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    Question

    Consider the following Python code for calculating the

    length of the LCS: def lcs_length(text1, text2):     m = len(text1)     n = len(text2)     dp = [[0] * (n + 1) for _ in range(m + 1)]     for i in range(1, m + 1):         for j in range(1, n + 1):             if text1[i - 1] == text2[j - 1]:                 dp[i][j] = 1 + dp[i - 1][j - 1]             else:                 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])     return dp[m][n] # Assume text1 = "AGGTAB" and text2 = "GXTXAYB" # And the dp table has been partially filled as follows (only relevant cells shown): #       ""  G   X   T   X   A   Y   B # ""    0   0   0   0   0   0   0   0 # A     0   0   0   0   0   1   1   1 # G     0   1   1   1   1   1   1   1 # G     0   1   1   1   1   1   1   1 # T     0   1   1   2   2   2   2   2 # A     0   1   1   2   2   3   3   3 # B     0   1   1   2   2   3   3   4 
    A 1 Correct Answer Incorrect Answer
    B 2 Correct Answer Incorrect Answer
    C 3 Correct Answer Incorrect Answer
    D 4 Correct Answer Incorrect Answer

    Solution

    Detailed Answer and Dry Run: We are calculating dp[4][4] for text1 = "AGGTAB" and text2 = "GXTXAYB". This corresponds to finding the LCS of text1[0...3] ("AGGT") and text2[0...3] ("GXTX"). Let's trace the execution for dp[4][4] using the provided Python code snippet and the partially filled table. The i loop is at i = 4 (for text1[3], which is 'T'). The j loop is at j = 4 (for text2[3], which is 'X'). 1.  Character Comparison:     text1[i - 1] is text1[3] which is 'T'.     text2[j - 1] is text2[3] which is 'X'.     Since 'T' != 'X', the else block will be executed. 2.  Applying the Recurrence Relation (Mismatch Case):     dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])     dp[4][4] = max(dp[3][4], dp[4][3])     We need the values of dp[3][4] and dp[4][3]. Let's look at the provided partial table:     dp[3][4] (LCS of "AGG" and "GXTX") is 1. (From the table: G matches G, so dp[3][4] is 1 from dp[2][3] which is 1 for "AG" and "GXT" - this is a simplification, the full table fill would show dp[3][4] as 1 because 'G' is common).     dp[4][3] (LCS of "AGGT" and "GXT") is 2. (From the table: "GT" is common, so dp[4][3] is 2).     Let's re-evaluate dp[3][4] and dp[4][3] based on the recurrence relation for clarity, assuming they were just computed:     For dp[3][4] (LCS of "AGG" and "GXTX"):         text1[2] is 'G', text2[3] is 'X'. They don't match.         dp[3][4] = max(dp[2][4], dp[3][3])         dp[2][4] (LCS of "AG" and "GXTX") is 1 (common 'G').         dp[3][3] (LCS of "AGG" and "GXT") is 1 (common 'G').         So, dp[3][4] = max(1, 1) = 1.     For dp[4][3] (LCS of "AGGT" and "GXT"):         text1[3] is 'T', text2[2] is 'T'. They match!         dp[4][3] = 1 + dp[3][2]         dp[3][2] (LCS of "AGG" and "GX") is 1 (common 'G').         So, dp[4][3] = 1 + 1 = 2.     Now, back to dp[4][4]:     dp[4][4] = max(dp[3][4], dp[4][3])     dp[4][4] = max(1, 2)     dp[4][4] = 2 Therefore, the value of dp[4][4] will be 2. The common subsequences for "AGGT" and "GXTX" are "GX" or "GT", both of length 2.

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