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Using a 4-bit version of the algorithm to save pages, divide 11ten by 4ten, or 00001011two by 0100 two. Use the algorithm outlined in the flowchart in figure 3.9 and fill in what numbers, in the table, would populate in the Quotient, Divisor, and Remainder in each step of each iteration.

Using a 4-bit version of the algorithm to save pages, divide 11ten by 4ten, or 00001011two by 0100 two. Use the algorithm outlined in the flowchart in figure 3.9 and fill in what numbers, in the table, would populate in the Quotient, Divisor, and Remainder in each step of each iteration.

Database System Concepts
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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not sure if my values are correct if yall could look through and correct me please I would appreciate it !

4. Using a 4-bit version of the algorithm to save pages, divide 11ten by 4ten, or 00001011two by 0100two. Use the algorithm outlined in the flowchart in figure 3.9 and fill in what numbers, in the table, would populate in the Quotient, Divisor, and Remainder in each step of each iteration. Remainder 0000 1011 0011 Iteration 0 1 2 3 4 divisor properly, relative to the dividend for the next iteration. These steps are repeated 33 times. 5 Step Initial Values 1: Rem Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div 1: Rem Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div 1: Rem = Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div 1: Rem Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div 1: Rem Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div Quotient 0000 0010 0010 0001 Divisor 0100 0000 1000 1.000 0006 2011 coll
Transcribed Image Text:4. Using a 4-bit version of the algorithm to save pages, divide 11ten by 4ten, or 00001011two by 0100two. Use the algorithm outlined in the flowchart in figure 3.9 and fill in what numbers, in the table, would populate in the Quotient, Divisor, and Remainder in each step of each iteration. Remainder 0000 1011 0011 Iteration 0 1 2 3 4 divisor properly, relative to the dividend for the next iteration. These steps are repeated 33 times. 5 Step Initial Values 1: Rem Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div 1: Rem Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div 1: Rem = Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div 1: Rem Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div 1: Rem Rem-Div 2b. Rem<0->+Div, ssl Q QO=0 3. Shift right Div Quotient 0000 0010 0010 0001 Divisor 0100 0000 1000 1.000 0006 2011 coll
1. Subtract the Divisor register from the Remainder register and place the result in the Remainder register Remainder 20 Start Test Remainder 2a. Shift the Quotient register to the left, setting the new rightmost bit to 1 Remainder <0 2b. Restore the original value by adding the Divisor register to the Remainder register and placing the sum in the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0 3. Shift the Divisor register right 1 bit 5th repetition? Done No: < 5 repetitions Yes: 5 repetitions FIGURE 3.9 A division algorithm, using the hardware in Figure 3.8. If the remainder is positive, the divisor did go into the dividend, so step 2a generates a I in the quotient. A negative remainder after
Transcribed Image Text:1. Subtract the Divisor register from the Remainder register and place the result in the Remainder register Remainder 20 Start Test Remainder 2a. Shift the Quotient register to the left, setting the new rightmost bit to 1 Remainder <0 2b. Restore the original value by adding the Divisor register to the Remainder register and placing the sum in the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0 3. Shift the Divisor register right 1 bit 5th repetition? Done No: < 5 repetitions Yes: 5 repetitions FIGURE 3.9 A division algorithm, using the hardware in Figure 3.8. If the remainder is positive, the divisor did go into the dividend, so step 2a generates a I in the quotient. A negative remainder after
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