Создание дешифратора на СИ! Студентам)))) Все кто любит поломать голову, приветствую вас))))) Нужно разработать программу, взламывающую шифр, основываясь на статистике естественного языка. Дан шифротекст и размер блока. Требуется получить открытый текст и ключ — последовательность перестановки символов (получить автоматически). Текст должен расшифровываться по методу двойной перестановки. Пример: 1 2 3 4 1 A З Ю Ж 2 E С Ш 3 Г T О О 4 И П E P Рассматривая маловероятные сочетания букв, легко найти истинную последовательность столбцов. Так, сочетание ГТ в 3 строке шифровки указывает на то, что после 1 столбца вряд ли следует 2 столбец. Рассчитаем статистически, какой столбец скорее всего следует за 1. Для этого воспользуемся таблицей логарифмов вероятностей биграмм русского текста, приведенной в приложении. Вероятность следования одного столбца за другим равна произведению вероятностей биграмм в строках этих столбцов. Поскольку в таблице даны логарифмы биграмм, то их достаточно суммировать, а потом выбрать сочетание столбцов с максимальной вероятностью. Для вероятностей следования за первым столбцом 2, 3 и 4 имеем выражения: р (1-2) =р(A3) р(Е ) р(ГТ) р(ИП)=7+9+0+5=21 р (1-3) =р(АЮ) р(ЕС) р(ГО) р(ИЕ)=6+8+8+8=30 р (1-4 )=р(АЖ) р(ЕШ) р(ГО) р(ЯР)=7+5+8+7=27 В нашем случае наиболее вероятно, что после столбца 1 следует столбец 3. Для такой небольшой таблицы шифрования, которую имеем, можно перебрать все варианты перестановок - их всего лишь 24. В случае большого числа столбцов целесообразно оценить вероятности пар сочетаний разных столбцов и решить оптимизационную задачу, которая укажет перестановку столбцов, дающую фрагменты естественного текста с большей вероятностью. В нашем случае наилучший результат достигается при расстановке столбцов (2413), что примерно вдвое по вероятностной оценке достовернее ближайшей к ней по вероятности расстановки (4132). После того, как столбцы шифровки расставлены, не составит труда правильно расставить и ее строки по смыслу фрагментов текста: 2 4 1 3 1 З Ж A Ю 2 Ш E С 3 T О Г О 4 П P И E Текст в ней уже читается и, расставив строки в порядке (4123), получим расшифровку ПРИЕЗЖАЮ ШЕСТОГО. 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Вы говорите "Текст должен расшифровываться по методу простой перестановки.", а сами рассказываете о двойной перестановке(когда помимо столбцов ещё и строки перемешиваются) Так что же конкретно нужно? P.S. "Статистика распределения частот символов" здесь не нужна, т.к. замена алфавита в этом методе шифрования не производится.
Ну нашу статистику мы помещаем в 2-у мерный массив arr[1][1]="Бо"; arr[1][2]="2"; arr[2][1]="Бы"; arr[1][2]="4"; (ну заносим конечно не таким методом, а более оптимальным) Ройте в направлении инициализации массивов. Ну а по поводу получения ключа у вас впринципе всё написано в шапке темы... Загоняем текст в двумерный массив и сравниваем между собой его элементы...
В общем вот такая рабочая программка: Unit1.cpp: #include <vcl.h> #pragma hdrstop #include "Unit1.h" #include "Unit2.h" #include "Unit3.h" //--------------------------------------------------------------------------- #pragma package(smart_init) #pragma resource "*.dfm" TForm1 *Form1; //--------------------------------------------------------------------------- __fastcall TForm1::TForm1(TComponent* Owner) : TForm(Owner) { } //--------------------------------------------------------------------------- void __fastcall TForm1::BitBtn1Click(TObject *Sender) { Memo1->Clear(); if (OpenDialog1->Execute()) Memo1->Lines->LoadFromFile(OpenDialog1->FileName); } //--------------------------------------------------------------------------- void __fastcall TForm1::BitBtn2Click(TObject *Sender) { struct{ int n; //вероятность бирграмы char bigrama[2];} //сама биграма str[811]={ {2," "}, {70," ("}, {74," -"}, {4," 1"}, {2," 2"}, {2," 4"}, {2," 6"}, {4," 9"}, {14," А"}, {6," Б"}, {74," В"}, {2," Г"}, {26," Д"}, {14," Е"}, {22," З"}, {42," И"}, {40," К"}, {8," Л"}, {14," М"}, {48," Н"}, {54," О"}, {56," П"}, {12," Р"}, {36," С"}, {18," Т"}, {12," У"}, {4," Ф"}, {10," Х"}, {2," Ц"}, {2," Ч"}, {6," Ш"}, {16," Э"}, {2," Ю"}, {4," Я"}, {162," а"}, {294," б"}, {652," в"}, {72," г"}, {284," д"}, {88," е"}, {36," ж"}, {204," з"}, {652," и"}, {472," к"}, {94," л"}, {228," м"}, {548," н"}, {556," о"}, {816," п"}, {218," р"}, {828," с"}, {372," т"}, {222," у"}, {20," ф"}, {76," х"}, {44," ц"}, {220," ч"}, {16," ш"}, {142," э"}, {6," ю"}, {26," я"}, {6,"! 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