Answer:
Link Aggregation Control Protocol
Explanation:
Link Aggregation Control Protocol can be used to assist in the formation of port channel bundles of physical links.
Link Aggregation Control Protocol is an IEEE standard defined in IEEE 802.3ad. LACP lets devices send Link Aggregation Control Protocol Data Units (LACPDUs) to each other to establish a link aggregation connection.
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:
Question was incomplete and continued the question
For each of the following scenarios, which of these choices would be best? Explain your answer.
BST
Sorted Array
Un-sorted Array
a) The records are guaranteed to arrive already sorted from lowest to highest (i.e., whenever a record is inserted, its key value will always be greater than that of the last record inserted). A total of 1000 inserts will be interspersed with 1000 searches.
b) The records arrive with values having a uniform random distribution (so the BST is likely to be well balanced). 1,000,000 insertions are performed, followed by 10 searches.
Explanation:
Answer for a: Un-sorted array or Un-sorted linked list : as mentioned in the question itself that the records are arriving in the sorted order and search will not be O(log n) and insert will be not be O(n).
Answer for b : Un-sorted array or Un-sorted linkedlist : Number of the items to be inserted is already known which is 1,000,000 but it is very high and at the same time search is low. Unsorted array or Unsorted linked list will be best option here.
Answer:
Computer professionals known as software engineers, or programmers use the software development life cycle to create software required for information systems.
Explanation:
Computer professionals are called software engineers and programmers because they develop and program software. Some additional titles for computer professionals are hardware engineers and iOS/Android developers.
MOHR-COULOMB FAILURE CRITERIA:
In 1900, MOHR-COULOMB states Theory of Rupture in Materials which defines as “A material fails due to because of a critical combination of normal and shear stress, not from maximum normal or shear stress”. Failure Envelope is approached by a linear relationship.
If you can not understand the below symbols see the attachment below
f f ()
Where: f = Shear Stress on Failure Plane
´= Normal Stress on Failure Plane
See the graph in the attachment
For calculating the shear stress, when Normal stress, cohesion and angle of internal friction are given. Use this formula: shear stress = f c tan
Where,
• f is Shear Stress on Failure Plane
• c is Cohesion
• is Normal Total Stress on Failure Plane
• is Friction Angle
