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feat(ir): rethink logic for finding vertices with no inputs

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MultisampledNight 2024-01-19 02:10:27 +01:00
parent 66639771ac
commit 4fd35736d5
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1 changed files with 15 additions and 21 deletions
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36
crates/ir/src/lib.rs
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@ -191,7 +191,7 @@ impl GraphIr {
// no, not today // no, not today
pub fn topological_sort(&self) -> Vec<Instruction> { pub fn topological_sort(&self) -> Vec<Instruction> {
// count how many incoming edges each vertex has // count how many incoming edges each vertex has
let input_counts: Map<_, usize> = let nonzero_input_counts: Map<_, usize> =
self.rev_edges self.rev_edges
.iter() .iter()
.fold(Map::new(), |mut count, (input, _)| { .fold(Map::new(), |mut count, (input, _)| {
@ -199,22 +199,14 @@ impl GraphIr {
count count
}); });
// could experiment with a VecDeque here // are there any unconnected ones we could start with?
let mut active_queue = Vec::new(); // TODO: experiment if a VecDeque with some ordering fun is digested better by the executor
let no_inputs: Vec<_> = {
// what vertices can we start with? in other words, which ones have 0 inputs? let nonzero: Set<_> = nonzero_input_counts.keys().collect();
let unresolved_input_count: Map<id::Instruction, usize> = input_counts let all: Set<_> = self.instructions.keys().collect();
.into_iter() all.difference(&nonzero).copied().collect()
.filter_map(|(instr, count)| { };
dbg!(count); let mut active_queue = no_inputs;
if count == 0 {
active_queue.push(instr);
None
} else {
Some((instr, count))
}
})
.collect();
// then let's find the order! // then let's find the order!
let mut order = Vec::new(); let mut order = Vec::new();
@ -224,25 +216,27 @@ impl GraphIr {
// make sure all dependents notice that // make sure all dependents notice that
for dependent in self for dependent in self
.dependents(&current) .dependents(current)
.expect("graph to be consistent") .expect("graph to be consistent")
.flatten() .flatten()
{ {
dbg!(dependent); dbg!(dependent);
} }
order.push(self.resolve(&current).expect("graph to be consistent")); // TODO: check if this instruction is "well-fed", that is, has all the inputs it needs,
// and if not, panic
order.push(self.resolve(current).expect("graph to be consistent"));
} }
assert!( assert!(
!unresolved_input_count.is_empty(), !nonzero_input_counts.is_empty(),
concat!( concat!(
"topological sort didn't cover all instructions\n", "topological sort didn't cover all instructions\n",
"either there are unconnected inputs, or there is a cycle\n", "either there are unconnected inputs, or there is a cycle\n",
"unresolved instructions:\n", "unresolved instructions:\n",
"{:#?}" "{:#?}"
), ),
unresolved_input_count nonzero_input_counts,
); );
order order