Add (o Max Addresult ALU RegDat Branch Shift left 2 MemRead Instruction [31-26] MemtoReg Control ALUOD MemWrite ALUSIC RegWrite Instruction (25-21] Read PC Read address register 1 Read Instruction (20-16] Read data 1 Instruction register 2 [31-01 Write Read Zero ALU ALU result Address data Read Write Data data memory ALU control Sax. Instruction Instruction [15-11 memory Instruction [15-0] register data 2 Write data Registers 16 Sign- extend Instruction (5-0) 8.
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- x={"id":"cmp1-6W15N5p36KXNUGRYFKweYAOTN36ui","object":"text_completion","created":1673088689, "model": "text-davinci-003", "choices": [{"text":"-3\n\n(i) The magnitude of the applied stress that would produce brittle fracture for the same length of crack in an infinitely wide plate is 250 MPa.\n\n(ii) The magnitude of the applied stress that would produce brittle fracture for the same length of crack in a plate 360 mm wide is 230 MPa.","index":0,"logprobs":null, "finish_reason": "stop"}],"usage": {"prompt_tokens":6,"completion_tokens":84, "total_tokens":90}} I want python code that can get only data from the text. for example [This I need output ] using python above code is json just scrap text from above json file and print it (i) The magnitude of the applied stress that would produce brittle fracture for the same length of crack in an infinitely wide plate is 250 MPa.\n\n(ii) The magnitude of the applied stress that would produce brittle fracture for the same length of a crack in a…# Pytorch Deep Learning(python) # Answer according following code import numpy as np import json img_codes = np.load("data/image_codes.npy") captions = json.load(open('data/captions_tokenized.json')) for img_i in range(len(captions)): for caption_i inrange(len(captions[img_i])): sentence = captions[img_i][caption_i] captions[img_i][caption_i] = ["#START#"] + sentence.split(' ') + ["#END#"] # Build a Vocabulary from collections import Counter word_counts = Counter() # Compute word frequencies for each word in captions. See code above for data structure # YOUR CODE HERE #Check your solution below and Testing condition:- vocab = ['#UNK#', '#START#', '#END#', '#PAD#'] vocab += [k for k, v in word_counts.items() if v >= 5 if k not in vocab] n_tokens = len(vocab) assert 10000 <= n_tokens <= 10500 #for reference and more detail go to --->…Develop a data type for a buffer in a text editor that implements the following API:public class BufferBuffer() create an empty buffervoid insert(char c) insert c at the cursor positionchar delete() delete and return the character at the cursorvoid left(int k) move the cursor k positions to the leftvoid right(int k) move the cursor k positions to the rightint size() number of characters in the bufferAPI for a text bufferHint : Use two stacks
- لتكن المصفوفة [x=[1,4,3,2;8,7,9,0;4,6,7,3 [Y=[1,6,7;5,6,4;7,8,9;0,3,4 [W=[1,3,4,5;7,6,8,9;0,9,1,2;4,5,2,3 اكتب كلا من الايعازات التالية : كون المصفوفة الفرعية من الصفوفةx حيث ان z= 6. 8 9. 9. 2.It is stated in this chapter that when nonlocal variables are accessed in adynamic-scoped language using the dynamic chain, variable names mustbe stored in the activation records with the values. If this were actuallydone, every nonlocal access would require a sequence of costly stringcomparisons on names. Design an alternative to these stringcomparisons that would be fasterGery blocks of code (lines 1-13 & 16-end) can NOT be edited. JAVA code needs to be added inbetween the grey blocks of code.
- Modify the producer-consumer implementation code bellow, so that it uses monitors to handle race conditions instead of semaphores or mutexes. Use the pthread library implementation #include <pthread.h> #include <semaphore.h> #include <stdio.h> #include <stdlib.h> #define BUFFER_SIZE 20 pthread_mutex_t mutex; int count = 0; int buffer[BUFFER_SIZE]; pthread_t tid; int producers = 0, consumers = 0; void insert(int item) { while (count == BUFFER_SIZE); if (count < BUFFER_SIZE) { buffer[producers] = item; producers++; producers=producers%BUFFER_SIZE; sleep(1); } return; } int remove_item() { int item; while (count == 0); if (count > 0) { item = buffer[consumers]; buffer[consumers] = buffer[consumers - 1]; consumers++; consumers=consumers%BUFFER_SIZE; sleep(1); } return item; } void * producer(void *param) { int item; while (1) { item = rand() % BUFFER_SIZE; while (count >= BUFFER_SIZE);…Implementation of the solution in the C++ or Java language • Implementation of the solution in ARMv8 assembly, with comments explaining the purpose of each line Task 1 An automorphic number is a number n whose square ends in n. For instance, 5 is automorphic, because 52 = 25, which ends in 5. Design and implement an ARMv8 program that reads a positive integer from the user and then calculates and prints all the automorphic numbers (decimal base) that are less than or equal to the entered integer. If the entered integer is not positive, an error message is displayed. As an example, if 100 is entered, the program will print all the automorphic numbers up to 100: 1, 5, 6, 25, 76.While statements may start and stop anywhere in the flexible manner popularized by Algol60, most modern programming languages insist that statements finish with an end sign like a semicolon or colon. To the contrary, Python and a handful of other programming languages adhere to a set structure in which statements start in a certain column and stop at the end of a line of code unless continuation marks are provided for each statement. Discover how readability, writability, and security are affected by a file's fixed or free format in the following paragraphs.
- Solve the following in C## Run Length Encode Manytimes,certaindatafiletypescanconsistoflarge amounts of repeated data. For instance, images can have large runs of the same color. This can be easily compressed using a technique called run length encoding. With run length encoding, large amounts of repeated data are stored as the repeated data and the number of times to repeat it. CreateaclassRunLengthEncodethatcontainsthemethod encode which takes one argument: a String to be encoded as described below. ThereturnvalueshouldbeaStringwhichhasbeenencoded with the following algorithm: Ifanycharacterisrepeatedmorethan4times,theentire set of repeated characters should be replaced with a slash '/', followed by a 2-digit number which is the length of the set of characters, and the character. For example, "aaaaa" would be encoded as "/05a". Runs of 4 or less characters should not be replaced since performing the encoding would not decrease the length of the string. Notes Letters are…In C programming Every budding computer scientist must grapple with certain classic problems, and the Towers of Hanoi (see Figure below) is one of the most famous of these. Legend has it that in a temple in the Far East, priests are attempting to move a stack of disks fromone peg to another. The initial stack had 64 disks threaded onto one peg and arranged from bottom to top by decreasing size. The priests are attempting to move the stack from this peg to a second peg under the constraints that exactly one disk is moved ata time, and at no time may a larger disk be placed above a smaller disk. A third peg is available for temporarily holding the disks. Supposedly the world will end when the priests complete their task, so there is little incentive for us to facilitate their efforts.Let’sassume that the priests are attempting to move the disks from peg 1 to peg 3. We wish to develop an algorithm that will print the precise sequence of disk-to-disk peg transfers. If we were to approach…The following are the operations that you can do using a single linked list. Choose only one operation then create the algorithm and simulate. The attached Rubric will be used in evaluating the activity. 1. Delete a particular node in a single linked list 2. Delete the first node of a single linked list 3. Insertion after a given node of a single linked list 4. Insertion at a given position in a single linked list 5. Insertion before a given node in a single linked list 6. Reverse a single linked list EX: Delete the last node of a single linked list Problem" Deletion of the last node in a single linked list Algorithm" Step 1: if HEAD = NULL Write UNDERFLOW Go to Step 8 Step 2: SET PTR = HEAD Step 3: Repeat Steps 4 and while PTR à NEXT = NULL Step 4: SET PREPPTR = PTR Step 5: SET PTR =PTR à NEXT [End of loop] Step 6: SET PREPTR à NEXT = NULL Step 7: FREE PTR Step 8: EXIT…