How do you filter to only return solvable == 0

[Affie]

[GearsAD]

As per #201:

Using solvable in this way means that it's a lower bound when filtering - filtering on solvable=0 means that you'll get both solvable=1 and solvable=0. That is correct? DF correct

At present there isn't a way to query unsolved nodes.

[Affie]

Ok, I missed the reasoning behind it. What about a compromise.

• use an Array{Int} = [] or
• use an UnitRange{Int64}

eg.

function _isSolvable(dfg::LightDFG, label::Symbol, ready::Array{Int})::Bool
	haskey(dfg.g.variables, label) && (return dfg.g.variables[label].solvable in ready)
	haskey(dfg.g.factors, label) && (return dfg.g.factors[label].solvable in ready)

...

[GearsAD]

Thinking about this - if filtering solvable==0 is needed, I'd suggest we don't use the lower limit then. Rather we switch it to an equality check than introduce the array?

[Affie]

I came across another case here #331. Maybe == will work better as you said. However, I don't understand the reason behind > so hesitant to change.

[dehann]

Think we resolved this in a previous discussion? The main idea is/was that there are levels of solvable. The way I find solvable == 0 is to use setdiff(ls(fg,solvable=0), ls(fg, solvable=1)). Its working well enough for the time being that we can close this issue?

[dehann]

understand the reason behind > so hesitant to change.

We need a way to search for multiple values of solvable at once, and rather than doing a ton of x in list it seemed more sensible to work with an Int and < or >. So ls(fg, solvable=0) will include all 0 <= val -- i.e. ==0, ==1, ==2, etc.

The easiest way then is to use setdiff for only getting ==0 but excluding ==1 or ==0 cases. However, this way is easier to get just solvables and skip unsolvables.