The longest common subsequence (LCS) for dynamic planning

The longest common sub-sequence (lcs,longest Common subsequence). It is defined as a sequence s, if it is a subsequence of two or more known sequences, and is the longest of all conforming to this condition sequence, then S is called the longest common subsequence of the known sequence. The longest common substring (which requires continuous) and the longest common subsequence is different.

Set X (M) ={x (1), X (2), X (3),...., X (M)} and Y (n) ={y (1), Y (2), Y (3),...., y (N)} The longest common subsequence Z (k) ={z (1), Z (2), Z (3),...., Z (k)}

First, the original problem is decomposed into sub-problems, a known conclusion: when m or n equals 0 o'clock, K equals 0, that is, the common subsequence length is 0

When M and n are not equal to 0 o'clock, there are three cases:

(1) x (m) = = Y (n) , at which point Z (k) = x (m) = Y (n), the element belongs to the last element of the longest common subsequence of the current. At this point Z (k-1) ={X (m-1) and Y (n-1) The longest common sub-sequence}

(2) x (m)! = Y (n) , and Z (k)! = x (m), at which point the longest common subsequence of z={x (m-1) and Y (n)}

(3) x (m)! = Y (n) , and Z (k)! = Y (n), at which point the longest common subsequence of z={X (m) and Y (n-1)}

where x (m-1) ={x (1), X (2), X (3),...., x (m-1)}, Y (n-1) ={y (1), Y (2), Y (3),...., y (n-1)},z (k-1) ={z (1), Z (2), Z (3),...., Z (k-1)}

The above three steps, each of which are based on the state of the current sequence, transform the problem into a sub-problem of a known solution (the first known sub-problem is when m==0 or n==0), thus solving the current problem.

The state transfer equation of the problem is obtained and the mathematical description is as follows:

Existing two sequence X={x1,x2,x3,...xi},y={y1,y2,y3,....,yi},

Set a c[i,j]: preserve the length of the LCS of Xi and YJ

Finally, the python implementation of the algorithm:

1 #Longest common sub-sequence problem2 __author__='Ice'3 4 5 #arr_x,arr_y [0 ~ length-1]6 #Subarr_len [0,1~x_length][0,1~y_length]7 #flag [0 ~ x_length-1][0 ~ y_length]8 9 Ten deflcs_length (arr_x, arr_y): OneX_length =Len (arr_x) AY_length =Len (arr_y) -Subarr_len = [[0 forJinchRange (Y_length + 1)] forIinchRange (X_length + 1)] -flag = [[0 forJinchRange (Y_length)] forIinchrange (x_length)] the      forIinchRange (1, x_length + 1): -          forJinchRange (1, y_length + 1): -             ifArr_x[i-1] = = Arr_y[j-1]: -SUBARR_LEN[I][J] = subarr_len[i-1][j-1] + 1 +FLAG[I-1][J-1] = 1 -             elifSUBARR_LEN[I-1][J] >= subarr_len[i][j-1]: +SUBARR_LEN[I][J] = subarr_len[i-1][j] AFLAG[I-1][J-1] = 2 at             Else: -SUBARR_LEN[I][J] = subarr_len[i][j-1] -FLAG[I-1][J-1] = 3 -     return{'Subarr_length': Subarr_len[x_length][y_length],'Flag': Flag} -  -  in defLCS (arr_x, x_i, Y_j, flag, result): -     ifX_i < 0orY_j <0: to         return +     ifFlag[x_i][y_j] = = 1: -LCS (arr_x, X_i-1, y_j-1, flag, result) the result.append (arr_x[x_i]) *     elifFlag[x_i][y_j] = = 2: $LCS (arr_x, x_i-1, Y_j, flag, result)Panax Notoginseng     elifFlag[x_i][y_j] = = 3: -LCS (arr_x, X_i, y_j-1, flag, result) the  +  Aarray_x = ['a','b','C','b','D','a','b'] theArray_y = ['b','D','C','a','b','a'] +Longest_common_subsequence = [] -Lcs_info =lcs_length (array_x, array_y) $LCS (Array_x, Len (array_x)-1, Len (array_y)-1, lcs_info['Flag'], longest_common_subsequence) $ Print(longest_common_subsequence)

The longest common subsequence (LCS) for dynamic planning