Question

我是算法编程的新手。我知道当涉及动态编程的序列对齐时，它应该遵循以下算法：

Alg: Compute C[i, j]: min-cost to align (the first i symbols of X) and (the j symbols of Y)(C1:cost for mismatch,C2:cost for gap alignment)
and def d[i, j] = { C1 if X[i] ≠ Y[j],0 otherwise}
Compute C[i, j]:
  case 1: align X[i] with Y[j]
    C[i, j] = C[i-1, j-1] + d[i, j]
  case 2: either X[i] or Y[j] is aligned to a gap
    C[i, j] = min{ (C[i-1, j] + C2), (C[i, j-1] + C2) }
    (C[i, j] = C[i-1, j] + C2 is case 2-1)
    (C[i, j] = C[i-1, j] + C2 is case 2-2)
Alg: C[i, 0] = iC2, ∀i
C[0, j] = jC2, ∀j
for i = 1 to m
  for j = 1 to n
    C[i, j] = min{ (case 1), (case 2-1), (case 2-2) }
return C[m, n]

然而，最后一部分：

for i = 1 to m
  for j = 1 to n
    C[i, j] = min{ (case 1), (case 2-1), (case 2-2) }
return C[m, n]

我有点困惑，因为从前一部分它只是一个问题。为什么要一次又一次地在多个i中迭代j？谢谢！

Answer 1

在这种情况下，C[i,j]是memo，它存储输入i和j的每个组合的计算值。您提到的嵌套循环只是遍历2-D矩阵中的每个单元格。实际上，相同的单元格从未计算过两次。

动态规划的序列对齐

1 个答案: