Name your assembly language source code file hw03_01.s . Let n be an integer tha
ID: 669932 • Letter: N
Question
Name your assembly language source code file hw03_01.s. Let n be an integer that is greater than 1. If n is
even, we change n to be n / 2; if n is odd, we change n to be 3n + 1. We repeat this procedure until n reaches 1. For
example, let n = 6,
6, 3, 10, 5, 16, 8, 4, 2, 1
The Collatz conjecture1 states for any natural number, this Collatz sequence will eventually reach 1 (this is called a
conjecture, because it has not been proven). The total stopping time is the number of iterations of applying the
rule (divide by 2 or multiply by 3 and add 1) until n reaches 1. For n = 6, we applied the rule 8 times, so for n = 6
the total stopping time is 8. For this exercise, you are to write a complete MIPS assembly language program that will
ask the user to enter n. The program shall generate and display the Collatz sequence, stopping when n reaches 1. The
program shall also display the total stopping time. Sample input and output,
Enter n (>= 1)? 27
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700
350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638
319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616
2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4
2 1
The total stopping time is 111
Here is the pseudocode,
Function main()
SysPrintStr("Enter n (>= 1)>? "
n = SysReadInt()
stop = 0
SysPrintInt(n)
SysPrintChar(' ')
While (n != 1) Do
If (n % 2 == 0) Then
n = n / 2
Else
n = 3n + 1
End If
++stop
SysPrintInt(n)
SysPrintChar(' ')
End While
SysPrintStr(" The total stopping time is ")
SysPrintInt(stop)
Explanation / Answer
hw03_01.s
HAILSTON CSECT
USING HAILSTON,R12
LR R12,R15
ST R14,SAVER14
BEGIN L R11,=F'100000' nmax
LA R8,27 n=27
LR R1,R8
MVI FTAB,X'01' ftab=true
BAL R14,COLLATZ
LR R10,R1 p
XDECO R8,XDEC n
MVC BUF1+10(6),XDEC+6
XDECO R10,XDEC p
MVC BUF1+18(5),XDEC+7
LA R5,6
LA R3,0 i
LA R4,BUF1+25
LOOPED L R2,TAB(R3) tab(i)
XDECO R2,XDEC
MVC 0(7,R4),XDEC+5
LA R3,4(R3) i=i+1
LA R4,7(R4)
C R5,=F'4'
BNE BCT
LA R4,7(R4)
BCT BCT R5,LOOPED
XPRNT BUF1,80 print hailstone(n)=p,tab(*)
MVC LONGEST,=F'0' longest=0
MVI FTAB,X'00' ftab=true
LA R8,1 i
LOOPI CR R8,R11 do i=1 to nmax
BH ELOOPI
LR R1,R8 n
BAL R14,COLLATZ
LR R10,R1 p
L R4,LONGEST
CR R4,R10 if longest<p
BNL NOTSUP
ST R8,IVAL ival=i
ST R10,LONGEST longest=p
NOTSUP LA R8,1(R8) i=i+1
B LOOPI
ELOOPI EQU * end i
XDECO R11,XDEC maxn
MVC BUF2+9(6),XDEC+6
L R1,IVAL ival
XDECO R1,XDEC
MVC BUF2+28(6),XDEC+6
L R1,LONGEST longest
XDECO R1,XDEC
MVC BUF2+36(5),XDEC+7
XPRNT BUF2,80 print maxn,hailstone(ival)=longest
B RETURN
* * * r1=collatz(r1)
COLLATZ LR R7,R1 m=n (R7)
LA R6,1 p=1 (R6)
LOOPP C R7,=F'1' do p=1 by 1 while(m>1)
BNH ELOOPP
CLI FTAB,X'01' if ftab
BNE NONOK
C R6,=F'1' if p>=1
BL NONOK
C R6,=F'3' & p<=3
BH NONOK
LR R1,R6 then
BCTR R1,0
SLA R1,2
ST R7,TAB(R1) tab(p)=m
NONOK LR R4,R7 m
N R4,=F'1' m&1
LTR R4,R4 if m//2=0 (if not(m&1))
BNZ ODD
EVEN SRA R7,1 m=m/2
B EIFM
ODD LA R3,3
MR R2,R7 *m
LA R7,1(R3) m=m*3+1
EIFM CLI FTAB,X'01' if ftab
BNE NEXTP
MVC TAB+12,TAB+16 tab(4)=tab(5)
MVC TAB+16,TAB+20 tab(5)=tab(6)
ST R7,TAB+20 tab(6)=m
NEXTP LA R6,1(R6) p=p+1
B LOOPP
ELOOPP LR R1,R6 end p; return(p)
BR R14 end collatz
*
RETURN L R14,SAVER14 restore caller address
XR R15,R15 set return code
BR R14 return to caller
SAVER14 DS F
IVAL DS F
LONGEST DS F
N DS F
TAB DS 6F
FTAB DS X
BUF1 DC CL80'hailstone(nnnnnn)=nnnnn : nnnnnn nnnnnn nnnnnn ...*
... nnnnnn nnnnnn nnnnnn'
BUF2 DC CL80'longest <nnnnnn : hailstone(nnnnnn)=nnnnn'
XDEC DS CL12
YREGS
END HAILSTON
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.