A Tutorial Introduction to the Lambda Calculus

Transcription

1 rxiv: v1 [cs.lo] 28 Mr 2015 A Tutoril Introduction to the Lmbd Clculus Rul Rojs Freie Universität Berlin Version 2.0, 2015 Abstrct This pper is concise nd pinless introduction to the λ-clculus. This formlism ws developed by Alonzo Church s tool for studying the mthemticl properties of effectively computble functions. The formlism becme populr nd hs provided strong theoreticl foundtion for the fmily of functionl progrmming lnguges. This tutoril shows how to perform rithmeticl nd logicl computtions using the λ-clculus nd how to define recursive functions, even though λ-clculus functions re unnmed nd thus cnnot refer explicitly to themselves. 1 Definition The λ-clculus cn be clled the smllest universl progrmming lnguge in the world. The λ-clculus consists of single trnsformtion rule (vrible substitution, lso clled β-conversion) nd single function definition Send corrections or suggestions to rojs@inf.fu-berlin.de 1

2 2 scheme. It ws introduced in the 1930s by Alonzo Church s wy of formlizing the concept of effective computbility. The λ-clculus is universl in the sense tht ny computble function cn be expressed nd evluted using this formlism. It is thus equivlent to Turing mchines. However, the λ-clculus emphsizes the use of symbolic trnsformtion rules nd does not cre bout the ctul mchine implementtion. It is n pproch more relted to softwre thn to hrdwre. The centrl concept in λ-clculus is tht of expression. A nme is n identifier which, for our purposes, cn be ny of the letters, b, c, etc. An expression cn be just nme or cn be function. Functions use the Greek letter λ to mrk the nme of the function s rguments. The body of the function specifies how the rguments re to be rerrnged. The identity function, for exmple, is represented by the string (λx.x). The frgment λx tell us tht the function s rgument is x, which is returned unchnged s x by the function. Functions cn be pplied to other functions. The function A, for exmple, pplied to the function B, would be written s AB. In this tutoril, cpitl letters re used to represent functions. In fct, nything of interest in λ- clculus is function. Even numbers or logicl vlues will be represented by functions tht cn ct on one nother in order to trnsform string of symbols into nother string. There re no types in λ-clculus: ny function cn ct on ny other. The progrmmer is responsible for keeping the computtions sensible. An expression is defined recursively s follows: < expression > := < nme > < function > < ppliction > < function > := λ < nme >. < expression > < ppliction > := < expression >< expression > An expression cn be surrounded by prenthesis for clrity, tht is, if E is n expression, (E) is the sme expression. Otherwise, the only keywords used in the lnguge re λ nd the dot. In order to void cluttering expressions with prenthesis, we dopt the convention tht function ppliction ssocites

3 3 from the left, tht is, the composite expression E 1 E 2 E 3... E n is evluted pplying the successive expressions s follows (... ( (E 1 E 2 )E 3 )... En ) As cn be seen from the definition of λ-expressions, well-formed exmple of function is the previously mentioned string, enclosed or not in prentheses: λx.x (λx.x) We use the equivlence symbol to indicte tht when A B, A is just synonym for B. As explined bove, the nme right fter the λ is the identifier of the rgument of this function. The expression fter the point (in this cse single x) is clled the body of the function s definition. Functions cn be pplied to expressions. A simple exmple of n ppliction is (λx.x)y This is the identity function pplied to the vrible y. Prenthesis help to void mbiguity. Function pplictions re evluted by substituting the vlue of the rgument x (in this cse the y being processed) in the body of the function definition. Fig. 1 shows how the vrible y is bsorbed by the function (red line), nd lso shows where it is used s replcement for x (green line). The result is reduction, represented by the right rrow, with the finl result y. Since we cnnot lwys hve pictures, s in Fig. 1, the nottion [y/x] is used to indicte tht ll occurrences of x re substituted by y in the function s body. We write, for exmple, (λx.x)y [y/x]x y. The nmes of the rguments in function definitions do not crry ny mening by themselves. They re just plce holders, tht is, they re used to indicte how to rerrnge the rguments of the function when it is evluted. Therefore ll the strings below represent the sme function: (λz.z) (λy.y) (λt.t) (λu.u) This kind of purely lphbeticl substitution is lso clled α-reduction.

4 4 (λx.x)y y (λ. )y y Figure 1: The sme reduction shown twice. The symbol for the function s rgument is just plce holder. 1.1 Free nd bound vribles If we only hd pictures of the plumbing of λ-expressions, we would not hve to cre bout the nmes of vribles. Since we re using letters s symbols, we hve to be creful if we repet them, s shown in this section. In λ-clculus ll nmes re locl to definitions (like in most progrmming lnguges). In the function λx.x we sy tht x is bound since its occurrence in the body of the definition is preceded by λx. A nme not preceded by λ is clled free vrible. In the expression (λx.x)(λy.yx) the x in the body of the first expression from the left is bound to the first λ. The y in the body of the second expression is bound to the second λ, nd the following x is free. Bound vribles re shown in bold fce. It is very importnt to notice tht this x in the second expression is totlly independent of the x in the first expression. This cn be more esily seen if we drw the plumbing of the function ppliction nd the consequent reduction, s shown in Fig. 2. In Fig. 2 we see how the symbolic expression (first row) cn be interpreted s kind of circuit, where the bound rgument is moved to new position inside the body of the function. The first function (the identity function) consumes the second one. The symbol x in the second function hs no connections with the rest of the expression, it is floting free inside the function definition.

5 5 (λx.x)(λy.yx) (λ. )(λ. x) (λ. x) Figure 2: In successive rows: The function ppliction, the plumbing of the symbolic expression, nd the resulting reduction. Formlly, we sy tht vrible <nme> is free in n expression if one of the following three cses holds: <nme> is free in <nme>. (Exmple: is free in ). <nme> is free in λ<nme 1 >. <exp> if the identifier <nme> <nme 1 > nd <nme> is free in <exp>. (Exmple: y is free in λx.y). <nme> is free in E 1 E 2 if <nme> is free in E 1 or if it is free in E 2. (Exmple: x is free in (λx.x)x). A vrible <nme> is bound if one of two cses holds: <nme> is bound in λ <nme 1 >. <exp> if the identifier <nme>=<nme 1 > or if <nme> is bound in <exp>. (Exmple: x is bound in (λy.(λx.x))). <nme> is bound in E 1 E 2 if <nme> is bound in E 1 or if it is bound in E 2. (Exmple: x is bound in (λx.x)x).

6 6 It should be emphsized tht the sme identifier cn occur free nd bound in the sme expression. In the expression (λx.xy)(λy.y) the first y is free in the prenthesized subexpression to the left, but it is bound in the subexpression to the right. Therefore, it occurs free s well s bound in the whole expression (the bound vribles re shown in bold fce). 1.2 Substitutions The more confusing prt of stndrd λ-clculus, when first pproching it,is the fct tht we do not give nmes to functions. Any time we wnt to pply function, we just write the complete function s definition nd then proceed to evlute it. To simplify the nottion, however, we will use cpitl letters, digits nd other symbols (sn serif) s synonyms for some functions. The identity function, for exmple, cn be denoted by the letter I, using it s shorthnd for (λx.x). The identity function pplied to itself is the ppliction II (λx.x)(λx.x). In this expression, the first x in the body of the first function in prenthesis is independent of the x in the body of the second function (remember tht the plumbing is locl). Just to emphsize the difference we cn in fct rewrite the bove expression s The identity function pplied to itself yields therefore tht is, the identity function gin. II (λx.x)(λz.z). II (λx.x)(λz.z) [(λz.z)/x]x λz.z I,

7 7 (λx.(λy.xy))y y bound in the subexpression (λx.(λy.xy))y dnger: free y should not be mixed with bound y s Figure 3: A free vrible should not be substituted in subexpression where it is bound, otherwise new plumbing, different to the originl, would be generted. When performing substitutions, we should be creful to void mixing up free occurrences of n identifier with bound ones. In the expression ( λx.(λy.xy) ) y the function on the left contins bound y, wheres the y on the right is free. An incorrect substitution would mix the two identifiers in the erroneous result (λy.yy). Simply by renming the bound y to t we obtin ( λx.(λt.xt) ) y (λt.yt) which is completely different result but nevertheless the correct one. Therefore, if the function λx. < exp > is pplied to E, we substitute ll free occurrences of x in < exp > with E. If the substitution would bring free vrible of E in n expression where this vrible occurs bound, we renme

8 8 the bound vrible before performing the substitution. For exmple, in the expression (λx.(λy.(x(λx.xy)))) y we ssocite the first x with y. In the body (λy.(x(λx.xy))) only the first x is free nd cn be substituted. Before substituting though, we hve to renme the vrible y to void mixing its bound with its free occurrence: [y/x] (λt.(x(λx.xt))) (λt(y(λx.xt))) In norml order reduction we reduce lwys the left most expression of series of pplictions first. We continue until no further reductions re possible. 2 Arithmetic A progrmming lnguge should be cpble of specifying rithmeticl clcultions. Numbers cn be represented in the λ-clculus strting from zero nd writing successor of zero, tht is suc(zero), to represent 1, suc(suc(zero)) to represent 2, nd so on. Since in λ-clculus we cn only define new functions, numbers will be defined s functions using the following pproch: zero cn be defined s λs.(λz.z) This is function of two rguments s nd z. We will bbrevite such expressions with more thn one rgument s λsz.z It is understood here tht s is the first rgument to be substituted during the evlution nd z the second. Using this nottion, the first nturl numbers cn be defined s 0 λsz.z 1 λsz.s(z) 2 λsz.s(s(z)) 3 λsz.s(s(s(z)))

9 9 nd so on. The big dvntge of defining numbers in this wy is tht we cn now pply function f to n rgument ny number of times. For exmple, if we wnt to pply f to three times we pply the function 3 to the rguments f nd yielding: 3f (λsz.s(s(sz)))f f(f(f)). This wy of defining numbers provides us with lnguge construct similr to n instruction such s FOR i=1 to 3 in other lnguges. The number zero pplied to the rguments f nd yields 0f (λsz.z)f. Tht is, pplying the function f to the rgument zero times leves the rgument unchnged. Our first interesting function, fter hving defined the nturl numbers, is the successor function. This cn be defined s S λnb.(nb). The definition looks wkwrd but it works. For exmple, the successor function pplied to our representtion for zero is the expression: S0 (λnb.(nb))0 In the body of the first expression we substitute the occurrence of n with 0 nd this produces the reduced expression: λb.(0b) λb.(b) 1 Tht is, the result is the representtion of the number 1 (remember tht bound vrible nmes re dummies nd cn be chnged). Successor pplied to 1 yields: S1 (λnb.(nb))1 λb.(1b) λb.(b) 2 Notice tht the only purpose of pplying the number 1 (λsz.sz) to the rguments nd b is to renme the vribles used internlly in the definition of our number.

10 Addition Addition cn be obtined immeditely by noting tht the body sz of our definition of the number 1, for exmple, cn be interpreted s the ppliction of the function s on z. If we wnt to dd sy 2 nd 3, we just pply the successor function two times to 3. Let us try the following in order to compute 2+3: 2S3 (λsz.s(sz)))s3 S(S3) 5 In generl m plus n cn be computed by the expression msn. 2.2 Multipliction The multipliction of two numbers x nd y cn be computed using the following function: (λxy.x(y)) The product of 3 by 3 is then which reduces to (λxy.x(y))33 (λ.3(3)) Using the definition of the number 3, we further reduce the bove expression to (λ.(λsb.s(s(sb)))(3)) (λb.(3)((3)((3)b))) In order to understnd why this function relly computes the product of 3 by 3, let us look t some digrms. The first ppliction (3) is computed on the left of Fig. 4. Notice tht the ppliction of 3 to hs the effect of producing new function which pplies three times to the function s rgument. Now, pplying the function 3 to the result of (3) produces three copies of the function obtined in Fig. 4, conctented s shown on the right in Fig. 4

11 11 (λsz.s(s(sz))) (λb.(3)((3)((3)b))) three pplic+ons of s pplied three +mes b pplied 3 by 3 +mes to b ((((((((b)))))))) Figure 4: Left: The number 3 pplied to n rgument produces new function. Right: The plumbing of the function 3 pplied to 3, nd the result to b. (where the result hs been pplied to b for clrity). Notice tht we hve tower of three times the sme function, ech one bsorbing the lower one s rgument for the ppliction of the function three times, for totl of nine pplictions. 3 Conditionls We introduce the following two functions which we cll the vlues true T λxy.x

12 12 nd flse F λxy.y The first function tkes two rguments nd returns the first one. The second function returns the second of two rguments. 3.1 Logicl opertions It is now possible to define logicl opertions using this representtion of the truth vlues. The AND function of two rguments cn be defined s λxy.xyf This definition works becuse given tht x is true, the truth vlue of the AND opertion depends on the truth vlue of y. If x is flse (nd selects thus the second rgument in xyf) the complete AND is flse, regrdless of the vlue of y. The OR function of two rguments cn be defined s λxy.xty Here, if x is true, the OR is true. If x is flse, it picks the second rgument y nd the vlue of the OR function depends now on the vlue of y. Negtion of one rgument cn be defined s λx.xft For exmple, the negtion function pplied to true is T (λx.xft)t which reduces to TFT (λcd.c)ft F

13 13 tht is, the truth vlue flse. Armed with this three logic functions we cn encode ny other logic function nd reproduce ny given circuit without feedbck (we look t feedbck when we del with recursion). 3.2 A conditionl test It is very convenient in progrmming lnguge to hve function which is true if number is zero nd flse otherwise. The following function Z fulfills this role: Z λx.xf F To understnd how this function works, remember tht 0f (λsz.z)f = tht is, the function f pplied zero times to the rgument yields. On the other hnd, F pplied to ny rgument yields the identity function F (λxy.y) λy.y I We cn now test if the function Z works correctly. The function pplied to zero yields Z0 (λx.xf F)0 0F F F T becuse F pplied 0 times to yields. The function Z pplied to ny other number N yields ZN (λx.xf F)N NF F The function F is then pplied N times to. But F pplied to nything is the identity (s shown before), so tht the bove expression reduces, for ny number N greter thn zero, to IF F

14 The predecessor function We cn now define the predecessor function combining some of the functions introduced bove. When looking for the predecessor of n, the generl strtegy will be to crete pir (n, n 1) nd then pick the second element of the pir s the result. A pir (, b) cn be represented in λ-clculus using the function (λz.zb) We cn extrct the first element of the pir from the expression pplying this function to T (λz.zb)t Tb, nd the second pplying the function to F (λz.zb)f Fb b. The following function genertes from the pir (n, n 1) (which is the rgument p in the function) the pir (n + 1, n): Φ (λpz.z(s(pt))(pt)) The subexpression pt extrcts the first element from the pir p. A new pir is formed using this element, which is incremented for the first position of the new pir nd just copied for the second position of the new pir. The predecessor of number n is obtined by pplying n times the function Φ to the pir (λ.z00) nd then selecting the second member of the new pir: P (λn.(nφ(λz.z00))f) Notice tht using this pproch the predecessor of zero is zero. This property is useful for the definition of other functions. 3.4 Equlity nd inequlities With the predecessor function s the building block, we cn now define function which tests if number x is greter thn or equl to number y: G (λxy.z(xpy))

15 15 If the predecessor function pplied x times to y yields zero, then it is true tht x y. If x y nd y x, then x = y. This leds to the following definition of the function E which tests if two numbers re equl: E (λxy. (Z(xPy))(Z(yPx))) In similr mnner we cn define functions to test whether x > y, x < y or x y. 4 Recursion Recursive functions cn be defined in the λ-clculus using function which clls function y nd then regenertes itself. This cn be better understood by considering the following function Y: Y (λy.(λx.y(xx))(λx.y(xx))) This function pplied to function R yields: which further reduced yields YR (λx.r(xx))(λx.r(xx)) R((λx.R(xx))(λx.R(xx)) but this mens tht YR R(YR), tht is, the function R is evluted using the recursive cll YR s the first rgument. An infinite loop, for exmple, cn be progrmmed s YI, since this reduces to I(YI), then to YI nd so d infinitum. A more useful function is one which dds the first n nturl numbers. We cn use recursive definition, since n i=0 i = n + n 1 i=0 i. Let us use the following definition for R: R (λrn.zn0(ns(r(pn))))

16 16 This definition tells us tht the number n is tested: if it is zero the result of the sum is zero. In n is not zero, then the successor function is pplied n times to the recursive cll (the rgument r) of the function pplied to the predecessor of n. How do we know tht r in the expression bove is the recursive cll to R, since functions in λ-clculus do not hve nmes? We do not know nd tht is precisely why we hve to use the recursion opertor Y. Assume for exmple tht we wnt to dd the numbers from 0 to 3. The necessry opertions re performed by the cll: YR3 R(YR)3 Z30(3S(YR(P3))) Since 3 is not equl to zero, the evlution is equivlent to 3S(YR(P3)) tht is, the sum of the numbers from 0 to 3 is equl to 3 plus the sum of the numbers from 0 to the predecessor of 3 (tht is, two). Successive recursive evlutions of YR will led to the correct finl result. Notice tht in the function defined bove the recursion will be stopped when the rgument becomes 0. The finl result will be tht is, the number 6. 3S(2S(1S0)) 5 Projects for the reder 1. Define the functions less thn nd greter thn of two numericl rguments. 2. Define the positive nd negtive integers using pirs of nturl numbers. 3. Define ddition nd subtrction of integers. 4. Define the division of positive integers recursively.

17 17 5. Define the function n! = n (n 1) 1 recursively. 6. Define the rtionl numbers s pirs of integers. 7. Define functions for the ddition, subtrction, multipliction nd division of rtionls. 8. Define dt structure to represent list of numbers. 9. Define function which extrcts the first element from list. 10. Define recursive function which counts the number of elements in list. 11. Cn you simulte Turing mchine using λ-clculus? References [1] P.M. Kogge, The Architecture of Symbolic Computers, McGrw-Hill, New york 1991, chpter 4. [2] G. Michelson, An Introduction to Functionl Progrmming through λ- clculus, Addison-Wesley, Wokinghm, [3] G. Revesz, Lmbd-Clculus Combintors nd Functionl Progrmming, Cmbridge University Press, Cmbridge, 1988, chpters 1-3.