially when one wants to break a particular \ cryptosystem, a probabilistic approach to describe a language is often \ already a powerful tool, as we shall see in ", ButtonBox["Section 2.2", BaseStyle->"Hyperlink", ButtonData:>"SectClass Incid Coin"], "." }], "Text"], Cell[TextData[{ "The person ", "Alice", " in ", ButtonBox["Figure 1.1", BaseStyle->"Hyperlink", ButtonData:>"FigIntro Classic CS"], " stands for a finite or infinite ", StyleBox["plaintext source", FontSlant->"Italic"], " \[GothicCapitalS] of text, that was called plaintext, from an alphabet \ \[ScriptCapitalA], e.g. ", Cell[BoxData[ FormBox[ SubscriptBox["\[DoubleStruckCapitalZ]", "q"], TraditionalForm]]], ". It can be described as a finite resp. infinite sequence of random \ variables ", Cell[BoxData[ FormBox[ SubscriptBox["M", "i"], TraditionalForm]]], ", so by sequences" }], "Text", CellTags->"DefIntro Plaint Source"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["M", "0"], ",", " ", SubscriptBox["M", "1"], ",", "\[Ellipsis]", " ", ",", " ", SubscriptBox["M", RowBox[{"n", "-", "1"}]]}], TraditionalForm]]], "\tfor some fixed value of ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "," }], "DisplayFormula"], Cell["resp.", "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["M", "0"], ",", " ", SubscriptBox["M", "1"], ",", SubscriptBox["M", "2"], ",", " ", "\[Ellipsis]", " "}], TraditionalForm]]], "," }], "DisplayFormula"], Cell[TextData[{ "each described by probabilities that events occur. So, for each letter \ combination (", Cell[BoxData[ FormBox["r", TraditionalForm]]], "-gram) ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{"(", RowBox[{ RowBox[{ SubscriptBox["m", RowBox[{"0", ","}]], SubscriptBox["m", "1"]}], ",", "\[Ellipsis]", " ", ","}]}], TraditionalForm], SubscriptBox["m", RowBox[{"r", "-", "1"}]]}], ")"}], TraditionalForm]]], " over \[ScriptCapitalA] and each starting point ", Cell[BoxData[ FormBox["j", TraditionalForm]]], " the probability" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ RowBox[{ SubscriptBox["M", "j"], "=", SubscriptBox["m", "0"]}], ",", " ", RowBox[{ SubscriptBox["M", RowBox[{"j", "+", "1"}]], "=", SubscriptBox["m", "1"]}], ",", "\[Ellipsis]", " ", ",", " ", RowBox[{ SubscriptBox["M", RowBox[{"j", "+", "r", "-", "1"}]], "=", SubscriptBox["m", RowBox[{"r", "-", "1"}]]}]}], ")"}], TraditionalForm]]]], "DisplayFormula"], Cell[TextData[{ "is well defined. In the case that ", Cell[BoxData[ FormBox[ RowBox[{"j", "=", "0"}], TraditionalForm]]], ", we shall simply write ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ SubscriptBox["m", "0"], ",", " ", SubscriptBox["m", "1"], ",", "\[Ellipsis]", " ", ",", " ", SubscriptBox["m", RowBox[{"r", "-", "1"}]]}], ")"}], TraditionalForm]]], ". Of course, the probabilities that describe the plaintext source \ \[GothicCapitalS] should satisfy the standard statistical properties, that we \ shall mention below but on which we shall not elaborate." }], "Text"], Cell[TextData[{ "i) ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ SubscriptBox["m", "0"], ",", " ", SubscriptBox["m", "1"], ",", "\[Ellipsis]", " ", ",", " ", SubscriptBox["m", RowBox[{"r", "-", "1"}]]}], ")"}], "\[GreaterEqual]", "0"}], TraditionalForm]]], " for all texts ", Cell[BoxData[ FormBox[ RowBox[{"(", RowBox[{ SubscriptBox["m", "0"], ",", " ", SubscriptBox["m", "1"], ",", "\[Ellipsis]", " ", ",", " ", SubscriptBox["m", RowBox[{"r", "-", "1"}]]}], ")"}], TraditionalForm]]], "." }], "DisplayFormula"], Cell[TextData[{ "ii) ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["\[Sum]", RowBox[{"(", RowBox[{ SubscriptBox["m", "0"], ",", " ", SubscriptBox["m", "1"], ",", "\[Ellipsis]", " ", ",", " ", SubscriptBox["m", RowBox[{"r", "-", "1"}]]}], ")"}]], RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ SubscriptBox["m", "0"], ",", " ", SubscriptBox["m", "1"], ",", "\[Ellipsis]", " ", ",", " ", SubscriptBox["m", RowBox[{"r", "-", "1"}]]}], ")"}]}], "=", "1"}], TraditionalForm]]], "." }], "DisplayFormula"], Cell[TextData[{ "iii) ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["\[Sum]", RowBox[{"(", RowBox[{ SubscriptBox["m", "r"], ",", " ", SubscriptBox["m", RowBox[{"r", "+", "1"}]], ",", "\[Ellipsis]", " ", ",", " ", SubscriptBox["m", RowBox[{"l", "-", "1"}]]}], ")"}]], RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ SubscriptBox["m", "0"], ",", " ", SubscriptBox["m", "1"], ",", "\[Ellipsis]", " ", ",", " ", SubscriptBox["m", RowBox[{"l", "-", "1"}]]}], ")"}]}], "=", RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ SubscriptBox["m", "0"], ",", " ", SubscriptBox["m", "1"], ",", "\[Ellipsis]", ",", " ", SubscriptBox["m", RowBox[{"r", "-", "1"}]]}], ")"}]}], TraditionalForm]]], ", for all ", Cell[BoxData[ FormBox[ RowBox[{"l", ">", "r"}], TraditionalForm]]], "." }], "DisplayFormula"], Cell[TextData[{ "The third property is called ", StyleBox["Kolmogorov's consistency condition", FontSlant->"Italic"], "." }], "Text", CellTags->"PropIntro Kolmogorov"], Cell[TextData[{ StyleBox["Example ", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold", FontSlant->"Plain"], StyleBox[".", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Example"], FontWeight->"Bold", FontSlant->"Plain"], "\nThe plaintext source \[GothicCapitalS] (Alice in ", ButtonBox["Figure 1.1", BaseStyle->"Hyperlink", ButtonData:>"FigIntro Classic CS"], ") generates individual letters (1-grams) from ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"a", ",", "b", ",", "\[Ellipsis]", ",", "z"}], "}"}], TraditionalForm]]], " with an independent but identical distribution, say ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"p", "(", "a", ")"}], ",", " ", RowBox[{"p", "(", "b", ")"}], ",", " ", "\[Ellipsis]", ",", " ", RowBox[{"p", "(", "z", ")"}]}], TraditionalForm]]], ". So," }], "Example", CellTags->"ExamIntro 1-grams"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ SubscriptBox["m", "0"], ",", " ", SubscriptBox["m", RowBox[{"1", " "}]], ",", "\[Ellipsis]", ",", SubscriptBox["m", RowBox[{"n", "-", "1"}]]}], ")"}], "=", RowBox[{ RowBox[{"p", "(", SubscriptBox["m", "0"], ")"}], RowBox[{"p", "(", SubscriptBox["m", "1"], ")"}], "\[CenterEllipsis]", " ", RowBox[{"p", "(", SubscriptBox["m", RowBox[{"n", "-", "1"}]], ")"}]}]}], TraditionalForm]]], ", ", Cell[BoxData[ FormBox[ RowBox[{"n", "\[GreaterEqual]", "1"}], TraditionalForm]]], "." }], "DisplayFormula"], Cell[TextData[{ "The distribution of the letters of the alphabet in normal English texts is \ given in ", ButtonBox["Table 1.1", BaseStyle->"Hyperlink", ButtonData:>"TableIntr Prob English"], " (see Table 12-1 in ", ButtonBox["[MeyM82]", BaseStyle->"Hyperlink", ButtonData:>"RefMey82"], "). In this model one has that" }], "Text", GeneratedCell->True, FontSlant->"Italic", FontColor->GrayLevel[0]], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", "run", ")"}], "=", RowBox[{ RowBox[{ RowBox[{"p", "(", "r", ")"}], RowBox[{"p", "(", "u", ")"}], RowBox[{"p", "(", "n", ")"}]}], "=", " ", RowBox[{ RowBox[{"0.0612", "\[Times]", "0.0271", "\[Times]", "0.0709"}], "\[TildeTilde]", RowBox[{"1.18", " ", SuperscriptBox["10", RowBox[{"-", "4"}]]}]}]}]}], TraditionalForm]]], "." }], "DisplayFormula"], Cell[TextData[{ "Note that in this model also ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", "nru", ")"}], "=", RowBox[{ RowBox[{"p", "(", "n", ")"}], RowBox[{"p", "(", "r", ")"}], RowBox[{"p", "(", "u", ")"}]}]}], TraditionalForm]]], ", etc., so, unlike in a regular English texts, all permutations of the \ three letters ", Cell[BoxData[ FormBox["r", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["u", TraditionalForm]]], ", and ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " are equally likely in \[GothicCapitalS]." }], "Text", FontSlant->"Italic", FontColor->GrayLevel[0]], Cell[TextData[{ "\n", Cell[BoxData[GridBox[{ {"a", "0.0804", "h", "0.0549", "o", "0.0760", "v", "0.0099"}, {"b", "0.0154", "i", "0.0726", "p", "0.0200", "w", "0.0192"}, {"c", "0.0306", "j", "0.0016", "q", "0.0011", "x", "0.0019"}, {"d", "0.0399", "k", "0.0067", "r", "0.0612", "y", "0.0173"}, {"e", "0.1251", "l", "0.0414", "s", "0.0654", "z", "0.0009"}, {"f", "0.0230", "m", "0.0253", "t", "0.0925", " ", " "}, {"g", "0.0196", "n", "0.0709", "u", "0.0271", " ", " "} }]]], "\n\nProbability distributions of 1-grams in English." }], "NumberedTable", CellTags->"TableIntr Prob English"], Cell[TextData[{ StyleBox["Example ", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold", FontSlant->"Plain"], StyleBox[".", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Example"], FontWeight->"Bold", FontSlant->"Plain"], "\n\[GothicCapitalS] generates 2-grams over the alphabet ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"a", ",", "b", ",", ",", "\[Ellipsis]", ",", "z"}], "}"}], TraditionalForm]]], " with an independent but identical distribution, say ", Cell[BoxData[ FormBox[ RowBox[{"p", "(", RowBox[{"s", ",", "t"}], ")"}], TraditionalForm]]], ", with ", Cell[BoxData[ FormBox[ RowBox[{"s", ",", RowBox[{"t", " ", "\[Element]", RowBox[{"{", RowBox[{"a", ",", "b", ",", "\[Ellipsis]", ",", "z"}], "}"}]}]}], TraditionalForm]]], ". So, for ", Cell[BoxData[ FormBox[ RowBox[{"n", "\[GreaterEqual]", "1"}], TraditionalForm]]] }], "Example"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ RowBox[{ SubscriptBox["m", RowBox[{"0", ","}]], SubscriptBox["m", "1"]}], ",", "\[Ellipsis]", ",", SubscriptBox["m", RowBox[{ RowBox[{"2", "n"}], "-", "1"}]]}], ")"}], "=", " ", RowBox[{ RowBox[{"p", "(", RowBox[{ SubscriptBox["m", "0"], SubscriptBox["m", "1"]}], ")"}], RowBox[{"p", "(", RowBox[{ SubscriptBox["m", "2"], ",", SubscriptBox["m", "3"]}], ")"}], "\[CenterEllipsis]"}]}], TraditionalForm]]], " ", Cell[BoxData[ FormBox[ RowBox[{"p", "(", RowBox[{ SubscriptBox["m", RowBox[{ RowBox[{"2", "n"}], "-", "2"}]], SubscriptBox["m", RowBox[{ RowBox[{"2", "n"}], "-", "1"}]]}], ")"}], TraditionalForm]]], ". 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Let ", Cell[BoxData[ FormBox[ StyleBox[ RowBox[{ StyleBox["p", FontSize->12, FontWeight->"Bold", FontVariations->{"StrikeThrough"->False, "Underline"->True}, FontColor->GrayLevel[0]], "=", RowBox[{"(", RowBox[{ RowBox[{"p", "(", "a", ")"}], ",", RowBox[{"p", "(", "b", ")"}], ",", "\[Ellipsis]", ",", RowBox[{"p", "(", "z", ")"}]}], ")"}]}], FontFamily->"Times", FontSlant->"Italic"], TraditionalForm]]], ", be the corresponding eigenvector (it is called the ", StyleBox["equilibrium distribution", FontSlant->"Italic"], " of the process).\nAssuming that the process is already in its equilibrium \ state at the beginning, one has" }], "Example", CellTags->"ExamIntro Markov"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ RowBox[{ SubscriptBox["m", RowBox[{"0", ","}]], SubscriptBox["m", "1"]}], ",", "\[Ellipsis]", ",", SubscriptBox["m", RowBox[{"n", "-", "1"}]]}], ")"}], "=", " ", RowBox[{ RowBox[{"p", "(", SubscriptBox["m", "0"], ")"}], RowBox[{"p", "(", RowBox[{ SubscriptBox["m", "1"], "|", SubscriptBox["m", "0"]}], ")"}], RowBox[{"p", "(", RowBox[{ SubscriptBox["m", "2"], "|", SubscriptBox["m", "1"]}], ")"}], "\[CenterEllipsis]"}]}], TraditionalForm]]], " ", Cell[BoxData[ FormBox[ RowBox[{"p", "(", RowBox[{ SubscriptBox["m", RowBox[{"n", "-", "1"}]], "|", SubscriptBox["m", RowBox[{"n", "-", "2"}]]}], ")"}], TraditionalForm]]], ". 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Then, one obtains the \ following, more realistic probabilities of occurrence:" }], "Text", GeneratedCell->True, FontSlant->"Italic", FontColor->GrayLevel[0]], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", "run", ")"}], "=", RowBox[{ RowBox[{"p", "(", "r", ")"}], RowBox[{"p", "(", RowBox[{"u", "|", "r"}], ")"}], RowBox[{"p", "(", RowBox[{"n", "|", "u"}], ")"}]}]}], TraditionalForm]]], " = 0.0751\[Times] 0.0192 \[Times] 0.1517 \[TildeTilde] 2.19 ", Cell[BoxData[ FormBox[ SuperscriptBox["10", RowBox[{"-", "4"}]], TraditionalForm]]], "," }], "DisplayFormula"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", "urn", ")"}], "=", RowBox[{ RowBox[{"p", "(", "u", ")"}], RowBox[{"p", "(", RowBox[{"r", "|", "u"}], ")"}], RowBox[{"p", "(", RowBox[{"n", "|", "r"}], ")"}]}]}], TraditionalForm]]], " = 0.0272\[Times] 0.1460 \[Times] 0.0325 \[TildeTilde] 1.29 ", Cell[BoxData[ FormBox[ SuperscriptBox["10", RowBox[{"-", "4"}]], TraditionalForm]]], "," }], "DisplayFormula"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["Pr", "plain"], "(", "nru", ")"}], "=", RowBox[{ RowBox[{"p", "(", "n", ")"}], RowBox[{"p", "(", RowBox[{"r", "|", "n"}], ")"}], RowBox[{"p", "(", RowBox[{"u", "|", "r"}], ")"}]}]}], TraditionalForm]]], " = 0.0814\[Times] 0.0011 \[Times] 0.0192 \[TildeTilde] 1.72 ", Cell[BoxData[ FormBox[ SuperscriptBox["10", RowBox[{"-", "6"}]], TraditionalForm]]], "," }], "DisplayFormula"], Cell[TextData[{ "By means of the ", StyleBox["Mathematica", FontSlant->"Italic"], " functions ", StyleBox["StringTake", FontVariations->{"Underline"->True}], ", ", StyleBox["ToCharacterCode", FontVariations->{"Underline"->True}], ", and ", StyleBox["StringLength", FontVariations->{"Underline"->True}], ", these probabilities can be computed in the following way (first enter the \ input ", ButtonBox["Table 1.2", BaseStyle->"Hyperlink", ButtonData:>"TableIntro Equil Distr"], " and the input before ", ButtonBox["Table 1.3,", BaseStyle->"Hyperlink", ButtonData:>"TableIntro Transition Prob"], ")" }], "Text", CellChangeTimes->{{3.423391330402695*^9, 3.4233913382933707`*^9}, { 3.4233914992006507`*^9, 3.4233915170445147`*^9}, {3.4233916045138245`*^9, 3.423391611138867*^9}}, FontSlant->"Italic", FontColor->GrayLevel[0]], Cell[BoxData[{ RowBox[{ RowBox[{"sourcetext", "=", "\"\\""}], ";"}], "\n", RowBox[{ RowBox[{"ed", "[", RowBox[{"StringTake", "[", RowBox[{"sourcetext", ",", RowBox[{"{", "1", "}"}]}], "]"}], "]"}], "*", RowBox[{ UnderoverscriptBox["\[Product]", RowBox[{"i", "=", "1"}], RowBox[{ RowBox[{"StringLength", "[", "sourcetext", "]"}], "-", "1"}]], RowBox[{"TrPr", "[", RowBox[{"[", "\t\t\t\t", RowBox[{ RowBox[{ RowBox[{"ToCharacterCode", "[", RowBox[{"StringTake", "[", RowBox[{"sourcetext", ",", RowBox[{"{", "i", "}"}]}], "]"}], "]"}], "-", "96"}], ",", RowBox[{ RowBox[{"ToCharacterCode", "[", RowBox[{"StringTake", "[", RowBox[{"sourcetext", ",", RowBox[{"{", RowBox[{"i", "+", "1"}], "}"}]}], "]"}], "]"}], "-", "96"}]}], "]"}], "]"}]}]}]}], "Input", GeneratedCell->True, CellTags->"FormIntro Markov Eval"], Cell[BoxData[ RowBox[{"{", RowBox[{"{", "0.00021873926399999997`", "}"}], "}"}]], "Output"], Cell["\<\ Better approximations of a language can be made, by considering transition \ probabilities that depend on more than one letter in the past.\ \>", "Text", FontSlant->"Italic", FontColor->GrayLevel[0]], Cell[TextData[{ "Note, that in the three examples above, the models are all ", StyleBox["stationary", FontSlant->"Italic"], ", which means that ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["Pr", "plain"], "(", RowBox[{ RowBox[{ SubscriptBox["M", "j"], "=", SubscriptBox["m", "0"]}], ",", " ", RowBox[{ SubscriptBox["M", RowBox[{"j", "+", "1"}]], "=", SubscriptBox["m", "1"]}], ",", "\[Ellipsis]", ",", " ", RowBox[{ SubscriptBox["M", RowBox[{"j", "+", "n", "-", "1"}]], "=", SubscriptBox["m", RowBox[{"n", "-", "1"}]]}]}], ")"}], TraditionalForm]]], " is independent of the value of ", Cell[BoxData[ FormBox["j", TraditionalForm]]], ". In the middle of a regular text one may expect this property to hold, but \ in other situations this is not the case. Think for instance of the date at \ the beginning of a letter." }], "Text", CellTags->"DefIntro stationary"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], "\tProblems" }], "Section", CellTags->"SectIntro Problems"], Cell[TextData[{ StyleBox["Problem ", FontWeight->"Bold"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold"], StyleBox[ CounterBox["Problem"], FontWeight->"Bold"], "\nWhat is the probability that the text \"apple'' occurs, when the \ plaintext source generates independent, identically distributed 1-grams, as \ described in ", ButtonBox["Example 1.1", BaseStyle->"Hyperlink", ButtonData:>"ExamIntro 1-grams"], ".\nAnswer the same question when the Markov model of ", ButtonBox["Example 1.3", BaseStyle->"Hyperlink", ButtonData:>"ExamIntro Markov"], " is used?" }], "Problem"], Cell[TextData[{ StyleBox["Problem ", FontWeight->"Bold"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold"], StyleBox[ CounterBox["Problem"], FontWeight->"Bold"], Cell[BoxData[ FormBox[ RowBox[{" ", SuperscriptBox["", "M"]}], TraditionalForm]]], "\nUse the ", StyleBox["Mathematica", FontSlant->"Italic"], " function ", ButtonBox["Permutations", BaseStyle->"Link", ButtonData->"paclet:ref/Permutations"], " and the ", ButtonBox["input formula", BaseStyle->"Hyperlink", ButtonData:>"FormIntro Markov Eval"], " at the end of Section 1.3 to determine for each of the 24 orderings of the \ four letters ", Cell[BoxData[ FormBox[ RowBox[{"e", ",", " ", "h", ",", " ", "l", ",", " ", "p"}], TraditionalForm]]], " the probability that it occurs in a language generated by the Markov model \ of ", ButtonBox["Example 1.3", BaseStyle->"Hyperlink", ButtonData:>"ExamIntro Markov"], "." }], "Problem"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], "\t", "Classical Cryptosystems" }], "Chapter", CellTags->"Chap Class"], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], "\tCaesar, Simple Substitution, Vigen\[EGrave]re" }], "Section", CellTags->"SectClass Caesar"], Cell[TextData[{ "In this chapter we shall discuss a number of classical cryptosystems. For \ further reading we refer the interested reader to (", ButtonBox["[BekP82]", BaseStyle->"Hyperlink", ButtonData:>"RefBek82"], ", ", ButtonBox["[Denn82]", BaseStyle->"Hyperlink", ButtonData:>"RefDen82"], ", ", ButtonBox["[Kahn67]", BaseStyle->"Hyperlink", ButtonData:>"RefKahn67"], ", ", ButtonBox["[Konh81]", BaseStyle->"Hyperlink", ButtonData:>"RefKonh81"], ", or ", ButtonBox["[MeyM82]", BaseStyle->"Hyperlink", ButtonData:>"RefMey82"], "). " }], "Text", GeneratedCell->True], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\t", "Caesar Cipher" }], "Subsection", CellTags->"SubsClass Caesar"], Cell[TextData[{ "One of the oldest cryptosystems is due to Julius Caesar. It shifts each \ letter in the text cyclicly over ", Cell[BoxData[ FormBox["k", TraditionalForm]]], " places. So, with ", Cell[BoxData[ FormBox[ RowBox[{"k", "=", "7"}], TraditionalForm]]], " one gets the following encryption of the word cleopatra (note that the \ letter ", Cell[BoxData[ FormBox["z", TraditionalForm]]], " is mapped to ", Cell[BoxData[ FormBox["a", TraditionalForm]]], "):" }], "Text"], Cell[TextData[{ StyleBox["cleopatra", FontColor->RGBColor[0.500008, 0.500008, 0]], " ", Cell[BoxData[ FormBox[ OverscriptBox["\[LongRightArrow]", RowBox[{"+", "1"}]], TraditionalForm]]], " dmfpqbusb ", Cell[BoxData[ FormBox[ OverscriptBox["\[LongRightArrow]", RowBox[{"+", "1"}]], TraditionalForm]]], " engqrcvtc ", Cell[BoxData[ FormBox[ OverscriptBox["\[LongRightArrow]", RowBox[{"+", "1"}]], TraditionalForm]]], " fohrsdwud ", Cell[BoxData[ FormBox[ OverscriptBox["\[LongRightArrow]", RowBox[{"+", "1"}]], TraditionalForm]]], " gpistexve ", Cell[BoxData[ FormBox[ OverscriptBox["\[LongRightArrow]", RowBox[{"+", "1"}]], TraditionalForm]]], " hqjtufywf ", Cell[BoxData[ FormBox[ OverscriptBox["\[LongRightArrow]", RowBox[{"+", "1"}]], TraditionalForm]]], " irkuvgzxg ", Cell[BoxData[ FormBox[ OverscriptBox["\[LongRightArrow]", RowBox[{"+", "1"}]], TraditionalForm]]], " ", StyleBox["jslvwhayh", FontColor->RGBColor[0, 0.500008, 0]] }], "Text"], Cell[TextData[{ "By using the ", StyleBox["Mathematica", FontSlant->"Italic"], " functions ", StyleBox["ToCharacterCode", FontVariations->{"Underline"->True}], " and ", StyleBox["FromCharacterCode", FontVariations->{"Underline"->True}], ", which convert symbols to their ASCI code and back (letter ", Cell[BoxData[ FormBox["a", TraditionalForm]]], " has value 97, letter ", Cell[BoxData[ FormBox["b", TraditionalForm]]], " has value 98, etc.), the Caesar cipher can be executed by the following \ function:" }], "Text", CellChangeTimes->{{3.4233911893080425`*^9, 3.423391236167717*^9}, { 3.4233917968588057`*^9, 3.4233918116870255`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"CaesarCipher", "[", RowBox[{"plaintext_", ",", " ", "key_"}], "]"}], ":=", "\t", RowBox[{"FromCharacterCode", "[", " ", RowBox[{ RowBox[{"Mod", "[", " ", RowBox[{ RowBox[{ RowBox[{"ToCharacterCode", "[", "plaintext", "]"}], " ", "-", " ", "97", " ", "+", "key"}], ",", " ", "26"}], "]"}], " ", "+", " ", "97"}], "]"}]}]], "Input", GeneratedCell->True, CellTags->"DefClass Caesar"], Cell["An example is given below.", "Text"], Cell[BoxData[{ RowBox[{ RowBox[{"plaintext", "=", "\"\\""}], ";"}], "\n", RowBox[{ RowBox[{"key", "=", "24"}], ";"}], "\n", RowBox[{"CaesarCipher", "[", RowBox[{"plaintext", ",", "key"}], "]"}]}], "Input", GeneratedCell->True], Cell[BoxData["\<\"rwncfcpcwmspnjyglrcvrglqkyjjjcrrcpq\"\>"], "Output"], Cell[TextData[{ "In the terminology of ", ButtonBox["Section 1.2", BaseStyle->"Hyperlink", ButtonData:>"DefIntro Crypto Syst"], ", the ", StyleBox["Caesar cipher", FontSlant->"Italic"], " is defined over the alphabet ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"{", RowBox[{"0", ",", "1", ",", "\[Ellipsis]", ",", "25"}], "}"}], " "}], TraditionalForm]]], " by:" }], "Text", CellTags->"DefClass Caesar Cipher"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ SubscriptBox["E", "k"], RowBox[{"(", "m", ")"}]}], " ", "=", " ", RowBox[{"(", RowBox[{ RowBox[{"(", RowBox[{"m", "+", "k"}], ")"}], " ", "mod", " ", "26"}], ")"}]}], ",", " ", RowBox[{"0", "\[LessEqual]", "m", "<", "26"}], ","}]], "DisplayFormula"], Cell["and", "Text"], Cell[BoxData[ RowBox[{" ", RowBox[{ RowBox[{"\[GothicCapitalE]", "=", " ", RowBox[{"{", RowBox[{ SubscriptBox["E", "k"], "|", RowBox[{"0", "\[LessEqual]", "k", "<", "26"}]}], "}"}]}], ","}]}]], "DisplayFormula"], Cell[TextData[{ "where ", Cell[BoxData[ FormBox[ RowBox[{"(", RowBox[{"i", " ", "mod", " ", "n"}], ")"}], TraditionalForm]]], " denotes the unique integer ", Cell[BoxData[ FormBox["j", TraditionalForm]]], " satisfying ", Cell[BoxData[ FormBox[ RowBox[{"j", "\[Congruent]", RowBox[{"i", " ", RowBox[{"(", RowBox[{"mod", " ", "n"}], ")"}]}]}], TraditionalForm]]], " and ", Cell[BoxData[ FormBox[ RowBox[{"0", "\[LessEqual]", "j", "<", "n"}], TraditionalForm]]], ". In this case, the key space \[ScriptCapitalK] is the set ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"0", ",", "1", ",", "\[Ellipsis]", ",", "25"}], "}"}], TraditionalForm]]], " and ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["D", "k"], "=", SubscriptBox["E", RowBox[{"q", "-", "1", "-", "k"}]]}], TraditionalForm]]], ". " }], "Text"], Cell[TextData[{ "An easy way to break the system is to try out all possible keys. This \ method is called ", StyleBox["exhaustive key search", FontSlant->"Italic"], ". In ", ButtonBox["Table 2.1", BaseStyle->"Hyperlink", ButtonData:>"TableClass Anal Caesar"], " one can find the cryptanalysis of the ciphertext \"", StyleBox["xyuysuyifvyxi", FontColor->RGBColor[0, 0.500008, 0]], "\"." }], "Text", CellTags->"DefClass exh key search"], Cell[TextData[{ Cell[BoxData[GridBox[{ { StyleBox["x", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0, 0.500008, 0]], StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0, 0.500008, 0]], StyleBox["u", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0, 0.500008, 0]], StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", 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"f", "c", "s", "v", "u", "f"}, { StyleBox["t", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["u", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["q", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["u", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["o", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["q", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["u", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["e", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["b", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["r", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["u", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["t", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox["e", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]]} }]]], "\n\nCryptanalysis of the Caesar cipher " }], "NumberedTable", CellTags->"TableClass Anal Caesar"], Cell[TextData[{ "To decrypt the ciphertext ", StyleBox["yhaklwpnw", FontColor->RGBColor[0.500008, 0.500008, 0]], "., one can easily check all keys with the ", ButtonBox["CaesarCipher", BaseStyle->"Hyperlink", ButtonData:>"DefClass Caesar"], " function 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",", "\<\"bkdnozsqz\"\>", ",", "\<\"ajcmnyrpy\"\>", ",", "\<\"ziblmxqox\"\>", ",", "\<\"yhaklwpnw\"\>"}], "}"}]], "Output"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\t", "Simple Substitution" }], "Subsection", CellTags->"SubsClass Simple sub"], Cell[CellGroupData[{ Cell["The System and its Main Weakness", "Subsubsection", CellTags->"SubsubsClass weakness simple"], Cell[TextData[{ "With the method of a ", StyleBox["simple substitution", FontSlant->"Italic"], " one chooses a fixed permutation \[Pi] of the alphabet ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"{", RowBox[{"a", ",", "b", ",", "\[Ellipsis]", ",", "z"}], "}"}], " "}], TraditionalForm]]], " and applies that to all letters in the plaintext." }], "Text", CellTags->"DefClass SimpleSubst"], Cell[TextData[{ StyleBox["Example ", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold", FontSlant->"Plain"], StyleBox[".", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Example"], FontWeight->"Bold", FontSlant->"Plain"], "\nIn the following example we only give that part of the substitution \[Pi] \ that is relevant for the given plaintext. We use the ", StyleBox["Mathematica", FontSlant->"Italic"], " function ", StyleBox["StringReplace", FontVariations->{"Underline"->True}], "." }], "Example", CellChangeTimes->{{3.423391904703246*^9, 3.4233919124532957`*^9}}], Cell[BoxData[ RowBox[{"StringReplace", "[", RowBox[{"\"\\"", ",", RowBox[{"{", RowBox[{ RowBox[{"\"\\"", "->", "\"\\""}], ",", " ", RowBox[{"\"\\"", "->", "\"\\""}], ",", " ", RowBox[{"\"\\"", "->", "\"\\""}], ",", " ", RowBox[{"\"\\"", "->", "\"\\""}], ",", "\t\t", RowBox[{"\"\\"", "->", "\"\\""}], ",", " ", RowBox[{"\"\\"", "->", "\"\\""}], ",", " ", RowBox[{"\"\\"", "->", "\"\\""}], ",", " ", RowBox[{"\"\\"", "->", "\"\\""}]}], "}"}]}], "]"}]], "Input", GeneratedCell->True], Cell[BoxData["\<\"vrkbaqzdq\"\>"], "Output"], Cell[TextData[{ "A more formal description of the simple substitution system is as follows: \ the key space \[ScriptCapitalK] is the set ", Cell[BoxData[ FormBox[ SubscriptBox["S", "q"], TraditionalForm]]], " of all permutations of ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"0", ",", "1", ",", "\[Ellipsis]", ",", RowBox[{"q", "-", "1"}]}], "}"}], TraditionalForm]]], " and the cryptosystem \[GothicCapitalE] is given by" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"\[ScriptCapitalE]", "=", RowBox[{"{", RowBox[{ SubscriptBox["E", "\[Pi]"], "|", RowBox[{"\[Pi]", "\[Element]", SubscriptBox["S", "q"]}]}], "}"}]}], ","}]], "DisplayFormula"], Cell["where", "Text"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ SubscriptBox["E", "\[Pi]"], RowBox[{"(", "m", ")"}]}], "=", RowBox[{"\[Pi]", " ", RowBox[{"(", "m", ")"}]}]}], ",", " ", RowBox[{"0", "\[LessEqual]", "m", "<", RowBox[{"q", "."}]}]}]], "DisplayFormula"], Cell[TextData[{ "The decryption function ", Cell[BoxData[ FormBox[ SubscriptBox["D", "\[Pi]"], TraditionalForm]]], " is given by ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["D", "\[Pi]"], "=", SubscriptBox["E", SuperscriptBox["\[Pi]", RowBox[{"-", "1"}]]]}], TraditionalForm]]], ", as follows from" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ SubscriptBox["D", "\[Pi]"], RowBox[{"(", RowBox[{ SubscriptBox["E", "\[Pi]"], RowBox[{"(", "m", ")"}]}], ")"}]}], "=", RowBox[{ RowBox[{"D", RowBox[{"(", RowBox[{"\[Pi]", RowBox[{"(", "m", ")"}]}], ")"}]}], "=", RowBox[{ RowBox[{ SubscriptBox["E", SuperscriptBox["\[Pi]", RowBox[{"-", "1"}]]], RowBox[{"(", RowBox[{"\[Pi]", RowBox[{"(", "m", ")"}]}], ")"}]}], "=", RowBox[{ RowBox[{ SuperscriptBox["\[Pi]", RowBox[{"-", "1"}]], RowBox[{"(", RowBox[{"\[Pi]", RowBox[{"(", "m", ")"}]}], ")"}]}], "=", "m"}]}]}]}], ",", " ", RowBox[{"0", "\[LessEqual]", "m", "<", RowBox[{"q", "."}]}]}]], "DisplayFormula"], Cell[TextData[{ "Unlike Caesar's cipher, this system does not have the drawback of a small \ key space. Indeed, ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"|", "\[ScriptCapitalK]", "|"}], "=", RowBox[{ RowBox[{"|", SubscriptBox["S", "26"], "|"}], "=", RowBox[{ RowBox[{"26", "!"}], "\[TildeTilde]"}]}]}], TraditionalForm]]], "4.03 ", Cell[BoxData[ FormBox[ SuperscriptBox["10", "26"], TraditionalForm]]], ". This system however does demonstrate very well that a large key space \ should not fool one into believing that a system is secure! On the contrary, \ by simply counting the letter frequencies in the ciphertexts and comparing \ these with the letter frequencies in ", ButtonBox["Table 1.1", BaseStyle->"Hyperlink", ButtonData:>"TableIntr Prob English"], ", one very quickly finds the images under \[Pi] of the most frequent \ letters in the plaintext. Indeed, the most frequent letter in the ciphertext \ will very likely be the image under \[Pi] of the letter ", Cell[BoxData[ FormBox["e", TraditionalForm]]], ". The next one is the image of the letter ", Cell[BoxData[ FormBox["n", TraditionalForm]]], ", etc. After having found the encryptions of the most frequent letters in \ the plaintext, it is not difficult to fill in the rest. Of course, the longer \ the cipher text, the easier the cryptanalysis becomes. In ", ButtonBox["Chapter 5", BaseStyle->"Hyperlink", ButtonData:>"Chap Shannon"], ", we come back to the cryptanalysis of the system, in particular how long \ the same key can be used safely." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Cryptanalysis by The Method of a Probable Word", "Subsubsection", CellTags->"DefClass Prob Word"], Cell["\<\ In the following example we have knowledge of a very long ciphertext. This is \ not necessary at all for the cryptanalysis of the ciphertext, but it takes \ that long to know the full key. Indeed, as long as two letters are missing in \ the plaintext, one does not know the full key, but the system is of course \ broken much earlier than that.\ \>", "Text"], Cell[TextData[{ "Apart from the ciphertext, given in ", ButtonBox["Table 2.2", BaseStyle->"Hyperlink", ButtonData:>"TableClass Anal Simple Subst"], ", we shall assume in this example that the plaintext discusses the concept \ of ''bidirectional communication theory''. Cryptanalysis will turn out to be \ very easy. " }], "Text"], Cell[BoxData[{ FormBox[ RowBox[{ StyleBox[GridBox[{ {"zhjeo", "ndize", "hicle", "osiol", "digic", "lmhzq", "zolyi", "zehdp", "zhjeo", "ndize"}, {"hycdh", "hlpvs", "uczyc", "dhzhj", "eondi", "zehge", "moylk", "zhjpm", "lhylg", "gidiz"}, {"gizyd", "ppsdo", "lylzr", "losye", "nnmhz", "ydize", "hicle", "osceu", "lrloq", "lgyoz"}, {"vlgic", "lneol", "flhlo", "dpydg", "lzhuc", "zyciu", "eeone", "olzhj", "eondi", "zehge"}, {"moylg", "zhjpm", "lhyll", "dycei", RowBox[{"clogi", " "}], "dizgi", "zydpp", "siclq", "zolyi", "zehej"}, {"iczgz", "hjpml", "hylzg", "lkaol", "gglqv", "sqzol", "yilqi", "odhgj", "eondi", "zehxm"}, {"dhizi", "zlguc", "zycyd", "hehps", "vlqlo", "zrlqz", "jiclp", "duejy", "dmgdp", "ziszg"}, {"evglo", "rlqqz", "gizhf", "mzgcz", "hficl", "ldopz", "loydm", "gljoe", "niclp", "dilol"}, {"jjlyi", "zhvze", "pefsd", "hqgey", "zepef", "syenn", "mhzyd", "izehi", "cleos", "gllng"}, {"iecdr", "luzql", "daapz", "ydize", "hgqml", "ieicl", "jdyii", "cdipz", "rzhfv", "lzhfg"}, {"dolvs", "iclzo", "dyize", "hggem", "oylge", "jzhje", "ondiz", "ehucz", "yczhj", "pmlhy"}, {"lldyc", "eiclo", "zhdpp", "aeggz", "vplqz", "olyiz", "ehgic", "laolg", "lhiad", "aloql"}, {"gyzvl", "gicly", "dglej", "vzqzo", "lyize", "hdpye", "nnmhz", "ydize", "hicle", "osdaa"}, {"pzlqi", "eiclg", "eyzdp", "vlcdr", "zemoe", "jneht", RowBox[{"lsg\[Ellipsis]", " "}], " ", " ", " "} }], FontFamily->"Courier New"], "\n"}], TextForm], "\n", FormBox[ FormBox[ RowBox[{"\n", RowBox[{ "Ciphertext", " ", "obtained", " ", "with", " ", "a", " ", "simple", " ", "substitution"}]}], TextForm], TextForm]}], "NumberedTable", CellMargins->{{17, 57}, {Inherited, Inherited}}, CellChangeTimes->{{3.4233920035945034`*^9, 3.423392006907025*^9}}, TextAlignment->Center, TextJustification->0, CellTags->"TableClass Anal Simple Subst"], Cell["\<\ Assuming that the word \"communication'' will occur in the plaintext, we look \ for strings of 13 consecutive letters, in which letter 1 = letter 8, letter 2 \ = letter 12, letter 3 = letter 4, letter 6 = letter 13 and letter 7 = letter \ 11. 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CellMargins->{{22, 23.0625}, {Inherited, Inherited}}, TextAlignment->Center, TextJustification->0, FontSize->9, CellTags->"TableClass Vigenere"], Cell[TextData[{ "Because of the redundancy in the English language one reduces the effective \ size of the key space tremendously by choosing an existing word as the key. \ Taking the name of a relative, as we have done above, reduces the security of \ the encryption more or less to zero.\nIn ", StyleBox["Mathematica", FontSlant->"Italic"], ", addition of two letters as defined by the Vigen\[EGrave]re Table can be \ realized in a similar way, as our earlier implementation of the ", ButtonBox["Caesar cipher", BaseStyle->"Hyperlink", ButtonData:>"DefClass Caesar"], ":" }], "Text", FontSlant->"Italic"], Cell[BoxData[ RowBox[{ RowBox[{"AddTwoLetters", "[", RowBox[{"a_", ",", "b_"}], "]"}], ":=", RowBox[{"FromCharacterCode", "[", RowBox[{ RowBox[{"Mod", "[", RowBox[{ RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"ToCharacterCode", "[", "a", "]"}], "-", 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Cell["with", "Text"], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["c", "i"], "=", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"(", RowBox[{ SubscriptBox["m", "i"], "+", SubscriptBox["k", RowBox[{"(", RowBox[{"i", " ", "mod", " ", "r"}], ")"}]]}], ")"}], " ", "mod", " ", "26"}], ")"}], "."}]}], TraditionalForm]], "NumberedEquation", CellTags->"FormClass Vigen Encr"], Cell[TextData[{ "Instead of using ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " Caesar ciphers periodically in the Vigen\[EGrave]re cryptosystem, one can \ of course also use ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " simple substitutions. Such a system is an example of a so-called ", StyleBox["polyalphabetic substitution", FontSlant->"Italic"], ". For centuries, no one had an effective way of breaking this system, \ mainly because one did not have a technique of determining the key length ", Cell[BoxData[ FormBox["r", TraditionalForm]]], ". Once one knows ", Cell[BoxData[ FormBox["r", TraditionalForm]]], ", one can find the ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " simple substitutions by grouping together the letters ", Cell[BoxData[ FormBox[ RowBox[{"i", ",", RowBox[{"i", "+", "r"}], ",", " ", RowBox[{"i", "+", RowBox[{"2", "r"}]}], ",", "\[Ellipsis]", ","}], TraditionalForm]], CellTags->"DefClass Poly subst"], " for each ", Cell[BoxData[ FormBox[ RowBox[{"i", ",", RowBox[{"0", "\[LessEqual]", "i", "<", "r"}], ","}], TraditionalForm]]], " and break each of these ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " simple substitutions individually. In 1863, the Prussian army officer, \ F.W. Kasiski, solved the problem of finding the key length ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " by statistical means. In the next section, we shall discuss this method." }], "Text", CellTags->"DefClass Poly Subst"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], "\tThe Incidence of Coincidences, Kasiski's Method" }], "Section", CellTags->"SectClass Incid Coin"], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\tThe Incidence of Coincidences" }], "Subsection", CellTags->"SubSClass incid of coinc"], Cell[TextData[{ "Consider a ciphertext ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["c", FontFamily->"Times", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "=", SubscriptBox["c", "0"]}], ",", SubscriptBox["c", "1"], ",", "\[Ellipsis]", ",", SubscriptBox["c", RowBox[{"n", "-", "1"}]]}], TraditionalForm]]], "which is the result of a ", ButtonBox["Vigen\[EGrave]re encryption", BaseStyle->"Hyperlink", ButtonData:>"DefClass Vigenere"], " of an English plaintext ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "=", SubscriptBox["m", "0"]}], ",", SubscriptBox["m", "1"], ",", "\[Ellipsis]", ",", SubscriptBox["m", RowBox[{"n", "-", "1"}]]}], TraditionalForm]]], "under the key ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["k", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "=", SubscriptBox["k", "0"]}], ",", SubscriptBox["k", "1"], ",", "\[Ellipsis]", ",", SubscriptBox["k", RowBox[{"r", "-", "1"}]]}], TraditionalForm]]], "(see also ", ButtonBox["(2.1)", BaseStyle->"Hyperlink", ButtonData:>"FormClass Vigen Encr"], "). As explained at the end of the previous section, the key to breaking the \ Vigen\[EGrave]re system is to determine the key length ", Cell[BoxData[ FormBox["r", TraditionalForm]]], "." }], "Text"], Cell[TextData[{ "In our analysis we are going to assume the very simple model of a plaintext \ source outputting independent, individual letters, each with probability \ distribution given by ", ButtonBox["Table 1.1", BaseStyle->"Hyperlink", ButtonData:>"TableIntr Prob English"], " (see ", ButtonBox["Example 1.1", BaseStyle->"Hyperlink", ButtonData:>"ExamIntro 1-grams"], "). We further assume that the letters ", Cell[BoxData[ FormBox[ SubscriptBox["k", "i"], TraditionalForm]]], " in the key are chosen with independent and uniform distribution from ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"a", ",", "b", ",", "\[Ellipsis]", ",", "z"}], "}"}], TraditionalForm]]], " (so, with probability 1/26)." }], "Text"], Cell[TextData[{ "Let ", Cell[BoxData[ FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "left", RowBox[{"(", "i", ")"}]], TraditionalForm]]], "and ", Cell[BoxData[ FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "right", RowBox[{"(", "i", ")"}]], TraditionalForm]]], " the substrings of ", Cell[BoxData[ FormBox["c", TraditionalForm]], FontWeight->"Bold"], " consisting of the ", Cell[BoxData[ FormBox["i", TraditionalForm]]], " left most resp. right most symbols of ", StyleBox["c", FontWeight->"Bold"], ", so:" }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "left", RowBox[{"(", "i", ")"}]], "=", SubscriptBox["c", "0"]}], ",", SubscriptBox["c", "1"], ",", "\[Ellipsis]", ",", SubscriptBox["c", RowBox[{"i", "-", "1"}]]}], TraditionalForm]]], "\t\tand\t\t", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "right", RowBox[{"(", "i", ")"}]], "=", SubscriptBox["c", RowBox[{"n", "-", "i"}]]}], ",", SubscriptBox["c", RowBox[{"n", "-", "i", "+", "1"}]], ",", "\[Ellipsis]", ",", SubscriptBox["c", RowBox[{"n", "-", "1"}]]}], TraditionalForm]]], "." }], "DisplayFormula"], Cell[TextData[{ "Let us now count the number of agreements between ", Cell[BoxData[ FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "left", RowBox[{"(", "i", ")"}]], TraditionalForm]]], "and ", Cell[BoxData[ FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "right", RowBox[{"(", "i", ")"}]], TraditionalForm]]], ", i.e. the number of coordinates ", Cell[BoxData[ FormBox["j", TraditionalForm]]], " where ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ RowBox[{"(", SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "left", RowBox[{"(", "i", ")"}]], ")"}], "j"], "=", SubscriptBox[ RowBox[{"(", FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "right", RowBox[{"(", "i", ")"}]], TraditionalForm], ")"}], "j"]}], TraditionalForm]]], ". We shall show in ", ButtonBox["Lemma 2.1", BaseStyle->"Hyperlink", ButtonData:>"LemClass Kasiski"], " that the expected value of this number divided by the string length ", Cell[BoxData[ FormBox["i", TraditionalForm]]], " will be ", Cell[BoxData[ FormBox["0.06875", TraditionalForm]]], " or ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"1", "/", "26"}], "\[TildeTilde]", "0.03846"}], TraditionalForm]]], ", depending on whether the (unknown) key length ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " divides ", Cell[BoxData[ FormBox[ RowBox[{"n", "-", "i"}], TraditionalForm]]], " or does not divide ", Cell[BoxData[ FormBox[ RowBox[{"n", "-", "i"}], TraditionalForm]]], "." }], "Text"], Cell[TextData[{ "Let us show by example how this difference in expected values can be used \ to determine the unknown key length ", Cell[BoxData[ FormBox["r", TraditionalForm]]], "." }], "Text"], Cell[TextData[{ StyleBox["Example ", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold", FontSlant->"Plain"], StyleBox[".", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Example"], FontWeight->"Bold", FontSlant->"Plain"], "\nIn this example we consider the ciphertext " }], "Example"], Cell["\<\ \"glrtnhklttbrxbxwnnhshjwkcjmsmrwnxqmvehuimnfxbzcwixbrnhxqhhclgcipcgimglrtnhkl\ ttbrshvilgwcmwyejqbxbmlywimbkhhjwkcjmsmrwnxqmplceiwkcjmehtpslmmlxowmylxbxflxee\ brahjwkcjmsmrwnxqm\".\ \>", "DisplayFormula", FontSlant->"Italic"], Cell[TextData[{ "By means of the ", StyleBox["Mathematica", FontSlant->"Italic"], " functions ", StyleBox["StringTake", FontVariations->{"Underline"->True}], ", ", StyleBox["StringLength", FontVariations->{"Underline"->True}], ", ", StyleBox["Characters", FontVariations->{"Underline"->True}], ", ", StyleBox["Count", FontVariations->{"Underline"->True}], ", and ", StyleBox["Table", FontVariations->{"Underline"->True}], ", we can easily compute the number of agreements between ", Cell[BoxData[ FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "left", RowBox[{"(", "i", ")"}]], TraditionalForm]]], "and ", Cell[BoxData[ FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "right", RowBox[{"(", "i", ")"}]], TraditionalForm]]], " in any range of values of ", Cell[BoxData[ FormBox["i", TraditionalForm]]], ":" }], "Text", CellChangeTimes->{{3.4233917193270597`*^9, 3.4233917232020845`*^9}, 3.423392796177701*^9, {3.423393297415284*^9, 3.423393300977807*^9}, { 3.4233940093417153`*^9, 3.423394015044877*^9}, {3.4233941832647038`*^9, 3.423394185499093*^9}}, FontSlant->"Italic"], Cell[BoxData[{ RowBox[{ RowBox[{ "ciphertext", "=", "\"\\"\ "}], ";"}], "\n", RowBox[{ RowBox[{"L", "=", RowBox[{"StringLength", "[", "ciphertext", "]"}]}], ";"}], "\n", RowBox[{ RowBox[{"Table", "[", " ", RowBox[{ RowBox[{"N", "[", " ", RowBox[{ RowBox[{ RowBox[{"Count", "[", " ", RowBox[{ RowBox[{ RowBox[{"Characters", "[", " ", RowBox[{"StringTake", "[", " ", RowBox[{"ciphertext", ",", " ", "i"}], "]"}], "]"}], " ", "-", " ", RowBox[{"Characters", "[", " ", RowBox[{"StringTake", "[", " ", RowBox[{"ciphertext", ",", " ", RowBox[{"-", "i"}]}], "]"}], "]"}]}], ",", " ", "0"}], "]"}], "/", "i"}], ",", " ", "1"}], "]"}], ",", " ", RowBox[{"{", RowBox[{"i", ",", " ", RowBox[{"L", "-", "20"}], ",", " ", RowBox[{"L", "-", "1"}]}], "}"}]}], " ", "]"}], " \t"}]}], "Input", GeneratedCell->True], Cell[BoxData[ RowBox[{"{", RowBox[{ "0.02857142857142857`", ",", "0.042704626334519574`", ",", "0.07801418439716312`", ",", "0.0176678445229682`", ",", "0.045774647887323945`", ",", "0.03859649122807018`", ",", "0.038461538461538464`", ",", "0.0313588850174216`", ",", "0.059027777777777776`", ",", "0.06920415224913495`", ",", "0.05517241379310345`", ",", "0.044673539518900345`", ",", "0.023972602739726026`", ",", "0.051194539249146756`", ",", "0.08163265306122448`", ",", "0.04406779661016949`", ",", "0.0472972972972973`", ",", "0.020202020202020204`", ",", "0.013422818791946308`", ",", "0.046822742474916385`"}], "}"}]], "Output"], Cell[TextData[{ "The (relative) higher values in this listing at places ", Cell[BoxData[ FormBox[ RowBox[{"-", "6"}], TraditionalForm]]], " and ", Cell[BoxData[ FormBox[ RowBox[{"-", "18"}], TraditionalForm]]], " indicate that the key length ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " is 6. Indeed, the key that has been used to generate this example is the \ word \"monkey'', which has 6 letters.\nThis can be checked with the following \ analogue of the Vigen\[EGrave]re encryption of ", ButtonBox["Example 2.3", BaseStyle->"Hyperlink", ButtonData:>"ExamClass Vigenere"], "." }], "Text", FontSlant->"Italic"], Cell[BoxData[ RowBox[{ RowBox[{"SubTwoLetters", "[", RowBox[{"a_", ",", "b_"}], "]"}], ":=", RowBox[{"FromCharacterCode", "[", RowBox[{ RowBox[{"Mod", "[", RowBox[{ RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"ToCharacterCode", "[", "a", "]"}], "-", "97"}], ")"}], "-", RowBox[{"(", RowBox[{ RowBox[{"ToCharacterCode", "[", "b", "]"}], "-", "97"}], ")"}]}], ",", "26"}], "]"}], "+", "97"}], "]"}]}]], "Input", GeneratedCell->True], Cell[BoxData[{ RowBox[{ RowBox[{ "ciphertext", "=", "\"\\"\ "}], ";"}], "\n", RowBox[{ RowBox[{"key", "=", "\"\\""}], ";"}], "\n", RowBox[{ RowBox[{"plaintext", "=", "\"\<\>\""}], ";"}], "\n", RowBox[{"Do", "[", RowBox[{ RowBox[{"plaintext", "=", RowBox[{"plaintext", "<>", RowBox[{"SubTwoLetters", "[", RowBox[{ RowBox[{"StringTake", "[", RowBox[{"ciphertext", ",", RowBox[{"{", "i", "}"}]}], "]"}], ",", "\t", RowBox[{"StringTake", "[", RowBox[{"key", ",", RowBox[{"{", RowBox[{ RowBox[{"Mod", "[", RowBox[{ RowBox[{"i", "-", "1"}], ",", RowBox[{"StringLength", "[", "key", "]"}]}], "]"}], "+", "1"}], "}"}]}], "]"}]}], "]"}]}]}], ",", "\t", RowBox[{"{", RowBox[{"i", ",", "1", ",", RowBox[{"StringLength", "[", "ciphertext", "]"}]}], "}"}]}], "]"}], "\n", "plaintext"}], "Input", CellChangeTimes->{{3.4233928068340197`*^9, 3.423392824787259*^9}}], Cell[BoxData["\<\"\ informationtheorytreatstheunidirectionalikformationchannelbywhichaninformation\ sourceinfluencesstatisticallyareceivercommunpcationtheoryhoweverdescribesthemo\ regeneralcaseinwhichtwoormoreinformationsourcesinfluenceeachotherstatistically\ thedirectionofthisinfluenceisexpressedbydsrectedtransinformationqu\"\>"], \ "Output"], Cell[TextData[{ StyleBox["Lemma ", FontWeight->"Bold"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold"], StyleBox[ CounterBox["Theorem"], FontWeight->"Bold"], "\nLet ", Cell[BoxData[ FormBox[ StyleBox["c", FontWeight->"Bold", FontSlant->"Italic"], TraditionalForm]]], " be a ciphertext which is the result of a Vigen\[EGrave]re encryption of a \ plaintext ", Cell[BoxData[ FormBox[ StyleBox["m", FontSize->12, FontWeight->"Bold", FontSlant->"Italic", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], TraditionalForm]]], " of length ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " with key ", Cell[BoxData[ FormBox[ StyleBox["k", FontSize->12, FontWeight->"Bold", FontSlant->"Italic", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], TraditionalForm]]], " of length ", Cell[BoxData[ FormBox["r", TraditionalForm]]], ". \nSuppose that ", StyleBox["m", FontWeight->"Bold", FontSlant->"Italic"], " is generated by the plaintext source of ", ButtonBox["Example 1.1", BaseStyle->"Hyperlink", ButtonData:>"ExamIntro 1-grams"], ". So, all the letters in ", StyleBox["m", FontWeight->"Bold", FontSlant->"Italic"], " are generated independently of each other, all with the frequency \ distribution ", Cell[BoxData[ FormBox[ RowBox[{"p", "(", "m", ")"}], TraditionalForm]]], " given by ", ButtonBox["Table 1.1", BaseStyle->"Hyperlink", ButtonData:>"TableIntr Prob English"], ". Suppose further that the letters ", Cell[BoxData[ FormBox[ SubscriptBox["k", "i"], TraditionalForm]]], " in the key are chosen with independent and uniform distribution from ", Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"a", ",", "b", ",", "\[Ellipsis]", ",", "z"}], "}"}], TraditionalForm]]], " (so, with probability 1/26).\nThen, for each ", Cell[BoxData[ FormBox[ RowBox[{"1", "\[LessEqual]", "i", "<", "j", "\[LessEqual]", "n"}], TraditionalForm]]], ",\n\n\t", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"Pr", "[", RowBox[{ SubscriptBox["c", "i"], "=", SubscriptBox["c", "j"]}], "]"}], "=", RowBox[{"{", GridBox[{ { RowBox[{ RowBox[{ RowBox[{ SubscriptBox["\[Sum]", "m"], SuperscriptBox[ RowBox[{"p", "(", "m", ")"}], "2"]}], "\[TildeTilde]", "0.06875"}], ","}]}, { RowBox[{ RowBox[{ RowBox[{ RowBox[{"1", "/", "26"}], "\[TildeTilde]", "0.03846"}], ","}], " "}]} }]}]}], TraditionalForm]], TextJustification->1], "\t\t", Cell[BoxData[ FormBox[GridBox[{ { RowBox[{ RowBox[{ RowBox[{ RowBox[{"if", " ", "r", " ", "divides", " ", "j"}], "-", "i"}], ","}], " "}]}, { RowBox[{ RowBox[{ "if", " ", "r", " ", "does", " ", "not", " ", "divide", " ", "j"}], "-", RowBox[{"i", "."}]}]} }], TraditionalForm]]] }], "Theorem", CellTags->"LemClass Kasiski"], Cell[TextData[{ StyleBox["Proof", FontWeight->"Bold"], ":\nIf ", Cell[BoxData[ FormBox[ RowBox[{"j", "-", "i"}], TraditionalForm]]], " is divisible by ", Cell[BoxData[ FormBox["r", TraditionalForm]]], ", then ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["c", "i"], "=", SubscriptBox["c", "j"]}], TraditionalForm]]], " if and only if ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["m", "i"], "=", SubscriptBox["m", "j"]}], TraditionalForm]]], ". This follows directly from formula ", ButtonBox["(2.1)", BaseStyle->"Hyperlink", ButtonData:>"FormClass Vigen Encr"], ", since ", Cell[BoxData[ FormBox[ RowBox[{"(", RowBox[{"j", " ", "mod", " ", "r"}], ")"}], TraditionalForm]]], " equals ", Cell[BoxData[ FormBox[ RowBox[{"(", RowBox[{"i", " ", "mod", " ", "r"}], ")"}], TraditionalForm]]], ". So, " }], "Text", FontColor->RGBColor[0.500008, 0, 0.500008]], Cell[TextData[Cell[BoxData[{ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ RowBox[{"Pr", "[", RowBox[{ SubscriptBox["c", "i"], "=", SubscriptBox["c", "j"]}], "]"}], "="}], TraditionalForm], RowBox[{"Pr", "[", RowBox[{ SubscriptBox["m", "i"], "=", SubscriptBox["m", "j"]}], "]"}]}], "="}], TraditionalForm], RowBox[{ SubscriptBox["\[Sum]", "m"], RowBox[{"Pr", "[", RowBox[{ SubscriptBox["m", "i"], "=", RowBox[{ SubscriptBox["m", "j"], "=", "m"}]}], "]"}]}]}], "=", "\n", "\t\t", AdjustmentBox[ RowBox[{ RowBox[{ SubscriptBox["\[Sum]", "m"], RowBox[{ RowBox[{"Pr", "[", RowBox[{ SubscriptBox["m", "i"], "=", "m"}], "]"}], RowBox[{"Pr", "[", RowBox[{ SubscriptBox["m", "j"], "=", "m"}], "]"}]}]}], "=", RowBox[{ RowBox[{ SubscriptBox["\[Sum]", "m"], SuperscriptBox[ RowBox[{"p", "(", "m", ")"}], "2"]}], "\[TildeTilde]", RowBox[{"0.06875", "."}]}]}], BoxBaselineShift->1.14286, BoxMargins->{{0, 0}, {-1.14286, 1.14286}}]}], TraditionalForm], "\n", FormBox[ RowBox[{"\t\t\t\t\t"}], TraditionalForm]}]]], "DisplayFormula", FontColor->RGBColor[0.500008, 0, 0.500008]], Cell[TextData[{ "If ", Cell[BoxData[ FormBox[ RowBox[{"j", "-", "i"}], TraditionalForm]]], " is not divisible by ", Cell[BoxData[ FormBox["r", TraditionalForm]]], ", then by ", ButtonBox["(2.1)", BaseStyle->"Hyperlink", ButtonData:>"FormClass Vigen Encr"], " ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["c", "i"], "=", SubscriptBox["c", "j"]}], TraditionalForm]]], " if and only if ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["m", "i"], "+", SubscriptBox["k", RowBox[{"(", RowBox[{"i", " ", "mod", " ", "r"}], ")"}]]}], "=", RowBox[{ SubscriptBox["m", "j"], "+", SubscriptBox["k", RowBox[{"(", RowBox[{"j", " ", "mod", " ", "r"}], ")"}]]}]}], TraditionalForm]]], ". Since ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", RowBox[{"j", " ", "mod", " ", "r"}], ")"}], " ", "\[NotEqual]"}], TraditionalForm]]], " ", Cell[BoxData[ FormBox[ RowBox[{"(", RowBox[{"i", " ", "mod", " ", "r"}], ")"}], TraditionalForm]]], ", it follows that ", Cell[BoxData[ FormBox[ SubscriptBox["k", RowBox[{"(", RowBox[{"j", " ", "mod", " ", "r"}], ")"}]], TraditionalForm]]], " takes on the value ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["m", "i"], "+", SubscriptBox["k", RowBox[{"(", RowBox[{"i", " ", "mod", " ", "r"}], ")"}]], "-", SubscriptBox["m", "j"]}], TraditionalForm]]], " with probability 1/26. We conclude that" }], "Text", FontColor->RGBColor[0.500008, 0, 0.500008]], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"Pr", "[", RowBox[{ SubscriptBox["c", "i"], "=", SubscriptBox["c", "j"]}], "]"}], "=", RowBox[{ RowBox[{"1", "/", "26"}], "\[TildeTilde]", "0.03846"}]}], TraditionalForm]]], "." }], "DisplayFormula", FontColor->RGBColor[0.500008, 0, 0.500008]], Cell["\[Square]", "Text", TextAlignment->Right, TextJustification->0, FontColor->RGBColor[0.500008, 0, 0.500008]], Cell[TextData[{ "It may be clear that with increasing length of the ciphertext, it is easier \ to determine the key length from the relative number of agreements between ", Cell[BoxData[ FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "left", RowBox[{"(", "i", ")"}]], TraditionalForm]]], "and ", Cell[BoxData[ FormBox[ SubsuperscriptBox[ StyleBox["c", FontSize->12, FontWeight->"Bold", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->GrayLevel[0]], "right", RowBox[{"(", "i", ")"}]], TraditionalForm]]], "." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\tKasiski's Method" }], "Subsection", CellTags->"SubSClass Kasiski"], Cell[TextData[{ "Kasiski based his cryptanalysis of the ", ButtonBox["Vigen", BaseStyle->"Hyperlink", ButtonData:>"DefClass Vigenere"], ButtonBox["\[EGrave]", BaseStyle->"Hyperlink", ButtonData:>"DefClass Vigenere"], ButtonBox["re cryptosystem", BaseStyle->"Hyperlink", ButtonData:>"DefClass Vigenere"], " on the fact that when a certain combination of letters (a frequent \ plaintext fragment) is encrypted more than once with the same segment of the \ key (because they occur at a multiple of the key length ", Cell[BoxData[ FormBox["r", TraditionalForm]]], "), one will see a repetition of the corresponding ciphertext at those \ places. " }], "Text"], Cell[TextData[{ "We quote an example from ", ButtonBox["[Baue97]", BaseStyle->"Hyperlink", ButtonData:>"RefBau97"], ":" }], "Text", GeneratedCell->True], Cell[TextData[{ StyleBox["Example ", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Chapter"], FontWeight->"Bold", FontSlant->"Plain"], StyleBox[".", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ CounterBox["Example"], FontWeight->"Bold", FontSlant->"Plain"], "\nConsider the following plaintext and ciphertext pair (where the key \ \"comet\" has been used): " }], "Example"], Cell[TextData[Cell[BoxData[GridBox[{ { StyleBox["plaintext", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0.500008, 0.500008, 0]], "t", "h", "e", "r", "e", "i", "s", "a", "n", "o", "t", "h", "e", "r", "f", "a", "m", "o", "u", "s", "p", "i", "a", "n", "o", "p", "l", "a", "y", "."}, { StyleBox["key", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}], "c", "o", "m", "e", "t", "c", "o", "m", "e", "t", "c", "o", "m", "e", "t", "c", "o", "m", "e", "t", "c", "o", "m", "e", "t", "c", "o", "m", "e", "."}, { StyleBox["ciphertext", FontFamily->"Courier", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontVariations->{"StrikeThrough"->False, "Underline"->False}, FontColor->RGBColor[0, 0.500008, 0]], "v", "v", "q", "v", "x", "k", "g", "m", "r", "h", "v", "v", "q", "v", "y", "c", "a", "a", "y", "l", "r", "w", "m", "r", "h", "r", "z", "m", "c", "."} }]]]], "Text"], Cell[TextData[{ "In the ciphertext one can find the substring \"vvqv\" (of length 4) \ repeated twice, namely starting at positions 1 and 11. This indicates that ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " divides 10. The substring \"mrh\" (of length 3) also occurs twice: at \ positions 8 and 23. So, it seems likely that ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " also divides 15. Combining these results, we conclude that ", Cell[BoxData[ FormBox[ RowBox[{"r", "=", "5"}], TraditionalForm]]], ", which is indeed the case." }], "Text", FontSlant->"Italic"], Cell[TextData[{ "See ", ButtonBox["[Baue97]", BaseStyle->"Hyperlink", ButtonData:>"RefBau97"], " for a further analysis of the Vigen\[EGrave]re cryptosystem." }], "Text", GeneratedCell->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], "\t", "Vernam, Playfair, Transpositions, Hagelin, Enigma" }], "Section", PageBreakAbove->True, CellTags->"SectClass Vernam"], Cell["\<\ In this section, we shall briefly discuss a few more cryptosystems, without \ going deep into their structure.\ \>", "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\t", "The One-Time Pad" }], "Subsection", CellTags->"SubsClas One-time"], Cell[TextData[{ "The ", StyleBox["one-time pad", FontSlant->"Italic"], ", also called the ", StyleBox["Vernam cipher", FontSlant->"Italic"], " (after the American A.T. & T. employee G.S. Vernam, who introduced the \ system in 1917), is a Vigen\[EGrave]re cipher with key length equal to the \ length of the plaintext. Also, the key must be chosen in a completely random \ way and can only be used once. In this way the system is unconditionally \ secure, as is intuitively clear and will be proved in ", ButtonBox["Chapter 5", BaseStyle->"Hyperlink", ButtonData:>"Chap Shannon"], ". The ''hot line'' between Washington and Moscow uses this system. The \ major drawback of this system is the length of the key, which makes this \ system impractical for most applications." }], "Text", TextAlignment->Left, TextJustification->0, CellTags->"DefClass Vernam"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\t", "The Playfair Cipher" }], "Subsection", CellTags->"SubsClass Playfair"], Cell[TextData[{ "The ", StyleBox["Playfair cipher", FontSlant->"Italic"], " (1854, named after the Englishman L. Playfair) was used by the British in \ World War I. It operates on 2-grams. First of all, one has to identify the \ letters ", Cell[BoxData[ FormBox["i", TraditionalForm]]], " and ", Cell[BoxData[ FormBox["j", TraditionalForm]]], ". The remaining 25 letters of the alphabet are put rowwise in a ", Cell[BoxData[ FormBox[ RowBox[{"5", "\[Times]", "5"}], TraditionalForm]]], " matrix ", Cell[BoxData[ FormBox["K", TraditionalForm]]], ", as follows. Put the first letter of a keyword in the top-left position. \ Continue rowwise from left to right. If a letter occurs more than once in the \ keyword, use it only once. The remaining letters of the alphabet are put into \ ", Cell[BoxData[ FormBox["K", TraditionalForm]]], " in their natural order. For instance, the keyword \"hieronymus'' gives \ rise to" }], "Text", TextAlignment->Left, TextJustification->0, CellTags->"DefClass Playfair"], Cell[TextData[Cell[BoxData[ RowBox[{"(", GridBox[{ {"h", "i", "e", "r", "o"}, {"n", "y", "m", "u", "s"}, {"a", "b", "c", "d", "f"}, {"g", "k", "l", "p", "q"}, {"t", "v", "w", "x", "z"} }], ")"}]]]], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "The 2-gram ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", RowBox[{"x", ",", "y"}], ")"}], "=", RowBox[{"(", RowBox[{ SubscriptBox["K", RowBox[{"i", ",", "j"}]], ",", SubscriptBox["K", RowBox[{"m", ",", "n"}]]}], ")"}]}], TraditionalForm]]], " with ", Cell[BoxData[ FormBox[ RowBox[{"x", "\[NotEqual]", "y"}], TraditionalForm]]], " will be encrypted into" }], "Text", TextAlignment->Left, TextJustification->0], Cell[TextData[Cell[BoxData[GridBox[{ { RowBox[{ RowBox[{"(", RowBox[{ SubscriptBox["K", RowBox[{"i", ",", "n"}]], ",", SubscriptBox["K", RowBox[{"m", ",", "j"}]]}], ")"}], ","}], RowBox[{ RowBox[{ RowBox[{"if", " ", "i"}], "\[NotEqual]", RowBox[{"m", " ", "and", " ", "j"}], "\[NotEqual]", "n"}], ","}]}, { RowBox[{ RowBox[{"(", RowBox[{ SubscriptBox["K", RowBox[{"i", ",", RowBox[{"j", "+", "1"}]}]], ",", SubscriptBox["K", RowBox[{"i", ",", RowBox[{"n", "+", "1"}]}]]}], ")"}], ","}], RowBox[{ RowBox[{ RowBox[{"if", " ", "i"}], "=", RowBox[{ RowBox[{"m", " ", "and", " ", "j"}], "\[NotEqual]", "n"}]}], ","}]}, { RowBox[{ RowBox[{"(", RowBox[{ SubscriptBox["K", RowBox[{ RowBox[{"i", "+", "1"}], ",", "j"}]], ",", SubscriptBox["K", RowBox[{ RowBox[{"m", "+", "1"}], ",", "j"}]]}], ")"}], ","}], RowBox[{ RowBox[{ RowBox[{ RowBox[{"if", " ", "i"}], "\[NotEqual]", RowBox[{"m", " ", "and", " ", "j"}]}], "=", "n"}], ","}]} }]]]], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "where the indices are taken modulo 5. If the symbols ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " and ", Cell[BoxData[ FormBox["y", TraditionalForm]]], " in the 2-gram ", Cell[BoxData[ FormBox[ RowBox[{"(", RowBox[{"x", ",", "y"}], ")"}], TraditionalForm]]], " are the same, one first inserts the letter ", Cell[BoxData[ FormBox["q", TraditionalForm]]], " and enciphers the text ", Cell[BoxData[ FormBox[ RowBox[{"\[Ellipsis]", " ", "x", " ", "q", " ", "y", " ", "\[Ellipsis]"}], TraditionalForm]]], " ." }], "Text", TextAlignment->Left, TextJustification->0] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\t", "Transposition Ciphers" }], "Subsection", CellTags->"SubsClas Transposition"], Cell[TextData[{ "A completely different way of enciphering is called ", StyleBox["transposition", FontSlant->"Italic"], ". This system breaks the text up into blocks of fixed length, say ", Cell[BoxData[ FormBox["n", TraditionalForm]]], ", and applies a fixed permutation \[Sigma] to the coordinates. For \ instance, with ", Cell[BoxData[ FormBox[ RowBox[{"n", "=", "5"}], TraditionalForm]]], " and \[Sigma] = (1, 4, 5, 2, 3), one gets the following encryption:" }], "Text", TextAlignment->Left, TextJustification->0, CellTags->"DefClass Transp cipher"], Cell[TextData[{ StyleBox["crypt ograp hical \[Ellipsis]", FontFamily->"Courier New", FontColor->RGBColor[0.500008, 0.500008, 0]], StyleBox[" ", FontFamily->"Courier New"], Cell[BoxData[ FormBox[ OverscriptBox["\[LongRightArrow]", "\[Sigma]"], TraditionalForm]], FontFamily->"Courier New"], StyleBox[" ", FontFamily->"Courier New"], StyleBox["ytrcp rpgoa cliha \[Ellipsis]", FontFamily->"Courier New", FontColor->RGBColor[0, 0.500008, 0]] }], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "Often the permutation is of a geometrical nature, as is the case with the \ so-called ", StyleBox["column transposition", FontSlant->"Italic"], ". The plaintext is written rowwise in a matrix of given size, but will be \ read out columnwise in a specific order depending on a keyword. For instance, \ after having identified letters ", Cell[BoxData[ FormBox[ RowBox[{"a", ",", "b", ",", "\[Ellipsis]", ",", "z"}], TraditionalForm]]], " with the numbers ", Cell[BoxData[ FormBox[ RowBox[{"1", ",", "2", ",", "\[Ellipsis]", ",", "26"}], TraditionalForm]]], " the keyword \"right'' will dictate you to read out column 3 first (being \ the alphabetically first of the 5 letters in \"right''), followed by columns \ 4, 2, 1 and 5. So, the plaintext" }], "Text", TextAlignment->Left, TextJustification->0, CellTags->"DefClass Column Transp"], Cell[TextData[StyleBox["computing science has had very little influence on \ computing practice", FontFamily->"Courier New", FontColor->RGBColor[0.500008, 0.500008, 0]]], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "when encrypted with a ", Cell[BoxData[ FormBox[ RowBox[{"5", "\[Times]", "5"}], TraditionalForm]]], " matrix and keyword ''right'' will first be filled in rowwise as depicted \ below" }], "Text", TextAlignment->Left, TextJustification->0], Cell[TextData[{ Cell[BoxData[GridBox[{ {"4", "3", "1", "2", "5"}, {"c", "o", "m", "p", "u"}, {"t", "i", "n", "g", "s"}, {"c", "i", "e", "n", "c"}, {"e", "h", "a", "s", "h"}, {"a", "d", "v", "e", "r"} }]]], "\t \t", Cell[BoxData[GridBox[{ {"4", "3", "1", "2", "5"}, {"y", "l", "i", "t", "t"}, {"l", "e", "i", "n", "f"}, {"l", "u", "e", "n", "c"}, {"e", "o", "n", "c", "o"}, {"m", "p", "u", "t", "i"} }]]], "\t\t", Cell[BoxData[GridBox[{ {"4", "3", "1", "2", "5"}, {"n", "g", "p", "r", "a"}, {"c", "t", "i", "c", "e"}, {".", ".", ".", " ", " "}, {" ", " ", " ", " ", " "}, {" ", " ", " ", " ", " "} }]]] }], "Text", TextAlignment->Center, TextJustification->0], Cell["\<\ and then read out (columnwise in the indicated order) to give the ciphertext:\ \ \>", "Text", TextAlignment->Left, TextJustification->0], Cell[TextData[{ StyleBox["mneav pgnse oiihd ctcea uschr iienu tnnct leuop yllem tfcoi \ \[Ellipsis]", FontFamily->"Courier New", FontColor->RGBColor[0, 0.500008, 0]], " ." }], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "Since transpositions do not change letter frequencies, but destroy \ dependencies between consecutive letters in the plaintext, while Vigen\ \[EGrave]re etc. do the opposite, one often combines such systems. Such a \ combined system is called a ", StyleBox["product cipher", FontSlant->"Italic"], ". Shannon used the words confusion and diffusion in this context." }], "Text", TextAlignment->Left, TextJustification->0, CellTags->"DefClass product cipher"], Cell[TextData[{ "Ciphersystems that encrypt the plaintext symbol for symbol in a way that \ depends on previous input symbols are often called ", StyleBox["stream ciphers", FontSlant->"Italic"], " (they will discussed in ", ButtonBox["Chapter 3", BaseStyle->"Hyperlink", ButtonData:>"Chap Shift"], "). Cryptosystems that encrypt blocks of symbols (of a fixed length) \ simultaneously but independent of previous encryptions, they are called ", StyleBox["block ciphers", FontSlant->"Italic"], " (see ", ButtonBox["Chapter 4", BaseStyle->"Hyperlink", ButtonData:>"Chap Block"], ")." }], "Text", TextAlignment->Left, TextJustification->0, CellTags->"DefClass stream cipher"], Cell["\<\ During World War II both sides used so called rotor machines for their \ encryption. Several variations of the machines described in the next two \ subsections were in use at that time. We shall give a rough idea of each one.\ \ \>", "Text", TextAlignment->Left, TextJustification->0] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Chapter"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\t", "Hagelin" }], "Subsection", CellTags->"SubSClass Hagelin"], Cell[GraphicsData["CompressedBitmap", "\<\ eJzsvT2PXMm2ppcn987P+mSR7K7u033v6Xtvzwx1pyUREAXUAISAAkQBBNQG DRplELggBM6AEChMDUCjBPAPtEmTf4Emzf4DMtqk2ZAns02aR/uNZ72xYu9M 8g5kyThoFLsqc+fO2BGxvt71rhX/67/8l//4v/3v//Jf/tPzf/nL//Kf/+X/ +I//6fn1Xx69+s/DS92fZrM//V+z2fzZ6Wz4/f+cdf/37O9Xb/46W978/uav b/6qf29+74cXOn6dzf378P85b/X8+frXZXNd75vc/M6/88mb9SbDz+L1r/6u WXvd8AWzNbe4+f31r7w6vDB846zjdb0/GlzvLx0umK30IV7Qb/qQH40vnDOw Bde0N3n9qwY1vL/QFzDA8eCG39ftE46miumZLXijfuBg+gGGVt5b6S997e7T vP610+g1iuHi4cX58L/DfDB/pj6UppgvHh5ttuS6fLfjV31FnxczFzurVl/Q t8QnFr5EY9cduDsf6D2cmFqmcrhSD7EaP7+GUIfoB6nr7mXSyx03KOs4n8y0 F7Lzs8WX1ecc7z+u78r18/GMdM2f7U6dbv3c9+Nbb9kt5V5/Gv6e1Tsvmo/3 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