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Insertion_Sort(num_list) length <-- length of num_list // At the start, the second element of the original list // is the first number in the set of "unsorted" numbers. number_index <-- 2 // We re done when we have looked at all positions in the list. while(number_index <= length) { // newNum is the no. being considered for sorting newNum <-- num_list[number_index] // sorted_index marks the end of previously sorted numbers. sorted_index <-- number_index - 1 // From high to low, look for the place for the new number. // If newNum is smaller than the previously sorted numbers, // move the previously sorted numbers up in the num_list. while newNum < num_list[sorted_index] AND sorted_index > 0 { num_list[sorted_index + 1] <-- num_list[sorted_index] sorted_index <-- sorted_index - 1 } // newNum is not smaller than the number at sorted_index. // We found the place for the new number, so insert it. num_list[sorted_index + 1] = newNum } end To repeat, the variable number_index keeps track of where the algorithm is in the unsorted set of numbers. The algorithm starts with the second number (number_index = 2). Then the algorithm compares the number to the largest number that has been sorted so far, num_list[sorted_index]. If the number is smaller than the previously sorted number, the algorithm moves the previously sorted number up one position in num_list, and checks the new number against the next largest number in the previously sorted elements of num_list. Finally, the algorithm will encounter a previously sorted number which is smaller than the number being inserted, or it will find itself past the starting position of num_list. At that point, the number can be inserted into the num_list. The algorithm completes when all of the positions in the num_list have been sorted. To analyze the running time of the insertion sort, we note first that the performance will be proportional to n, the number of elements to be sorted. We also note that each element to be sorted must be compared one or many times with the elements already sorted. In the best case, the elements will be sorted already, and each element will require only a single comparison, so the best-case performance of the insertion sort is (n). In the worst case, the elements to be sorted will be in reverse order, so that every element will require comparison with every element already sorted. The second number will be compared with the first, the third with the second and first, the fourth with the third, second, and first, etc. If there were four numbers in reverse order, the number of comparisons would be six. In general, the number of comparisons in the worst case for the insertion sort will be: n2/2 - n/2 The number of comparisons will grow as the square of the number of elements to be sorted. The negative term of -n/2, and the division of n2 by the constant 2, mean that the rate of growth in number of comparisons will not be the full rate that n2 would imply. However, for very large values of n, those terms other than
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n2 become relatively insignificant. Imagine the worst case of sorting a million numbers. The n2 term will overwhelm the other terms of the equation. Since one usually reports the order of growth for an algorithm as the worst-case order of growth, the insertion sort has a theta of n2, or (n2). If one computes the average case order of growth for the insertion sort, one also finds a quadratic equation; it s just somewhat smaller, since on average each new element will be compared with only half of the elements already sorted. So we say the performance of the insertion sort is (n2). Merge sort An example of order of growth of n(lg n) Q(n lg n) Another algorithm for sorting numbers uses recursion, a technique we will discuss in more detail shortly, to divide the problem into many smaller problems before recombining the elements of the full solution. First, this solution requires a routine to combine two sets of sorted numbers into a single set. Imagine two piles of playing cards, each sorted from smallest to largest, with the cards face up in two piles, and the two smallest cards showing. The merge routine compares the two cards that are showing, and places the smaller card face down in what will be the merged pile. Then the routine compares the two cards showing after the first has been put face down on the merged pile. Again, the routine picks up the smaller card, and puts it face down on the merged pile. The merge routine continues in this manner until all the cards have been moved into the sorted merged pile. Here is pseudocode for the merge routine. It expects to work on two previously sorted lists of numbers, and it merges the two lists into one sorted list, which it returns. The variable index keeps track of where it is working in sorted_list. The routine compares the first (top) numbers in the two original lists, and puts the smaller of the two into sorted_list. Then it discards the number from the original list, which means that the number that used to be the second one in the original list becomes the first number in that list. Again the routine compares the first numbers in the two lists, and again it moves the smaller to sorted_list. The routine continues this way until one of the original lists becomes empty. At that point, it adds the remaining numbers (which were in sorted order originally, remember) to sorted_list, and returns sorted_list. merge(list_A, list_B) // index keeps track of where we are in the // sorted list index <-- 1 // Repeat as long as there are numbers in both // original lists. while list_A is not empty AND list_B is not empty // Compare the 1st elements of the 2 lists. // Move the smaller to the sorted list. // "<" means "smaller than." if list_A[1] < list_B[1] sorted_list[index] <-- list_A[1] discard list_A[1] else sorted_list[index] <-- list_B[1] discard list_B[1] index <-- index + 1 // If numbers remain only in list_A, move those // to the sorted list while list_A is not empty sorted_list[index] <-- list_A[1] discard list_A[1] index <-- index + 1
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