COMBINATORS I is
http://www.escribe.com/science/theory/m5913.html
COMBINATORS II is
http://www.escribe.com/science/theory/m5942.html
COMBINATORS III is
http://www.escribe.com/science/theory/m5946.html

Resume:

Kxy = x
Sxyz = xz(yz)

You are supposed to remember also that
abc is an abbreviation of ((ab)c), and a(bc) is an abbreviation for (a(bc)), so
Kxy is put for ((K x) y), and Sxyz is put for (((S x) y) z), and xz(yz) is put for
((x z)(y z)). We just don't write the left parentheses).


We have seen:

Ix = x,   a solution for I is I = SKK
Mx = xx,  a solution for M is M = SII = S(SKK)(SKK)
Bxyz = x(yz), a solution for B is B = S(KS)K

I recall the last exercises:
Find combinators W, L, T, C and INFINITY such that

Wxy = xyy
Lxy = x(yy)
Txy = yx
Cxyz = xzy
INFINITY gives INFINITY (by the dynamical rules for K and S).

SOLUTIONS:

1) Wxy = xyy
           = (xy)y      because xyy abbreviates xyy
           = (xy)(Iy)    because y = Iy
           = SxIy        dynamics of S
           = (SxI)y        abbrev.
           = Sx(KIx)y    because I = KIx
           = (Sx(KIx))y  abbrev.
           = (SS(KI)x)y    dynamics of S
           = SS(KI)xy    abbrev.

Thus W = SS(KI) = SS(K(SKK))

2) Lxy = x(yy)
          = x(My)   because My = yy (Mx = xx).
          = BxMy  dynamics of B
          = Bx(KMx)y  'cause KMx = M
          = SB(KM)xy  dyn. of S, on ((Bx)((KM)x)

Thus L = SB(KM) = S(S(KS)K)(KS(SKK)(SKK))

Od course SB(KM) is a better presentation, giving that we know already B and M.
But I give the (abbreviated) combinator (in term of S and K) to be sure you see it is
indeed a combinator (a combination of S and K).


3) Txy = yx
= y(Kxy) 'cause x = Kxy
= (Iy)(Kxy) 'cause y = (Iy)
= SI(Kx)y dyn. of S (which explain the transformation just above btw)
= B(SI)Kxy dyn. of B


Thus T = B(SI)K = S(KS)K(S(SKK))K

4) Cxyz = xzy
            = xz(Kyz)
            = Sx(Ky)z
            = B(Sx)Kyz
            = BBSxKyz
            = BBSx(KKx)yz
            = (BBS)x(KKx)yz
            = S(BBS)(KK)xyz

Thus C = S(BBS)(KK)
            = S((S(KS)K)(S(KS)K)S)(KK)    cf B = S(KS)K

5) INFINITY   Well this one is easy!

Mx = xx, thus MM gives MM which gives MM, etc.

Thus INFINITY = MM = SII(SII) = S(SKK)(SKK)(S(SKK)(SKK))

It is a simple example of a perpetually unstable molecule.

All right so far? To find those programs, the heuristic has been
to use K and I to change some combination of variables (like xzy) into
a form such that a known dynamic can be used (in general the one of
S, or of B). B, for example, is useful to shift parentheses on the left.
Such exercises need a bit of training and a taste for programming or
puzzles. I hope that you are able to verify (at least) a solution.
Let us verify that L admits another solution based on the use of C:

L = CBM  (by which I mean CBM behaves like it is aked for L to behave,
i.e. Lxy = x(yy)

Indeed CBMxy = BxMy (because Cabc=acb, or Cxyz = xzy)
                       = x(My)
                       = x(yy).

Such a sequence of application of the dynamical rules will be what I mean
by the term "computation". This will be justify when we will see that the
combinators are Turing-equivalent. Any program can be matched by a
combinator.

Note also this. We see that SB(KM) and CBM are two different programs
computing L. But SB(KM), that is  S(S(KS)K)(KS(SKK)(SKK)), and
CBM, that is S((S(KS)K)(S(KS)K)S)(KK)(S(KS)K)(S(SKK)(SKK)) are
different combinators doing the same things. We will use the (extensionality)
rule which identifies the combinators which have the same behavior. When two
combinators are syntactically identical, we will use the expression "strictly
identical" (written ==). A (more simple) example is:

 I = SKK = SKS = SK(KK) = ....

They all give an identity combinator. But none can be obtained from this
other by a reduction, i.e. an application of the dynamical rule from left to right (like
in a computation).


Finally, because the combinators we have met so far will play some
persisting role in the sequel, they deserve names.
I give you the (birdy) terminology of Smullyan:

K is the kestrel. It is "the" eliminator. K(KK)(SSS) eliminates (SSS) for example.
S is the starling. S does many things at once: it duplicates, composes, and permutates
its arguments: Sxyz = xz(yz). Look carefully.
T is the trush. It is a crude permutator. It permutates its arguments: Txy = yx.
M is the mocking bird itself! It is just a crude duplicator: Mx = xx.
C is the cardinal. C is called the elementary or regular permutator.
It is less crude than the trush. Much more easy to use, like in general the regular
combinators, which by definition are those combinators which leaves unchanged they
first argument. All combinators seen so far are regular except the trush.
W is the warbler. It is the elementary duplicator (less crude thanb the mocking bird!).
B is the elementary compositor. Bxyz = x(yz). Suppose that f = log, and g = sin,
then Bfgx = log(sin x), so that Bfg gives f g. That is the composition of the functions
f and g. I forget to say that B is called the blue bird.
L is the lark: it is an hybrid of a warbler or a mocking bird and a compositor.


Exercise:
We have seen how to program the blue bird B, the cardinal C and the Warbler W
with the kestrel K and the starling S. Could you define the starling S from B, W and C?
I give you the first line: Sxyz = xz(yz)
= Cx(yz)z
= ...


Bruno

http://iridia.ulb.ac.be/~marchal/




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