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digitalmars.D.learn - Ranges help

reply Xinok <xinok live.com> writes:
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
Oct 12 2011
next sibling parent reply Dmitry Olshansky <dmitry.olsh gmail.com> writes:
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();
 }

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

sort infinite range? hasSlicing is needed though. So my take on this would be: isRandomAccessRange && hasSlicing -- Dmitry Olshansky
Oct 12 2011
parent Xinok <xinok live.com> writes:
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

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.
Oct 12 2011
prev sibling parent reply travert phare.normalesup.org (Christophe) writes:
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
Oct 14 2011
parent Xinok <xinok live.com> writes:
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.
Oct 14 2011