IN210 − lecture 2
Review
How to solve the information-processing
problems efficiently.
;
: abstraction, formalisation
Problems
solutions
efficiency
;
;
;
I/O pairs,
functions,
“interesting
problems”
algorithms
resources,
upper/lower
bounds
;
;
;
formal
languages
Turing
machines
complexity
classes
high-level information
low-level information
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IN210 − lecture 2
Theorem 1 There are more problems than
solutions.
Corollarly 1 There are (many) problems that
cannot be solved by algorithms.
The “egg of all problems” which we want to
divide into (complexity) classes:
Unsolvable
Intractable
P
Today
• Turing machines as an algorithm model
• Turing’s theorem which gives us a
(provably) unsolvable problem.
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Algorithm
11
397
46
443
397 + 46 =
input
443
output
computation
rules
Turing machine – intuitive description
b
... b 0 1 0 b b ...
(input/output)
tape
read/write head
"processor" or
s finite state control
...
q2
q1
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steps of
computation
...
states
"loaded
program"
or rules
δ (s,o) =( q1, b, R)
δ (q1 ,1)=(q2, b , R)
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Turing machine – formal description
A Turing machine (TM) is M = (Σ, Γ, Q, δ)
where
Σ , the input alphabet is a finitive set of input
symbols
Γ , the tape alphabet is a finite set of tape
symbols which includes Σ, a special blank
symbol b ∈ Γ \ Σ, and possibly other
symbols
Q is a finite set of states which includes a
start state s and a halt state h
δ , the transition function is
δ : (Q \ {h}) × Γ → Q × Γ × {L, R}
NB! “Almost” every textbook has its own
unique definition of a Turing machine, but
they are all equivalent in a certain sense.
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Computation – formal definition
A configuration of a Turing machine M is a
triple C = (q, wl , wr ) where q ∈ Q is a state and
wl and wr are strings over the tape alphabet.
We say that a configuration (q, wl , wr ) yields in
one step configuration (q 0, wl0, wr0 ) and write
(q, wl , wr )` (q 0, wl0, wr0 ) if (and only if) for some
M
a, b, c ∈ Γ and x, y ∈ Γ∗ either
wl = xa
wl0 = x
wl
... b b
and
δ(q, b) = (q 0, c, L)
wr = by
wr0 = acy
x
wr
a bc
w l’
y
b b ...
w r’
or
wl = x
wl0 = xb
wl
... b b
x
w l’
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and
δ(q, a) = (q 0, b, R)
wr = acy
wr0 = cy
wr
ab c
y
b b ...
w r’
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Note:
wl is the written portion of the tape to the left
of the read/write head.
wr is the written portion of the tape to the
right of the read/write head, inluding the
square the head is currently scanning.
We say that a configuration C = (q, wl , qr )
yields configuration C 0 = (q 0, wl0, qr0 ) and write
∗
C ` C 0 if (and only if) there is a sequence of
M
configurations C = C1, C2, · · · , Cn = C 0 of M
such that
Ci ` Ci+1
M
for i = 1, · · · , n − 1
We say that Turing machine M computes
function f if (and only if) for all w1, w1 ∈ Σ∗
∗
(s, , w1) ` (h, , w2) ⇔ f (w1) = w2
M
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We say that Turing machine M decides
language L if (and only if) M computes the
function
f : Σ∗ → {Y, N } and for each x ∈ L : f (x) = Y
for each x ∈
/ L : f (x) = N
Language L is (Turing) decidable if (and only
if) there is a Turing machine which decides it.
We say that Turing machine M accepts
language L if M halts if and only if its input is
an string in L.
Language L is (Turing) acceptable if (and
only if) there is a Turing machine which
accepts it.
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Example
A Turing machine M which decides
L = {010}.
... b 0 1 0 b b ...
s
h
qe
q2
q3
q1
M = (Σ, Γ, Q, δ)
Γ = {0, 1, b, Y, N }
Σ = {0, 1}
Q = {s, h, q1, q2, q3, qe}
δ:
s
q1
q2
q3
qe
0
1
b
(q1, b, R)
(qe, b, R)
(q3, b, R)
(qe, b, R)
(qe, b, R)
(qe, b, R)
(q2, b, R)
(qe, b, R)
(qe, b, R)
(qe, b, R)
(h, N, −)
(h, N, −)
(h, N, −)
(h, Y, −)
(h, N, −)
(’−’ means “don’t move the read/write head”)
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Church’s thesis
’Turing machine’ ∼
= ’algorithm’
Turing machines can compute every function
that can be computed by some algorithm or
program or computer.
’Expressive power’ of PL’s
Turing complete programming languages.
’Universality’ of computer models
Neural networks are Turing complete (Mc
Cullok, Pitts).
Uncomputability
If a Turing machine cannot compute f , no
computer can!
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Uncomputability
What algorithmic can and cannot do.
Strategy
1. Show that H ALTING (the Halting problem)
is unsolvable
LH
Unsolvable
Solvable
R
2. Use reductions 7−→ to show that other
problems are unsolvable
R1
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Step 1: H ALTING is unsolvable
Def. 1 (H ALTING)
LH = {(M, x)|M halts on input x}
Lemma 1 Every Turing decidable language is
Turing acceptable.
Proof (by reduction): Given a Turing
machine M that decides L we can construct a
Turing machine M 0 that accepts L as follows:
YES
input
halt
M
NO
M’
Lemma 2 The complement of every decidable
language is decidable.
(The complement of language L is LC , Σ∗ \ L.)
Proof: Given a Turing machine M that
decides L . . .
YES
YES
NO
NO
M
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M’
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