Fundamental Algorithms

Transcription

1 Techische Uiversität Müche Fakultät für Iformatik Lehrstuhl für Effiziete Algorithme Dmytro Chibisov Sadeep Sadaada Witer Semester 2007/08 Solutio Sheet 6 November 30, 2007 Fudametal Algorithms Problem 1 (10 Poits) A biary tree is full if all of its vertices have either zero or two childre. Let B deote the umber of full biary trees with vertices. 1. By drawig out all full biary trees with 3, 5, or 7 vertices, determie the eact values of B 3, B 5, ad B 7. Why have we left out eve umbers of vertices, like B 4? 2. For geeral, derive a recurrece relatio for B. Solutio 1. By drawig out all full biary trees with 3, 5, or 7 odes, determie the eact values of B 3, B 5, ad B 7. Why have we left out eve umbers of vertices, like B 4? The figure shows all the full biary trees with 3, 5 or 7 odes. The the umber of trees are 1, 2 ad 5 respectively. There are o eve umber of odes because, a tree with eve umber of odes caot be a full tree.

2 B 3 = 1 B 5 = 2 B 7 = 5 2. For geeral, derive a recurrece relatio for B. ( ) 2 B 2 + B 4 B B 2 B = ( B 2 1 ) 2 B 2 + B 4 B B 2 B 2 B 2 B 2 if = 4k + 1 if = 4k + 3 Problem 2 (10 Poits) Review all the sort algorithms take i the class. Compare their compleities. If possible, try to eplai them with day-to-day eamples. Prove that the lower boud for sortig is lg 2

3 Solutio Sort Average Best Worst Remarks Bubble sort Selectio sort Isertio sort 2 2 I best case, isert requires costat time Merge sort lg lg lg Heap sort lg lg lg Quick sort lg lg 2 Proof: For a iput of size, the decisio tree has! leaves. Which leaves the tree with a height h lg(!) h lg(!) ( ( ) ) 2 lg 2 = (lg() 1) ( 2 lg 4) Problem 3 Stacks ad Queues. 1. Write pseudo code for push(), pop(), add(), delete(). 2. How ca oe simulate a queue with two stacks! (o coutig) What is a circular queue? Solutio 1. Stack #defie STACKSIZE 1000 usiged it stack[stacksize]; it top; void push(it data) 3

4 if (top < STACKSIZE) stack[top++] = data; else pritf("stack Full"); it pop() if(top!= 0) retur stack[--top]; else prit("stack Empty"); retur -1; 2. Simulate Queue with Stacks stack Stack1, Stack2; void add(it data) Stack1.push(data); it del() while possible to pop from Stack1 Stack2.push(Stack1.pop()); retur Stack2.pop(); while possible to pop from Stack2 Stack1.push(Stack2.pop()); 3. Circular Queue A circular queue is a queue which has a maimum capacity at a givem poit of time. It acts as if its head ad tail are coected. It is usually implemeted with a ormal array. Oce the head/tail reaches the ed of the array, the cout starts agai from the begiig. 4

5 Problem 4 Desig the fuctios isert(data), search(data) ad delete(data) i a biary search tree Recursively. Compare the compleity with the iterative implemetatios. Solutio 1. isert(data) ode * isert(ode * tree, it data) if(tree == NULL) retur ewode(data); if (data < tree->data) tree->left = isert(tree->left, data); if (data > tree->data) tree->right = isert(tree->right, data); if (data == tree->data) tree->cout++; retur tree; 2. search(data) is eactly like isert(data) - so, left as eercise. 3. delte(data) void delete(ode * tree, ode * vater, it data) if (tree == NULL) retur; // othig to delete if(data < tree->data) // happes to be i the left tree delete(tree->left, tree, data); if(data > tree->data) // let s delete it from the right subtree. delete(tree->right, tree, data); // ow we are o the tree NODE to be deleted. if(tree == vater) // happes to be the root ode. if(isleaf(tree)) // the oly ode i the tree 5

6 else free(tree); retur ; if(isleaf(tree)) if(vater->left == tree) vater->left = NULL; else // if (vater->right == tree) vater->right = NULL; retur; // if tree has oly oe child, we ca replace tree by it s kid. if((olykid = sigle_kid(tree))!= NULL) if(vater->left == tree) vater->left = olykid; else // if (vater->right == tree) vater->right = olykid; retur; // ot a leaf, or the father of oly oe child - // hece replace tree with leftmost child of right child or // rightmost child of left child // radom == 1 --> left child s rightmost child ad // radom == 2 --> right child s leftmost child radom = replace(tree, vater); // does the radom replacemet. if(radom == 1) delete(tree->left, tree, data); else // (radom == 2) delete(tree->right, tree, data); The umber of recursive calls is the same as the umber of iteratios i the iterative loops. Hece the compleities of both the methods are the same. Ad it is O(lg ), where is the umber of odes i the tree. 6