The series below, (four to be precise) of pseudo-code segments that represent typical code and data structures that one might implement with a language such as Java or Python. Devise a strategy to...


The series below, (four to be precise) of pseudo-code segments that represent typical code and data structures that one might implement with a language such as Java or Python. Devise a strategy to implement these code and data structures using only the Hack assembler.    You will be able to use the following assembler codes or symbols to complete this assignment.


The following 4 problems are the pseudocode routines that you must implement in the Hack assembler.  You should develop the code for your assembler program, compile and test it using the CPU Simulator (Please suggest a link for the CPU Simulator).  Kindly show the assembler code by pasting it directly.     You should document your code using the // symbol as a comment to describe what part of the pseudo code is being implemented by your assembler code in the assignment so that I can understand how you implemented the code in the assembler. The four problems that you must code include.   Comments should not be included in your code when you compile and test it.


A couple of tips that might be useful:




Comparisons:
Your program will need to compare two numbers to determine if one is larger than the other.  In assembler, we do not have an ‘if’ expression that allows us to test the equality or inequality of two variables.  The way that this capability can be implemented is by subtracting the second number from the first number and then jumping based upon the result.    For example, if A and B are our two numbers then the rules for A-B are as follows:


If A-B = 0, then A=B.
If A-B > 0, then A > B.
If A-B < 0,="" then="" a=""><>


If the result of the operation is 0 then we can execute a JEQ which means jump if the output of the operation is equal to zero (see the jump table above).


In the algorithm, we need to be able to test if a variable is less than, greater than, or equal to either another variable or a constant value.  Each of these tests can be accomplished by subtracting the two numbers and then using the appropriate jump instructions such as:


 JLT – jump less than 0
JGT – jump greater than 0
JEQ – jump equal to 0


Here, you should develop your solution to the exercise and test it using the CPU emulator.


(uhen -o)<br>(when a1)<br>el<br>c2<br>03 c4 e5 06<br>Cmp emnic<br>10 1 0<br>11 1 1<br>1 1 1 0 10<br>0 0 1 1 0 0<br>1 !0 0 . 0<br>O 0 1 1 1<br>1 10 0 o 1<br>0 0 11 1 1<br>1 10 0 1 1<br>0 1 1 11 1<br>1 101 1 1<br>0 011 10<br>1 10 0 10<br>0 0 0 0 10<br>010 o11<br>0 0 0 1 1 1<br>0 00 0 DO<br>010 LDI<br>-1<br>A<br>ID<br>IA<br>IM<br>-A<br>D+1<br>D-1<br>A-1<br>N-1<br>D-A<br>D-<br>A-D<br>M-D<br>DAA<br>DlA<br>In addition you can take advantage of the jump and dest functionality that is represented in the fallowing two tables<br>j2<br>(out < 0) (oat =0) (ut ><br>Mnenie<br>Elfecet<br>No jump<br>If or>0 jump<br>If at jump<br>null<br>If or 20 jump<br>If ar <0 jump<br>If at0 jump<br>If at S0 jump<br>Jump<br>1.<br>JLT<br>JHE<br>JLE<br>INP<br>Figure 45 The ny fiedd of dhe Citstnxtien. Oat sekes e Ihe ALU output jeesulting from<br>the inaruction's coxp parn, and ep inples

Extracted text: (uhen -o) (when a1) el c2 03 c4 e5 06 Cmp emnic 10 1 0 11 1 1 1 1 1 0 10 0 0 1 1 0 0 1 !0 0 . 0 O 0 1 1 1 1 10 0 o 1 0 0 11 1 1 1 10 0 1 1 0 1 1 11 1 1 101 1 1 0 011 10 1 10 0 10 0 0 0 0 10 010 o11 0 0 0 1 1 1 0 00 0 DO 010 LDI -1 A ID IA IM -A D+1 D-1 A-1 N-1 D-A D- A-D M-D DAA DlA In addition you can take advantage of the jump and dest functionality that is represented in the fallowing two tables j2 (out < 0)="" (oat="0)" (ut=""> Mnenie Elfecet No jump If or>0 jump If at jump null If or 20 jump If ar <0 jump="" if="" at0="" jump="" if="" at="" s0="" jump="" jump="" 1.="" jlt="" jhe="" jle="" inp="" figure="" 45="" the="" ny="" fiedd="" of="" dhe="" citstnxtien.="" oat="" sekes="" e="" ihe="" alu="" output="" jeesulting="" from="" the="" inaruction's="" coxp="" parn,="" and="" ep="" inples="" "cotinas="" evaution="" uah="" the="" instnuction="" adressod="" by="" the="" a="" register."="" di="" d2="" 43="" macnic="" destiwaniow="" (="" where="" te="" store="" the="" compwted="" ralve)="" the="" value="" is="" not="" stored="" anywhere="" memoryja]="" (memory="" register="" addressed="" by="" a)="" d="" register="" memory(a)="" and="" d="" register="" a="" register="" a="" register="" and="" memorya)="" a="" register="" and="" d="" register="" a="" register,="" memorylal="" and="" d="" register="" null="" md="" 100="" 10="" 1="" am="" ad="" amd="" figure="" 44="" the="" dat="" field="" of="" the="">
/ Problem 1<br>for loap<br>J=5<br>forfi=1; i«5: ir+)<br>/ Problem 2<br>/- then else<br>i=4<br>if i < 5) then<br>j= 3<br>else<br>j= 2<br>/ Problem 3<br>/while loap<br>i = 0<br>i=0<br>whileli--0)<br>j++<br>if j= 5 then<br>i=j<br>/ Problem 4<br>/ load and traverse an array<br>A0 =5<br>A11-4<br>A2]=3<br>A31=2<br>A4=1<br>ASI-0<br>for (i=0, ie=5; i++)<br>if A == 0 then<br>Alil = 5:<br>

Extracted text: / Problem 1 for loap J=5 forfi=1; i«5: ir+) / Problem 2 /- then else i=4 if i < 5) then j= 3 else j= 2 / problem 3 /while loap i = 0 i=0 whileli--0) j++ if j= 5 then i=j / problem 4 / load and traverse an array a0 =5 a11-4 a2]=3 a31=2 a4=1 asi-0 for (i=0, ie=5; i++) if a == 0 then alil = 5: 5)="" then="" j="3" else="" j="2" problem="" 3="" while="" loap="" i="0" i="0" whileli--0)="" j++="" if="" j="5" then="" i="j" problem="" 4="" load="" and="" traverse="" an="" array="" a0="5" a11-4="" a2]="3" a31="2" a4="1" asi-0="" for="" (i="0," ie="5;" i++)="" if="" a="=" 0="" then="" alil="">
Jun 11, 2022
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