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3 years ago

A(n) _________________________ conveys a visual representation of data.

Computers and Technology

1 answer:

Komok [63]3 years ago

3 0

I believe that the answer to fit this sentence is chart. Hope that this answer helped! ☺

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You are planning to depart on a flight from area 2 to area 4 in 12 hours. What weather is forecast to occur along your route?

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It depends on where you live!

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2 years ago

Which type of operating system is usually used in personal computers?

Vitek1552 [10]

Answer:

I think A modern operating system is usually used in personal computers

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3 years ago

Coment on this if your user starts with dida

ratelena [41]

Answer:

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Explanation:

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2 years ago

Suppose you have a program P and 90 percent of P can be parallelized, but 10 percent of P is inherently sequential and cannot be

EleoNora [17]

Answer:

Means no matter how many processors you use, speed up never increase from 10 times.

Explanation:

If a problem of size W has a serial component Ws,then performance using parallelism:

Using Amdahl's Law:

Tp = (W - Ws )/ N + Ws

Here, Ws = .1,

W - Ws = .9

Performance Tp = (.9 / N) + .1

---------------------------------------------------------

Speed Up = 1 / ( (.9 / N) + .1)

If N -> infinity, Speed Up <= 10

Means no matter how many processors you use, speed up never increase from 10 times.

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2 years ago

Show the array that results from the following sequence of key insertions using a hashing system under the given conditions: 5,

sergejj [24]

Answer:

a) Linear probing is one of the hashing technique in which we insert values into the hash table indexes based on hash value.

Hash value of key can be calculated as :

H(key) = key % size ;

Here H(key) is the index where the value of key is stored in hash table.

----------

Given,

Keys to be inserted are : 5 , 205, 406,5205, 8205 ,307

and size of the array : 100.

First key to be inserted is : 5

So, H(5) = 5%100 = 5,

So, key 5 is inserted at 5th index of hash table.

-----

Next key to inserted is : 205

So, H(205) = 205%100 = 5 (here collision happens)

Recompute hash value as follows:

H(key) =(key+i) % size here i= 1,2,3...

So, H(205) =( 205+1)%100 = 206%100 = 6

So, key 205 is inserted in the 6th index of the hash table.

----------

Next Key to be inserted : 406

H(406) = 406%100 = 6(collision occurs)

H(406) =(406+1) %100 = 407%100 = 7

So, the value 406 is inserted in 7the index of the hash table.

-----------------

Next key : 5205

H(5205) = 5205%100 = 5(collision)

So, H(5205) = (5205+1)%100 = 6( again collision)

So, H(5205) = 5205+2)%100 = 7(again collision)

So, H(5205) = (5205+3)%100 = 8 ( no collision)

So, value 5205 is inserted at 8th index of the hash table.

-------------

Similarly 8205 is inserted at 9th index of the hash table because , we have collisions from 5th to 8th indexes.

-------------

Next key value is : 307

H(307) = 307%100 = 7(collision)

So, (307+3)%100 = 310%100 = 10(no collision)

So, 307 is inserted at 10th index of the hash table.

So, hash table will look like this:

Key index

5 5

205 6

406 7

5205 8

8205 9

307 10

b) Quadratic probing:

Quadratic probing is also similar to linear probing but the difference is in collision resolution. In linear probing in case of collision we use : H(key) = (key+i)%size but here we use H(key) =( key+i^2)%size.

Applying Quadratic probing on above keys:.

First key to be inserted : 5.

5 will go to index 5 of the hash table.

-------

Next key = 205 .

H(205) = 205%100 = 5(collision)

So. H(key)= (205+1^2)%100 = 6(no collision)

So, 205 is inserted into 6th index of the hash table.

--------

Next key to be inserted 406:

So, 406 %100 = 6 (collision)

(406+1^2)%100 = 7(no collision)

So, 406 is moved to 7th index of the hash table.

----------

Next key is : 5205

So, 5205%100 = 5 (collision)

So, (5205+1^2)%100 = 6 ( again collision)

So, (5205+2^2)%100 = 9 ( no collision)

So, 5205 inserted into 9th index of hash table.

-----------

Next key is 8205:

Here collision happens at 5the , 6the , 9th indexes,

So H(8205) = (8205+4^2)%100 = 8221%100 = 21

So, value 8205 is inserted in 21st index of hash table.

--------

Next value is 307.

Here there is collision at 7the index.

So, H(307) = (307+1^2)%100 = 308%100= 8.

So, 307 is inserted at 8the index of the hash table.

Key Index

5 5

205 6

406 7

5205 9

8205 21

307 8

3 0

3 years ago