pf1e-simulator/tests/test_move_footprint_terrain.py

"""Multi-cell creatures move at the difficult rate under MIXED terrain (SRD).

SRD "Movement, Position, and Distance" (movement-position-and-distance-2.md):
"If you occupy squares with different kinds of terrain, you can move only as
fast as the most difficult terrain you occupy will allow." A 2x2 creature whose
ANCHOR square is plain floor but whose TRAILING footprint square sits in rubble
is hampered exactly like a 1x1 creature standing in it.

The charge, the run and the 5-foot step already test the whole footprint
(``_footprint_touches_difficult``); the ``move``/withdraw DESCENTS previously
priced every step by the anchor cell alone (``grid.step_cost``), so a Large
creature straddling rubble crossed the board at full speed. The descent now
prices a step by the most difficult square under the footprint
(``_footprint_step_cost`` -> ``grid.step_cost_with_terrain``), keeping the
5-10-5 + elevation formula in ``grid.py``. For a 1x1 creature the footprint is
the anchor cell, so the cost is bit-identical to the old anchor-only pricing.
"""

from __future__ import annotations

from pf1e_simulator.combat import CombatantState, CombatEngine
from pf1e_simulator.grid import Grid
from pf1e_simulator.map import MapSpec, TerrainType
from pf1e_simulator.models import (
    AbilityScores,
    ACProfile,
    AttackSpec,
    Combatant,
    DamageComponent,
    Saves,
)
from pf1e_simulator.rng import ScriptedRng


def _combatant(cid: str, *, size: str, speed: int = 30) -> Combatant:
    attack = AttackSpec(
        id=f"{cid}-w",
        name="club",
        kind="melee",
        attack_bonus=0,
        damage=[DamageComponent(formula="1d6", types=["bludgeoning"])],
    )
    return Combatant(
        id=cid,
        name=cid,
        level=1,
        size=size,
        abilities=AbilityScores(
            str_score=10, dex_score=10, con_score=10, int_score=10, wis_score=10,
            cha_score=10,
        ),
        hp_max=30,
        ac=ACProfile(total=10, touch=10, flat_footed=10),
        bab=0,
        initiative_mod=0,
        speed_land_ft=speed,
        saves=Saves(fort=0, ref=0, will=0),
        attacks=[attack],
    )


def _state(c: Combatant, side: str, pos: tuple[int, int]) -> CombatantState:
    return CombatantState(combatant=c, side=side, pos=pos, hp=c.hp_max)


def _rubble_at(cells: set[tuple[int, int]], size: int = 12) -> Grid:
    """Open size x size grid; the listed cells are rubble (cost 2)."""
    legend = {
        ".": TerrainType(type="floor", move_cost=1),
        "~": TerrainType(type="rubble", move_cost=2),
    }
    rows = [list("." * size) for _ in range(size)]
    for r, c in cells:
        rows[r][c] = "~"
    return Grid.from_spec(
        MapSpec(name="rubble", terrain=tuple("".join(r) for r in rows), legend=legend),
    )


def _rubble_cols(cols: set[int], size: int = 12) -> Grid:
    """Open grid with FULL-HEIGHT rubble in the given columns (no escape
    around the difficult lane: every footprint touching those columns is
    hampered)."""
    return _rubble_at(
        {(r, c) for r in range(size) for c in cols},
        size=size,
    )


# ── the descent step cost itself ──────────────────────────────────────────────


def test_large_step_priced_by_trailing_rubble_square() -> None:
    """Large anchor at (0,0) covers rows 0-1, cols 0-1. Rubble at (1,1) is a
    TRAILING footprint square: the anchor (0,1) is floor, so the old
    anchor-only pricing charged 1 for the step (0,0)->(0,1); the SRD 'most
    difficult terrain you occupy' rule charges the difficult rate (2)."""
    large = _state(_combatant("ogre", size="Large"), "players", (0, 0))
    engine = CombatEngine(ScriptedRng([]), _rubble_at({(1, 1)}), [large])
    cost = engine._footprint_step_cost(large, (0, 0), (0, 1), 0)
    assert cost == 2


def test_large_step_clean_footprint_stays_normal() -> None:
    """Same creature, rubble far off every footprint: the step is priced
    normally (1) - the difficult rate only applies when the footprint touches
    rubble."""
    large = _state(_combatant("ogre", size="Large"), "players", (0, 0))
    engine = CombatEngine(ScriptedRng([]), _rubble_at({(9, 9)}), [large])
    cost = engine._footprint_step_cost(large, (0, 0), (0, 1), 0)
    assert cost == 1


def test_medium_step_bit_identical_to_anchor_pricing() -> None:
    """1x1 regression: the footprint IS the anchor cell, so the footprint-aware
    cost equals ``grid.step_cost`` exactly - into floor and into rubble."""
    med = _state(_combatant("orc", size="Medium"), "players", (0, 0))
    engine = CombatEngine(ScriptedRng([]), _rubble_at({(0, 2)}), [med])
    assert engine._footprint_step_cost(med, (0, 0), (0, 1), 0) == engine._grid.step_cost(
        (0, 0), (0, 1), 0
    )
    assert engine._footprint_step_cost(med, (0, 1), (0, 2), 0) == engine._grid.step_cost(
        (0, 1), (0, 2), 0
    )


def test_large_diagonal_step_priced_by_trailing_rubble() -> None:
    """Diagonal step: a footprint straddling rubble prices the diagonal at the
    difficult-diagonal rate (3), matching grid.step_cost's own diagonal-into-
    difficult rule (SRD: 'Each diagonal move into a difficult terrain square
    counts as 3 squares')."""
    large = _state(_combatant("ogre", size="Large"), "players", (0, 0))
    engine = CombatEngine(ScriptedRng([]), _rubble_at({(1, 2)}), [large])
    # (0,0)->(1,1): footprint rows 1-2, cols 1-2 covers the rubble cell (1,2).
    cost = engine._footprint_step_cost(large, (0, 0), (1, 1), 0)
    assert cost == 3


# ── the observable effect: the move budget ────────────────────────────────────


def test_large_straddling_rubble_cannot_cross_at_full_speed() -> None:
    """A Large creature (30 ft = 6 squares) advancing east through a full-
    height rubble lane is hampered the whole way: it covers only 3 squares on a
    6-square move (cost 2 per square), where the anchor-only pricing let it
    cross 6. The foe far east keeps the move auto-directed straight down the
    lane (no south escape: the rubble spans every row)."""
    large = _state(_combatant("ogre", size="Large"), "players", (0, 0))
    foe = _state(_combatant("archer", size="Medium"), "monsters", (0, 11))
    engine = CombatEngine(
        ScriptedRng([15, 1]),
        _rubble_cols({0, 1, 2, 3, 4, 5, 6, 7}),
        [large, foe],
    )
    engine._move(large, foe)
    # 6-square budget / cost-2 squares = 3 squares of advance (cols 0 -> 3).
    assert large.pos[1] == 3


def test_large_clean_lane_still_crosses_at_full_speed() -> None:
    """Same geometry with the rubble removed: the Large creature crosses its
    full 6-square speed (the difficult rate never applies)."""
    large = _state(_combatant("ogre", size="Large"), "players", (0, 0))
    foe = _state(_combatant("archer", size="Medium"), "monsters", (0, 11))
    engine = CombatEngine(ScriptedRng([15, 1]), _rubble_at(set()), [large, foe])
    engine._move(large, foe)
    assert large.pos[1] == 6


def test_large_straddling_rubble_withdraw_is_hampered() -> None:
    """The withdraw ascent shares the footprint-aware cost: a Large creature
    retreating through a full-height rubble lane spends double the budget per
    square, so its 2x-speed flee covers FEWER squares than the same flee on
    clean ground (the difficult rate bites on every step)."""
    large = _state(_combatant("ogre", size="Large"), "players", (5, 8))
    threat = _state(_combatant("brute", size="Medium"), "monsters", (5, 10))
    rubble_engine = CombatEngine(
        ScriptedRng([]),
        _rubble_cols(set(range(16)), size=16),
        [large, threat],
    )
    rubble_path = rubble_engine._flee_path(large, threat)

    large2 = _state(_combatant("ogre", size="Large"), "players", (5, 8))
    threat2 = _state(_combatant("brute", size="Medium"), "monsters", (5, 10))
    clean_engine = CombatEngine(
        ScriptedRng([]), _rubble_at(set(), size=16), [large2, threat2]
    )
    clean_path = clean_engine._flee_path(large2, threat2)

    # Same 12-square (2x30ft) budget; the rubble lane halves the squares.
    assert len(rubble_path) < len(clean_path)
    assert len(rubble_path) == 4  # 4 diagonal steps x cost-3 = 12 budget