Introduction To Computing Systems
Introduction To Computing Systems
3rd Edition
ISBN: 9781260150537
Author: PATT, Yale N., Patel, Sanjay J.
Publisher: Mcgraw-hill,
Question
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Chapter 2, Problem 33E

a.

Program Plan Intro

OR operation:

  • OR function needs two inputs and produces one output.
  • It is also known as binary logical function.
  • If one of the inputs or both the inputs are “1”, then one-bit OR operation produces the output as “1”.
  • If both the inputs are “0”, then OR operation produces the output as “0”.
  • The following diagram depicts the one-bit OR operation,

Introduction To Computing Systems, Chapter 2, Problem 33E

b.

Explanation of Solution

101 OR 110:

The truth table for OR operation is as follows,

XYZ=X OR Y
000
011
101
111

Explanation:

  • In the above table, “X” and “Y” are the inputs, and “Z” is the output.
  • In the above table, when “X=0”, and “Y=0”, the output “Z” is “0”, because both the inputs “X” and “Y” contains the value “0”.
  • When “X=0”, and “Y=1”, the output “Z” is “1”, because one of the input “Y” contains the value “1”.
  • When “X=1”, and “Y=0”, the output “Z” is “1”, because one of the input “X” contains the value “1”...

c.

Explanation of Solution

11100000 OR 10110100:

The truth table for OR operation is as follows,

XYZ=X OR Y
000
011
101
111

Explanation:

  • In the above table, “X” and “Y” are the inputs, and “Z” is the output.
  • In the above table, when “X=0”, and “Y=0”, the output “Z” is “0”, because both the inputs “X” and “Y” contains the value “0”.
  • When “X=0”, and “Y=1”, the output “Z” is “1”, because one of the input “Y” contains the value “1”.
  • When “X=1”, and “Y=0”, the output “Z” is “1”, because one of the input “X” contains the value “1”...

d.

Explanation of Solution

00011111 OR 10110100:

The truth table for OR operation is as follows,

XYZ=X OR Y
000
011
101
111

Explanation:

  • In the above table, “X” and “Y” are the inputs, and “Z” is the output.
  • In the above table, when “X=0”, and “Y=0”, the output “Z” is “0”, because both the inputs “X” and “Y” contains the value “0”.
  • When “X=0”, and “Y=1”, the output “Z” is “1”, because one of the input “Y” contains the value “1”.
  • When “X=1”, and “Y=0”, the output “Z” is “1”, because one of the input “X” contains the value “1”...

e.

Explanation of Solution

(0101 OR 1100) OR 1101:

The truth table for OR operation is as follows,

XYZ=X OR Y
000
011
101
111

Explanation:

  • In the above table, “X” and “Y” are the inputs, and “Z” is the output.
  • In the above table, when “X=0”, and “Y=0”, the output “Z” is “0”, because both the inputs “X” and “Y” contains the value “0”.
  • When “X=0”, and “Y=1”, the output “Z” is “1”, because one of the input “Y” contains the value “1”.
  • When “X=1”, and “Y=0”, the output “Z” is “1”, because one of the input “X” contains the value “1”.
  • When “X=1”, and “Y=1”, the output “Z” is “1”, because both the inputs “X” and “Y” contains the value “1”...

f.

Explanation of Solution

0101 OR (1100 OR 1101):

The truth table for OR operation is as follows,

XYZ=X OR Y
000
011
101
111

Explanation:

  • In the above table, “X” and “Y” are the inputs, and “Z” is the output.
  • In the above table, when “X=0”, and “Y=0”, the output “Z” is “0”, because both the inputs “X” and “Y” contains the value “0”.
  • When “X=0”, and “Y=1”, the output “Z” is “1”, because one of the input “Y” contains the value “1”.
  • When “X=1”, and “Y=0”, the output “Z” is “1”, because one of the input “X” contains the value “1”.
  • When “X=1”, and “Y=1”, the output “Z” is “1”, because both the inputs “X” and “Y” contains the value “1”...

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Here are two diagrams. Make them very explicit, similar to Example Diagram 3 (the Architecture of MSCTNN). graph LR subgraph Teacher_Model_B [Teacher Model (Pretrained)] Input_Teacher_B[Input C (Complete Data)] --> Teacher_Encoder_B[Transformer Encoder T] Teacher_Encoder_B --> Teacher_Prediction_B[Teacher Prediction y_T] Teacher_Encoder_B --> Teacher_Features_B[Internal Features F_T] end subgraph Student_B_Model [Student Model B (Handles Missing Labels)] Input_Student_B[Input C (Complete Data)] --> Student_B_Encoder[Transformer Encoder E_B] Student_B_Encoder --> Student_B_Prediction[Student B Prediction y_B] end subgraph Knowledge_Distillation_B [Knowledge Distillation (Student B)] Teacher_Prediction_B -- Logits Distillation Loss (L_logits_B) --> Total_Loss_B Teacher_Features_B -- Feature Alignment Loss (L_feature_B) --> Total_Loss_B Partial_Labels_B[Partial Labels y_p] -- Prediction Loss (L_pred_B) --> Total_Loss_B Total_Loss_B -- Backpropagation -->…
Please provide me with the output  image of both of them . below are the diagrams code I have two diagram : first diagram code  graph LR subgraph Teacher Model (Pretrained) Input_Teacher[Input C (Complete Data)] --> Teacher_Encoder[Transformer Encoder T] Teacher_Encoder --> Teacher_Prediction[Teacher Prediction y_T] Teacher_Encoder --> Teacher_Features[Internal Features F_T] end subgraph Student_A_Model[Student Model A (Handles Missing Values)] Input_Student_A[Input M (Data with Missing Values)] --> Student_A_Encoder[Transformer Encoder E_A] Student_A_Encoder --> Student_A_Prediction[Student A Prediction y_A] Student_A_Encoder --> Student_A_Features[Student A Features F_A] end subgraph Knowledge_Distillation_A [Knowledge Distillation (Student A)] Teacher_Prediction -- Logits Distillation Loss (L_logits_A) --> Total_Loss_A Teacher_Features -- Feature Alignment Loss (L_feature_A) --> Total_Loss_A Ground_Truth_A[Ground Truth y_gt] -- Prediction Loss (L_pred_A)…
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