COMP2300/assignment2/topics.txt

/* ---------------------------------------------------------------------

             Name: Andrew Pollock
   Student Number: 4137129
             Unit: COMP2300
Assignment Number: 2
Name of this file: topics.txt
        Lab Group: Tuesday 0900


I declare that the material I am submitting in this file is entirely
my own work.  I have not collaborated with anyone to produce it, nor
have I copied it, in part or in full, from work produced by someone else.

   --------------------------------------------------------------------- */

For all machines, it takes 6 bits to reference one of the 65536 memory cells.

Question 1:

Machine 1:      ; dc is your friend

PUSH Mb 
PUSH Mc
PUSH Md
MUL             ; Top of stack now contains result of Mc * Md
ADD             ; Top of stack now contains result of (Mc * Md) + Mb
PUSH Mb
PUSH Me
PUSH Mf
DIV             ; Top of stack now contains result of Me / Mf
SUB             ; Top of stack now contains result of Mb - (Me / Mf)
DIV             ; Top of stack now contains result of formula
POP Ma          ; Save top of stack in Ma

Machine 2:

LOAD Mc         ; This is just like PeANUt
MUL Md
ADD Mb
STORE Mc
LOAD Me
DIV Mf
STORE Mf
LOAD Mb
SUB Mf
STORE Mf
LOAD Mc
DIV Mf
STORE Ma

Machine 3:

LOAD R0,Mb      ; Load Mb into register 0
LOAD R1, Mc     ; Load Mc into register 1
LOAD R2, Md     ; Load Md into register 2
MUL R1, R2, R3  ; Multiply register 1 (Mc) and register 2 (Md) store in 3
ADD R3, R0, R1  ; Add register 3 (Mc*Md) and register 0 (Mb), store in register 1
LOAD R2, Me     ; Load Me into register 2
LOAD R3, Mf     ; Load Mf into register 3
DIV R2, R3, R4  ; Divide register 2 (Me) by register 3 (Mf), store in 4
SUB R0, R4, R2  ; Subtract register 4 (Me/Mf) from register 0, store in 2
DIV R1, R2, R0  ; Divide register 1 by register 2, store in register 0
STORE Ma, R0    ; Store register 0 in Ma

Machine 4:

MUL Mc, Md      ; Mc contains Mc * Md
ADD Mc, Mb      ; Mc contains Mc + Mb [ ((Mc * Md) + Mb ]
DIV Me, Mf      ; Me contains Me / Mf
SUB Mb, Me      ; Mb contains Mb - Me [ Mb - (Me / Mf) ]
DIV Mc, Mb      ; Mc contains Mc / Mb
MOV Mc, Ma      ; Store Mc in Ma

Question 2:

Machine 1:

I = instruction length
M = memory address size
P = pad to get to a 4 bit boundary

                I   M        P
PUSH Mb         8 + 6 = 14 + 2 = 16 
PUSH Mc         8 + 6 = 14 + 2 = 16
PUSH Md         8 + 6 = 14 + 2 = 16
MUL             8
ADD             8
PUSH Mb         8 + 6 = 14 + 2 = 16
PUSH Me         8 + 6 = 14 + 2 = 16
PUSH Mf         8 + 6 = 14 + 2 = 16
DIV             8
SUB             8
DIV             8
POP Ma          8 + 6 = 14 + 2 = 16

Machine 1 total memory used: 152 bytes

Machine 2:

LOAD Mc         8 + 6 = 14 + 2 = 16
MUL Md          8 + 6 = 14 + 2 = 16
ADD Mb          8 + 6 = 14 + 2 = 16
STORE Mc        8 + 6 = 14 + 2 = 16
LOAD Me         8 + 6 = 14 + 2 = 16
DIV Mf          8 + 6 = 14 + 2 = 16
STORE Mf        8 + 6 = 14 + 2 = 16
LOAD Mb         8 + 6 = 14 + 2 = 16
SUB Mf          8 + 6 = 14 + 2 = 16
STORE Mf        8 + 6 = 14 + 2 = 16
LOAD Mc         8 + 6 = 14 + 2 = 16
DIV Mf          8 + 6 = 14 + 2 = 16
STORE Ma        8 + 6 = 14 + 2 = 16

Machine 3 total memory used: 208 bytes

Machine 3:

LOAD R0,Mb      8 + 4 + 6 = 18 + 2 = 20
LOAD R1, Mc     8 + 4 + 6 = 18 + 2 = 20
LOAD R2, Md     8 + 4 + 6 = 18 + 2 = 20
MUL R1, R2, R3  8 + 4 + 4 + 4 = 20
ADD R3, R0, R1  8 + 4 + 4 + 4 = 20
LOAD R2, Me     8 + 4 + 6 = 18 + 2 = 20
LOAD R3, Mf     8 + 4 + 6 = 18 + 2 = 20
DIV R2, R3, R4  8 + 4 + 4 + 4 = 20
SUB R0, R4, R2  8 + 4 + 4 + 4 = 20
DIV R1, R2, R0  8 + 4 + 4 + 4 = 20
STORE Ma, R0    8 + 6 + 4 + 2 = 20

Machine 3 total memory used: 242 bytes

Machine 4:

MUL Mc, Md      8 + 6 + 6 = 20
ADD Mc, Mb      8 + 6 + 6 = 20
DIV Me, Mf      8 + 6 + 6 = 20
SUB Mb, Me      8 + 6 + 6 = 20
DIV Mc, Mb      8 + 6 + 6 = 20
MOV Mc, Ma      8 + 6 + 6 = 20

Machine 4 total memory used: 120 bytes
1	/* ---------------------------------------------------------------------
2
3	Name: Andrew Pollock
4	Student Number: 4137129
5	Unit: COMP2300
6	Assignment Number: 2
7	Name of this file: topics.txt
8	Lab Group: Tuesday 0900
9
10
11	I declare that the material I am submitting in this file is entirely
12	my own work. I have not collaborated with anyone to produce it, nor
13	have I copied it, in part or in full, from work produced by someone else.
14
15	--------------------------------------------------------------------- */
16
17	For all machines, it takes 6 bits to reference one of the 65536 memory cells.
18
19	Question 1:
20
21	Machine 1: ; dc is your friend
22
23	PUSH Mb
24	PUSH Mc
25	PUSH Md
26	MUL ; Top of stack now contains result of Mc * Md
27	ADD ; Top of stack now contains result of (Mc * Md) + Mb
28	PUSH Mb
29	PUSH Me
30	PUSH Mf
31	DIV ; Top of stack now contains result of Me / Mf
32	SUB ; Top of stack now contains result of Mb - (Me / Mf)
33	DIV ; Top of stack now contains result of formula
34	POP Ma ; Save top of stack in Ma
35
36	Machine 2:
37
38	LOAD Mc ; This is just like PeANUt
39	MUL Md
40	ADD Mb
41	STORE Mc
42	LOAD Me
43	DIV Mf
44	STORE Mf
45	LOAD Mb
46	SUB Mf
47	STORE Mf
48	LOAD Mc
49	DIV Mf
50	STORE Ma
51
52	Machine 3:
53
54	LOAD R0,Mb ; Load Mb into register 0
55	LOAD R1, Mc ; Load Mc into register 1
56	LOAD R2, Md ; Load Md into register 2
57	MUL R1, R2, R3 ; Multiply register 1 (Mc) and register 2 (Md) store in 3
58	ADD R3, R0, R1 ; Add register 3 (Mc*Md) and register 0 (Mb), store in register 1
59	LOAD R2, Me ; Load Me into register 2
60	LOAD R3, Mf ; Load Mf into register 3
61	DIV R2, R3, R4 ; Divide register 2 (Me) by register 3 (Mf), store in 4
62	SUB R0, R4, R2 ; Subtract register 4 (Me/Mf) from register 0, store in 2
63	DIV R1, R2, R0 ; Divide register 1 by register 2, store in register 0
64	STORE Ma, R0 ; Store register 0 in Ma
65
66	Machine 4:
67
68	MUL Mc, Md ; Mc contains Mc * Md
69	ADD Mc, Mb ; Mc contains Mc + Mb [ ((Mc * Md) + Mb ]
70	DIV Me, Mf ; Me contains Me / Mf
71	SUB Mb, Me ; Mb contains Mb - Me [ Mb - (Me / Mf) ]
72	DIV Mc, Mb ; Mc contains Mc / Mb
73	MOV Mc, Ma ; Store Mc in Ma
74
75	Question 2:
76
77	Machine 1:
78
79	I = instruction length
80	M = memory address size
81	P = pad to get to a 4 bit boundary
82
83	I M P
84	PUSH Mb 8 + 6 = 14 + 2 = 16
85	PUSH Mc 8 + 6 = 14 + 2 = 16
86	PUSH Md 8 + 6 = 14 + 2 = 16
87	MUL 8
88	ADD 8
89	PUSH Mb 8 + 6 = 14 + 2 = 16
90	PUSH Me 8 + 6 = 14 + 2 = 16
91	PUSH Mf 8 + 6 = 14 + 2 = 16
92	DIV 8
93	SUB 8
94	DIV 8
95	POP Ma 8 + 6 = 14 + 2 = 16
96
97	Machine 1 total memory used: 152 bytes
98
99	Machine 2:
100
101	LOAD Mc 8 + 6 = 14 + 2 = 16
102	MUL Md 8 + 6 = 14 + 2 = 16
103	ADD Mb 8 + 6 = 14 + 2 = 16
104	STORE Mc 8 + 6 = 14 + 2 = 16
105	LOAD Me 8 + 6 = 14 + 2 = 16
106	DIV Mf 8 + 6 = 14 + 2 = 16
107	STORE Mf 8 + 6 = 14 + 2 = 16
108	LOAD Mb 8 + 6 = 14 + 2 = 16
109	SUB Mf 8 + 6 = 14 + 2 = 16
110	STORE Mf 8 + 6 = 14 + 2 = 16
111	LOAD Mc 8 + 6 = 14 + 2 = 16
112	DIV Mf 8 + 6 = 14 + 2 = 16
113	STORE Ma 8 + 6 = 14 + 2 = 16
114
115	Machine 3 total memory used: 208 bytes
116
117	Machine 3:
118
119	LOAD R0,Mb 8 + 4 + 6 = 18 + 2 = 20
120	LOAD R1, Mc 8 + 4 + 6 = 18 + 2 = 20
121	LOAD R2, Md 8 + 4 + 6 = 18 + 2 = 20
122	MUL R1, R2, R3 8 + 4 + 4 + 4 = 20
123	ADD R3, R0, R1 8 + 4 + 4 + 4 = 20
124	LOAD R2, Me 8 + 4 + 6 = 18 + 2 = 20
125	LOAD R3, Mf 8 + 4 + 6 = 18 + 2 = 20
126	DIV R2, R3, R4 8 + 4 + 4 + 4 = 20
127	SUB R0, R4, R2 8 + 4 + 4 + 4 = 20
128	DIV R1, R2, R0 8 + 4 + 4 + 4 = 20
129	STORE Ma, R0 8 + 6 + 4 + 2 = 20
130
131	Machine 3 total memory used: 242 bytes
132
133	Machine 4:
134
135	MUL Mc, Md 8 + 6 + 6 = 20
136	ADD Mc, Mb 8 + 6 + 6 = 20
137	DIV Me, Mf 8 + 6 + 6 = 20
138	SUB Mb, Me 8 + 6 + 6 = 20
139	DIV Mc, Mb 8 + 6 + 6 = 20
140	MOV Mc, Ma 8 + 6 + 6 = 20
141
142	Machine 4 total memory used: 120 bytes