
Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Exercise 3: Multiple Scatter plots
- Generate scatter plots between every pairs of numerical columns from the "iris" dataframe as a set of subplots
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- Let N be an unordered array of integers. The maximum number of compares required to find the minimum value is N. Select one True or Falsearrow_forwardWhat is the run-time complexity for inserting an item when the insertion point is the logical size of the array.arrow_forwardYou can see in the above display, we first sort each row of the 2D array; we then take the transpose of a two D array, i.e., all the row elements becoming the column elements; we then sort each row of the 2D again. If you read the final array, each row is sorted; each column is also sorted. The smallest element obviously is the 1st element of the two D array and the last element is the largest element of a two D array. Let us now look at the following UML diagram: (Note that additional methods are allowed; proposed methods and instance variable cannot be changed) Main method firstly constructs a 2D array of certain sizes and then construct a TwoD object and drive the task according to the above runtime interactions and displays. TwoD class has only one instance variable which is a two D array of numbers ( int or double). The constructor must do some “deep” copying. A copy constructor. The other three methods are obvious in definition: to sort each row, to rotate the 2D array (i.e.,…arrow_forward
- Exercise 2.3. What is wrong with this procedure, which is supposed to return True if the element x occurs in the list items, and False otherwise? def member (x, items): for i in items: if x == 1: return True else: return Falsearrow_forwardThe algorithm: –In an array of n elements, go to index [n/2] –If the record there is the one you want, you are done –If the record value there is smaller than your search value, all records less than the current record can be ignored – set your search range of elements to [n/2+1…n] and return to step 1 –Otherwise, set your range of elements to [0…(n/2)-1] and return to step 1 –Repeat this loop until you have 0 elements (record is not found) or record is found Short answer Another approach to the update algorithm is to perform use the delete function for the old value and if it is successful, call the insert function using the new value. Explain in your own words if you think this approach is significantly better, worse, or in the same category as the algorithm discussed in the slides, and why.arrow_forwardIn matlab, Write a function that performs the data saturation operation, that is, if the values found within the vector exceed a certain limit value (in module), these values are equal to this limit value.arrow_forward
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