Add test module exercising formal verification.
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Cesar Strauss 2024-11-20 17:39:00 -03:00
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// SPDX-License-Identifier: LGPL-3.0-or-later
// See Notices.txt for copyright information
//! Formal tests in Fayalite
use fayalite::{
cli::FormalMode,
clock::{Clock, ClockDomain},
expr::{CastTo, HdlPartialEq},
firrtl::ExportOptions,
formal::{any_seq, formal_reset, hdl_assert, hdl_assume},
hdl_module,
int::{Bool, UInt},
module::{connect, connect_any, reg_builder, wire},
reset::ToReset,
testing::assert_formal,
};
/// Test hidden state
///
/// Hidden state can cause problems for induction, since the formal engine
/// can assign invalid values to the state registers, making it traverse
/// valid but unreachable states.
///
/// One solution is to go sufficiently in the past so the engine is forced
/// to eventually take a reachable state. This may be hampered by
/// existence of loops, then assumptions may be added to break them.
///
/// Another solution is to "open the black box" and add additional
/// assertions involving the hidden state, so that the unreachable states
/// become invalid as well.
///
/// Both approaches are taken here.
///
/// See [Claire Wolf's presentation] and [Zipcpu blog article].
///
/// [Claire Wolf's presentation]: https://web.archive.org/web/20200115081517fw_/http://www.clifford.at/papers/2017/smtbmc-sby/
/// [Zipcpu blog article]: https://zipcpu.com/blog/2018/03/10/induction-exercise.html
mod hidden_state {
use super::*;
/// Test hidden state by shift registers
///
/// The code implement the ideas from an article in the [Zipcpu blog]. Two
/// shift registers are fed from the same input, so they should always have
/// the same value. However the only observable is a comparison of their
/// last bit, all the others are hidden. To complicate matters, an enable
/// signal causes a loop in state space.
///
/// [Zipcpu blog]: https://zipcpu.com/blog/2018/03/10/induction-exercise.html
#[test]
fn shift_register() {
enum ConstraintMode {
WithExtraAssertions,
WithExtraAssumptions,
}
use ConstraintMode::*;
#[hdl_module]
fn test_module(constraint_mode: ConstraintMode) {
#[hdl]
let clk: Clock = m.input();
#[hdl]
let cd = wire();
connect(
cd,
#[hdl]
ClockDomain {
clk,
rst: formal_reset().to_reset(),
},
);
// input signal for the shift registers
#[hdl]
let i: Bool = wire();
connect(i, any_seq(Bool));
// shift enable signal
#[hdl]
let en: Bool = wire();
connect(en, any_seq(Bool));
// comparison output
#[hdl]
let o: Bool = wire();
// shift registers, with enable
#[hdl]
let r1 = reg_builder().clock_domain(cd).reset(0u8);
#[hdl]
let r2 = reg_builder().clock_domain(cd).reset(0u8);
#[hdl]
if en {
connect_any(r1, (r1 << 1) | i.cast_to(UInt[1]));
connect_any(r2, (r2 << 1) | i.cast_to(UInt[1]));
}
// compare last bits of both shift registers
connect(o, r1[7].cmp_eq(r2[7]));
// what we want to prove: last bits are always equal
hdl_assert(clk, o, "");
// additional terms below are only needed to assist with the induction proof
if let WithExtraAssertions = constraint_mode {
// "Open the box": add assertions about hidden state.
// In this case, the hidden bits are also always equal.
hdl_assert(clk, r1.cmp_eq(r2), "");
}
if let WithExtraAssumptions = constraint_mode {
// Break the loop, do not allow "en" to remain low forever
#[hdl]
let past_en_reg = reg_builder().clock_domain(cd).reset(false);
connect(past_en_reg, en);
hdl_assume(clk, past_en_reg | en, "");
}
}
// we need a minimum of 16 steps so we can constrain all eight shift register bits,
// given that we are allowed to disable the shift once every two cycles.
assert_formal(
"shift_register_with_assumptions",
test_module(WithExtraAssumptions),
FormalMode::Prove,
16,
None,
ExportOptions::default(),
);
// here a couple of cycles is enough
assert_formal(
"shift_register_with_assertions",
test_module(WithExtraAssertions),
FormalMode::Prove,
2,
None,
ExportOptions::default(),
);
}
}