Implement INTERCAL's Binary Operators

Question

The Compiler Language With No Pronounceable Acronym, abbreviated INTERCAL, is a very unique programming language. Among its unreproducible qualities are its binary operators.

INTERCAL's two binary operators are interleave (also known as mingle), and select. Interleave is represented with a change (¢), and select is represented with a sqiggle (~).

Interleave works by taking two numbers in the range 0-65535 and alternating their bits. For instance:

234 ¢ 4321
234   = 0000011101010
4321  = 1000011100001
Result: 01000000001111110010001001
Output: 16841865

Select works by taking two numbers in the range 0-65535, taking the bits in the first operand which are in the same position as 1s in the second operand, and right packing those bits.

2345 ~ 7245
2345  = 0100100101001
7245  = 1110001001101
Taken : 010   0  10 1
Result: 0100101
Output: 37

In this challenge, you will be given a binary expression using either the interleave or select operation. You must calculate the result, using the fewest possible bytes.

The expression will be given as a space separated string, consisting of an integer in 0-65535, a space, either ¢ or ~, a space, and an integer in 0-65535.

Input and output may be through any standard system (STDIN, function, command line, etc.). Standard loopholes banned.

Examples:

5 ¢ 6
54

5 ~ 6
2

51234 ¢ 60003
4106492941

51234 ~ 60003
422

This is code golf - fewest bytes wins. Good luck.

EDIT: Since some languages do not support INTERCAL's change (¢) symbol, you may use the big money ($) symbol instead, at a 5 byte penalty.

Answer 1

Pyth, 32 31 29 bytes

isummFdG}\~zCm.[Z16jvd2%2cz)2

Try it online: Regular Input / Test Suite

Thanks to @isaacg for golfing off one byte.

Explanation:

                         cz)   split input at spaces
                       %2      only take every second item (the numbers)
             m                 map each number d to:
                    vd           convert d to int
                   j  2          convert to base 2
              .[Z16              pad zeros at the left
            C                  zip
  u     }\~z                   apply the following function ("~" in input) times:
   m   G                         map each pair d to:
    mFd                          convert [x,0] to [] and [x,1] to [x]
 s                             take sum (unfold all lists)
i                           2  convert back from base 2 and print

Answer 2

Python 2, 115 112 bytes

x,y,z=input().split()
d=y<""
f=lambda a,b:a+b and(b%2+5&4-d)*f(a/2,b/2)+(a%2*2+b%2)/3**d
print f(int(x),int(z))

The string on the second line contains a single unprintable character \x7d, the next char after ~.

All hopes of a nice, single lambda get crushed by the input format. There's probably a better way to read in input. Input like "51234 ¢ 60003" via STDIN.

The function f combines the following two recursive functions:

g=lambda a,b:a+b and 4*g(a/2,b/2)+a%2*2+b%2    # ¢
h=lambda a,b:a+b and(b%2+1)*h(a/2,b/2)+a*b%2   # ~

(-3 bytes with the help of @xnor)

Answer 3

CJam, 31 bytes

rrc\r]{i2bF0e[}%(7=\zf{_)*?~}2b

Try it online in the CJam interpreter.

How it works

rr                              e# Read two tokens from STDIN.
  c\                            e# Cast the second to char and swap with the first.
    r                           e# Read a third token from STDIN.
     ]                          e# Wrap everything in an array.
      {       }%                e# For all three elements:
       i2b                      e#   Cast to int and convert to base 2.
          F0e[                  e#   Left-pad with zeroes to complete 15 digits.
                (               e# Shift out the first base 2 array.
                 7=             e# Select its eighth MSB (1 for '¢', 0 for '~').
                   \            e# Swap with the array of base 2 arrays.
                    z           e# Zip to transpose rows with columns.
                     f{     }   e# For each pair of base 2 digits:
                                e#   Push the bit, then the pair.
                       _        e#   Copy the pair.
                        )       e#   Pop the second digit.
                         *      e#   Repeat the first digit that many times.
                          ?     e#   Ternary if. Select the pair if the bit is
                                e#    truthy, the repeated first bit if it's falsy.
                           ~    e#   Dump the selected array on the stack.
                             2b e# Convert from base 2 to integer.

Answer 4

JavaScript (ES6), 103 117 119 124

Edit now working with numbers instead of strings

(not counting leading spaces, newlines and comments)

Test running the snippet on any EcmaScript 6 compliant browser (notably not Chrome not MSIE. I tested on Firefox, Safari 9 could go)

I=s=>
  (i=>{
    for(m=r=0,[a,o,b]=s.split` `;i>0;i<<=1) // loop until bit 31 of i is set
      o>'~'?r+=(b&i)*i+(a&i)*2*i:b&i?r+=(a&i)>>m:++m
  })(1)||r


// TEST
out=x=>O.innerHTML+=x+'\n\n';

[ ['234 ¢ 4321', 16841865], ['2345 ~ 7245', 37]
, ['5 ¢ 6', 54], ['5 ~ 6', 2]
, ['51234 ¢ 60003',4106492941], ['51234 ~ 60003', 422]]
.forEach(([i,o,r=I(i)])=>{
  out('Test '+ (o==r?'OK':'Fail')+'\nInput:    '+ i+'\nResult:   '+r+'\nExpected: '+o)})

<pre id=O></pre>

Answer 5

Python 3, 174 166 148 126

Pretty straightforward, string operations, then conversion back to integer.

Limited to numbers which in binary have 99 digits (max 2^99-1 = 633825300114114700748351602687).

Thanks, Sp3000 and Vioz!

a,o,b=input().split()
print(int(''.join([(i+j,i[:j>'0'])[o>'~']for i,j in zip(*[bin(int(j))[2:].zfill(99)for j in(a,b)])]),2))

Or 165 chars, without limit:

a,o,b=input().split()
a,b=[bin(int(j))[2:]for j in(a,b)]
print(int(''.join([(i if j=='1'else'')if o=='~'else i+j for i,j in zip(a.zfill(len(b)),b.zfill(len(a)))]),2))

Ungolfed:

a, op, b = input().split()
a, b = [bin(int(j))[2:] for j in(a,b)] #convert to int (base 10), then to binary, remove leading '0b'
m = max(len(a), len(b))
a = a.zfill(m) #fill with leading zeroes
b = b.zfill(m)
if op == '~':
    ret = [i if j=='1' else'' for i, j in zip(a, b)]
else:
    ret = [i + j for i, j in zip(a, b)]
ret = ''.join(ret) #convert to string
ret = int(ret, 2) #convert to integer from base 2
print(ret)

Answer 6

Matlab, 119 113 bytes

function f(s)
t=dec2bin(str2double(strsplit(s,{'¢' '~'}))');u=any(s>'~');[~u u]*bin2dec({t(1,t(2,:)==49) t(:)'})

Ungolfed:

function f(s)                                     % input s is a string
t = dec2bin(str2double(strsplit(s,{'¢' '~'}))');  % get the two numbers and convert to
                                                  % two-row char array of zeros of ones
u = any(s>'~');                                   % 1 indicates '¢'; 0 indicates '~'
[~u u]*bin2dec({t(1,t(2,:)==49) t(:)'})           % compute both results and display
                                                  % that indicated by u

Examples:

>> f('234 ¢ 4321')
ans =
    16841865

>> f('2345 ~ 7245')
ans =
    37

Answer 7

R, 145 bytes

s=scan(,"");a=as.double(c(s[1],s[3]));i=intToBits;cat(packBits(if(s[2]=="~")c(i(a[1])[i(a[2])>0],i(0))[1:32] else c(rbind(i(a[2]),i(a[1]))),"i"))

Ungolfed + explanation:

# Read a string from STDIN and split it on spaces
s <- scan(, "")

# Convert the operands to numeric
a <- as.double(c(s[1], s[3]))

o <- if (s[2] == "~") {
    # Get the bits of the first operand corresponding to ones in
    # the second, right pad with zeros, and truncate to 32 bits
    c(intToBits(a[1])[intToBits(a[2]) == 1], intToBits(0))[1:32]
} else {
    # Interleave the arrays of bits of the operands
    c(rbind(intToBits(a[2]), intToBits(a[1])))
}

# Make an integer from the raw bits and print  it to STDOUT
cat(packBits(o, "integer"))

Answer 8

Pyth, 43 bytes

Part of me feels nervous posting such a long Pyth answer on isaacg's question... :oP

J.(Kczd1Am.BvdKiu?qJ\~u+G?qeH\1hHk+VGHk.iGH2

Explaination:

                                               Implicit: z=input(), k='', d=' '
   Kczd                                        Split z on spaces, store in K
J.(    1                                       Remove centre element from K, store in J
         m    K                                For each d in K
          .Bvd                                 Evaluate as int, convert to binary string
        A                                      Store pair in G and H
                                               ~ processing:
                                 +VGH          Create vectorised pairs ([101, 110] -> [11, 01, 10])
                     u               k         Reduce this series, starting with empty string
                        ?qeH\1                 If 2nd digit == 1...
                              hHk              ... take the 1st digit, otherwise take ''
                      +G                       Concatenate
                                      .iGH     ¢ processing: interleave G with H
                ?qJ\~                          If J == ~, take ~ processing, otherwise take ¢
               i                          2    Convert from binary to decimal

Answer 9

CJam, 61 50 46 41 34 bytes

Thanks @Dennis for pointing out a 4 bytes golf.

rrc'~=:X;r]{i2bF0e[}/.{X{{;}|}&}2b

Try it online.

Answer 10

C, 127 123 bytes + 5 penalty = 128

scanf counts the unicode symbol as more than one character which complicates things a lot, so I'm applying the 5-byte penalty for using $.

a,b,q,x,i;main(){scanf("%d %c %d",&a,&q,&b);for(i=65536;i/=2;)q%7?x=x*4|a/i*2&2|b/i&1:b/i&1&&(x=x*2|a/i&1);printf("%u",x);}

The changes from the original version are:

-The test for $ or ~ has been revised from q&2 to q%7. This reverses the true/false values, allowing the code for $ operator to go before the : which means a set of parentheses can be eliminated.

-The i loop now counts down in powers of 2 which is longer, but permits >> to be substituted by / and saves some parentheses.

Original version 127 bytes

a,b,q,x,i;
main(){
  scanf("%d %c %d",&a,&q,&b);
  for(i=16;i--;)
    q&2?
      b>>i&1&&(x=x*2|a>>i&1):    // ~ operator. && used as conditional: code after it is executed only if code before returns truthy.
      (x=x*4|(a>>i&1)*2|b>>i&1); // $ operator
  printf("%u",x);
}

I went with a single loop with the conditionals inside to avoid the overhead of two loops. In both cases I rightshift the bits of the operands down to the 1's bit, and build up the result from the most significant to least significant bit, leftshifting the result (multiplying by 2 or 4) as I go.

Answer 11

K5, 53 52 bytes

{b/({,/x,'y};{x@&y})[*"~"=y][b\.x;(b:20#2)\.z]}." "\

53-byte version:

{b/({,/x,'y};{x@&y})[*"¢~"?y][b\.x;(b:20#2)\.z]}." "\

Still needs a bit more golfing.

Answer 12

Haskell, 77

g=(`mod`2)
h=(`div`2)
0¢0=0
a¢b=g a+2*b¢h a
a?0=0
a?b=g a*g b+(1+g b)*h a?h b

input is given by applying the input to the functions/operators ? and ¢ defined in the code (Haskell can't define an operator ~ for technical reasons).

basically works the old recursive approach.

Answer 13

Python 3, 157 bytes

a,x,y=input().split()
i=int
b=bin
print(i(''.join(([c for c,d in zip(b(i(a)),b(i(y)))if d=='1'],[c+d for c,d in zip(b(i(a))[2:],b(i(y))[2:])])['¢'==x]),2))

Full and explanatory version can be found on my pastebin.

Answer 14

J, 173

f=:|."1@(>@(|.&.>)@(#:@{:;#:@{.))
m=:2&#.@:,@:|:@:|.@:f
s=:2&#.@#/@:f
a=:{&a.@-.
(1!:2)&2(s@".@:a&126)^:(126 e.i)((m@".@:a&194 162)^:(1 e.194 162 E.i)i=._1}.(a.i.((1!:1)3)))

expects one line of input

input expected to terminate after new line with EOF

Answer 15

Javascript ES6 (3 arguments) 141 138 136 121 119 bytes

b=x=>(65536|x).toString`2`
f=(x,o,y)=>+eval(`'0b'+(b(y)+b(x)).replace(/^1|${o=='~'?1:'(.)'}(?=.{16}(.)())|./g,'$2$1')`)

Test:

;[f(234,'¢',4321),f(2345,'~',7245)]=="16841865,37"

Javascript ES6 (1 argument) 135 133 bytes

b=x=>(65536|x).toString`2`
f=s=>([x,o,y]=s.split` `)|eval(`'0b'+(b(y)+b(x)).replace(/^1|${o=='~'?1:'(.)'}(?=.{16}(.)())|./g,'$2$1')`)

Test:

;[f('234 ¢ 4321'),f('2345 ~ 7245')]=="16841865,37"

PS: New line is counted as 1 byte as it can be replaced by ;.

Answer 16

Mathematica, 155 bytes

f=IntegerDigits[#,2,16]&;
g=#~FromDigits~2&;
¢=g[f@#~Riffle~f@#2]&;
s=g@Cases[Thread@{f@#,f@#2},{x_,1}->x]&;
ToExpression@StringReplace[#,{" "->"~","~"->"s"}]&

Evaluates to an anonymous function taking the string as input. Line breaks added for clarity.

f and g convert to/from base 2. Riffle does exactly what interleave is supposed to. I wanted to use Select for select but Cases is better unfortunately. The last line is a bit of trickery; spaces are changed to ~ which is Mathematica's infix operator, then the string is eval'd.