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Problem with categories LSTM network

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Marek Ziak
Marek Ziak 2019-11-29
评论： Marek Ziak 2019-11-29
Hi everyone,
I am using LSTM network to break ciphers.Input is password encoded to matrix 94x6. I am using ASCII table from where I picked these 94 valid characters . Password will be always 6 characters. My categories are 94. I generetad train data and it worked quite good. Now I would like to add Validation data. They should't be the same which can be handled. But when I generate them NN won't because it thinks it does nt ahve the same categories. I checked that and look like totally the same . Any help would be appriecated.
k = 5; % posun znakov pre Cezarovu sifru
dlzka_Hesla = 6; % dlzka nahodne generovanych hesiel
N = 4000; % pocet vygenerovanych hesiel (velkost trenovacej mnoziny)
max_Epochs = 100; % maximalny pocet epoch trenovania siete
sifra = 1;
% vytvorime si vektor 94 znakov pre mozne vytvorenie hesiel
uniqueCharacters = char(33:126);
NumValid = 1000;
NumTest = 1000;
[textData,ValidationData,testData] = GenPasswords(uniqueCharacters,dlzka_Hesla,N,NumValid,NumTest);
% pre kazdy vstup do siete (jedno heslo) vytvor kategoricke premenne
[XTrain,YTrain,vec_cipher2] = GenTrainData(uniqueCharacters,sifra,N,textData,k);% konvertovanie stringu na one hot vectory
[XValidation,YValidation,vec_cipher3] = GenTrainData(uniqueCharacters,sifra,length(ValidationData),ValidationData,k);
Cat_Train = sort(categories([YTrain{:}]));
Cat_Val = sort(categories([YValidation{:}]));
CAT = [ Cat_Train Cat_Val];
if (sum (strcmp(Cat_Train,Cat_Val)) ==94)
disp ('Kategorie su identicke')
end
function [XTrain,YTrain,vec_cipher2] = GenTrainData(uniqueCharacters,sifra,N,textData,k)
% ulozime si ich pocet uniaktnych zakov
numUniqueCharacters = numel(uniqueCharacters);
c = 1;
for i = 1:N
% vyber jeden heslo
characters = textData{i};
% zisti pocet znakov
sequenceLength = numel(characters);
% zisti indexy kazdeho unikatneho znaku v riadku
[~,idx] = ismember(characters,uniqueCharacters);
% vytvor vektor reprezentujuci, ci sa dany znak nachadza v tomto vstupe
% teda, kategoricke premenne
X = zeros(numUniqueCharacters,sequenceLength);
% pre vsetky znaky, ktore sa nachadzaju v texte nastav vlajku na 1
for j = 1:sequenceLength
X(idx(j),j) = 1;
end
% vytvorime ocakavane mapovanie vystupu pouzitim Cezarovej sifry
if (sifra ==1)
cipher = caesar_cipher(textData{i}, k);
end
if (sifra ==2)
cipher = vignerie_cipher(textData{i},[25, 14, 17, 10]);
end
vec_cipher2(i,1:6) = cipher;
% ak sifra obsahuje znaky, ktore nie su v slovniku, tak ignorujeme
b = false;
for s=1:length(cipher)
if(sum(ismember(uniqueCharacters, cipher(s)))<=0)
b = true;
break;
end
end
% dont add this example to the training data
if(b)
cipher;
continue;
end
% vytvorime zasifrovany vektor vstupneho textu, ktory vznikol pouzitim sifry
charactersOutput = cellstr(cipher')';
% konvertujeme vystupny vektor na kategoricku premennu (pole znakov)
Y = categorical(charactersOutput);
% priradime vytvorene vstupno-vystupne pary do datasetu
% XTrain{c} = X;
% YTrain{c} = Y;
XTrain{c} = X;
YTrain{c} = Y;
c = c + 1;
end
end
function [textData,validationData,testData] = GenPasswords(uniqueCharacters,dlzka_Hesla,N,NumValid,NumTest)
% vygenerujeme nahodne indexy pre tvorbu hesiel
r = randi([1 length(uniqueCharacters)],dlzka_Hesla,N);
% vygenerujeme hesla v textovej podobe na zaklade indexov
textData = cell(N,1);
for e=1:N
textData{e} = uniqueCharacters(r(:,e));
end
r = randi([1 length(uniqueCharacters)],dlzka_Hesla,10000);
k = 1;
validationData = cell(NumValid,1);
% vygenerovanie validacnej mnoziny
for e = 1:N
validationData{k,1} = uniqueCharacters(r(:,e));
if( sum(ismember(validationData{k,1},textData{e,1}))~= 6)
k = k+1;
if k ==1001
break;
end
end
end
testData = cell(NumTest,1);
for i = 1:NumTest
% vygenerujeme heslo v textovej podobe na zaklade indexov
testData{i} = uniqueCharacters(r(:,i));
end
end

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