%%
%?Purpose:?
%?
%?This?algorithm?decrypts?and?encrypts?images?based?on?keys
%?
%?INPUTS:
%?
%?
%?
%?imMat:?If?the?nargin?is?one?imMat?is?the?filename?of?the?image?which
%?you?would?like?to?encrypt.?If?the?nargin?is?4?imMat?is?the?encrypted
%?matrix?
%
%?key1?key2?key3:?These?are?number?values?used?to?encrypt?and?decrypt
%?images.
%?key3?and?key2?and?key1?can?be?left?blank?if?the?nargin?is?one?and??you?would?like
%?the?computer?to?make?the?keys?for?you.?if?the?nargin?is?4?key1?is
%?a?vector?at?baseto?that?is?the?encrypted?vector?of?the?original?random?numbers.
%?key2?is?the?base?of?the?original?vector?（baseKey）?**Please
%?read?the?help?section?of?baseVecExpand.
%?
%?OUTPUTS:
%
%?cryptPic:?If?the?nargin?=?1?cryptPic?is?a?matrix?of?encrypted
%?numbers?representing?an?encrypted?image.?If?the?nargin?=?4?imMat
%?is?a?uint8?array?which?is?a?decrypted?image.?This?function?will?also
%?display?the?decrypted?image.?
%
%?keyVec:?If?the?nargin?=?1?the?computer?will?use?the?random?number?generator?to
%?make?three?keys.?These?three?keys?are?then?changed?to?a?random?base?baseto
%?using?baseVecExpand.?The?starting?base?is?baseKey.?
%?keyVec?is?a?vector?containing?the??key?vector?at?baseto.?if?nargin?=?4
%?keyVec?is?not?returned
%
%?baseKey:?If?the?nargin?=?1?baseKey?is?the?base?of?the
%?original?vector.?please?read?the?help?section?of?baseVecExpand.?baseKey
%?is?not?returned?if?the?nargin?=?4.?
%
%?COMMENTS:?
%
%?Please?change?the?formulas?used?to?encrypt?and?decrypt?the?images?into
%?your?own?unique?formulas?so?that?nobody?can?see?your?images.?
function?[cryptPic?keyVec?baseto?baseKey]?=?picCrypt（imMatkeykey2key3）
if?nargin?==?1
???key?=?rand;
???key2?=?rand;
???key3?=?rand;
???baseto?=?rand*1000;
???baseKey?=?max（floor（1000000*[key?key2?key3]））?+?rand*100;
???keyVec?=?baseVecExpand（floor（1000000*[key?key2?key3]）baseKeybaseto）;
???pic?=?10000*im2double（imread（imMat））;
???cryptPic?=?（（floor（pic）/key2）.^key）+key3;
???imagesc（im2uint8（cryptPic））;
???axis?equal;
???axis?tight;
elseif?nargin?==?4
???key10?=?baseVecExpand（keykey3key2）/1000000;
???cryptPic?=?im2uint8（（（imMat?-?key10（3））.^（1/key10（1））*key10（2））/10000）;?%?CHANGE?THIS?FORMULA?（this?is?the?opposite?of?the?formula?above
???imagesc（cryptPic）;
???axis?equal;
???axis?tight;??
else
????error（‘You?must?enter?the?correct?number?of?parameters‘）;
end

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????2403??2011-02-17?05:02??picCrypt.m
?????文件????????1316??2014-02-12?13:26??license.txt