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Transcript
```5 COMPUTABLE FUNCTIONS
Computable functions are deﬁned on the set of nonnegative integers. Some texts use N to denote this set. We
use N to denote the set of positive integers, so we will use the notation
N0 = {0, 1, 2, 3, . . .}
Throughout this section, the terms number, integer, and nonnegative integer are used synonymously. The preceding
section described the way a Turing machine M manipulates and recognizes character data. Here we show how
M manipulates numerical data. First, however, we need to be able to represent our numbers by our tape set A.
We will write 1 for the tape symbol a1 and 1n for 111 . . . 1, where 1 occurs n times.
Deﬁnition 12: Each number n will be represented by the tape expression )n* where )n* = 1n+1 . Thus:
)4* = 11111 = 15 , )0* = 1, )2* = 111 = 13 .
Deﬁnition 13: Let E be an expression. Then [E] will denote the number of times 1 occurs in E. Thus
[11Bs 2 a3 111Ba 4 ] = 5,
[a4 s2 Ba 2 ] = 0, [)n*] = n + 1.
Deﬁnition 14: A function f : N0 → N0 is computable if there exists a Turing machine M such that, for every
integer n, M halts on )n* and
f (n) = [term(α()n*)]
We then say that M computes f .
That is, given a function f and an integer n, we input )n* and apply M . If M always halts on )n* and the
number of 1’s in the ﬁnal picture is equal to f (n), then f is a computable function and we say that M computes f .
EXAMPLE 4 The function f (n) = n + 3 is computable. The input is W = 1n+1 . Thus we need only add
two 1’s to the input. A Turing machine M which computes f follows:
M = {q1 , q2 , q3 } = {s0 1s0 L, s0 B 1s1 L, s1 B 1sH L}
Observe that:
(1) q1 moves the machine M to the left.
(2) q2 writes 1 in the blank square B , and moves M to the left.
(3) q3 writes 1 in the blank square B , and halts M .
Accordingly, for any positive integer n,
s0 1n+1 → s0 B 1n+1 → s1 B 1n+2 → sH B 1n+3
Thus M computes f (n) = n + 3. It is clear that, for any positive integer k, the function f (n) = n + k is
computable.
The following theorem applies.
Theorem 4: Suppose f : N0 → N0 and g : N0 → N0 are computable. Then the composition function
h = g ◦ f is computable.
We indicate the proof of this theorem here. Suppose Mf and Mg are the Turing machines which compute
f and g, respectively. Given the input )n*, we apply Mf to )n* to ﬁnally obtain an expression E with [E] = f (n).
We then arrange that E = s0 1f (n) . We next add 1 to E to obtain E = s0 11f (n) and apply Mg to E . This yields
E where [E ] = g(f (n)) = (g ◦ f )(n), as required.
Functions of Several Variables
This subsection deﬁnes a computable function f (n1 , n2 , . . . , nk ) of k variables. First we need to represent
the list m = (n1 , n2 , . . . , nk ) in our alphabet A.
Deﬁnition 15: Each list m = (n1 , n2 , . . . , nk ) of k integers is represented by the tape expression
)m* = )n1 *B )n2 *B · · · B )nk *
For example, )(2, 0, 4)* = 111B 1B 11111 = 13 B 11 B 15 .
Deﬁnition 16: A function f (n1 , n2 , . . . , nk ) of k variables is computable if there is a Turing machine M such
that, for every list m = (n1 , n2 , . . . , nk ), M halts on )m* and
f (m) = [term(α()m*))]
We then say that M computes f .
The deﬁnition is analogous to Deﬁnition 14 for one variable.
EXAMPLE 5 The addition function f (m, n) = m + n is computable. The input is W = 1m+1 B 1n+1 .
Thus we need only erase two of the 1s. A Turing machine M which computes f follows:
M = {q1 , q2 , q3 , q4 } = {s0 1Bs 1 R, s1 1Bs H R, s1 BBs 2 R, s2 1Bs H R}
Observe that:
(1) q1 erases the ﬁrst 1 and moves M to the right.
(2) If m = 0, then q2 erases the second 1 and halts M .
(3) If m = 0, q3 moves M to the right past the blank space B .
(4) q4 erases the 1 and halts M .
Accordingly, if m = 0 we have:
s0 1m+1 B 1n+1 → s1 1m B 1n+1 → sH 1m−1 B 1n+1
but if m = 0 and m + n = n, we have
s0 1B 1n+1 → s1 B 1n+1 → s2 1n+1 → s H 1n
Thus M computes f (m, n) = m + n.
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