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# CodeQ: an online programming tutor.
# Copyright (C) 2015 UL FRI
#
# This program is free software: you can redistribute it and/or modify it under
# the terms of the GNU Affero General Public License as published by the Free
# Software Foundation, either version 3 of the License, or (at your option) any
# later version.
#
# This program is distributed in the hope that it will be useful, but WITHOUT
# ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS
# FOR A PARTICULAR PURPOSE. See the GNU Affero General Public License for more
# details.
#
# You should have received a copy of the GNU Affero General Public License
# along with this program. If not, see <http://www.gnu.org/licenses/>.
import time
from prolog.util import tokenize, stringify, parse, interesting_ranges, rename_vars_list
from .util import PQueue
# Starting from [code], find a sequence of edits that transforms it into a
# correct predicate for [name]. The [test] function is used to test generated
# programs.
# Return (solution, edits, time spent, #programs checked). If no solution is
# found within [timeout] seconds, return solution='' and edits=[].
def fix(code, edits, test, timeout=30, debug=False):
def step(program, path=None):
tokens = program.leaves()
for part, range_path in interesting_ranges(program):
names = {}
part_normal = tuple(rename_vars_list(part, names))
for (path, a, b), (p, uids) in edits.items():
if path == range_path and a == part_normal:
reverse_names = {v: k for k, v in names.items()}
b_real = tuple(rename_vars_list(b, reverse_names))
new_tokens = []
idx = None
for i, t in enumerate(tokens):
if t.pos >= part[0].pos and t.pos+len(t.val) <= part[-1].pos+len(part[-1].val):
if idx is None:
idx = i
new_tokens.extend(b_real)
else:
new_tokens.append(t)
new_code = stringify(new_tokens)
new_program = parse(new_code)
new_step = (idx, tuple(part), b_real)
if new_program is not None:
yield (new_program.freeze(), new_step, p)
# Main loop: best-first search through generated programs.
todo = PQueue() # priority queue of candidate solutions
done = set() # programs we have already visited
n_tested = 0
start_time = time.monotonic()
total_time = 0
# Each program gets a task with the sequence of edits that generated it and
# the associated cost. First candidate with cost 1 is the initial program.
program = parse(code)
if program is None:
return '', [], total_time, n_tested
todo.push((program.freeze(), (), 1.0), -1.0)
while total_time < timeout:
# Get next candidate.
task = todo.pop()
if task == None:
break
program, path, path_cost = task
# If we have already seen this code, skip it.
code = stringify(program)
if code in done:
continue
done.add(code)
# Print some info about the current task.
if debug:
print('Cost {:.12f}'.format(path_cost))
for idx, a, b in path:
print('{}: {} → {}'.format(idx, stringify(a), stringify(b)))
# If the code is correct, we are done.
n_correct, n_all = test(code)
if n_correct == n_all:
return code, path, total_time, n_tested
n_tested += 1
# Otherwise generate new solutions.
for new_program, new_step, new_cost in step(program, path):
new_path_cost = path_cost * new_cost
if len(todo) > 500 and new_path_cost < 0.001:
continue
new_path = path + (new_step,)
todo.push((new_program, new_path, new_path_cost), -new_path_cost)
total_time = time.monotonic() - start_time
return '', [], total_time, n_tested
def min_diff(a, b):
first = 0
while first < len(a) and first < len(b) and a[first] == b[first]:
first += 1
last = 0
while first-last < len(a) and first-last < len(b) and a[last-1] == b[last-1]:
last -= 1
return first, last
def fix_hints(code, steps):
hints = []
tokens = tokenize(code)
for idx, a, b in steps:
start, end = min_diff(a, b)
if start >= len(a)+end:
hint_id = 'monkey_insert'
elif start >= len(b)+end:
hint_id = 'monkey_remove'
else:
hint_id = 'monkey_change'
pos_start = tokens[idx+start].pos
pos_end = tokens[idx+len(a)+end-1].pos + len(tokens[idx+len(a)+end-1].val)
pos_end = max(pos_end, pos_start+1)
hints += [{'id': hint_id, 'start': pos_start, 'end': pos_end}]
tokens[idx:idx+len(a)] = [t.clone(pos=tokens[idx].pos) for t in b]
return hints
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