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Dovator [93]
3 years ago
14

We investigated a program which is probably used as one component of a bigger password breaking algorithm. We determined that th

e program can input arbitrary N-bit queue and for actual N-bit input also the program output will be always N bits long. Additionally we noticed that the longer program input is, the longer will be the output calculating time. After performing some repeating tests we also determined that the program working time depends only and exactly on input length, not on the input itself.
Finally we fixed some actual working times:



-for N=10 - 10.576 seconds;

-for N=20 - 11.087 seconds;

-for N=25 - 13.544 seconds;

-for N=30 - 27.442 seconds;

-for N=35 - 1 minute 46.059 seconds;

-for N=40 - 9 minutes 10.784 seconds.



Task:



a) Find the program working time for N=50.

b) Please derive the mathematical formula using which is possible to calculate actual working time for arbitrary N.
Computers and Technology
1 answer:
Svetlanka [38]3 years ago
3 0

Answer:

i dont know

Explanation:

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Because of Moore's Law, widely-used encryption programs currently employed by banks and ecommerce sites are now easily penetrate
melomori [17]

Answer:

False

Explanation:

Moore's law observes the principle that the computing speed and capability of computers doubles after every two years, because of increases in the number of transistors a microchip can contain. This increase in coputational power will rather help banks and e-commerce sites (the defenders) much more than it will help the attackers. This is so because cryptographic systems are designed so that a small amount of extra work by the defender leads to an much much greater amount of work for the attacker.

6 0
3 years ago
Write a script that will read from a file x and y data points in thefollowing format:
worty [1.4K]

Answer /Explanation

load hw92.dat

FID = fopen(file, 'r');

if FID == -1  

   fprintf('ERROR CANNOT OPEN FILE TO READ!');  

else

   fclose(file);

The we have:

datacell = textscan(FID, 'x%fy%f', 'CollectData', 1);

xycoords = datacell{1};

x 0 y 1

x 1.3 y 2.2

x 2.2 y 6

x 3.4 y 7.4

x 4.2 y 5.5

x 4.4 y 4.5

x 6.2 y 7.8

x 7.7 y 11.1

x 8.2 y 11.5

x 9.9 y 15.2

x 7.2 y 9.5

x 8.9 y 12.5

end

8 0
4 years ago
)duplicate hash code
mash [69]

Answer:A) Create collisions

Explanation: Hash codes are the numeric codes present in the programming infrastructures which are responsible for focusing on the object while the equality testing is in process. Hash codes does not display the property of having a permanent value.

When there is the creation of the duplicate hash codes then ,chances of collision arise because they search for the particular object and cannot have the same numeric value.

5 0
3 years ago
Consider the following method intended to modify the parameter names by removing all instances of the String n.
Anastasy [175]

Answer:

int i = 0; i < names.size(); i++

Explanation:

The ArrayList must be read in the forward direction, and it is going to start from 0 certainly. Also, the iteration is going to end when i is exactly one less than the size of the ArrayList. And this is possible only if we choose the option mentioned in the Answer section. In this, i starts from 0 and iterates till i is one less than name.size() which is the size of the ArrayList.

7 0
3 years ago
Write the code for the method getNewBox. The method getNewBox will return a GiftBox that has dimensions that are m times the dim
liraira [26]

Answer:

public class GiftBox{

private int length;

private int width;

private int height;

 

public GiftBox(double length, double width, double height) {

this.length = length;

this.width = width;

this.height = height;

}

 

public static GiftBox getNewBox(GiftBox giftBox, double m) {

return new GiftBox(m * giftBox.length, m * giftBox.width, m * giftBox.height);

}

 

private boolean fitsInside(GiftBox giftBox) {

if(giftBox.length < this.length && giftBox.width <this.width

&& giftBox.height < this.height) {

return true;

}

return false;

}

 

public static void main(String []args){

GiftBox giftBox = new GiftBox(3.0 , 4.0, 5.0);

GiftBox newGiftBox = getNewBox(giftBox, 0.5);

System.out.println("New Box length: " + newGiftBox.length);

System.out.println("New Box width: " + newGiftBox.width);

System.out.println("New Box height: " + newGiftBox.height);

 

GiftBox gift = new GiftBox(3.0 , 4.0, 5.0);

GiftBox other = new GiftBox(2.1 , 3.2, 4.3);

GiftBox yetAnother = new GiftBox(2.0 , 5.0, 4.0);

 

System.out.println(gift.fitsInside(other));

System.out.println(gift.fitsInside(yetAnother));

}

}

Explanation:

The getNewBox is a public method in the GiftBox class in the Java source code above. It returns the GiftBox object instance increasing or multiplying the dimensions by the value of m of type double.

3 0
3 years ago
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