Implement multiplication by a constant with addition and bit shifts

Question

While looking at the ARM instruction set, you notice that the ADD instruction has the so-called "Flexible second operand", which can be abused for quick multiplication. For example, the following instruction multiplies register r1 by 17 (shifting it left by 4 bits and adding to itself):

ADD r1, r1, r1, LSL #4
; LSL means "logical shift left"

The subtraction instruction has that feature too. So multiplication by 15 is also easy to implement:

RSB r1, r1, r1, LSL #4
; RSB means "reverse subtract", i.e. subtract r1 from (r1 << 4)

So this is an excellent optimization opportunity! Certainly, when you write in a C program x *= 15, the compiler will translate that to some inefficient code, while you could replace it by x = (x << 4) - x, and then the compiler will generate better code!

Quickly, you come up with the following cunning plan:

1. Write a program or a subroutine, in any language, that receives one parameter m, and outputs a C optimizing macro, like the following (for m = 15):

   #define XYZZY15(x) (((x) << 4) - (x))
   

   or (which is the same)

   #define XYZZY15(x) (-(x) + ((x) << 4))
   

2. ???

3. Profit!

Notes:

• The macro's name is essential: magic requires it.
• Parentheses around x and the whole expression are required: ancient C arcana cannot be taken lightly.
• You can assume m is between 2 and 65535.
• The input and output can be passed through function parameters, stdin/stdout or in any other customary way.
• In C, addition and subtraction have tighter precedence than bit shift <<; use parentheses to clarify order of calculations. More parentheses than necessary is OK.
• It is absolutely vital that the C expression be optimal, so e.g. x+x+x is not acceptable, because (x<<1)+x is better.
• Expressions x + x and x << 1 are equivalent, because both require one ARM instruction to calculate. Also, (x << 1) + (x << 2) (multiplication by 6) requires two instructions, and cannot be improved.
• You are amazed to discover that the compiler detects that (x << 0) is the same as x, so x + (x << 0) is an acceptable expression. However, there is no chance the compiler does any other optimizations on your expression.

• Even though multiplication by 17 can be implemented in one instruction, multiplication by 17*17 cannot be implemented in two instructions, because there is no C expression to reflect that:

  #define XYZZY289(x) x + (x << 4) + (x + (x << 4) << 4) // 3 operations; optimal
  

  In other words, the compiler doesn't do common sub-expression elimination. Maybe you will fix that bug in version 2...

This is code-golf: the shortest code wins (but it must be correct, i.e. produce correct and optimal expressions for all m)

Please also give example outputs for m = 2, 10, 100, 14043 and 65535 (there is no requirement on the length of the expression - it can contain extra parentheses, whitespace and shift left by 0 bits, if that makes your code simpler).

Answer 1

Python - 414 393 331

This is my entry, which finds the least possible addition/subtraction operations, and subsequently evaluates to a C expression. Not sure it is optimal as multiplication of subexpressions may offer less operations.

import itertools as l
def w(N,s=''):
 for k,v in enumerate(min([t for t in l.product((1,0,-1),repeat=len(bin(N))-1)if sum([v*2**k for k,v in enumerate(t)])==N],key=lambda t:sum(abs(k)for k in t))):
  if v!=0:s+='{:+d}'.format(v)[0];s+='((x)<<{})'.format(k)if k>0 else'(x)'
 return '#define XYZZY{}(x) ({})'.format(N,s.lstrip('+'))

For the cases 2, 10, 100, 14043, 65535:

#define XYZZY2 (((x)<<1))
#define XYZZY10 (((x)<<1)+((x)<<3))
#define XYZZY100 (((x)<<2)+((x)<<5)+((x)<<6))
#define XYZZY14043 (-(x)-((x)<<2)-((x)<<5)-((x)<<8)-((x)<<11)+((x)<<14))
#define XYZZY65535(x) (-(x)+((x)<<16))

The final 65535 took around 11 minutes to compute. Suggestions for improvement (also speedwise) are welcome :-). Thanks to mbomb007 and plg for golfing improvements.

Answer 2

Python 508

Slightly more characters than my other answer, but this code runs in roughly one second for all cases 2, 10, 100, 14043, 65535.

def l(n,k=0,o=0):
 while 1:
  if k==0:
   for j in range(1,n):
    yield(j,);yield(-j,)               
   k+=1
  else:
   for j in range(1,n):
    for u in l(j,k-1,1):
     yield(j,)+u;yield(-j,)+u 
   k+=1
  if o:break 
r=lambda t:sum((1,-1)[k<0]*2**(abs(k)-1)for k in t)
def w(N,s=''):
 g = loop(len(bin(N))) 
 while 1:
  a = g.next()
  if r(a)==N:
   for k in a:
    s+='{:+d}'.format(k)[0];s+='((x)<<{})'.format(abs(k-1)) if abs(k)>1 else'(x)'
   return '#define XYZZY{}(x) ({})'.format(N,s.lstrip('+'))""")