Re: [PATCH v2] sched/topology: Check average distances to remote packages

[Chen, Yu C]

On 2/25/2026 8:30 PM, Peter Zijlstra wrote:
> On Tue, Feb 24, 2026 at 07:43:10PM -0600, Kyle Meyer wrote:

[ ... ]

> And it also shows that using REMOTE_DISTANCE (20) was completely random
> and 'wrong'.
>
> So per 4d6dd05d07d0 ("sched/topology: Fix sched domain build error for GNR, CWF in SNC-3 mode")
>
> Tim's original crazy SNC-3 SLIT table was:
>
> node distances:
> node     0    1    2    3    4    5
>      0:   10   15   17   21   28   26
>      1:   15   10   15   23   26   23
>      2:   17   15   10   26   23   21
>      3:   21   28   26   10   15   17
>      4:   23   26   23   15   10   15
>      5:   26   23   21   17   15   10
>
> And per:
>
>    https://lore.kernel.org/lkml/20250825075642.GQ3245006@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx/
>
> My suggestion was to average the off-trace clusters to restore sanity.

In above example, the node distances become:
"Eg. since (21+28+26+23+26+23+26+23+21)/9 ~ 24, you end up with:

 node     0    1    2    3    4    5
     0:   10   15   17   24   24   24
     1:   15   10   15   24   24   24
     2:   17   15   10   24   24   24
     3:   24   24   24   10   15   17
     4:   24   24   24   15   10   15
     5:   24   24   24   17   15   10
"

> +
> +static int slit_cluster_distance(int i, int j)
> +{
> +	static int u = 0;
> +	long d = 0;
> +	int x, y;
> +
> +	if (!u)
> +		u = slit_cluster_size();
> +
> +	/*
> +	 * Is this a unit cluster on the trace?
> +	 */
> +	if ((i / u) == (j / u))
> +		return node_distance(i, j);

the u is 3 in above example, because slit_cluster_size()
found that node0, node1 and node2 are in the same biggest
symmetric cluster. Not sure if I understand it correctly,
here we will treat node4 and node5 as the same cluster,
but without checking whether node_distance(4, 5) and
node_distance(5,4) are the same. If node_dist(4,5)!=node_dist(5,4),
will we keep it as it is?

thanks,
Chenyu