# Solving of tasks using two-dimensional arrays. Examples

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##### 1. Creation of a third matrix based on two matrices

Task. Given two matrices of numbers A and B. Dimension of matrices m×n (m – number of rows, n – number of columns). The numbers of matrices are formed randomly. Form a matrix C in which each element is determined by the formula

where i = 1, 2, …, m; j = 1, 2, …, n.

Solution.

```// Two-dimensional arrays. Solution of tasks.

// Creating a third matrix based on two

// 1. Declaring matrices and creating them
var A = new Array(), B = new Array() // original matrices
var C = [] // resulting matrice

// 2. Enter the dimension of matrices m, n
var m = prompt("m = ")
var n = prompt("n = ")

// 3. Allocate memory for matrix cells
// 3.1. Allocate memory for m rows of matrices
A.length = m
B.length = m

// 3.2. Allocate memory for n columns in each row
for (var i=0; i<A.length; i++)
{
// Create the array A[i] and allocate memory for it
A[i] = new Array()
A[i].length = n

// Create the array B[i] and allocate memory for it
B[i] = new Array()
B[i].length = n
}

// 4. Fill matrices A, B with random numbers in the interval [min; max]
var min = -10 // lower bound of interval
var max = 10 // upper bound of the interval

for (var i=0; i<A.length; i++)
for (var j=0; j<A[i].length; j++)
{
// The array sizes are the same, so you can form two arrays at once
A[i][j] = parseInt(min + Math.random()*(max - min + 1))
B[i][j] = parseInt(min + Math.random()*(max - min + 1))
}

// 5. Display matrix A
document.write("A:<br>")
for (var i=0; i<A.length; i++)
{
for (var j=0; j<A[i].length; j++)
document.write(A[i][j] + " ")
document.write("<br>")
}
document.write("<br>")

// 6. Display matrix B
document.write("B:<br>")
for (var i=0; i<B.length; i++)
{
for (var j=0; j<B[i].length; j++)
document.write(B[i][j] + "\t")
document.write("<br>")
}

document.write("<br>")

// 7. Form the resulting matrix C
// 7.1. Allocate memory for the matrix C with size m*n
C.length = A.length // number of rows

for (var i=0; i<C.length; i++)
{
C[i] = [] // create the array C[i]
C[i].length = n // set the number of columns
}

// 7.2. Fill in the values of the matrix cells according to the task
for (var i=0; i<C.length; i++)
for (var j=0; j<C[i].length; j++)
C[i][j] = A[i][j]*A[i][j] + B[i][j]

// 8. Display the resulting matrix C
document.write("C:<br>")
for (var i=0; i<C.length; i++)
{
for (var j=0; j<C[i].length; j++)
document.write(C[i][j] + " ")
document.write("<br>")
}```

Result

```A:
-1 5 -8 5 3
5 8 -3 -5 0
-5 10 0 -5 9

B:
9 -4 -9 1 -4
0 2 5 -1 6
1 -4 5 0 8

C:
10 21 55 26 5
25 66 14 24 6
26 96 5 25 89```

### ⇑

##### 2. Processing a matrix consisting of rows. Determining the number of rows according to the criterion

Task. A matrix A of words (rows) of size m×n is given. Determine how many rows of the matrix have length greater than the given k. Matrix elements A and m, n, k are entered from the keyboard.

Solution.

```// Two-dimensional arrays. Solution of tasks.

// Determining the number of matrix rows according to the criterion

// 1. Matrix declaration
var A = new Array()

// 2. Input the dimension of the matrix m*n
var m = prompt("m = ")
var n = prompt("n = ")

// 3. Input k
var k = prompt("k = ")

// 4. Allocate memory for matrix
A.length = m

for (var i=0; i<m; i++)
{
A[i] = new Array()
A[i].length = n
}

// 5. The cycle of forming a matrix of rows
for (var i=0; i<m; i++)
for (var j=0; j<n; j++)
{
A[i][j] = prompt("A[" + i + "][" + j + "] = ")
}

// 6. Display matrix A
document.write("A:<br>")
for (var i=0; i<A.length; i++)
{
for (var j=0; j<A[i].length; j++)
document.write(A[i][j] + " ")
document.write("<br>")
}
document.write("<br><br>")

// 7. Loop for calculating the number of rows with length no more than k.
var count = 0 // result

for (var i=0; i<A.length-1; i++)
for (var j=0; j<A[i].length; j++)
if (A[i][j].length <= k)
count++

// 7. Display the result
document.write("-----------------------------<br>")
document.write("count = " + count)```

Result

```A:
abcd jklmn
jprstq 123456

-----------------------------
count = 2```

### ⇑

##### 3. Create a matrix based on the other two according to the formula

Task. Given natural numbers m, n and a matrix of real numbers Aij, Bij, where i = 1, 2, …, m; j = 1, 2, …, n. Calculate the value of matrix elements Cij, if:

Solution.

```// Two-dimensional arrays. Solution of tasks.

// Formation of the third matrix based on two given ones

// 1. Ввод m, n
var m = prompt("m = ")
var n = prompt("n = ")

// 2. Declaring matrices A, B and filling them with values.
// 2.1. Create arrays A, B
var A = []
var B = []

// 2.2. Set the number of rows m
A.length = m
B.length = m

// 2.3. Create subarrays and customize them.
for (var i=0; i<A.length; i++)
{
// Create subarrays in arrays A, B.
A[i] = []
A[i].length = n
B[i] = []
B[i].length = n
}

// 2.4. Input numbers into arrays A, B
for (var i=0; i<A.length; i++)
for (var j=0; j<A[i].length; j++)
A[i][j] = prompt("A["+i+"]["+j+"] = ")

for (var i=0; i<B.length; i++)
for (var j=0; j<B[i].length; j++)
B[i][j] = prompt("B["+i+"]["+j+"] = ")

// 3. Display arrays A and B
document.write("A:<br>")
for (var i=0; i<A.length; i++)
{
for (var j=0; j<A[i].length; j++)
document.write(A[i][j]+" ")
document.write("<br>")
}
document.write("<br>")

document.write("B:<br>")
for (var i=0; i<B.length; i++)
{
for (var j=0; j<B[i].length; j++)
document.write(B[i][j]+" ")
document.write("<br>")
}
document.write("<br>")

// 4. Creating a C matrix and setting the dimensions of the matrix
var C = []
C.length = m
for (var i=0; i<C.length; i++)
{
C[i] = []
C[i].length = n
}

// 5. Filling the matrix C with the resulting values according to the formula
for (var i=1; i<=C.length; i++)
for (var j=1; j<=C[i-1].length; j++)
{
if (i<j)
C[i-1][j-1] = i*i + j - 5;
else
if (i==j)
C[i-1][j-1] = 1 / ((i+j)*(i+j)*(i+j));
else
C[i-1][j-1] = Math.sin(A[i-1][j-1])*Math.sin(B[i-1][j-1]) +
Math.cos(B[i-1][j-1])
}

// 6. Display the result
document.write("C:<br>")
for (var i=0; i<C.length; i++)
{
for (var j=0; j<C[i].length; j++)
document.write(C[i][j]+" ")
document.write("<br>")
}```

Result

```A:
1 2 3
4 5 6

B:
1 1 1
1 1 1

C:
0.125 -2 -1
-0.09652503516369604 0.015625 2```

### ⇑

##### 4. Determine the contents of a two-dimensional table according to a condition

A natural number n is given. Determine the number of positive and the number of negative elements of table A with dimension n×n, if

where i, j = 1, 2, … , n.

Solution.

```// Two-dimensional arrays. Solving of tasks.

// A natural number n is given. Determine the number
// of positive and the number of negative elements of table A.

// 1. Input n
var n = prompt("n = ")

// 2. Table A declaration
var A = []

// 3. Set the dimensions of table A
A.length = n
for (var i=0; i<A.length; i++)
{
A[i] = []
A[i].length = n
}

// 4. Form table A according to the formula
for (var i=0; i<A.length; i++)
for (var j=0; j<A[i].length; j++)
A[i][j] = Math.sin(i*i-j)

// 5. Print table A with 2 decimal places.
document.write("A:<br>")
for (var i=0; i<A.length; i++)
{
for (var j=0; j<A[i].length; j++)
document.write(A[i][j].toFixed(2)+" ")
document.write("<br>")
}
document.write("<br>")

// 6. Calculate the number of positive and negative elements.
var n1 = 0 // positive elements
var n2 = 0 // negative elements

for (var i=0; i<A.length; i++)
for (var j=0; j<A.length; j++)
{
if (A[i][j]<0)
n1++
else
n2++
}

// 7. Display the result
document.write("n1 = " + n1 + "<br>")
document.write("n2 = " + n2 + "<br>")```

Result

```A:
0.00 -0.84 -0.91 -0.14
0.84 0.00 -0.84 -0.91
-0.76 0.14 0.91 0.84
0.41 0.99 0.66 -0.28

n1 = 7
n2 = 9```

### ⇑

##### 5. Formation of a one-dimensional array based on a two-dimensional matrix. Determining the arithmetic mean column of a matrix

Task. The matrix Aij is given, where i=1, 2, …, m, j = 1, 2, …, n. Create a new one-dimensional array Bi of size m, in which the elements are equal to the arithmetic mean of each column of matrix A.

Solution.

```// Two-dimensional arrays. Solution of tasks.

// Form an array of average arithmetic columns of a given matrix.
// 1. Input m, n
var m = prompt("m = ")
var n = prompt("n = ")

// 2. Declaring the A matrix
var A = []

// 3. Set the dimensions of matrix A.
A.length = m
for (var i=0; i<A.length; i++)
{
A[i] = []
A[i].length = n
}

// 4. Input matrix A
for (var i=0; i<A.length; i++)
for (var j=0; j<A[i].length; j++)
A[i][j] = prompt("A["+i+"]["+j+"] = ")

// 5. Display m, n
document.write("m = " + m + "<br>")
document.write("n = " + n + "<br>")

// 6. Display matrix A
document.write("A:<br>")
for (var i=0; i<A.length; i++)
{
for (var j=0; j<A[i].length; j++)
document.write(A[i][j]+" ")
document.write("<br>")
}

// 7. Create the resulting array B and set its dimensions
var B = []
B.length = n // number of columns, one-dimensional array

// 8. The cycle of calculating the arithmetic mean and filling the matrix B
var avg // additional variables - average
for (var i=0; i<n; i++) // outer loop over columns
{
avg = 0;
for (var j=0; j<m; j++) // calculate the sum of column i
avg += parseFloat(A[j][i])

B[i] = avg / m
}

// 9. Return the result
document.write("B:<br>")
for (var i=0; i<B.length; i++)
document.write(B[i].toFixed(2) + " ")
document.write("<br>")```

Result

```m = 2
n = 3
A:
1 2 3
4 5 6
B:
2.50 3.50 4.50```