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**Then**

(A1∗A2)∗A3

Total number to be multiplied in this bracketing (10×100×5) + (10×5×50) = 7500 times

A1∗(A2∗A3)

In this bracketing, the total number to be multiplied is (100×5×50) + (10×100×50) = 75,000 times.

(Here multiplication refers to the number of times the scalar is multiplied inside the matrix multiplication, see condition number 3)

The amount of calculation increased by 10 times in the 2nd way! Using the matrix chain multiplication method, we will find a “bracketing” that minimizes number multiplication. You A1,A2……..An

Many such matrices will have only given dimensions, say the minimum number of times A1∗A2∗………..An

can be extracted We assume the dimensions are valid, i.e. satisfy the first condition.

We will solve it by divide and conquer method. Dynamic programming will be needed because the same subproblem will come up again and again. Let n be

value is 5, then there are 5 matrices A1,A2,A3,A4,A5, you A1∗A2∗A3∗A4∗A5

Find out how many numbers it takes to multiply. Now see that we can bisect matrices in different ways, one way is as follows:

(A1∗A2)∗(A3∗A4∗A5)

That means we have a bracketing Aleft=A1∗A2

I will extract and extract Aright=A3∗A4∗A5 in any bracketing. We don’t know how to figure them out, but if someone tells us Aleft and Aright

To find out how many numbers need to be multiplied then we can tell how many numbers need to be multiplied in total. The total number multiplied by the number will be:

Total number of times = Aleft

Number of product to determine + Aright Number of product to determine + Aleft∗Aright

Number of multiplications to be determined

Or we can write the last term from the 3rd term: Aleft

Number of Rows×Aleft Number of Columns×Aright

Its column number

Total number of times = Aleft

Number of multiplications to determine + Aright Number of multiplications to determine + Number of rows of Aleft×Number of columns of Aleft×Aright

Its column number

That is, if someone is magically Aleft

Aright of

Only then we can find out the total number.

But we could have divided it in many other ways, such as:

(A1)∗(A2∗A3∗A4∗A5)

(A1∗A2∗A3)∗(A4∗A5)

(A1∗A2∗A3∗A4)∗(A5)

We will divide each way and magically Aleft after dividing

and Aright

To find out how many numbers need to be multiplied, I will find out the total number. My answer is that the total number is minimum if you divide it!

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If you have a good understanding of recursion then by now you know how to magically do the work. We will solve the left and right sides recursively in the same way, that is, we will divide them again in many ways and bring out the optimal answer! So our algorithm is very simple:

Share as many ways as possible

Recursively solve for smaller parts

Find the total number of times by merging the left and right sides

Take the best of all means

So think what will be the parameter or state of our recursion? What information can I give to solve the problem?

I just need to know which part of the input matrices I’m working with now. That means we will keep the start point and end point as states.

int f(int beg, int end)

Such will be the parameter or state of the function. f

The function will tell you the total number of multiplications needed to multiply the matrices from beg to end optimally! When does recursion stop? That is, what is the base case? When we see that there is one or fewer matrices (b>=e)

Then we know we don’t need to do any multiplication, we will return zero.

So now we write a code. It’s best if you try it yourself first, the joy of solving it without help is unmatched! Even if you can’t, there’s nothing to be upset about. As input you take the number of rows and columns of each matrix. The output will be the number of operations.

After getting the state, our task will be to find out how to go from one state to another state, that is, the recurrence relation. In this problem we will recursively solve the problem for the left and right sides by selecting one midpoint and merging them. The recurrence will then be: MCM

Now we convert it to code:

#define MAX 100

int row[MAX], col[MAX];

int dp[MAX][MAX];

bool visited[MAX][MAX];

int f(int beg,int end)

{

if(beg>=end) return 0;

if(visited[beg][end])return dp[beg][end];

int ans=1<<30; // Taking 2^30 to infinity

for(int mid=beg; mid<end;mid++) //Dividing in two

{

int opr_left = f(beg, mid); //opr = multiplication operation

int opr_right = f(mid+1, end);

int opr_to_multiply_left_and_right = row[beg]*col[mid]*col[end];

int total = opr_left + opr_right + opr_to_multiply_left_and_right;

ans = min(ans, total);

}

visited[beg][end] = 1;

dp[beg][end] = ans;

return dp[beg][end];

}

int main()

{

int n;

cin>>n;

rep(i,n)cin>>row[i]>>col[i];

cout<<f(0,n-1)<<endl;

}

#define MAX 100

int row[MAX], col[MAX];

int dp[MAX][MAX];

bool visited[MAX][MAX];

int f(int beg,int end)

{

if(beg>=end) return 0;

if(visited[beg][end])return dp[beg][end];

int ans=1<<30; // Taking 2^30 to infinity

for(int mid=beg; mid<end;mid++) //Dividing in two

{

int opr_left = f(beg, mid); //opr = multiplication operation

int opr_right = f(mid+1, end);

int opr_to_multiply_left_and_right = row[beg]*col[mid]*col[end];

int total = opr_left + opr_right + opr_to_multiply_left_and_right;

ans = min(ans, total);

}

visited[beg][end] = 1;

dp[beg][end] = ans;

return dp[beg][end];

}

int main()

{

int n;

cin>>n;

rep(i,n)cin>>row[i]>>col[i];

cout<<f(0,n-1)<<endl;

}

A very simple code, dividing it into two parts and solving the smaller part. I don’t want to solve for the same part again and again, so I am saving it in DP array, if I see that any state has been visited before, then I am returning the old result!

Complexity:

beg and end can have values of 1

from to n. Then there are about n2 different states. The loop can run up to n times in each state. Then the time complexity is O(n^3). Memory required is O(n2).

.

Related problems:

You may not get direct matrix chain multiplication problems in the contest, but you can solve many problems with the idea of dividing both sides by states in different ways, which is why it’s so important to learn.

1. http://www.spoj.com/problems/MIXTURES/

2. Cutting Sticks

Happy Coding

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