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https://github.com/Jaxan/satuio.git
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376 lines
12 KiB
Python
376 lines
12 KiB
Python
"""
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Script for finding UIO sequences in a Mealy machine. Uses SAT solvers
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(in pysat) to search efficiently. The length is incremented each
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iteration, so that short UIOs are found fast. It also uses simple
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techniques to extend UIOs to more states if an UIO is found. This
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can result in UIOs which are longer than the specified bound. For
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the usage, please run
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python3 uio-incr.py --help
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If you want to read this script, my suggestion is to first read
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and understand uio.py. That file is simpler and perhaps better
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structured.
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© Joshua Moerman, Open Universiteit, 2022
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"""
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# Import the solvers and utilities
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from pysat.solvers import Solver
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from pysat.formula import IDPool
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from pysat.card import CardEnc, EncType
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from argparse import ArgumentParser # Command line options
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from rich.console import Console # Import colorized output
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from tqdm import tqdm # Import fancy progress bars
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from utils.parser import read_machine
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from utils.utils import *
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# We set up some things for nice output
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console = Console(markup=False, highlight=False)
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# *****************
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# Reading the input
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# *****************
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# command line options
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parser = ArgumentParser()
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parser.add_argument('filename', help='File of the mealy machine (dot format)')
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parser.add_argument('bound', help='Upper bound (incl.) for the UIO solving', type=int)
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parser.add_argument('--solver', help='Which solver to use (default g3)', default='g3')
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parser.add_argument('--bases', help='For which states to compute an UIO (leave empty for all states)', nargs='*')
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args = parser.parse_args()
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# reading the automaton
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(alphabet, outputs, states, delta, labda) = read_machine(args.filename)
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# if the base states are not specified, take all
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if args.bases == None:
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# Has to be a copy, because we will modify bases!
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bases = states.copy()
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else:
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bases = args.bases
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bound = args.bound
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measure_time('Constructed automaton with', len(states), 'states and', len(alphabet), 'symbols')
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# ****************
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# UIO implications
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# And length = 1
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# ****************
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# We are going to look for transitions s --i/o--> t such that if
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# t has an UIO, then automatically s has an UIO. For this, we need
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# to consider the predecessors of t with unique input/output. We
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# use a dictionary to sort this out.
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# incoming :: state -> (input, output) -> [predecessor state]
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incoming = {}
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# And we are going to look for UIOs of length 1. Again we use a
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# dictionary for this.
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# (input, output) -> [origin state]
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input_output_pairs = {}
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for s in states:
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for i in alphabet:
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t = delta[(s, i)]
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o = labda[(s, i)]
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# First, we record input/output pair
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# for the UIOs of length = 1.
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if (i, o) not in input_output_pairs:
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input_output_pairs[(i, o)] = []
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input_output_pairs[(i, o)].append(s)
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# For predecessors, we can skip self-loops, because if t has
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# an UIO we already know that s = t has an UIO...
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if t == s:
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continue
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if t not in incoming:
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incoming[t] = {}
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if (i, o) not in incoming[t]:
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incoming[t][(i, o)] = []
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# Record predecessors of t
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incoming[t][(i, o)].append(s)
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# Extract edges from `incoming`, if there is only a single
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# state for the given input/output pair. We only store the
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# input word, not the output, as we don't use it.
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# uio_implication_graph :: state -> [(predecessor, input)]
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uio_implication_graph = {}
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for (s, m) in incoming.items():
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uio_implication_graph[s] = []
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for ((i, o), pred) in m.items():
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if len(pred) == 1:
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uio_implication_graph[s].append((pred[0], i))
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incoming.clear()
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# Extract unique input/output from the `input_output_pairs`.
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# We only store the input word, as we don't use the output.
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# new_uios :: state -> word
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new_uios = {}
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for ((i, o), ls) in input_output_pairs.items():
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if len(ls) == 1:
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new_uios[ls[0]] = [i]
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input_output_pairs.clear()
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# By using a bfs on the uio-implication-graph, we can extend known UIOs
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# to other states. We define a function to do this. It operates on global
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# variables. We do this now, for the length = 1 UIOs, but also after each
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# iteration of the solver.
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# uios :: state -> word
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uios = {}
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def extend(new_uios):
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global uios, bases
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queue = list(new_uios.items())
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while queue:
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(t, uio) = queue.pop(0)
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# If we already have an UIO for s, skip
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# (unless it is a smaller one)
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if t in uios and len(uio) >= len(uios[t]):
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continue
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# Store the UIO
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uios[t] = uio
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# We don't have to search anymore for t, so remove it
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if t in bases:
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bases.remove(t)
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# And follow the graph to extend to other states.
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# Note: these may not be the smallest UIO for that state.
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# But it is an UIO nevertheless.
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if t in uio_implication_graph:
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for (s, i) in uio_implication_graph[t]:
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queue.append((s, [i] + uio))
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# We extend our 1-letter UIOs
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extend(new_uios)
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new_uios = {}
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measure_time('Found', len(uios), 'simple uios already')
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# ********************
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# Seting up the solver
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# And the variables
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# ********************
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for length in range(2, bound + 1):
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# If there are no more states for which we want an UIO,
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# we stop the search.
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if not bases:
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break
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# Otherwise, we will setup a solver and search for them
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print('*** Solving for length', length, 'for', len(bases), 'remaining states out of', len(states))
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with Solver(name=args.solver) as solver:
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# Since the solver can only deal with variables x_i, we need
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# a mapping of variabeles: x_whatever -> x_i.
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# We use the IDPool of pysat for this. It generates variables
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# on the fly.
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vpool = IDPool()
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def var(x):
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return(vpool.id(('uio', x)))
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# Variables for the guessed word
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# avar(i, a) means: on place i there is symbol a
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def avar(i, a):
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return var(('a', i, a))
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# Each state has its own path, and on this path we encode
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# states and outputs
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# svar(s, i, t) means: on path s, at place i, we are in state t
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def svar(s, i, t):
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return var(('s', s, i, t))
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# ovar(s, i, o) means: on path s, on place i, there is output o
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def ovar(s, i, o):
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return var(('o', s, i, o))
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# We use extra variables to encode the fact that there is
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# a difference in output (a la Tseytin transformation)
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# evar(s, i) means: on path s, on place i, there is a difference
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# in output. Note: the converse (if there is a difference
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# evar(s, i) is true) does not hold!
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def evar(s, i):
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return var(('e', s, i))
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# In order to re-use parts of the formula, we introduce
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# enabling variables. These indicate the fixed state for which
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# we want to compute an UIO. By changing these variables only, we
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# can keep most of the formula the same and incrementally solve it.
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# The fixed state is called the "base".
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# bvar(s) means: s is the base.
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def bvar(s):
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return var(('base', s))
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# We often need to assert that exacly one variable in a list holds.
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# For that we use pysat's cardinality encoding. This might introduce
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# additional variables. But that does not matter for us.
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def unique(lits):
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cnf = CardEnc.equals(lits, 1, vpool=vpool, encoding=EncType.seqcounter)
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solver.append_formula(cnf.clauses)
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# ********************
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# Constructing the CNF
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# ********************
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# Guessing the word:
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for i in range(length):
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unique([avar(i, a) for a in alphabet])
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# We should only enable one base state.
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# (This allows for an optimisation later.)
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unique([bvar(base) for base in bases])
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# For each state s, we construct a path of possible successor states,
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# following the guessed word. This path should be consistent with delta,
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# and we also record the outputs along this path. The outputs are later
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# used to decide whether we found a different output.
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possible_outputs = {}
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for s in tqdm(states, desc="CNF paths"):
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# current set of possible states we're in
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current_set = set([s])
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# set of successors for the next iteration of i
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next_set = set()
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for i in range(length):
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# Only one successor state should be enabled.
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# For i == 0, this is a single state (s).
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unique([svar(s, i, t) for t in current_set])
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# We keep track of the possible outputs
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possible_outputs[(s, i)] = set()
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for t in current_set:
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sv = svar(s, i, t)
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for a in alphabet:
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av = avar(i, a)
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output = labda[(t, a)]
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possible_outputs[(s, i)].add(output)
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# Constraint: on path s, when in state t and input a, we output o
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# x_('s', s, i, t) /\ x_('in', i, a) => x_('o', i, labda(t, a))
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# == -x_('s', s, i, t) \/ -x_('in', i, a) \/ x_('o', i, labda(t, a))
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solver.add_clause([-sv, -av, ovar(s, i, output)])
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# when i == length-1 we don't need to consider successors
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if i < length-1:
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next_t = delta[(t, a)]
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next_set.add(next_t)
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# Constraint: on path s, when in state t and input a, we go to next_t
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# x_('s', s, i, t) /\ x_('in', i, a) => x_('s', s, i+1, delta(t, a))
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# == -x_('s', s, i, t) \/ -x_('in', i, a) \/ x_('s', s, i+1, delta(t, a))
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solver.add_clause([-sv, -av, svar(s, i+1, next_t)])
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# Only one output should be enabled
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unique([ovar(s, i, o) for o in possible_outputs[(s, i)]])
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# Next iteration with successor states
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current_set = next_set
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next_set = set()
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# If the output of a state is different than the one from our base state,
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# then, we encode that in a new variable. This is only needed when the base
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# state is active, so the first literal in these clauses is -bvar(base).
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# Also note, we only encode the converse: if there is a difference claimed
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# and base has a certain output, than the state should not have that output.
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# This means that the solver doesn't report all differences, but at least one.
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for s in tqdm(states, desc="CNF diffs"):
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# Constraint: there is a place, such that there is a difference in output
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# \/_i x_('e', s, i)
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# If s is our base, we don't care (this can be done, because only
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# a single bvar is true).
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if s in bases:
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solver.add_clause([bvar(s)] + [evar(s, i) for i in range(length)])
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else:
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solver.add_clause([evar(s, i) for i in range(length)])
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# Now we actually encode when the difference occurs
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for base in bases:
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if s == base:
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continue
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bv = bvar(base)
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for i in range(length):
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outputs_base = possible_outputs[(base, i)]
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outputs_s = possible_outputs[(s, i)]
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# We encode: if the base is enabled and there is a difference,
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# then the outputs should actually differ. (We do not have to
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# encode the other implication!)
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# x_('b', base) /\ x_('e', s, i) /\ x_('o', base, i, o) => -x_('o', s, i, o)
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# Note: when o is not possible for state s, then the clause already holds
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for o in outputs_base:
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if o in outputs_s:
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solver.add_clause([-bv, -evar(s, i), -ovar(base, i, o), -ovar(s, i, o)])
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measure_time('Constructed CNF with', solver.nof_clauses(), 'clauses and', solver.nof_vars(), 'variables, solving with', args.solver)
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# ******************
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# Solving and output
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# ******************
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new_uios = {}
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# We want to find an UIO for each base. We have already constructed
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# the CNF. So it remains to add assumptions to the solver, this is
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# called "incremental solving" in SAT literature.
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for base in tqdm(bases, desc='solving'):
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# Solve with bvar(base) being true
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b = solver.solve(assumptions=[bvar(base)])
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# if there is no UIO, we go to the next state
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# we might find one later, with the length increased.
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if not b:
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continue
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# Get the set of true variables
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truth = get_truth(solver)
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# We extract the word
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uio = []
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for i in range(length):
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for a in alphabet:
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if avar(i, a) in truth:
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uio.append(a)
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new_uios[base] = uio
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print('found', len(new_uios), 'new uios, total =', len(new_uios) + len(uios))
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extend(new_uios)
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measure_time('after extending, we have', len(uios), 'uios')
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new_uios = {}
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for (s, uio) in uios.items():
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console.print(s, style='bold black', end=' => ')
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console.print(uio, style='bold green')
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print('')
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# Report some final stats
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measure_total_time('\nDone')
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print('With UIO:', len(uios))
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print('without: ', len(states) - len(uios))
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