61A Lecture 23Friday, October 21SetsOne more built-in Python container type•Set literals are enclosed in braces•Duplicate elements are removed on construction•Sets are unordered, just like dictionary entries2>>> s = {3, 2, 1, 4, 4}>>> s{1, 2, 3, 4}>>> 3 in sTrue>>> len(s)4>>> s.union({1, 5}){1, 2, 3, 4, 5}>>> s.intersection({6, 5, 4, 3}){3, 4}Implementing SetsThe interface for sets•Membership testing: Is a value an element of a set?•Union: Return a set with all elements in set1 or set2•Intersection: Return a set with any elements in set1 and set2•Adjunction: Return a set with all elements in s and a value v3Union13423513425Intersection1342353Adjunction13421342Sets as Unordered SequencesProposal 1: A set is represented by a recursive list that contains no duplicate items4 def empty(s): return s is Rlist.empty def set_contains(s, v): if empty(s): return False elif s.first == v: return True return set_contains(s.rest, v)DemoR(n)=Θ(n)k1· n ≤ R(n) ≤ k2· nReview: Order of GrowthFor a set operation that takes "linear" time, we say that5n: size of the setR(n): number of steps required to perform the operationwhich means that there are constants k1 and k2 such thatfor sufficiently large values of n.DemoΘ(n)Θ(n2)Sets as Unordered Sequences6 def adjoin_set(s, v): if set_contains(s, v): return s return Rlist(v, s) def intersect_set(set1, set2): f = lambda v: set_contains(set2, v) return filter_rlist(set1, f) def union_set(set1, set2): f = lambda v: not set_contains(set2, v) set1_not_set2 = filter_rlist(set1, f) return extend_rlist(set1_not_set2, set2)Θ(n2)Time order of growthThe size of the setThe size of the larger setSets as Ordered SequencesProposal 2: A set is represented by a recursive list with unique elements ordered from least to greatest7 def set_contains2(s, v): if empty(s) or s.first > v: return False elif s.first == v: return True return set_contains2(s.rest, v)Θ(n)Order of growth?Set Intersection Using Ordered SequencesThis algorithm assumes that elements are in order.8def intersect_set2(set1, set2): if empty(set1) or empty(set2): return Rlist.empty e1, e2 = set1.first, set2.first if e1 == e2: rest = intersect_set2(set1.rest, set2.rest) return Rlist(e1, rest) elif e1 < e2: return intersect_set2(set1.rest, set2) elif e2 < e1: return intersect_set2(set1, set2.rest)DemoΘ(n)Order of growth?Tree SetsProposal 3: A set is represented as a Tree. Each entry is:•Larger than all entries in its left branch and•Smaller than all entries in its right branch9731 59117315 911531 7911Membership in Tree SetsSet membership tests traverse the tree•The element is either in the left or right sub-branch•By focusing on one branch, we reduce the set by about half10531 7911 def set_contains3(s, v): if s is None: return False elif s.entry == v: return True elif s.entry < v: return set_contains3(s.right, v) elif s.entry > v: return set_contains3(s.left, v)9If 9 is in the set, it is in this branchAdjoining to a Tree Set11531 791187911878Right! Left! Right!None None8NoneStop!87879118531 79118DemoWhat Did I Leave Out?Sets as ordered sequences:•Adjoining an element to a set•Union of two setsSets as binary trees:•Intersection of two sets•Union of two setsThat's homework 8!12No lecture on MondayMidterm 2 on Monday, 7pm-9pmGood
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