b, 1:11b, 0:10a, 1:01a, 0:00b, z; lza, z; Ozb, 1; λa, 0;z, z; zb, z; Ab0;λc, 1; 1a, z; Ac, 0:0a, 1;A90q192What is the next configuration after (q0, abca, OZ)A. 92,λ, ezB. q1,bca, 00Zc.q1,bca,ozD. q2,Z,1ZE. q2,bca,00

Given is (q0,aacb,Z), Currently stack has Z as top.When we read the first alphabet a, 0 will be…

Answered: b, 1:11 b, 0:10 a, 1:01 a, 0:00 b, z;…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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explain this program bus1:mov ax,400add amount, axmov dx,0mov bx,10mov cx,0l222:div bxpush dxmov dx,0mov ah,0inc cxcmp ax,0jne l222l333:pop dxadd dx,48mov ah,2int 21hloop l333;mov am3,amount inc countinc bjmp while_jmp end_ recrd procmov dx,offset msg7mov ah,9int 21h ; print here the whole amountmov ax, amount mov dx,0mov bx,10mov cx,0
Up Counter Finite State Machine Construct the Finite State Machine repressentation for an up counter with a cycle length of 4 - i.e. a circuit that counts 0 – 1– 2 – 3 (output as a binary value, obviously) with successive clock pulses, and then starts over. The external output is the 2-bit count. The only external input is R, a reset pulse: when R = 1 it resets the next count to 1 (not zero, as in the practice exam), no matter what the current state; when R = 0 it keeps counting (i.e. the system transitions to the next state in sequence).
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b, 1:11 b, 0:10 a, 1:01 a, 0:00 b, z; lz a, z; Oz b, 1; λ a, 0; z, z; z b, z; A b0;λ c, 1; 1 a, z; A c, 0:0 a, 1;A 90 q1 92 What is the next configuration after (q0, abca, OZ) A. 92,λ, ez B. q1,bca, 00Z c. q1,bca,oz D. q2,Z,1Z E. q2,bca,00