采用NLJ隨機搜索的方法辨識一個以狀態(tài)方法表示的非線性系統(tǒng)。選其初值 a1(0) =50 , a2(0) =100 , a3(0) =100 , a4(0) =50 , a5(0) =10 , 選范圍為 r(1)(i)=0.5 a(0)(i) , 取數(shù)據(jù)長度 L ＝40, t =0.005 , 性能指標 J＝ 。迭代計算結(jié)果得 a 的估計值 1=17.6043243, 1=17.5977, 2=72.9573, 3=51.3014, 4=22.9889, 5=5.99965, J = 0.000000916 。
標簽：
100
50
NLJ
10
上傳時間：
2013-12-20
上傳用戶：weiwolkt
Instead of finding the longest common
subsequence, let us try to determine the
length of the LCS.
Then tracking back to find the LCS.
Consider a1a2…am and b1b2…bn.
Case 1: am=bn. The LCS must contain am,
we have to find the LCS of a1a2…am-1 and
b1b2…bn-1.
Case 2: am≠bn. Wehave to find the LCS of
a1a2…am-1 and b1b2…bn, and a1a2…am and
b b b
b1b2…bn-1
Let A = a1 a2 … am and B = b1 b2 … bn
Let Li j denote the length of the longest i,g g
common subsequence of a1 a2 … ai and b1 b2
… bj.
Li,j = Li-1,j-1 + 1 if ai=bj
max{ L L } a≠b i-1,j, i,j-1 if ai≠j
L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
標簽：
the
subsequence
determine
Instead
上傳時間：
2013-12-17
上傳用戶：evil
