mirror of
https://github.com/Jaxan/satuio.git
synced 2025-04-27 06:37:45 +02:00
235 lines
6.8 KiB
Python
235 lines
6.8 KiB
Python
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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import time
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import argparse
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from tqdm import tqdm
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solver_name = 'g3'
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verbose = True
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start = time.time()
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def measure_time(*str):
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global start
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now = time.time()
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print('***', *str, "in %.3f seconds" % (now - start))
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start = now
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# Automaton
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parser = argparse.ArgumentParser()
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parser.add_argument('filename', help='File of the mealy machine (dot format)')
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parser.add_argument('length', help="Length of the uio", type=int)
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args = parser.parse_args()
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length = args.length
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alphabet = set()
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outputs = set()
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states = set()
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bases = set(['s0', 's4', 's37', 's555'])
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delta = {}
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labda = {}
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# Quick and dirty .dot parser
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with open(args.filename) as file:
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for line in file.readlines():
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asdf = line.split()
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if len(asdf) > 3 and asdf[1] == '->':
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s = asdf[0]
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t = asdf[2]
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rest = ''.join(asdf[3:])
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label = rest.split('"')[1]
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[i, o] = label.split('/')
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states.add(s)
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states.add(t)
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alphabet.add(i)
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outputs.add(o)
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delta[(s, i)] = t
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labda[(s, i)] = o
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measure_time('Constructed automaton with', len(states), 'states and', len(alphabet), 'symbols')
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# Solver setup
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vpool = IDPool()
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solver = Solver(name=solver_name)
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# mapping van variabeles: x_... -> x_i
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def var(x):
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return(vpool.id(('uio', x)))
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# On place i we have 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
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# 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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# On path s, we are in state t on place i
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def svar(s, i, t):
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return var(('s', s, i, t))
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# Extra variable (a la Tseytin transformation)
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# On path s, there is a difference on place i
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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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def bvar(s):
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return var(('base', s))
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# maakt de constraint dat precies een van de literals waar moet zijn
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def unique(lits):
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# deze werken goed: pairwise, seqcounter, bitwise, mtotalizer, kmtotalizer
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# anderen geven groter aantal oplossingen
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# alles behalve pairwise introduceert meer variabelen
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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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measure_time('Setup solver')
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# Construction
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# Guessing the word:
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# variable x_('in', i, a) says: on place i there is an input a
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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 (this allows for an optimisation later)
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unique([bvar(base) for base in bases])
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for s in tqdm(states, desc="simple stuff"):
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for i in range(length):
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# variable x_('out', s, i, a) says: on place i there is an output o of the path s
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unique([ovar(s, i, o) for o in outputs])
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if i == 0:
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# The paths start in the corresponding state
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# This could be used to reduce some variables, but I'm lazy now
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solver.add_clause([svar(s, 0, s)])
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else:
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# variable x_('state', s, i, t) denotes the path through the automaton starting in s
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unique([svar(s, i, t) for t in states])
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# The path is consistent with the delta function
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# The outputs correspond to the output along the path
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# I have merged these loops, it was slightly faster
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for s in tqdm(states, desc="paths & outputs"):
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for i in range(length):
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for t in states:
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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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# We couple i with i+1, and so skip the last iteration
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if i < length-1:
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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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next_t = delta[(t, a)]
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solver.add_clause([-sv, -av, svar(s, i+1, next_t)])
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# x_('s', state, i, t) /\ x_('in', i, a) => x_('o', i, labda(t, a))
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# == -x_('s', state, i, t) \/ -x_('in', i, a) \/ x_('o', i, labda(t, a))
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output = labda[(t, a)]
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solver.add_clause([-sv, -av, ovar(s, i, output)])
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# If(f) 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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for s in tqdm(states, desc="diff1"):
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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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for o in outputs:
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# x_('o', state, i, o) /\ -x_('o', s, i, o) => x_('e', s, i)
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# == -x_('o', state, i, o) \/ x_('o', s, i, o) \/ -x_('e', s, i)
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solver.add_clause([-bv, -ovar(base, i, o), ovar(s, i, o), evar(s, i)])
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# We also need the other direction, we can do this:
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# x_('e', s, i) /\ x_('o', state, i, o) => -x_('o', s, i, o)
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# == -x_('e', s, i) \/ -x_('o', state, i, o) \/ -x_('o', s, i, o)
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solver.add_clause([-bv, -evar(s, i), -ovar(base, i, o), -ovar(s, i, o)])
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# Now we have to say that the other state have some different output on their path
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for s in tqdm(states, desc="diff2"):
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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
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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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measure_time('Constructed CNF with', solver.nof_clauses(), 'clauses and', solver.nof_vars(), 'variables')
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# Solving
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for s in bases:
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print('*** UIO for state', s)
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b = solver.solve(assumptions=[bvar(s)])
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measure_time('Solver finished')
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if b:
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m = solver.get_model()
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truth = set()
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for l in m:
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if l > 0:
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truth.add(l)
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print('! word')
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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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print(a, end=' ')
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print('')
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if verbose:
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print('! paths')
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for s in states:
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print(s, '=>', end=' ')
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for i in range(length):
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for t in states:
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if svar(s, i, t) in truth:
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print(t, end=' ')
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print('')
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print('! outputs')
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for s in states:
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print(s, '=>', end=' ')
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for i in range(length):
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for o in outputs:
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if ovar(s, i, o) in truth:
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print(o, end=' ')
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print('')
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print('! differences')
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for s in states:
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if s == base:
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continue
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print(s, '=>', end=' ')
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for i in range(length):
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if evar(s, i) in truth:
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print('x', end='')
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else:
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print('.', end='')
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print('')
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else:
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print('! no word')
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core = solver.get_core()
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print(core)
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