OuterReference.steps_out uses a 0-based convention, but Substrait requires steps_out >= 1

[nielspardon]

Summary

substrait-python builds and resolves OuterReference.steps_out with a 0-based convention (0 = the immediately enclosing query), but Substrait specifies steps_out must be >= 1 (1-based). See the v0.89.0 release note ("... this is not a real breaking change as Substrait always required non-zero value", #188) and proto/substrait/algebra.proto (steps_out ... Must be >= 1).

Details

• Builder default is 0: outer_reference(field, steps_out=0) (src/substrait/builders/extended_expression.py) and sub.outer(name, steps_out=0) (src/substrait/dataframe/expr.py), documented as "0 = the immediately enclosing query".
• Inference treats 0 as immediate: infer_expression_type resolves stack[len(stack) - 1 - steps], so steps_out=0 selects the immediately enclosing schema (src/substrait/type_inference.py, the steps_out branch).
• Tests encode the 0-based convention: tests/dataframe/test_frame.py::test_correlated_exists asserts steps_out == 0 for an immediate correlation; test_nested_correlation_steps_out uses steps_out=1 for "two levels out".

Impact

Plans emitted by the builder / DataFrame APIs carry steps_out=0, which the spec disallows, and are off-by-one for deeper nesting levels. Correlated subqueries may therefore not round-trip correctly with spec-conformant producers/consumers.

Suggested fix

Make the public API 1-based: default steps_out=1, emit and resolve accordingly, and update the outer_schemas indexing and the associated tests. This is a breaking change to the correlated-subquery API, so it warrants its own PR.

Context

Noticed while implementing lateral-join id-based references in #228. The lateral path uses OuterReference.rel_reference (id-based), not steps_out, so it is unaffected by this bug.