Merge pull request gzc#214 from ravijoshiBITS/master

made FA code more readable, added English solution to a question of exercise 15.4

C15-Dynamic-Programming/15.4.md

@@ -32,11 +32,11 @@ Give a memoized version of LCS-LENGTH that runs in O(mn) time.
 ### `Answer`
 
 	LCS-LENGTH(X, Y)
-		m <- length[X]
-		n <- length[Y]
-		for i <- 1 to m do
-			for j <- 1 to n do
-				c[i,j] <- -1
+		m ← length[X]
+		n ← length[Y]
+		for i ← 1 to m do
+			for j ← 1 to n do
+				c[i,j] ← -1
 			end for
 		end for
 		return LOOKUP-LENGTH(X,Y,m,n)
@@ -46,12 +46,12 @@ Give a memoized version of LCS-LENGTH that runs in O(mn) time.
 			return c[i,j]
 		end if
 		if i = 0 or j = 0 then
-			c[i,j] <- 0
+			c[i,j] ← 0
 		else
 			if X[i] = Y[j] then
-				c[i,j] <- LOOKUP-LENGTH(X,Y,i-1,j-1)+1
+				c[i,j] ← LOOKUP-LENGTH(X,Y,i-1,j-1)+1
 			else
-				c[i,j] <- max(LOOKUP-LENGTH(X,Y,i,j-1),LOOKUP-LENGTH(X,Y,i-1,j))
+				c[i,j] ← max(LOOKUP-LENGTH(X,Y,i,j-1),LOOKUP-LENGTH(X,Y,i-1,j))
 			end if
 		end if
 		return c[i,j]
@@ -65,11 +65,16 @@ Show how to compute the length of an LCS using only 2 · min(m, n) entries in th
 
 那如何做到只利用min(m, n) entries plus O(1) additional space呢? 其实我们只需要用一行保存状态即可. c[i,j-1]已经保存在该行中. 然后用一个常量维护c[i-1,j-1]即可.每次更新c[i,j]的时候将old valuy保存下来,因为下次要用到.
 
+Since we need only c[i-1,j-1], c[i,j-1], c[i-1,j] to compute c[i,j], we just need to save the previous row and the current row of the dp table. We will maintain the row parallel to the shorter one of the X and Y strings. So we can run the algorithm with 2 · min(m, n) space.
+
+In fact, we only need to save only one row. c[i,j-1] is already stored in this row. Then, we use an extra variable to maintain c[i-1,j-1]. Every time c[i,j] is updated, the value of c[i-1,j] is saved into the extra variable because it will be used next time.
+
 ### Exercises 15.4-5
 ***
 Give an O(n^2)-time algorithm to find the longest monotonically increasing subsequence of a sequence of n numbers.
 
 ### `Answer`
+Method 1:
 Given a sequence X = <x1,x2,...,xn> we wish to find the longest monotonically increasing subsequence. 
 
 1. First we sort the string X which produces sequence X'.
@@ -78,6 +83,20 @@ Given a sequence X = <x1,x2,...,xn> we wish to find the longest monotonically in
 
 The running time is O(n^2) since sorting can be done in O(nlgn) and the call to LCS-LENGTH is O(n^2). 
 
+Method 2:
+	LONGEST-INC-SEQUENCE(Arr, n)
+		len [1...n] be a new array
+		for i←1 to n
+			len[i] ← 1	// length of longest increasing sequence ending at i.
+		for i←2 to n do
+			for j←1 to i-1 do
+				if Arr[j] < Arr[i]
+					len [i] ← max(len[i], len[j] +1)
+				end if
+			end for
+		end for
+		return len [n]
+
 ### Exercises 15.4-6 *
 ***
 Give an O(n lg n)-time algorithm to find the longest monotonically increasing sub-sequence of a sequence of n numbers. (Hint: Observe that the last element of a candidate subsequence of length i is at least as large as the last element of a candidate subsequence of length i - 1. Maintain candidate subsequences by linking them through the input sequence.)

C32-String-Matching/FA.c

@@ -57,7 +57,7 @@ int FA_match(char *text,int *array,int numchars,int receive)
 	int i;
 	for(i = 0;i < n;i++)
 	{
-		int index = numchars*q+text[i]-97;    
+		int index = numchars*q+text[i]-'a';
 		q = array[index]; 
 		if(q == receive)
 			return i+1-receive;
@@ -67,23 +67,32 @@ int FA_match(char *text,int *array,int numchars,int receive)
 
 int main()
 {
-	char *text = "abababacaba";
-	char *pattern = "ababaca";
+	char *text = "aaababaabaababaab";
+	char *pattern = "aabab";
+	const int num_chars_alphabet = 2;
 	int length = strlen(pattern);
-	int *array = (int*)malloc(sizeof(int)*3*(length+1));
+	int *array = (int*)malloc(sizeof(int)*num_chars_alphabet*(length+1));
 
-	compute_transition_function(pattern,array,3);
+	compute_transition_function(pattern,array,num_chars_alphabet);
 
-	int i;
-	printf("This is our chart\n");
+
+	//This prints a chart showing present state and next states given
+	//the corresponding inputs.
+	printf("This is our chart\nstate\t");
+	for(char j = 'a'; j < 'a'+num_chars_alphabet; j++)
+		printf("%c\t", j);
+    printf("\n");
+
+    int i;
 	for(i = 0;i <= length; i++)
 	{
+	    printf("%d\t",i);
 		int j;
-		for(j = 0; j < 3; j++)
-			printf("%d  ",array[i*3+j]);
+		for(j = 0; j < num_chars_alphabet; j++)
+			printf("%d\t",array[i*num_chars_alphabet+j]);
 		printf("\n");
 	}
-	int offset = FA_match(text,array,3,length);
+	int offset = FA_match(text,array,num_chars_alphabet,length);
 	printf("offset is %d\n",offset);
 	free(array);
 	return 0;