QUESTION ONE
MOV AL, 5A      #This is hex for 'Z'
                #Move this value to the A Register
MOV BL, D3      #It says VDU starts at D3 in the Q

LOOP:           #Begin LOOP
MOV [BL], AL    #Move the A Reg value into memory (B Register tells us where in mem)
DEC AL          #Decrement A Reg as its reverse order
                #A now has hex value for 'Y'
INC BL          #Increment B Reg as next character must be displayed on next VDU mem addr
                #B now has hex value D4
CMP AL, 40      #This is hex for character BEFORE 'A'
                #See if A register is at charcter before alphabet
JNZ LOOP        #If flag is not set go back to start of loop
                #i.e A register is still alphabetical charcater
JMP END         #Otherwise we'll execute this line that jumps to end label

END:
END             #End label calls end


QUESTION TWO
#Similar to one but must check if VDU is full
#FF is 255 in decimal. 255 - 25 (26 letters in the alphabet)
#This will give us the last possible memory adress we should use
#255 - 25 = 230
#230 in hex is E6

MOV AL, 5A
MOV BL, D3

LOOP:
CMP BL, E7      #Is BL in our unacceptable range? (E6 + 1)
                #S flag will be set if in acceptable range (less than or equal)
JNS CHANGEVAL   #If S flag not set, fix BL
MOV [BL], AL
DEC AL
INC BL
CMP AL, 40
JNZ LOOP
JMP END

CHANGEVAL:
MOV BL, E6      #Put BL in acceptable range
JMP LOOP        #Go back to loop

END:
END

QUESTION THREE
MOV AL, 04
ADD AL, 30      #30 is the ascii hex code for the character '0'
                #We are using a numeric four and we need to convert it to ascii 4
MOV BL, 03
ADD BL, 30      #30 is the ascii hex code for the character '0'
                #We are using a numeric four and we need to convert it to ascii 3
ADD AL, BL      #Result is 7 which is less than 9
MOV [C0], AL

#Algorithm explanation:
# By getting the number modulus 10 (i.e., 0a hex) the units portion of the number can be determined.
# This can be written to the
# appropriate place in the VDU after adding 30hex to determine
# its corresponding character value.
# By dividing the number by 10 (i.e., 0a hex) the 10s portion of the number can be determined. This can be written to the
# appropriate place in the VDU after adding 30hex to determine its corresponding character value.

#Write:
#push al ;          place the number on the stack for save keeping since the al value will change
#mod al, 0a ;       determine the units portion of the number
#add al, 30 ;       convert to numeric character
#mov [c1], al;      place in appropriate place in VDU
#pop al ;           retrieve original number from stack
#div al, 0a ;       determine the 10s portion of the number
#add al, 30 ;       convert ot numeric character11
#mov [c0], al ;     place in appropriate place on the stack
#end

QUESTION FOUR
MOV AL, 0C      #12
MOV BL, 03      #3
MUL AL, BL
MOV CL, AL
PUSH CL
MOV AL, 08
MOV BL, 06
MUL AL, BL
MOV DL, AL
DIV CL, 04
ADD CL, DL
POP DL
SUB CL, DL
PUSH CL
MOD CL, 0a
ADD CL, 30
MOV [c1], a
POP CL
DIV CL, 0a
ADD CL, 30
MOV [c0], CL