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Sort by two criteria: x and index, in that order, so if there's two points on the same X, the index controls order. Indices are unique so such 2 criteria are enough.

$O(n^2)$ - for each vert, entire geometry is duplicated. I don't know how it compares to convex hull algorithm, which I'd consider a hacky solution - you're relying on undocumented implementation details - still, quite awesome, so keep it coming!

(left Transfer Attribute could be replaced with Field at Index node)

Accumulate Field counts for each vertex how many other vertices are either before it on X, or exactly on the same X but have lower index. That's the number of vertices that should have lower indices, and so (in 0-based indexing) it's the desired index of the currently evaluated vertex - which is saved as ID:

For clarity, this is how you would use this data to reposition the vertices:

Quadratic Sort (Pre B3.4)

This is one of outdated answers that would unnecessarily bury (currently) objectively the best answer by quellenform. You can still access this answer by reading the previous revision.

Sort by two criteria: x and index, in that order, so if there's two points on the same X, the index controls order. Indices are unique so such 2 criteria are enough.

$O(n^2)$ - for each vert, entire geometry is duplicated. I don't know how it compares to convex hull algorithm, which I'd consider a hacky solution - you're relying on undocumented implementation details - still, quite awesome, so keep it coming!

(left Transfer Attribute could be replaced with Field at Index node)

Accumulate Field counts for each vertex how many other vertices are either before it on X, or exactly on the same X but have lower index. That's the number of vertices that should have lower indices, and so (in 0-based indexing) it's the desired index of the currently evaluated vertex - which is saved as ID:

Sort by two criteria: x and index, in that order, so if there's two points on the same X, the index controls order. Indices are unique so such 2 criteria are enough.

$O(n^2)$ - for each vert, entire geometry is duplicated. I don't know how it compares to convex hull algorithm, which I'd consider a hacky solution - you're relying on undocumented implementation details - still, quite awesome, so keep it coming!

(left Transfer Attribute could be replaced with Field at Index node)

Accumulate Field counts for each vertex how many other vertices are either before it on X, or exactly on the same X but have lower index. That's the number of vertices that should have lower indices, and so (in 0-based indexing) it's the desired index of the currently evaluated vertex - which is saved as ID:

For clarity, this is how you would use this data to reposition the vertices:

Sort by two criteria: x and index, in that order, so if there's two points on the same X, the index controls order. Indices are unique so such 2 criteria are enough.

$O(n^2)$ - for each vert, entire geometry is duplicated. I don't know how it compares to convex hull algorithm, which I'd consider quite a hacky solution - you're relying on undocumented implementation details - still, quite awesome, so keep it coming!

Accumulate Field counts for each vertex how many other vertices are either before it on X, or exactly on the same X but have lower index. That's the number of vertices that should have lower vertices, and so (in 0-based indexing) it's the desired index of the currently evaluated vertex - which is saved as ID:

Sort by two criteria: x and index, in that order, so if there's two points on the same X, the index controls order. Indices are unique so such 2 criteria are enough.

$O(n^2)$ - for each vert, entire geometry is duplicated. I don't know how it compares to convex hull algorithm, which I'd consider a hacky solution - you're relying on undocumented implementation details - still, quite awesome, so keep it coming!

(left Transfer Attribute could be replaced with Field at Index node)

Accumulate Field counts for each vertex how many other vertices are either before it on X, or exactly on the same X but have lower index. That's the number of vertices that should have lower indices, and so (in 0-based indexing) it's the desired index of the currently evaluated vertex - which is saved as ID: