diff --git a/qsort.py b/qsort.py index b09a5ac..3609eec 100644 --- a/qsort.py +++ b/qsort.py @@ -6,24 +6,32 @@ def qsort(xs): right = [x for x in xs[1:] if x >= pivot] return [qsort(left)] + [pivot] + [qsort(right)] +def _sorted(tree, acc): + if tree == []: return + + _sorted(tree[0], acc) + acc.append(tree[1]) + _sorted(tree[2], acc) + def sorted(tree): - if tree == []: return [] - return sorted(tree[0]) + [tree[1]] + sorted(tree[2]) + acc = [] + _sorted(tree, acc) + return acc def search(tree, x): - return _sorted(tree, x) != [] + return _search(tree, x) != [] def insert(tree, x): - node = _sorted(tree, x) + node = _search(tree, x) if node == []: node.append([]) node.append(x) node.append([]) -def _sorted(tree, i): +def _search(tree, i): if tree == []: return tree pivot = tree[1] if pivot == i: return tree - elif i < pivot: return _sorted(tree[0], i) - else: return _sorted(tree[2], i) + elif i < pivot: return _search(tree[0], i) + else: return _search(tree[2], i) diff --git a/report.txt b/report.txt index 4296844..af3c04f 100644 --- a/report.txt +++ b/report.txt @@ -12,7 +12,11 @@ A: Quicksort has the worst-case complexity of O(n^2). This is because in the wor On average, Quicksort is also O(n*log(n)). It's quite difficult to consistently pick a pivot that is either the smallest or the largest. I am unfamilliar with proof - techniques that help formalize this. + techniques that help formalize this, but we can think of a case in which + some non-half fraction (say j/k) of the elements + is on the left of the pivot. In this case, the depth ends up being a multiple + of log_k(n), meaning that the depth is still logarithmic and the complexity is + still O(n*log(n)). Q: What's the best-case, worst-case, and average-case time complexities? Briefly explain. A: For the same reason as quicksort, in the worst case, the complexity is O(n^2). @@ -25,19 +29,17 @@ A: For the same reason as quicksort, in the worst case, the complexity is O(n^2) is n(1-r^k)/(1-r). This simplifies to 2n(1-r^k). Since 1-2^k < 1, n*(1+1/2+1/4+...) < 2n. This means the complexity is O(n). - For similarly hand-wavey reasons to those in Q0, the average case complexity aligns - with the best-case complexity rather than worst-case complexity. + Similarly to quicksort, we can assume j/k elements are on the left + of the pivot. Then, the the longest possible computation will end up + looking at nj/k elements, then nj^2/k^2, and so on. This is effectively + n times the sum of the geometric series with r=j/k. This means + the sum is n * c, and thus, the complexity is O(n). Q: What are the time complexities for the operations implemented? -A: The complexity of sorted is O(n*log(n)) in best, and O(n^2) in worst case. - This is because of the way in which it implements - "flattening" the binary search tree - it recursively calls itself, creating - a new array from the results of the two recursive calls and the "pivot" between them. - Since creating a new array from arrays of length m and n is an O(m+n) operation. - Just like with qsort, in the best case, the tree is balanced with a depth of log(n). - Since concatenation at each level will effectively take n steps, the best case complexity - is O(n*log(n)). On the other hand, in the case of a tree with only right children, - the concatenation will take 1+2+...+n steps, which is in the order O(n^2). +A: The complexity of sorted is O(n). + Since I use an accumulator array, array append is O(1). Then, all + that's done is an in-order traversal of the tree, which is O(n), + since it visits every element of the tree. Since insert and search both use _search, and perform no steps above O(1), they are of the same complexity as _search. _search itself is O(logn) in the average case,