フィルターのクリア

can someone help me in coding hash 512 algorithm?

2 ビュー (過去 30 日間)
yogya
yogya 2014 年 11 月 9 日
コメント済み: Jan 2014 年 11 月 9 日
hello i need a hash value that provide me as decimal value from 512 bits binary value. can someone please correct my code and help me. Where my code is
data = input('enter the data:','s');
pawrd = input('enter the password:','s');
bin = double(data);
display(bin);
n = dec2base(bin,16);
display(n);
len_n = length(n);
display(len_n);
k = dec2bin(bin,8);
display(k);
pwd = dec2bin(pawrd,8)';
pwd = pwd(:)'-'0';
ln_pwd = length(pwd);
disp(ln_pwd);
pad_pwd = [pwd , ones(1,64-ln_pwd)];
A = dec2bin(data,8)';
A = A(:)'-'0';
le_a = length(A);
pad2 = dec2bin(le_a,8)';
pad2 = pad2(:)'-'0';
j = length(pad2);
dap2 = [zeros(1,128-j), pad2];
display(pad2);
display(A);
display(le_a);
x = 896-le_a;
display(x);
if x == 0
display('seq need not to be padded');
else
padarray = [ones(1,1), zeros(1,x-1)];
paddedseq = [A, padarray];
finalpad = [paddedseq, dap2];
l_paddedseq = length(finalpad);
display(l_paddedseq);
display(paddedseq);
display(finalpad);
x1 = '6A09E667F3BCC908';
x2 = 'BB67AE8584CAA73B';
x3 = '3C6EF372FE94F82B';
x4 = 'A54FF53A5F1D36F1';
x5 = '510E527FADE682D1';
x6 = '9B05688C2B3E6C1F';
x7 = '1F83D9ABFB41BD6B';
x8 = '5BE0CD19137E2179';
bx1 = base2dec(x1,16);
bx2 = base2dec(x2,16);
bx3 = base2dec(x3,16);
bx4 = base2dec(x4,16);
bx5 = base2dec(x5,16);
bx6 = base2dec(x6,16);
bx7 = base2dec(x7,16);
bx8 = base2dec(x8,16);
a = dec2bin(bx1,64);
display(a);
b = dec2bin(bx2,64);
display(b);
c = dec2bin(bx3,64);
display(c);
d = dec2bin(bx4,64);
display(d);
e = dec2bin(bx5,64);
display(e);
f = dec2bin(bx6,64);
display(f);
g = dec2bin(bx7,64);
display(g);
h = dec2bin(bx8,64);
display(h);
end
w =reshape(finalpad,16,64);
display(w);
wi = w(i, 1:64);
disp(wi);
ki = circshift(pad_pwd,1);
disp(ki);
and1 = e&f;
and2 = (~e)&g;
and3 = a&b;
and4 = a&c;
and5 = b&c;
Ch = bitxor(and1,and2);
display(and1);
display(and2);
display(Ch);
Maj1 = bitxor(and3,and4);
display(Maj1);
maj = bitxor(Maj1,and5);
display(maj);
r1 = circshift(e,[0 +14])-'0' ;
disp(r1);
r2 = circshift(e,[0 +18])-'0';
disp(r2);
r3 = circshift(e,[0 +41])-'0';
r4 = circshift(a,[0 +28])-'0';
r5 = circshift(a,[0 +34])-'0';
r6 = circshift(a,[0 +39])-'0';
r12 = bitxor(r1,r2);
r45 = bitxor(r4,r5);
sum1e = bitxor(r12,r3);
sum0a = bitxor(r45,r6);
T1 = h|Ch|wi|ki|sum1e;
disp(T1);
T2 = maj|sum0a;
disp(T2);
h = g;
disp(h);
g = f;
disp(g);
f = e;
disp(f);
e = d|T1;
disp(e);
d = c;
disp(d);
c = b;
disp(c);
b = a;
disp(b);
a = T1|T2;
disp(a);
a=a(a~='0');
disp(a);
  1 件のコメント
Jan
Jan 2014 年 11 月 9 日
If you want a correction of your code, you have to mention, what you assume to be a bug. It is impossible to find a bug based on working code only.

サインインしてコメントする。

回答 (0 件)

カテゴリ

Help Center および File ExchangeChemistry についてさらに検索

タグ

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by