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OTTER
(Organized Techniques
for Theorem-proving and Effective Research)
Heuristics computing
2017/12/18 9:25 – 11:00
Keio University, SFC
http://www.mcs.anl.gov/research/projects/AR/otter/
Example2: Truth teller, Liar and Spy
(Normal)
Example 2: Truth tellers, liars and spy
On the fabled Island of Knights and Knaves, we meet three people, A, B, and C, one of whom
is a knight, one a knave, and one a spy. The knight always tells the truth, the knave always
lies, and the spy can either lie or tell the truth.
A says: "C is a knave."
B says: "A is a knight."
C says: "I am the spy."
Who is the knight, who the knave,
and who the spy?
set(hyper_res).
list(usable).
-P(T(x)) | -P(Says(x,y)) | P(y).
-P(L(x)) | -P(Says(x,y)) | -P(y).
P(T(x)) | P(L(x)) | P(N(x)).
-P(T(x)) | -P(L(x)).
-P(T(x)) | -P(N(x)).
-P(L(x)) | -P(N(x)).
-P(T(x)) | P(L(x)) | P(N(x)).
-P(L(x)) | P(T(x)) | P(N(x)).
end_of_list.
list(sos).
P(Says(A,L(C))).
P(Says(B,T(A))).
P(Says(C,N(C))).
end_of_list.
Always true (恒
真)
Always false(恒
偽)
Satisfiable(証明可能) Unsatisfiable(証明不可
能)
私は正直者だと言っているスパイ
私は嘘つきだと言っているスパイ
https://github.com/RuoAndo/otter-book/tree/master/takefuji-lab/2017
Proof and satisfiability
A
B C
a
A ∩ B → empty set
B∪¬A → not Ф
b
c
fermi particle model bose particle model
X
Y
Z
X
Y
Z
a
ab
c
c
a
X(a) | Y(b)
either, exclusive
electron, proton
X(a) or Y(b)
inclusive
photon
Always true Always false
satisfiable
unsatisfiable
unsatisfiable
satisfiable
Generalization: consistency is non-deterministic
(∃y) (x) ~ Dem (x, y) ⊃ (x) ~ Dem(x, sub(n, A, n))
Dem(x,y) : yの言っていることをxが確かめる。
Island is consistent if there are only truth tellers and liars.
島に正直者と嘘つきしかいない場合、確かめる事のできない言明がある。
その言明とは、「私は嘘好きである」というものである。
(∃y) (x) ~ Dem (x, y)
Dem(x, sub(n,A,n)): 自分について言っていること(sub(n,A,n)をxが確かめることができる。
If the island is consistent, no one in the island know if there are some spy.
「私は嘘つきである」という言明を確かめることができない場合、誰も島にスパイがいることを発見できない。
しかし、OTTERは発見できる。
set(para_into).
list(usable).
EQUAL(l(hole,l(n(x),y)),l(n(x),l(hole,y))).
EQUAL(l(hole,l(x,l(y,l(z,l(u,l(n(w),v)))))),l(n(w), l(x,l(y,l(z,l(u,l(hole,v))))))).
-STATE(l(n(1),l(n(2),l(n(3),l(n(4),l(end,l(n(5), l(n(6),l(n(7),l(n(8),l(end,l(n(9),l(n(10),l(n(11),
l(n(12),l(end,l(n(13),l(n(14),l(n(15), l(hole,end)))))))))))))))))))).
end_of_list.
list(sos).
STATE(l(n(1),l(n(6),l(n(2),l(n(4),l(end,l(n(5), l(hole,l(n(3),l(n(8),l(end,l(n(9),l(n(10),l(n(7),
l(n(11),l(end,l(n(13),l(n(14),l(n(15), l(n(12),end)))))))))))))))))))).
end_of_list.
1 6 2 4
5 3 8
9 10 7 11
13 14 15 12
Prg4-8.in : SCORE 437
1 6 2 4
5 3 8
9 10 7 11
13 14 15 12
Prg4-8-7.in : SCORE 891
11 2 6 9
4 14 3 5
1 12 7 15
13 10 6
Prg4-8-2.in : SCORE ????
Example 3: 15 puzzle
Paramodulation
1 |set(para_into).
2 |assign(stats_level,1).
3 |
4 |list(usable).
5 | father(ken)=jim.
6 | mother(jim)=fay.
7 |end_of_list.
8 |
9 |list(sos).
10 | Grandmother(mother(father(x)),x).
11 |end_of_list.
Grandmother
Fay
Jim
Ken
Jim
father
mother
SoS List:
list(list)
Usable List:
list(usable)
• Condition 1: For all a and x, b exists
where Parity(b, x) == Parity(a, x)
• Condition 2: For all b and x, a exists
where Parity(a,x) != Parity(b,x)
Result: For all x,
H exists Parity (x[i], H) != Parity (x[j], H)
Curious property of parity
Example 4: Two-inverter puzzle
A
B
C
-A
-B
-C
The 2-NOTs problem: Synthesize a black box which computes NOT-
A, NOT-B, and NOT-C from A, B, and C, using an arbitrary number
of ANDs and ORs, but only 2 NOTs.
NOT
AND
AND
AND
AND
AND
AND
AND
AND
AND
AND
OR
OR
OR
OR
OR
OR
OR
OR
NOT
Example 4: Two-inverter puzzle
Primitive recursive function
• The basic primitive recursive functions are given by
these axioms:
• Constant function: The 0-ary constant function 0 is
primitive recursive. => AND
• Successor function: The 1-ary successor function S,
which returns the successor of its argument (see Peano
postulates), is primitive recursive. That is, S(k) = k + 1. =>
OR
• Projection function: For every n≥1 and each i with 1≤i≤n,
the n-ary projection function Pi
n, which returns its i-th
argument, is primitive recursive. => NOT*NOT
Backup slides
Usable Set+
+
++
+
-
-
lightest
+
-
EMPTY
Usable Set
SoS Set
SoS Set
--
prg2-4.out.lined
-----> EMPTY CLAUSE at 0.00 sec
--> 8 [hyper,7,4,6] $F.
Size of Usable set (nonliear)
Main Loop
A
B
C
C ⊃ B
B ⊃ A
There exists
A[i] where
B∪¬A → Ф
(empty)
A’
Clause Sets and representation
SoS
(Set of Support)
Demodulators
Usable
Passive
Term
Rewriting
Resolution
and paramodulation
conflict
a b
a b c
a b
c
IF A THEN B : -A | B
IF A & B THEN B : -A | -B | C
IF A THEN B or C : -A | B | C
Strategies
Restriction Strategies
Set of Support
Weighting
Direction Strategies
Ratio
Resonance
Look-ahead Strategies
Hot List
Redundancy control
Subsumption
DemodulationParamodulation
Dynamic Hot List
Subtautology
Resonance restriction
Level SaturationRecursive Tail
Hints
First-in Last-out
Proof checking filter
http://www.mcs.anl.gov/research/projects/AR/strategies.html
Set of Support

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Otter 2017-12-18-01-ss

  • 1. OTTER (Organized Techniques for Theorem-proving and Effective Research) Heuristics computing 2017/12/18 9:25 – 11:00 Keio University, SFC http://www.mcs.anl.gov/research/projects/AR/otter/
  • 2. Example2: Truth teller, Liar and Spy (Normal)
  • 3. Example 2: Truth tellers, liars and spy On the fabled Island of Knights and Knaves, we meet three people, A, B, and C, one of whom is a knight, one a knave, and one a spy. The knight always tells the truth, the knave always lies, and the spy can either lie or tell the truth. A says: "C is a knave." B says: "A is a knight." C says: "I am the spy." Who is the knight, who the knave, and who the spy? set(hyper_res). list(usable). -P(T(x)) | -P(Says(x,y)) | P(y). -P(L(x)) | -P(Says(x,y)) | -P(y). P(T(x)) | P(L(x)) | P(N(x)). -P(T(x)) | -P(L(x)). -P(T(x)) | -P(N(x)). -P(L(x)) | -P(N(x)). -P(T(x)) | P(L(x)) | P(N(x)). -P(L(x)) | P(T(x)) | P(N(x)). end_of_list. list(sos). P(Says(A,L(C))). P(Says(B,T(A))). P(Says(C,N(C))). end_of_list. Always true (恒 真) Always false(恒 偽) Satisfiable(証明可能) Unsatisfiable(証明不可 能) 私は正直者だと言っているスパイ 私は嘘つきだと言っているスパイ https://github.com/RuoAndo/otter-book/tree/master/takefuji-lab/2017
  • 4. Proof and satisfiability A B C a A ∩ B → empty set B∪¬A → not Ф b c fermi particle model bose particle model X Y Z X Y Z a ab c c a X(a) | Y(b) either, exclusive electron, proton X(a) or Y(b) inclusive photon Always true Always false satisfiable unsatisfiable unsatisfiable satisfiable
  • 5. Generalization: consistency is non-deterministic (∃y) (x) ~ Dem (x, y) ⊃ (x) ~ Dem(x, sub(n, A, n)) Dem(x,y) : yの言っていることをxが確かめる。 Island is consistent if there are only truth tellers and liars. 島に正直者と嘘つきしかいない場合、確かめる事のできない言明がある。 その言明とは、「私は嘘好きである」というものである。 (∃y) (x) ~ Dem (x, y) Dem(x, sub(n,A,n)): 自分について言っていること(sub(n,A,n)をxが確かめることができる。 If the island is consistent, no one in the island know if there are some spy. 「私は嘘つきである」という言明を確かめることができない場合、誰も島にスパイがいることを発見できない。 しかし、OTTERは発見できる。
  • 6. set(para_into). list(usable). EQUAL(l(hole,l(n(x),y)),l(n(x),l(hole,y))). EQUAL(l(hole,l(x,l(y,l(z,l(u,l(n(w),v)))))),l(n(w), l(x,l(y,l(z,l(u,l(hole,v))))))). -STATE(l(n(1),l(n(2),l(n(3),l(n(4),l(end,l(n(5), l(n(6),l(n(7),l(n(8),l(end,l(n(9),l(n(10),l(n(11), l(n(12),l(end,l(n(13),l(n(14),l(n(15), l(hole,end)))))))))))))))))))). end_of_list. list(sos). STATE(l(n(1),l(n(6),l(n(2),l(n(4),l(end,l(n(5), l(hole,l(n(3),l(n(8),l(end,l(n(9),l(n(10),l(n(7), l(n(11),l(end,l(n(13),l(n(14),l(n(15), l(n(12),end)))))))))))))))))))). end_of_list. 1 6 2 4 5 3 8 9 10 7 11 13 14 15 12 Prg4-8.in : SCORE 437 1 6 2 4 5 3 8 9 10 7 11 13 14 15 12 Prg4-8-7.in : SCORE 891 11 2 6 9 4 14 3 5 1 12 7 15 13 10 6 Prg4-8-2.in : SCORE ???? Example 3: 15 puzzle
  • 7. Paramodulation 1 |set(para_into). 2 |assign(stats_level,1). 3 | 4 |list(usable). 5 | father(ken)=jim. 6 | mother(jim)=fay. 7 |end_of_list. 8 | 9 |list(sos). 10 | Grandmother(mother(father(x)),x). 11 |end_of_list. Grandmother Fay Jim Ken Jim father mother SoS List: list(list) Usable List: list(usable)
  • 8. • Condition 1: For all a and x, b exists where Parity(b, x) == Parity(a, x) • Condition 2: For all b and x, a exists where Parity(a,x) != Parity(b,x) Result: For all x, H exists Parity (x[i], H) != Parity (x[j], H) Curious property of parity
  • 9. Example 4: Two-inverter puzzle A B C -A -B -C The 2-NOTs problem: Synthesize a black box which computes NOT- A, NOT-B, and NOT-C from A, B, and C, using an arbitrary number of ANDs and ORs, but only 2 NOTs. NOT AND AND AND AND AND AND AND AND AND AND OR OR OR OR OR OR OR OR NOT
  • 11. Primitive recursive function • The basic primitive recursive functions are given by these axioms: • Constant function: The 0-ary constant function 0 is primitive recursive. => AND • Successor function: The 1-ary successor function S, which returns the successor of its argument (see Peano postulates), is primitive recursive. That is, S(k) = k + 1. => OR • Projection function: For every n≥1 and each i with 1≤i≤n, the n-ary projection function Pi n, which returns its i-th argument, is primitive recursive. => NOT*NOT
  • 13. Usable Set+ + ++ + - - lightest + - EMPTY Usable Set SoS Set SoS Set -- prg2-4.out.lined -----> EMPTY CLAUSE at 0.00 sec --> 8 [hyper,7,4,6] $F. Size of Usable set (nonliear) Main Loop A B C C ⊃ B B ⊃ A There exists A[i] where B∪¬A → Ф (empty) A’
  • 14. Clause Sets and representation SoS (Set of Support) Demodulators Usable Passive Term Rewriting Resolution and paramodulation conflict a b a b c a b c IF A THEN B : -A | B IF A & B THEN B : -A | -B | C IF A THEN B or C : -A | B | C
  • 15. Strategies Restriction Strategies Set of Support Weighting Direction Strategies Ratio Resonance Look-ahead Strategies Hot List Redundancy control Subsumption DemodulationParamodulation Dynamic Hot List Subtautology Resonance restriction Level SaturationRecursive Tail Hints First-in Last-out Proof checking filter http://www.mcs.anl.gov/research/projects/AR/strategies.html Set of Support