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勢函數的matlab實現

代碼片段和文件信息
clc
clear?all
start?=?[00;
?????????????01;
?????????????10;
?????????????11];
w?=?[1122];
Nc?=?0;
r?=?[];?
G?=?0;?
k?=?1;
N?=?4;?
m?=?1;?
n?=?0;
while(Nc?4?)
????r(1)?=?0;
????if?(?w(m)?==?1)?%屬于第一類
????????if?(G?>?0)
????????????Nc?=?Nc?+?1;
????????????r(k+1)?=?0;
????????end
????????if?(?G?<=?0)
????????????r?(?k+1?)?=?1;
????????????Nc?=?0;
????????end
????end
????if?(w(m)?==?2)
????????if?(G?0)
????????????r(k+1)?=?0;
????????????Nc?=?Nc?+?1;
????????end
????????if?(?G?>=?0)
????????????r(k+1)?=?-1;
????????????Nc?=?0;
????????end
????end
????k?=?k+1;
????m?=?mod(k-1N)?+?1;
????G?=?0;
????for?i=1:k
????????n?=?mod(i-1N)?+?1;
????????G?=?G?+?r(i)*KA(?start(m?1)start(n1)?start(m2)start(n2)?);
????end
end
X?=?input(‘please?input?X:‘)
G?=?0;
k?=?k-1;
for?i?=?1:k
????n?=?mod(i-1N)?+?1;
????G?=?G?+?r(i)*KA(X(1)start(n1)X(2)start(n2)?);
end
if?(G?>?0)
????fprintf(‘X?is?first?kind\n‘);
else
????fprintf(‘X?is?secont?kind\n‘);
end
????????
????????
????????
????????
????????
????????
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????1092??2016-03-23?22:45??勢函數\demoka.m
?????文件?????????68??2016-03-23?22:35??勢函數\KA.m
?????目錄??????????0??2016-03-23?22:35??勢函數
-----------?---------??----------?-----??----
?????????????????1160????????????????????3
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