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Solnce55 [7]
3 years ago
15

The sum of a number and forty-six

Mathematics
1 answer:
ad-work [718]3 years ago
4 0

\textsf{Hey there!}

\mathsf{\star \ The\ sum\ of\ a\ number\ \&\  forty-six}

\frak{Let's\ lable\ the\ keypoints\ so\ it\ can\ be\ easier\ to\ solve}

\bullet \ \textsf{The word \bf{\underline{SUM}}}\textsf{\ means\ add/addition}}

\bullet\textsf{ \underline{"A number"} is an unknown number so we can lable it as \bf{x}}

\bullet\textsf{ \underline{forty-six} is 46}

\textsf{The sum (+) does in the middle while \underline{x} \& \underline{46} will either be on left/right}

- \ \textsf{We can say that x is on your left}

-\  \textsf{  + can be in your middle}

- \ \textsf{\& lastly 46 can be on your right}

\boxed{\textsf{Thus, your answer SHOULD LOOK like: \boxed{\huge\text{\bf{x + 46}}}}}\checkmark

\textsf{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

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You are given 4 matrices M1, M2, M3, M4 and you are asked to determine the optimal schedule for the product M1 ×M2 × M3 ×M4 that
alexandr1967 [171]

Answer:

Step-by-step explanation:

first method is to try out all possible combinations and pick out the best one which has the minimum operations but that would be infeasible method if the no of matrices increases  

so the best method would be using the dynamic programming approach.

A1 = 100 x 50

A2 = 50 x 200

A3 = 200 x 50

A4 = 50 x 10

Table M can be filled using the following formula

Ai(m,n)

Aj(n,k)

M[i,j]=m*n*k

The matrix should be filled diagonally i.e., filled in this order

(1,1),(2,2)(3,3)(4,4)

(2,1)(3,2)(4,3)

(3,1)(4,2)

(4,1)

<u>                  Table M[i, j]                                             </u>

             1                      2                  3                    4

4    250000          200000        100000                0  

3      

750000        500000            0

2      1000000             0

1            

0

Table S can filled this way

Min(m[(Ai*Aj),(Ak)],m[(Ai)(Aj*Ak)])

The matrix which is divided to get the minimum calculation is selected.

Table S[i, j]

           1          2         3        

4

4          1           2         3

3          

1          2

2            1

1

After getting the S table the element which is present in (4,1) is key for dividing.

So the matrix multiplication chain will be (A1 (A2 * A3 * A4))

Now the element in (4,2) is 2 so it is the key for dividing the chain

So the matrix multiplication chain will be (A1 (A2 ( A3 * A4 )))

Min number of multiplications: 250000

Optimal multiplication order: (A1 (A2 ( A3 * A4 )))

to get these calculations perform automatically we can use java

code:

public class MatrixMult

{

public static int[][] m;

public static int[][] s;

public static void main(String[] args)

{

int[] p = getMatrixSizes(args);

int n = p.length-1;

if (n < 2 || n > 15)

{

System.out.println("Wrong input");

System.exit(0);

}

System.out.println("######Using a recursive non Dyn. Prog. method:");

int mm = RMC(p, 1, n);

System.out.println("Min number of multiplications: " + mm + "\n");

System.out.println("######Using bottom-top Dyn. Prog. method:");

MCO(p);

System.out.println("Table of m[i][j]:");

System.out.print("j\\i|");

for (int i=1; i<=n; i++)

System.out.printf("%5d ", i);

System.out.print("\n---+");

for (int i=1; i<=6*n-1; i++)

System.out.print("-");

System.out.println();

for (int j=n; j>=1; j--)

{

System.out.print(" " + j + " |");

for (int i=1; i<=j; i++)

System.out.printf("%5d ", m[i][j]);

System.out.println();

}

System.out.println("Min number of multiplications: " + m[1][n] + "\n");

System.out.println("Table of s[i][j]:");

System.out.print("j\\i|");

for (int i=1; i<=n; i++)

System.out.printf("%2d ", i);

System.out.print("\n---+");

for (int i=1; i<=3*n-1; i++)

System.out.print("-");

System.out.println();

for (int j=n; j>=2; j--)

{

System.out.print(" " + j + " |");

for (int i=1; i<=j-1; i++)

System.out.printf("%2d ", s[i][j]);

System.out.println();

}

System.out.print("Optimal multiplication order: ");

MCM(s, 1, n);

System.out.println("\n");

System.out.println("######Using top-bottom Dyn. Prog. method:");

mm = MMC(p);

System.out.println("Min number of multiplications: " + mm);

}

public static int RMC(int[] p, int i, int j)

{

if (i == j) return(0);

int m_ij = Integer.MAX_VALUE;

for (int k=i; k<j; k++)

{

int q = RMC(p, i, k) + RMC(p, k+1, j) + p[i-1]*p[k]*p[j];

if (q < m_ij)

m_ij = q;

}

return(m_ij);

}

public static void MCO(int[] p)

{

int n = p.length-1;     // # of matrices in the product

m    =    new    int[n+1][n+1];        //    create    and    automatically initialize array m

s = new int[n+1][n+1];

for (int l=2; l<=n; l++)

{

for (int i=1; i<=n-l+1; i++)

{

int j=i+l-1;

m[i][j] = Integer.MAX_VALUE;

for (int k=i; k<=j-1; k++)

{

int q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j];

if (q < m[i][j])

{

m[i][j] = q;

s[i][j] = k;

}

}

}

}

}

public static void MCM(int[][] s, int i, int j)

{

if (i == j) System.out.print("A_" + i);

else

{

System.out.print("(");

MCM(s, i, s[i][j]);

MCM(s, s[i][j]+1, j);

System.out.print(")");

}

}

public static int MMC(int[] p)

{

int n = p.length-1;

m = new int[n+1][n+1];

for (int i=0; i<=n; i++)

for (int j=i; j<=n; j++)

m[i][j] = Integer.MAX_VALUE;

return(LC(p, 1, n));

}

public static int LC(int[] p, int i, int j)

{

if (m[i][j] < Integer.MAX_VALUE) return(m[i][j]);

if (i == j) m[i][j] = 0;

else

{

for (int k=i; k<j; k++)

{

int   q   =   LC(p,   i,   k)   +   LC(p,   k+1,   j)   +   p[i-1]*p[k]*p[j];

if (q < m[i][j])

m[i][j] = q;

}

}

return(m[i][j]);

}

public static int[] getMatrixSizes(String[] ss)

{

int k = ss.length;

if (k == 0)

{

System.out.println("No        matrix        dimensions        entered");

System.exit(0);

}

int[] p = new int[k];

for (int i=0; i<k; i++)

{

try

{

p[i] = Integer.parseInt(ss[i]);

if (p[i] <= 0)

{

System.out.println("Illegal input number " + k);

System.exit(0);

}

}

catch(NumberFormatException e)

{

System.out.println("Illegal input token " + ss[i]);

System.exit(0);

}

}

return(p);

}

}

output:

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have a good day :)

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