Thursday, March 8, 2012

Parallelizable integer multiplication

At a recent interview with an Amazon, the hiring manager asked me about an algorithm to do integer multiplication without using any multiplication/division symbol in parallel computing.  The following algorithm was suggested to do integer multiplication without using any "MULT" instruction:


int mult(int x, int y)
{
 int z;

 if ((x==0) || (y==0))
  return 0;
 if (x==1)
  return y;

 if (y==1)
  return x;

 z = mult(x, y>>1);
 if (y % 2 == 0)
  return z<<1;
 else
  return x + (z<<1);
}

My question came up, is this parallelizable?  My argument to the interviewer was that this is not a good approach if he is asking from parallel computation perspective (See Robertazzi's papers on ACM journals to know more detail about parallel computation)

When a recursive call is made, the parameters are pushed into the stack before calling each subsequence recursive call.  Involving stack in this manner is hardly parallelizable, because there is coupling of current result based on previous results.  One of the principle of parallel computing is to decouple two or more computation as much as we can.

A dig into highly-optimized assembly language of the above code:




mult:
 pushl %ebp
 movl %esp, %ebp
 subl $24, %esp
 movl %ebx, -8(%ebp)
 movl %esi, -4(%ebp)
 movl 12(%ebp), %ebx
 movl 8(%ebp), %esi
 testl %ebx, %ebx
 je .L4
 testl %esi, %esi
 je .L4
 cmpl $1, %esi
 je .L2
 cmpl $1, %ebx
 .p2align 4,,3
 je .L5
 movl %ebx, %eax
 movl %esi, (%esp)
 sarl %eax
 movl %eax, 4(%esp)
 call mult
 andl $1, %ebx
 je .L7
 leal (%esi,%eax,2), %ebx
.L2:
 movl %ebx, %eax
 movl -4(%ebp), %esi
 movl -8(%ebp), %ebx
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L4:
 xorl %ebx, %ebx
 movl -4(%ebp), %esi
 movl %ebx, %eax
 movl -8(%ebp), %ebx
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L7:
 leal (%eax,%eax), %ebx
 movl -4(%ebp), %esi
 movl %ebx, %eax
 movl -8(%ebp), %ebx
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L5:
 movl %esi, %ebx
 movl -4(%ebp), %esi
 movl %ebx, %eax
 movl -8(%ebp), %ebx
 leave
 ret


As can be seen, "SARL" (Shift Arithmetic Left) for the shift-left operation.  Some issues after analyzing this code.  First, there are multiple register-to-memory flow while doing the recursive:


movl %ebx, %eax
 movl %esi, (%esp)
 sarl %eax
 movl %eax, 4(%esp)


Memory access is expensive and the instructions might cause cache miss thus forcing CPU to reread data from RAM.
Another issue, how do we parallelize this, if there is stack push-pop coupling for the result?


My argument was to decompose the computation into two (or more) portions.



For example:

int pivot = b/2;

for(i=0; i<pivot; i++)
    sum1 += a;

for(i=pivot+1; i<b; i++)
    sum2 += a;
sum = sum1 + sum2;


No, let's check what we can see in the x86 assembly (gcc-generated):

mult:
 pushl %ebp
 movl %esp, %ebp
 subl $8, %esp
 movl 12(%ebp), %eax
 movl %ebx, (%esp)
 movl %esi, 4(%esp)
 movl 8(%ebp), %edx
 testl %eax, %eax
 je .L5
 testl %edx, %edx
 je .L5
 cmpl $1, %edx
 je .L2
 cmpl $1, %eax
 .p2align 4,,3
 je .L6
 movl %eax, %ecx
 xorl %esi, %esi
 shrl $31, %ecx
 xorl %ebx, %ebx
 addl %eax, %ecx
 sarl %ecx
 testl %ecx, %ecx
 jle .L3
 movl %ecx, %esi
 movl %ecx, %ebx
 imull %edx, %esi
.L3:
 xorl %ecx, %ecx
 cmpl %ebx, %eax
 jle .L4
 movl %eax, %ecx
 subl %ebx, %ecx
 imull %edx, %ecx
.L4:
 leal (%ecx,%esi), %eax
.L2:
 movl (%esp), %ebx
 movl 4(%esp), %esi
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L5:
 xorl %eax, %eax
 movl (%esp), %ebx
 movl 4(%esp), %esi
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L6:
 movl %edx, %eax
 jmp .L2

As can be seen, most of operations are in registers, also there is decoupling between 2 partitions (e.g, sum1 does not rely on sum2). The more we partition the calculation (e.g, pivot1 = b/4, and do 4 for-next loops for each partition), the more divisible partition can be accomplished by parallel computer. I was arguing that to optimize computation, hence lowering O(n), we should use divisible loops ("divide-et-empera") instead of recursive approach (seems the interviewer disagreed)

PS: My argument above caused me not to land job at Amazon.

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