Universal Lambda Machine
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You are encouraged to solve this task according to the task description, using any language you may know.
One of the foundational mathematical constructs behind computer science is the universal machine, as a machine that can emulate the behaviour of any other machine, given a description of it. Alan Turing introduced the idea of a universal Turing machine in 1936–1937.
The lambda calculus is an even older, and in many ways simpler, model of computation. That simplicity is reflected in the Binary Lambda Calculus (BLC for short), which describes lambda terms with binary tokens 00 for lambda, 01 for application, and 1^n0 for variable n (which binds to the n'th enclosing lambda).
BLC also specifies a way to represent bits and lists as lambda terms, which provides the following I/O convention:
The lambda universal machine parses the binary encoding of a lambda term from the start of its input, applies that term to the remainder of input, and outputs the result interpreted as a list of bits or bytes.
BLC as a programming language has its own entry on Rosetta Code at https://rosettacode.org/wiki/Category:Binary_Lambda_Calculus which links to more detailed descriptions of the language.
- Task
Simulate the universal lambda machine Or in other words, write a BLC interpreter. Support either bit-mode or byte-mode, or preferably both (with byte-mode as the default, and a -b command line flag for bit mode).
To test your universal lambda machine, you should execute the following BLC programs.
For bit-mode, one should reproduce the BLC Rosetta Code solutions of
- https://rosettacode.org/wiki/Quine#Binary_Lambda_Calculus
- https://rosettacode.org/wiki/Sieve_of_Eratosthenes#Binary_Lambda_Calculus (
producing as much output as possible before running out of stack/heap space).
Also, the 342 bit program
010100011010000000011000010110011110000010010111110111100001010110000000011000011111000000101111110110010111111011001011110100111010111100010000001011100101010001101000000000010110000101011111101111100000010101111011111011111100001011000000101111111010110111000000111111000010110111101110011110100000010110000011011000100000101111000111001110
should produce output
11010
For byte-mode, one should reproduce the BLC Rosetta Code solutions of
- https://rosettacode.org/wiki/Hilbert_curve#Binary_Lambda_Calculus
- https://rosettacode.org/wiki/Execute_Brain****#Binary_Lambda_Calculus
Existing solutions may be found at https://rosettacode.org/wiki/Hello_world/Newbie#Binary_Lambda_Calculus
Binary Lambda Calculus
The following self interpreters are taken from the IOCCC entry at https://www.ioccc.org/2012/tromp/hint.html
Bit-wise:
01010001
10100000
00010101
10000000
00011110
00010111
11100111
10000101
11001111
000000111
10000101101
1011100111110
000111110000101
11101001 11010010
11001110 00011011
00001011 11100001
11110000 11100110
11110111 11001111
01110110 00011001
00011010 00011010
Byte-wise (showing https://www.ioccc.org/2012/tromp/uni8.Blc in hex):
19468
05580
05f00
bfe5f
85f3f
03c2d
b9fc3f8
5e9d65e5f
0decb f0fc3
9befe 185f7
0b7fb 00cf6
7bb03 91a1a
C
#ifndef M
#define M 50000
#endif
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
enum {I,O,V,A,L};
long n=44,i,c,T[M]={L,A,8,A,2, V,0,L,L,V,
A,30,L,A,2,V,0,L,A,5,A,7,L,V,0,O,
A,14,L,A,2,V,0,L,A,5,A,2, V,0,O,O,A},b,s;
long nc = 0, nf = 0, na = 0; // number of cells, number freed, number of allocs
typedef struct _{long t,r; struct _*e,*n;} C;C*e,*f,*l,*S[M];
void x(long l,long u){for(;l<=u;T[n++]=T[l++]);}
long g(){i--||(i=b,c=getchar());return c>>i&1;}
void d(C*l){!l||--l->r||(d(l->e),d(l->n),l->n=f,nf++,f=l);}
long p(long m){if(g()){for(T[n++]=V;g();T[n]++){}n++;}else
T[m]=n++&&g()?(T[m+1]=p(++n),A):L,p(n);return n-m;}
int main(int t,char **_){char o;
b=t>1?0:7;T[43]=p(n);i=0;
for(t=b?10:26;;)switch(T[t]){
case I: g();i++;assert(n<M-99);if(~c&&b){x(0,6);for(T[n-5]=96;i;T[n++]=!g())
x(0,9);}x(c<0?7:b,9);T[n++]=!b&&!g();break;
case O: t=b+t>42?(o=2*o|t&1,28):(putchar
(b?o:t+8),fflush(stdout),b?12:28);break;
case V: l=e;for(t=T[t+1];t--;e=e->n){}
t=e->t;(e=e->e)&&e->r++;d(l);break;
case A: t+=2;
nc++;
f||(na++,f=calloc(1,sizeof(C)));
assert(f&&s<M);S[s++]=l=f;f=l->n;
l->r=1;l->t=t+T[t-1];(l->e=e)&&e->r++;break;
case L: if(!s--){printf("\n%ld cells\n%ld freed\n%ld allocated\n", nc, nf, na);return 0;};S[s]->n=e;e=S[s];t++;break;
}
printf("%ld cells %ld allocated\n", nf, na);
return T[t+2];
}
Ruby
#!/usr/bin/ruby
def bit2lam(bit)
return lambda { |x0| lambda { |x1| bit==0 ? x0 : x1 } }
end
def byte2lam(bits,n)
return n==0 ? lambda { |_| lambda { |y| y } }
: lambda { |z| z.call(bit2lam((bits>>(n-1))&1)).call(byte2lam(bits,n-1)) }
end
def input(n) # input from n'th character onward
if n >= $inp.length()
c = STDIN.getbyte
$inp.append(c==nil ? (lambda { |_| lambda { |y| y } })
: lambda { |z| z.call($bytemode ? byte2lam(c,8) : bit2lam(c&1)).call(input(n+1)) } )
end
return $inp[n]
end
def lam2bit(lambit)
return lambit.call(lambda { |_| 0 }).call(lambda { |_| 1 }).call(0) # force suspension
end
def lam2byte(lambits, x)
return lambits.call(lambda { |lambit| lambda { |lamtail| lambda { |_| lam2byte(lamtail, 2*x+lam2bit(lambit)) } } }).call(x)
end
def output(prog)
return prog.call(lambda { |c| putc($bytemode ? lam2byte(c,0) : (lam2bit(c)==0 ? '0' : '1')) and (lambda { |tail| lambda { |_| output(tail) } }) }).call(0)
end
def getbit()
if ($nbit==0)
$progchar = STDIN.getbyte
$nbit = $bytemode ? 8 : 1
end
return ($progchar >> $nbit -= 1) & 1
end
def program()
if getbit()==1 # variable
i = 0
while (getbit()==1) do i += 1 end
return lambda { |*args| args[i] }
elsif getbit()==1 # application
p = program()
q = program()
return lambda { |*args| p.call(*args).call(lambda { |arg| q.call(*args).call(arg) }) } # suspend argument
else
p = program()
return lambda { |*args| lambda { |arg| p.call(arg, *args) } } # extend environment with one more argument
end
end
$inp = []
$nbit = $progchar = 0
$bytemode = ARGV.length <= 0
$prog = program().call(0)
output($prog.call(input(0))) # run program with empty env on input