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Similar one I enjoyed in the Naive Bayes article from yesterday https://blog.floydhub.com/naive-bayes-for-machine-learning/ Multiple Choice: If you choose an answer to this question at random, what is the chance you will be correct? A) 25% B) 50% C) 60% D) 25%
Here is a single python statement that solves the problem: print([ q for q in itertools.product((True, False), repeat=6) if q == ( all(q[1:]), # 1. All of the below. not any(q[2:]), # 2. None of the below. all(q[:2]), # 3. All of the above. any(q[:3]), # 4. One of the above. not any(q[:4]), # 5. None of the above. not any(q[:5]), # 6. None of the above. ) ]) https://gist.github.com/lovasoa/f2b4ed93e755bf4172583d28f20…
import itertools
def below(n):
return slice(n+1, None)
def above(n):
return slice(None, n)
print([
q for q in itertools.product((True, False), repeat=6)
if q == tuple(
f(i) for i, f in enumerate([
lambda n: all(q[below(n)]), # 1. All of the below.
lambda n: not any(q[below(n)]), # 2. None of the below.
lambda n: all(q[above(n)]), # 3. All of the above.
lambda n: any(q[above(n)]), # 4. One of the above.
lambda n: not any(q[above(n)]), # 5. None of the above.
lambda n: not any(q[above(n)]), # 6. None of the above.
])
)
])Do you agree it can be rephrased as "Which answers in this list are the correct answers to this question?", so as to make it more difficult, while keeping the same answer?
Here is a single python statement that solves the problem: print([ q for q in itertools.product((True, False), repeat=6) if q == ( all(q[1:]), # 1. All of the below. not any(q[2:]), # 2. None of the below. all(q[:2]), # 3. All of the above. any(q[:3]), # 4. One of the above. not any(q[:4]), # 5. None of the above. not any(q[:5]), # 6. None of the above. ) ]) https://gist.github.com/lovasoa/f2b4ed93e755bf4172583d28f20…
Run it online here: https://onlinegdb.com/rkZhWVniH
itertools.product((True, False), repeat=6)
The site made it easy for me to spend 30 seconds figuring it out. My interpretation is that that line generates what I'll call a truth table (not sure if that's what it formally is) then brute force searches for rows that satisfy all of the listed constraints.
I'd also hazard that the fact that there's only one answer is engineered into the question and not a natural product of this search.
Similar one I enjoyed in the Naive Bayes article from yesterday https://blog.floydhub.com/naive-bayes-for-machine-learning/ Multiple Choice: If you choose an answer to this question at random, what is the chance you will be correct? A) 25% B) 50% C) 60% D) 25%
What's the correct answer?
Here is a single python statement that solves the problem: print([ q for q in itertools.product((True, False), repeat=6) if q == ( all(q[1:]), # 1. All of the below. not any(q[2:]), # 2. None of the below. all(q[:2]), # 3. All of the above. any(q[:3]), # 4. One of the above. not any(q[:4]), # 5. None of the above. not any(q[:5]), # 6. None of the above. ) ]) https://gist.github.com/lovasoa/f2b4ed93e755bf4172583d28f20…
import z3
answers = [z3.Bool(f"answer{i}") for i in range(1,7)]
implications = [
z3.And(answers[1:]), # All of the below
z3.Not(z3.Or(answers[2:])), # None of the below
z3.And(answers[:2]), # All of the above
z3.Or(answers[:3]), # Any of the above
z3.Not(z3.Or(answers[:4])), # None of the above
z3.Not(z3.Or(answers[:5]))] # None of the above
constraints = [z3.Implies(ans, impl) for ans,impl in zip(answers, implications)]
z3.solve(constraints)By definition:
#1: xTTTTT
#2: xxFFFF
#3: TTxxxx
#4: tttxxx
#5: FFFFxx
#6: FFFFFx
Let's assume 6 is T so we have xxxxxT. In that case 5 need to be F. But if 5 is F, then any from [1,2,3,4] need to be T. But if any from [1,2,3,4] is T then 6 cannot be T. Thus 6 cannot be T, so it is F if correct answer exists.
Now we have: xxxxxF
Next, we try to set 5 to T: xxxxTF. 1 cannot be T, 2 cannot be T, since their last elements are different (TT!=TF, FF!=TF), 3 cannot be T because it requires 1 and 2 be both T , 4 cannot be T because (1 or 2 or 3) are F.
So we have: FFFFTF
Thus only 5 is True.
SELF REFERENTIAL QUIZ http://faculty.uml.edu/jpropp/srat-Q.txt
Soltion(s): https://faculty.uml.edu/jpropp/srat.html
Earlier quoted context omitted.
Using the clpb library [0]: solution([A1,A2,A3,A4,A5,A6]) :- sat(A1 =:= A2*A3*A4*A5*A6), sat(A2 =:= ~(A3+A4+A5+A6)), sat(A3 =:= A1*A2), sat(A4 =:= A1+A2+A3), sat(A5 =:= ~(A1+A2+A3+A4)), sat(A6 =:= ~(A1+A2+A3+A4+A5)). [0] https://www.metalevel.at/prolog/puzzles
Alternatively using z3 [0]: (declare-const a1 Bool) (declare-const a2 Bool) (declare-const a3 Bool) (declare-const a4 Bool) (declare-const a5 Bool) (declare-const a6 Bool) (assert (= a1 (and a2 (and a3 (and a4 (and a5 a6)))))) (assert (= a2 (not (or a3 (or a4 (or a5 a6)))))) (assert (= a3 (and a1 a2))) (assert (= a4 (or a1 (or a2 a3)))) (assert (= a5 (not (or a1 (or a2 (or a3 a4)))))) (assert (= a6 (not (or a1 (or a2…