@@ -50,7 +50,7 @@ static int levenshtein_internal(text *s, text *t,
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* array.
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*
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* If max_d >= 0, we only need to provide an accurate answer when that answer
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- * is less than or equal to the bound. From any cell in the matrix, there is
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+ * is less than or equal to the bound. From any cell in the matrix, there is
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* theoretical "minimum residual distance" from that cell to the last column
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* of the final row. This minimum residual distance is zero when the
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* untransformed portions of the strings are of equal length (because we might
@@ -141,7 +141,7 @@ levenshtein_internal(text *s, text *t,
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stop_column = m + 1 ;
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/*
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- * If max_d >= 0, determine whether the bound is impossibly tight. If so,
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+ * If max_d >= 0, determine whether the bound is impossibly tight. If so,
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* return max_d + 1 immediately. Otherwise, determine whether it's tight
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* enough to limit the computation we must perform. If so, figure out
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* initial stop column.
@@ -168,7 +168,7 @@ levenshtein_internal(text *s, text *t,
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* need to fill in. If the string is growing, the theoretical
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* minimum distance already incorporates the cost of deleting the
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* number of characters necessary to make the two strings equal in
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- * length. Each additional deletion forces another insertion, so
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+ * length. Each additional deletion forces another insertion, so
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* the best-case total cost increases by ins_c + del_c. If the
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* string is shrinking, the minimum theoretical cost assumes no
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* excess deletions; that is, we're starting no further right than
@@ -246,7 +246,7 @@ levenshtein_internal(text *s, text *t,
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/*
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* The main loop fills in curr, but curr[0] needs a special case: to
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* transform the first 0 characters of s into the first j characters
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- * of t, we must perform j insertions. However, if start_column > 0,
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+ * of t, we must perform j insertions. However, if start_column > 0,
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* this special case does not apply.
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*/
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if (start_column == 0 )
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