Introduction To Algorithms, Third Edition (international Edition)
3rd Edition
ISBN: 9780262533058
Author: Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein
Publisher: TRILITERAL
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Chapter 4.3, Problem 6E
Program Plan Intro
To show that the solution of the recurrence relation
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Chapter 4 Solutions
Introduction To Algorithms, Third Edition (international Edition)
Ch. 4.1 - Prob. 1ECh. 4.1 - Prob. 2ECh. 4.1 - Prob. 3ECh. 4.1 - Prob. 4ECh. 4.1 - Prob. 5ECh. 4.2 - Prob. 1ECh. 4.2 - Prob. 2ECh. 4.2 - Prob. 3ECh. 4.2 - Prob. 4ECh. 4.2 - Prob. 5E
Ch. 4.2 - Prob. 6ECh. 4.2 - Prob. 7ECh. 4.3 - Prob. 1ECh. 4.3 - Prob. 2ECh. 4.3 - Prob. 3ECh. 4.3 - Prob. 4ECh. 4.3 - Prob. 5ECh. 4.3 - Prob. 6ECh. 4.3 - Prob. 7ECh. 4.3 - Prob. 8ECh. 4.3 - Prob. 9ECh. 4.4 - Prob. 1ECh. 4.4 - Prob. 2ECh. 4.4 - Prob. 3ECh. 4.4 - Prob. 4ECh. 4.4 - Prob. 5ECh. 4.4 - Prob. 6ECh. 4.4 - Prob. 7ECh. 4.4 - Prob. 8ECh. 4.4 - Prob. 9ECh. 4.5 - Prob. 1ECh. 4.5 - Prob. 2ECh. 4.5 - Prob. 3ECh. 4.5 - Prob. 4ECh. 4.5 - Prob. 5ECh. 4.6 - Prob. 1ECh. 4.6 - Prob. 2ECh. 4.6 - Prob. 3ECh. 4 - Prob. 1PCh. 4 - Prob. 2PCh. 4 - Prob. 3PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6P
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- 4. Suppose we have a perfect binary tree with height h 0 representing a heap, meaning it = has n 2+1 1 keys indexed from 1 to 2+1 1. When we run convertomaxheap we run maxheapify in reverse order on every key with children. Let's examine the worst-case - In the worst-case every single key gets swapped all the way to the leaf level. (a) For each level in the tree there are a certain number of nodes and each of those nodes [10 pts] requires a certain number of swaps. Fill in the appropriate values/expressions in the table: Level Number of Keys Number of Swaps per Key 0 2 .. (b) Write down a sum for the total number of swaps required. This should involve h, not n. [10 pts] Totalarrow_forwardThe next problem concerns the following C code: /copy input string x to buf */ void foo (char *x) { char buf [8]; strcpy((char *) buf, x); } void callfoo() { } foo("ZYXWVUTSRQPONMLKJIHGFEDCBA"); Here is the corresponding machine code on a Linux/x86 machine: 0000000000400530 : 400530: 48 83 ec 18 sub $0x18,%rsp 400534: 48 89 fe mov %rdi, %rsi 400537: 48 89 e7 mov %rsp,%rdi 40053a: e8 di fe ff ff callq 400410 40053f: 48 83 c4 18 add $0x18,%rsp 400543: c3 retq 400544: 0000000000400544 : 48 83 ec 08 sub $0x8,%rsp 400548: bf 00 06 40 00 mov $0x400600,%edi 40054d: e8 de ff ff ff callq 400530 400552: 48 83 c4 08 add $0x8,%rsp 400556: c3 This problem tests your understanding of the program stack. Here are some notes to help you work the problem: ⚫ strcpy(char *dst, char *src) copies the string at address src (including the terminating '\0' character) to address dst. It does not check the size of the destination buffer. • You will need to know the hex values of the following characters:arrow_forward1234 3. Which line prevents compiler optimization? Circle one: 1234 Suggested solution: Store strlen(str) in a variable before the if statement. ⚫ Remove the if statement. Replace index 0 && index < strlen(str)) { 5 } } = str [index] = val;arrow_forward
- Character Hex value | Character Hex value Character Hex value 'A' 0x41 'J' Ox4a 'S' 0x53 'B' 0x42 'K' 0x4b "T" 0x54 0x43 'L' Ox4c 'U' 0x55 0x44 'M' 0x4d 'V' 0x56 0x45 'N' Ox4e 'W' 0x57 0x46 '0' Ox4f 'X' 0x58 0x47 'P' 0x50 'Y' 0x59 0x48 'Q' 0x51 'Z' Ox5a 'T' 0x49 'R' 0x52 '\0' 0x00 Now consider what happens on a Linux/x86 machine when callfoo calls foo with the input string "ZYXWVUTSRQPONMLKJIHGFEDCBA". A. On the left draw the state of the stack just before the execution of the instruction at address Ox40053a; make sure to show the frames for callfoo and foo and the exact return address, in Hex at the bottom of the callfoo frame. Then, on the right, draw the state of the stack just after the instruction got executed; make sure to show where the string "ZYXWVUTSRQPONMLKJIHGFEDCBA" is placed and what part, if any, of the above return address has been overwritten. B. Immediately after the ret instruction at address 0x400543 executes, what is the value of the program counter register %rip?…arrow_forward1 typedef struct node* { 2 struct node* next; 3 char* key; 4 char* val; 5} node_t; 6 7 char* find_node (node_t* node, char* key_to_find) { while(strcmp (node->key, key_to_find ) != 0 ) { node = node->next; 8 9 10 } 11 return node->val; 12 }arrow_forwardMatch each of the assembler routines on the left with the equivalent C function on the right. Write the name of the label (e.g., foo) to the right of the corresponding function. Note: shrq is the logical right shift instruction, and sarq is the arithmetic right shift instruction. foo1: leaq 0(,%rdi, 8), %rax long choice1 (long x) { ret return x - 8 >8; foo3: } movq sarq %rdi, %rax $8, %rax long choice4 (long x) ret { return x*256; } foo4: long choice5 (long x) leaq -8 (%rdi), %rax { ret return x-8; } long choice6 (long x) foo5: { leaq -8 (%rdi), %rax return x+8; shrq $63, %rax } retarrow_forward
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