CSCI 370 Computer Architecture: Homework 2

Due date: On or before Monday, March 31, 2025 Absolutely no copying others’ works

Name: Professor Hu

• Upload the completed homework to the section of “COVID-19 Exams, Homeworks, & Programming Exercises” of Blackboard.
• There are four algorithms discussed totally for multiplication and division.   Make sure you are using the correct ones for the following two questions.  
• The purpose of homeworks is for students to practice for the exams without others’ help, so the penalty of mistakes will be minor.
• Without practicing for the exams properly, students will not be able to do well on the exams.

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1. (Refined multiplication: 50%) Using a table similar to that shown in the Slide 8.6, calculate the product of the octal unsigned 6-bit integers 53₈ (or 101011₂) and 65₈ (or 110101₂) using the hardware and algorithm described in the figures of Slide 8.6. You should show the contents of each register on each step.
Ans>

Iteration		Multiplicand	Carry	Product = HI, LO
0	Initialize (LO=Multiplier, HI=0)	101011		000000 110101
1	LO[0]=1 ⇒ Add.		0	000000 110101 + 101011 000000    101011 110101
	Shift product right by 1 bit.	101011		010101 111010
2	LO[0]=0 ⇒ Do nothing.
	Shift product right by 1 bit.	101011		001010 111101
3	LO[0]=1 ⇒ Add.		0	001010 111101 + 101011 000000    110101 111101
	Shift product right by 1 bit.	101011		011010 111110
4	LO[0]=0 ⇒ Do nothing.
	Shift product right by 1 bit.	101011		001101 011111
5	LO[0]=1 ⇒ Add.		0	001101 011111 + 101011 000000    111000 011111
	Shift product right by 1 bit.	101011		011100 001111
6	LO[0]=1 ⇒ Add.		1	011100 001111 + 101011 000000    000111 001111
	Shift product right by 1 bit.	101011		100011 100111


^†You may double-check the answer as follows:

538×658 = 1010112×1101012 = 43×53 = 2279 = 100011 1001112


2. (First-version division: 50%) Using a table similar to that shown in the Slide 8.10, calculate the octal unsigned 6-bit integer 76₈ (or 111110₂) divided by another octal unsigned 6-bit integer 13₈ (or 001011₂) using the hardware and algorithm described in the figures of Slide 8.10. You should show the contents of each register on each step.
   ^†Note that you have to actually show the differences in the procedures, not just the signs.
   
Ans>

Iteration		Remainder	Divisor	Difference	Quotient
0	Initialize	000000 111110	001011 000000		000000
1	1: SHR, SHL, Difference	000000 111110	000101 100000	000000 111110 – 000101 100000 ⇓    000000 111110 + 111010 100000    111011 011110	000000
	2: Diff<0 ⇒ Do Nothing
2	1: SHR, SHL, Difference	000000 111110	000010 110000	000000 111110 – 000010 110000 ⇓    000000 111110 + 111101 010000    111110 001110	000000
	2: Diff<0 ⇒ Do Nothing
3	1: SHR, SHL, Difference	000000 111110	000001 011000	000000 111110 – 000001 011000 ⇓    000000 111110 + 111110 101000    111111 100110	000000
	2: Diff<0 ⇒ Do Nothing
4	1: SHR, SHL, Difference	000000 111110	000000 101100	000000 111110 – 000000 101100 ⇓    000000 111110 + 111111 010100  1 000000 010010	000000
	2: Diff≥0 ⇒ Rem=Diff, set lsb Quotient	000000 010010			000001
5	1: SHR, SHL, Difference	000000 010010	000000 010110	000000 010010 – 000000 010110 ⇓    000000 010010 + 111111 101010    111111 111100	000010
	2: Diff<0 ⇒ Do Nothing
6	1: SHR, SHL, Difference	000000 010010	000000 001011	000000 010010 - 000000 001011 ⇓    000000 010010 + 111111 110101  1 000000 000111	000010
	2: Diff≥0 ⇒ Rem=Diff, set lsb Quotient	000000 000111			000101


^†You may double-check the answer as follows:

768÷138 = 1111102÷0010112 = 62÷11 ⇒ Q: 5 = 0001012 and R: 7 = 000000 0001112