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

[Peter Zijlstra]

On Wed, Feb 25, 2026 at 05:32:46PM +0100, Peter Zijlstra wrote:
> +static bool slit_cluster_symmetric(int i, int j, int n)
>  {
> +	WARN_ON_ONCE((i % n) || (j % n));
>  
> +	for (int k = i; k < i + n; k++) {
> +		for (int l = k; l < j + n; l++) {
> +			if (node_distance(k, l) != node_distance(k, l))
> +				return false;
>  		}
>  	}
>  
> +	return true;
> +}
> +
> +static bool slit_cluster_match(int i, int j, int x, int y, int n)
> +{
> +	WARN_ON_ONCE((i % n) || (j % n) || (x % n) || (y % n));
> +
> +	for (int k = 0; k < n; k++) {
> +		for (int l = k; l < n; l++) {
> +			if (node_distance(i + k, j + l) != node_distance(x + k, y + l))
> +				return false;
> +		}
> +	}
> +
> +	return true;
> +}
> +
> +/*
> + * Find the largest symmetric,repeating cluster in an attempt to identify the
> + * unit size.
> + */
> +static int slit_cluster_size(void)
> +{
> +	int nodes = num_possible_nodes();
> +
> +	/*
> +	 * There are at least 2 packages; so half-nodes is the largest
> +	 * possible unit, go down from that.
> +	 */
> +	for (int u = nodes / 2; u; u--) {

nodes / topology_max_packages() might also do. And I worry about
num_possible_nodes(), if that includes CXL or other such nonsense then
we're up a creek.

But as stated before, ideally the hardware can actually just tell us the
right number.

> +		/*
> +		 * If u doesn't divide nodes, it can't be a unit.
> +		 */
> +		if (nodes % u)
> +			continue;
> +
> +		/*
> +		 * Unit must be symmetric,
> +		 */
> +		if (!slit_cluster_symmetric(0, 0, u))
> +			continue;
> +
> +		/*
> +		 * and repeating.
> +		 */
> +		if (slit_cluster_match(0, 0, u, u, u))
> +			return u;
> +	}
> +
> +	return nodes;
> +}
> +
> +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);
> +
> +	/*
> +	 * Off-trace cluster, return average of the cluster to force symmetry.
> +	 */
> +	x = i - (i % u);
> +	y = j - (j % u);
> +
> +	for (i = x; i < x + u; i++) {
> +		for (j = y; j < y + u; j++) {
> +			d += node_distance(i, j);
> +			d += node_distance(j, i);
> +		}
> +	}
> +
> +	return d / (2*u*u);
>  }

Note that I changed this to average over the symmetric pair of off-trace
clusters. Because if some BIOS monkey is really taking bong-hits, there
is absolutely no guarantee the just the one cluster average will result
in a symmetric system in the end.