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.
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Answer:
#include <iostream>//including libraries
using namespace std;
int main()
{
int arr[6] = { 0,1,2,3,4,5 };//make sure size of arr is 1 less than secArr
int secArr[7];//second array (1 element bigger)
for (int i = 0;i < 6;i++)//looping through each element (6 times)
{
secArr[i + 1] = arr[i];//transferring elements to second array and shifting by 1 cell
cout << secArr[i + 1] << endl;//printing elements of second array
}
return 0;//terminating program
}
Explanation:
The array size can range from any number. just make sure to keep arr one less than secArr. This is because we need the room for the extra element. This task is to help you understand how array work and how to parse through them using loops. For loops are the best for this task because even if you think intuitively, they work for as long as there are items in the array. and you can define the size yourself.
Hi,
JVM - Java Virtual Machine
Hope this helps.
r3t40
