Applied Test Series: Live Mock

Question

A 30*30  upper triangular matrix of integers is stored optimally in a 1-D array with starting address of 3000 assume the matrix indexing starts at (1,1) and array indexing starts at 0 what would be the address of element with index (25,25)), assume one integer takes up two bytes of memory.___

Need help in this one. Didn’t get the solution properly provided by them. Anyone!

Comments

Hii @ramakrushna please check this approach

Memory Address of (25,25) element should be =base address+no_of_element*(size of each element)

memory address=3000+444*2=3888

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@lalitver10 Your approach is correct but I have a doubt, since index 444 of the array will represent the (25, 25) index of the matrix therefore size of 444 elements will be added to the base address not of 445 elements otherwise it will represent (25, 26) index of the matrix.

 

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Yes you are correct memory address should be 3888 not 3890 for (25,25)^th element.

Thanks.

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Getting a general formula for n*n matrix as :

$\frac{{n}({n+1})}{2} - \frac{({n+2-i})({n+1-i})}{2} + j-i$

Putting n = 30; i=j=25;

444.

Can someone verify the formula?

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Where did you find this formula. I have never seen this one before. Any source ?

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Nah I derived it.

Answer 1

Since the matrix is upper triangular matrix number of elements in :

1^st row = 30

2^nd row = 29

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24^th row = 30 – (n-1) = 30 – 23 = 7

Since till 24^th row we have to consider all the elements.

∴ Total #elements = 30 + 29 + 28 + … + 7  (which is in AP)

= $\frac{n}{2}(2*a + (n-1)*d  =  \frac{24}{2}(2*30 + (24-1)*-1)$

= $444$

Now since matrix is upper triangular matrix index (25, 25) will correspond to 1^st element of the 25^th row.

∴ Address of (25, 25) = 3000 + (number of elements)* size of each element

 = 3000 + 444*2 = 3000+888 = 3888

Answer: 3888

Comments

Very well explained!