This is in relation to my sorting algorithm. This is what I need to accomplish with ranges in the most efficient way possible:

1. Merge sort - This involves copying elements to a temporary buffer, which can simply be an array, then merging the two lists together. The important thing is that it may merge left to right, or right to left, which requires a bidirectional range.

c[] = a[0..$/2];
foreach(a; arr) if(!b.empty && !c.empty) if(b.front <= c.front){
	a = b.front; b.popFront();
} else{
	a = c.front; c.popFront();
}

2. Range swap - First, I need to do a binary search, which requires a random access range. Then I need to swap two ranges of elements.

while(!a.empty && !b.empty){
	swap(a.front, b.front);
	a.popFront(); b.popFront();
}


That's the best I can come up with. I'm wondering if there's a more efficient way to accomplish what I have above.

I also need to figure out the template constraints. Would this be correct? Or would this be too much?

isRandomAccessRange && !isFiniteRange && isBidirectionalRange && hasSlicing

On 12.10.2011 22:23, Xinok wrote:
> This is in relation to my sorting algorithm. This is what I need to
> accomplish with ranges in the most efficient way possible:
>
> 1. Merge sort - This involves copying elements to a temporary buffer,
> which can simply be an array, then merging the two lists together. The
> important thing is that it may merge left to right, or right to left,
> which requires a bidirectional range.
>
> c[] = a[0..$/2];
> foreach(a; arr) if(!b.empty && !c.empty) if(b.front <= c.front){
> a = b.front; b.popFront();
> } else{
> a = c.front; c.popFront();
> }
How about:

if(b.empty)
	copy(c, a);
else if(c.empty)
	copy(b, a);
foreach(a; arr)
 if(b.front <= c.front){
   a = b.front;
   b.popFront();
   if(b.empty){
      	copy(c, a);
	break;
    }
  }
  else{
  a = c.front; c.popFront();
  if(c.empty){
	copy(b, a);
	break;
  }
 }

no need to check c if it hasn't changed from the last time, same about b.
>
> 2. Range swap - First, I need to do a binary search, which requires a
> random access range. Then I need to swap two ranges of elements.
>
> while(!a.empty && !b.empty){
> swap(a.front, b.front);
> a.popFront(); b.popFront();
> }
>

If your ranges have equal lengths (or you assume it)
you can skip one of !a.empty or !b.empty in while clause.

Otherwise :
for(;;){
	swap(a.front, b.front);
	a.popFront();
	if(a.empty)
		break;
	b.popFront();
	if(b.empty)
		break;
}
might save you a couple of ops in case a is shorter then b, and with sorting every bit counts isn't it?

>
> That's the best I can come up with. I'm wondering if there's a more
> efficient way to accomplish what I have above.
>
> I also need to figure out the template constraints. Would this be
> correct? Or would this be too much?
>
> isRandomAccessRange && !isFiniteRange && isBidirectionalRange && hasSlicing
isRandomAccessRange should be enough. Also why !isFinite how would one sort infinite range? hasSlicing is needed though. So my take on this would be:
isRandomAccessRange && hasSlicing

-- 
Dmitry Olshansky

On 10/12/2011 4:04 PM, Dmitry Olshansky wrote:
> On 12.10.2011 22:23, Xinok wrote:
>> I also need to figure out the template constraints. Would this be
>> correct? Or would this be too much?
>>
>> isRandomAccessRange && !isFiniteRange && isBidirectionalRange &&
>> hasSlicing
> isRandomAccessRange should be enough. Also why !isFinite how would one
> sort infinite range? hasSlicing is needed though. So my take on this
> would be:
> isRandomAccessRange && hasSlicing
>

Sorry, typo. That should be !isInfiniteRange. But I can drop !isInfinteRange anyways, so:

isRandomAccessRange && isBidirectionalRange && hasSlicing

isRandomAccessRange can be a bidirectional range or an infinite forward range, so isBidirectionalRange is still required.

Xinok , dans le message (digitalmars.D.learn:30054), a écrit :
> This is in relation to my sorting algorithm. This is what I need to accomplish with ranges in the most efficient way possible:
> 
> 1. Merge sort - This involves copying elements to a temporary buffer, which can simply be an array, then merging the two lists together. The important thing is that it may merge left to right, or right to left, which requires a bidirectional range.
> 
> c[] = a[0..$/2];
> foreach(a; arr) if(!b.empty && !c.empty) if(b.front <= c.front){
> 	a = b.front; b.popFront();
> } else{
> 	a = c.front; c.popFront();
> }
> 
> 2. Range swap - First, I need to do a binary search, which requires a random access range. Then I need to swap two ranges of elements.
> 
> while(!a.empty && !b.empty){
> 	swap(a.front, b.front);
> 	a.popFront(); b.popFront();
> }
> 
> 
> That's the best I can come up with. I'm wondering if there's a more efficient way to accomplish what I have above.
> 
> I also need to figure out the template constraints. Would this be correct? Or would this be too much?
> 
> isRandomAccessRange && !isFiniteRange && isBidirectionalRange && hasSlicing


You should look at:
std.algorithm.SetUnion
std.algorithm.swapRanges

-- 
Christophe

Thanks. I'll run a benchmark with swapRanges, see how it compares to my own code. But it would be better if I coded the merge function myself, since I can do it in-place using very little memory.