pablo23
asked on
my bubblesort for 100 ints
Hi I have been trying to implent this sorting algorithm for 100 ints.. I used the bubble sort one as well as quicksort...
I have spent nearly two weeks trying to code quicksort and gladly changed to bubble seen that qsort nearly gave me a nervous breakdown!
bubble sort is this:
Bubblesort (int data[],int n) {
int tmp,i,j;
for (i=0; i<n-1; i++) {
for (j=0; j<n-i-1; j++)
if (data[j] > data[j+1]) {
tmp = data[j];
data[j] = data[j+1];
data[j+1] = tmp;
}
}
}
and here is my implementation in bubblesort........
.data
array: .word 0x11,0x111,0x1111111,0x111
.text
dadd r1, r0, r0 ;i
dadd r2, r0, r0 ;j
daddi r6, r2, 8 ;j+1
ld r3, array(r0)
dsll r3, r3, 3 ;n
daddi r7, r3, -8 ;n-1
label: slt r12 ,r1, r7 ;check i< n-1
beqz r12, done ;if not halt
dsub r13, r3, r1 ; else set up for j loop....(ie) n-i
daddi r14, r13, -8 ; n-i-1
here: slt r15, r2, r14 ; if !r2 < n-i-1
beqz r15,inci ; go to next i
ld r19, array(r2) ; else do the swap
ld r20, array(r6)
slt r21, r19, r20
bnez r21, incj
ld r18, array(r6)
ld r16, array(r2)
sd r16, array(r6)
sd r18, array(r2)
incj: daddi r2, r2, 8 ; next iteration of j
daddi r6, r2, 8
j here
inci: dadd r2, r0, r0 ; inc i and reset j and j+ 1
daddi r1, r1, 8
daddi r6, r2, 8
j label ; go again
done: halt
the thing is i have an application from that shows the stats of the reg and mem etc.. a kind of simulator! can be downloaded from here http://www.computing.dcu.ie/~mike/winmips64.html installs in seconds
but this sort is taking way to many clock cycles maybe it can be optimized better also it will not sort the last four of the array...please help
I have spent nearly two weeks trying to code quicksort and gladly changed to bubble seen that qsort nearly gave me a nervous breakdown!
bubble sort is this:
Bubblesort (int data[],int n) {
int tmp,i,j;
for (i=0; i<n-1; i++) {
for (j=0; j<n-i-1; j++)
if (data[j] > data[j+1]) {
tmp = data[j];
data[j] = data[j+1];
data[j+1] = tmp;
}
}
}
and here is my implementation in bubblesort........
.data
array: .word 0x11,0x111,0x1111111,0x111
.text
dadd r1, r0, r0 ;i
dadd r2, r0, r0 ;j
daddi r6, r2, 8 ;j+1
ld r3, array(r0)
dsll r3, r3, 3 ;n
daddi r7, r3, -8 ;n-1
label: slt r12 ,r1, r7 ;check i< n-1
beqz r12, done ;if not halt
dsub r13, r3, r1 ; else set up for j loop....(ie) n-i
daddi r14, r13, -8 ; n-i-1
here: slt r15, r2, r14 ; if !r2 < n-i-1
beqz r15,inci ; go to next i
ld r19, array(r2) ; else do the swap
ld r20, array(r6)
slt r21, r19, r20
bnez r21, incj
ld r18, array(r6)
ld r16, array(r2)
sd r16, array(r6)
sd r18, array(r2)
incj: daddi r2, r2, 8 ; next iteration of j
daddi r6, r2, 8
j here
inci: dadd r2, r0, r0 ; inc i and reset j and j+ 1
daddi r1, r1, 8
daddi r6, r2, 8
j label ; go again
done: halt
the thing is i have an application from that shows the stats of the reg and mem etc.. a kind of simulator! can be downloaded from here http://www.computing.dcu.ie/~mike/winmips64.html installs in seconds
but this sort is taking way to many clock cycles maybe it can be optimized better also it will not sort the last four of the array...please help
Hi pablo23,
I'm not familliar with this language so forgive me if these are not of use.
There's some optimisations you can do:
1. You use -8 and 8 as values. Have you got two more registers you can use to hold them?
2. You load both array elements into registers before comparison. Can you compare a register with a memory location?
3. ld r18, array(r6). Surely, that value is in r20 already.
Are there any instructions to read a number of values into sequential registers in one hit?
Paul
I'm not familliar with this language so forgive me if these are not of use.
There's some optimisations you can do:
1. You use -8 and 8 as values. Have you got two more registers you can use to hold them?
2. You load both array elements into registers before comparison. Can you compare a register with a memory location?
3. ld r18, array(r6). Surely, that value is in r20 already.
Are there any instructions to read a number of values into sequential registers in one hit?
Paul
ASKER
No probs Paul-. Maybe I will try that with the r20 op (use a dadd)instead of ld, but the cc is still very poor and it won't sort the last five data regs.... and thats when its just sort 20 ints as you can see. Its up around 2000 and more! I want it down well past 1000 ;(!
ASKER CERTIFIED SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
>> for (i=0; i<n-1; i++) {
You dont need to stop i at n-1 since j is stopping at n-i-1. Use:
for (i=0; i<n; i++) {
I'll take a look at the rest in a bit.
Paul