# from math import floor,log

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1 Homework What follows is all the python code that you ll need for the homework assignment. # from math import floor,log # Trees # To begin we ll represent our binary trees with dictionaries. def left(x): 2*x def right(x): 2*x+1 def parent(x): x//2 # Tree walks def inorder_tree_walk(t,x=1): inorder_tree_walk(t,left(x)) print(t[x]) inorder_tree_walk(t,right(x)) def preorder_tree_walk(t,x=1): print(x,t[x]) preorder_tree_walk(t,left(x)) preorder_tree_walk(t,right(x)) def postorder_tree_walk(t,x=1): postorder_tree_walk(t,left(x)) postorder_tree_walk(t,right(x)) print(t[x]) # Tree search def tree_search(t,val,x=1): try: while T[x]!= val: if val<t[x]: x=left(x) x=right(x) x # T[x] == val except KeyError:

2 def tree_minimum(t,x=1): x = left(x) parent(x) # While x is in T, move down to the # left child until fallen out of tree. # Then go back up one step. def tree_maximum(t,x=1): x = right(x) parent(x) # While x is in T, move down to the # right child until fallen out of tree. # Then go back up one step. def tree_successor(t,x): if T.get(right(x))!= None: tree_minimum(t,right(x)) if T.get(parent(x)) == None: y=parent(x) while T.get(y)!=None and x==right(y): x,y = y,parent(y) if T.get(y) == None: # x is in T and has a right subtree, # so the successor is the minimum in # this subtree. # x is the root of T and has no right # child, so has no successor. # x is in T, has a parent but has no # right subtree, so the successor is # the root of the smallest subtree, if # any, containing x in its left subtree. # No subtree of T contains x in any # left subtree. y

3 def tree_predecessor(t,x): if T.get(left(x))!= None: tree_maximum(t,left(x)) if T.get(parent(x)) == None: y=parent(x) while T.get(y)!=None and x==left(y): x,y = y,parent(y) if T.get(y) == None: # x is in T and has a left subtree, so # the predecessor is the maximum in this # subtree. # x is the root of T and has no left # child, so has no predecessor. # x is in T, has a parent but has no # left subtree, so the predecessor is # the root of the smallest subtree, if # any, containing x in its right subtree. # No subtree of T contains x in any # right subtree. y def tree_insert(t,val): if T.get(1) = None: T[1] = val # If T is empty put val in T[1] and # exit the function. x = 1 # Otherwise start with x the root of T. # Walk x down T looking for a spot for y = x # val keeping y as the parent of x. Move if val < T[x]: # left if val is smaller than T[x] and x = left(x) # right otherwise. When the loop is left # y will be a leaf of T. x = right(x) if val < T[y]: T[left(y)] = val T[right(y)] = val # We ll store val in the left or right # child of y. To the left if val<t[y]; # to the right otherwise.

4 def tree_delete(t,z): if T.get(z) == None: if T.get(left(z))==None and T.get(right(z))==None: T.pop(z) # If T[z] is empty do nothing # and exit. # If z is a leaf # remove it # and exit. if T.get(left(z))==None: # If z has a right child but new_root = z # no left child we will move old_root = right(z) # the right subtree up to the elif T.get(right(z))==None: # position of z. Otherwise we new_root = z # will move the left subtree old_root = left(z) # up to the position of z. new_root = tree_successor(t,z) old_root = right(new_root) T[z] = T.pop(new_root) subtree = {} def sub_cut(y): if T.get(y)!= None: q = 2**floor(log(y/old_root, 2)) s = y % (q*old_root) subtree[(q,s)] = T.pop(y) sub_cut(left(y)) sub_cut(right(y)) sub_cut(old_root) for (q,s) in subtree: T[q*new_root+s] = subtree[(q,s)] # If z has left and right child # we move successor value to # position z and will move right # subtree of successor up to # successor s position. # We extract a subtree of T and # store it in a tree T with keys # of the form (q,s). Let r be the # root of the subtree of T. For # any y under r we find q, a power # of 2, so that y=qr+s with 0<=s<q. # We store T[y] in T [(q,s)]. # We insert the tree T into T by # choosing a new root r and putting # T [(q,s)] into T[qr +s]. # Build Tree def list_2_tree(l): T = {} for x in L: tree_insert(t,x) T

5 # Homework problem def check(l,val): T = list_2_tree(l[:3]) K1 = list(t.keys()) K1.sort() z = tree_search(t,val) tree_delete(t,z) tree_insert(t,l[3]) K2 = list(t.keys()) K2.sort() tuple(k1),tuple(k2) def check_all(): L = [(1,2,3,4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3,2), (4,1,2,3), (3,1,2,4), (4,1,3,2), (2,1,3,4), (3,1,4,2), (2,1,4,3), (3,4,1,2), (4,3,1,2), (2,4,1,3), (4,2,1,3), (2,3,1,4), (3,2,1,4), (2,3,4,1), (2,4,3,1), (3,2,4,1), (3,4,2,1), (4,2,3,1), (4,3,2,1)] K1 = {(1,2,3):0, (1,2,4):0, (1,2,5):0, (1,3,6):0, (1,3,7):0} K2 = {(1,2,3):0, (1,2,4):0, (1,2,5):0, (1,3,6):0, (1,3,7):0} for l in L: for v in l[:3]: k1,k2 = check(l,v) K1[k1] += 1 K2[k2] += 1 K1,K2

Binary search trees. Support many dynamic-set operations, e.g. Search. Minimum. Maximum. Insert. Delete ...

Binary search trees. Support many dynamic-set operations, e.g. Search. Minimum. Maximum. Insert. Delete ... Binary search trees Support many dynamic-set operations, e.g. Search Minimum Maximum Insert Delete... Can be used as dictionary, priority queue... you name it Running time depends on height of tree: 1