# Multiply with restricted operations

Goal

Your goal is to multiply two numbers using only a very limited set of arithmetic operations and variable assignment.

1. Addition `x,y -> x+y`
2. Reciprocal `x -> 1/x` (not division `x,y -> x/y`)
3. Negation `x -> -x` (not subtraction `x,y -> x-y`, though you can do it as two operations `x + (-y)`)
4. The constant `1` (no other constants allowed, except as produced by operations from `1`)
5. Variable assignment `[variable] = [expression]`

Scoring: The values start in variables `a` and `b`. Your goal is to save their product `a*b` into the variable `c` using as few operations as possible. Each operation and assignment `+, -, /, =` costs a point (equivalently, each use of (1), (2), (3), or (4)). Constants `1` are free. The fewest-point solution wins. Tiebreak is earliest post.

Allowance: Your expression has to be arithmetically correct for "random" reals `a` and `b`. It can fail on a measure-zero subset of R2, i.e. a set that has no area if plotted in the `a`-`b` Cartesian plane. (This is likely to be needed due to reciprocals of expressions that might be `0` like `1/a`.)

Grammar:

This is an . No other operations may be used. In particular, this means no functions, conditionals, loops, or non-numerical data types. Here's a grammar for the allowed operations (possibilities are separated by `|`). A program is a sequence of `<statement>`s, where a `<statement>` is given as following.

``````<statement>: <variable> = <expr>
<variable>: a | b | c | [string of letters of your choice]
<expr>: <arith_expr> | <variable> | <constant>
<arith_expr>: <addition_expr> | <reciprocal_expr> | <negation_expr>
<reciprocal_expr>: 1/(<expr>)
<negation_expr>: -<expr>
<constant>: 1
``````

You don't actually have to post code in this exact grammar, as long as it's clear what you're doing and your operation count is right. For example, you can write `a-b` for `a+(-b)` and count it as two operations, or define macros for brevity.

(There was a previous question Multiply without Multiply, but it allowed a much looser set of operations.)

I will give a bounty of 300 rep to the first person to post an optimal solution with proof of optimality. The proof must show that no lower-operation solution exists. If the current best 23-operation solution is optimal, a proof of that would suffice and be given the bounty. A brute force search can be a proof if you explain how it checks everything needed.

An optimal result restricted not to use variable assignment will be given 150 rep. Other partial bounties may be given for partial results.

To avoid wasting rep if the bounty isn't collected, I'm announcing it unofficially. If someone meets the criteria, I'll post the bounty and award to them. Redeemable until Jan 1, 2016 for as long as I'm active.

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Is this even possible? – Ypnypn Sep 1 '14 at 20:19
@Ypnypn Yes, and I've written an example to make sure. – xnor Sep 1 '14 at 20:24
This feels like a challenge where an optimal solution is likely to be found (once any solution has been found). So what's the tie breaker in that case? – Martin Ender Sep 1 '14 at 20:27
@MartinBüttner Tiebreak is earliest posting in that case. I think there's a good amount of room for optimizations, so I don't think it will just be a race to find one that works and write it cleanly. At least, that's what I found in trying it; maybe someone will find a clearly minimal solution. – xnor Sep 1 '14 at 20:30
Ok since not everyone thought my anwer was as funny as I did, I deleted it and comment here: The rule about the measure zero set is not very wisely chosen since rational numbers are a measure zero set regarding the lebesgue measure, I'd suggest using a certain percentage instead. (Or another kind) But I totally like the idea of this challenge! – flawr Sep 1 '14 at 21:43

## 22 operations

Phew, this took a while to get!

``````# 22 ops
itx = 1/(1+a+b)     #4
nx = -1/(itx+itx)   #4
c = -( 1/(itx + itx + 1/(1+nx)) + 1/(1/(a+nx) + 1/(b+nx)) ) #14
``````

The ops are 10 additions, 7 inverses, 2 negations, and 3 assignments.

So, how did I get this? I started with the promising-looking template of the sum of two double-decker fractions, a motif that had appeared in many previous attempts.

`c = 1/(1/x + 1/y) + 1/(1/z + 1/w)`

When we restrict the sum to `x+y+z+w=0`, a beautiful cancellations occur, giving:

`c = (x+z)*(y+z)/(x+y)`,

which contains a product. (It's often easier to get `t*u/v` rather than `t*u` because the first has degree 1.)

There's a more symmetric way to think about this expression. With the restriction `x+y+z+w=0`, their values are specified by three parameters `p,q,r` of their pairwise sums.

`````` p = x+y
-p = z+w
q = x+z
-q = y+w
r = x+w
-r = y+z
``````

and we have `c=-q*r/p`. The sum `p` is distinguished as being in the denominator by corresponding to the pairs `(x,y)` and `(z,w)` of variables that are in the same fraction.

This is a nice expression for `c` in `p,q,r`, but the double-decker fraction is in `x,y,z,w` so we must express the former in terms of the latter:

``````x = ( p + q + r)/2
y = ( p - q - r)/2
z = (-p + q - r)/2
w = (-p - q + r)/2
``````

Now, we want to choose `p,q,r` so that `c=-q*r/p` equals `a*b`. One choice is:

``````p = -4
q = 2*a
r = 2*b
``````

Then, the doubled values for `q` and `r` are conveniently halved in:

``````x = -2 + a + b
y = -2 - a - b
z =  2 + a - b
w =  2 - a + b
``````

Saving `2` as a variable `t` and plugging these into the equation for `c` gives a 24-op solution.

``````#24 ops
t = 1+1   #2
c = 1/(1/(-t+a+b) + 1/-(t+a+b))  +  1/(1/(-b+t+a) + 1/(-a+b+t)) #1, 10, 1, 10
``````

There's 12 additions, 6 inverses, 4 negations, and 2 assignments.

A lot of ops are spent expressing `x,y,z,w` in terms of `1,a,b`. To save ops, instead express `x` in `p,q,r` (and thus `a,b,1`) and then write `y,z,w` in terms of `x`.

``````y = -x + p
z = -x + q
w = -x + r
``````

Choosing

``````p = 1
q = a
r = b
``````

and expressing `c` with a negation as `c=-q*r/p`, we get

``````x = (1+a+b)/2
y = -x + 1
z = -x + a
w = -x + b
``````

Unfortunately, halving in `x` is costly. It needs to be done by inverting, adding the result to itself, and inverting again. We also negate to produce `nx` for `-x`, since that's what `y,z,w` use. This gives us the 23-op solution:

``````#23 ops
itx = 1/(1+a+b)     #4
nx = -1/(itx+itx)   #4
c = -( 1/(1/(-nx) + 1/(1+nx))  +  1/(1/(a+nx) + 1/(b+nx)) ) #15
``````

`itx` is 1/(2*x) and `nx` is `-x`. A final optimization of expressing `1/x` as `itx+itx` instead of the templated`1/(-nx)` cuts a character and brings the solution down to 22 ops.

-

# 23 operations

``````z = 1/(1/(1/(1/(a+1)+1/(b+1))-1+1/(a+b+1+1))-(1/a+1/b))
res = z+z
``````

proof by explosion:

``````z = 1/(1/(1/(1/(a+1)+1/(b+1))-1+1/(a+b+1+1))-(1/a+1/b))
1/(a+1)+1/(b+1)                            == (a+b+2) / (ab+a+b+1)
1/(1/(a+1)+1/(b+1))                           == (ab+a+b+1) / (a+b+2)
1/(1/(a+1)+1/(b+1))-1                         == (ab - 1) / (a+b+2)
1/(1/(a+1)+1/(b+1))-1+1/(a+b+1+1)             == ab / (a+b+2)
1/(1/(1/(a+1)+1/(b+1))-1+1/(a+b+1+1))            == (a+b+2) / ab
1/a+1/b   == (a+b) / ab
1/(1/(1/(a+1)+1/(b+1))-1+1/(a+b+1+1))-(1/a+1/b)  == 2 / ab
1/(1/(1/(1/(a+1)+1/(b+1))-1+1/(a+b+1+1))-(1/a+1/b)) == ab / 2

z = ab / 2 and therefore z+z = ab
``````

I abused wolfram alpha to get this beautiful image (wolfram alpha tried to get me to subscribe to pro to save it, but then ctrl-c ctrl-v ;-)):

score (with added `+` on subtraction):

``````z = ////++/++-+/++++-/+/
res = +
``````
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Congrats on the shortest solution! – xnor Sep 11 '14 at 4:04
@xnor thanks for giving me my first accepted answer and my first bounty! – proud haskeller Sep 11 '14 at 5:33
Not to be nit-picky, but shouldn't ...(b+1))-1+1... and ...1))-(1/a+... be ...(b+1))+-1+1... and ...1))+-(1/a+... respectively? – tfitzger May 2 '15 at 17:51
@tfitzger I think it's easier that way. The question does say that it doesn't matter. Note I do count the score correctly (Every minus is a two) – proud haskeller May 2 '15 at 17:53
Wolfram Alpha has a 7-day free trial, fyi. – ghosts_in_the_code Nov 27 '15 at 18:05

# 29 operations

Does not work for the set { (a,b) ∈ R2 | a+b = 0 or a+b = -1 or a-b = 0 or a-b = -1 }. That's probably measure zero?

``````sum = a+b
nb = -b
diff = a+nb
rfc = 1/(1/(1/sum + -1/(sum+1)) + -1/(1/diff + -1/(diff+1)) + nb + nb)  # rfc = 1/4c
c = 1/(rfc + rfc + rfc + rfc)

# sum  is  2: =+
# nb   is  2: =-
# diff is  2: =+
# rfc  is 18: =///+-/++-//+-/+++
# c    is  5: =/+++
# total = 29 operations
``````

The structure of `rfc` (Reciprocal-Four-C) is more evident if we define a macro:

``````s(x) = 1/(1/x + -1/(x+1))              # //+-/+ (no = in count, macros don't exist)
rfc = 1/(s(sum) + - s(diff) + nb + nb) # =/s+-s++ (6+2*s = 18)
``````

Let's do the math:

• `s(x)`, mathematically, is `1/(1/x - 1/(x+1))` which is after a bit of algebra is `x*(x+1)` or `x*x + x`.
• When you sub everything into `rfc`, it's really `1/((a+b)*(a+b) + a + b - (a-b)*(a-b) - a + b + (-b) + (-b))` which is just `1/((a+b)^2 - (a-b)^2)`.
• After difference of squares, or just plain expansion, you get that `rfc` is `1/(4*a*b)`.
• Finally, `c` is the reciprocal of 4 times `rfc`, so `1/(4/(4*a*b))` becomes `a*b`.
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+1, I was in the middle of finishing up this identical calculation – Eric Tressler Sep 1 '14 at 21:19
That's definitely measure zero; it's a union of lines. – xnor Sep 1 '14 at 21:54
Not gonna make a comment about union of lines... @algorithmshark Can you tell us more how you did come up with this identity? How did you aproach the problem? – flawr Sep 2 '14 at 12:31
@flawr I recalled that the properties of `s(x)` fit the requirements of the question, from calculus, so that meant I had a square function. After some faffing about, I found I could get an `a*b` term with the difference of squares trick. Once I had that, it was a matter of trying out which assignments saved operations. – algorithmshark Sep 2 '14 at 15:22
Since you use `-1` three times in `rfc`, couldn't you golf a character out by assigning it to a variable? – isaacg Sep 2 '14 at 18:23

# 27 operations

``````tmp = 1/(1/(1+(-1/(1/(1+(-a))+1/(1+b))))+1/(1/(1/b+(-1/a))+1/(a+(-b))))
res = tmp+tmp+(-1)

# tmp is 23: =//+-//+-+/++///+-/+/+-
# res is 4: =++-
``````

There is no theory behind this. I just tried to get `(const1+a*b)/const2` and started with `(1/(1-a)+1/(1+b))` and `(-1/a+1/b)`.

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Your `tmp` is actually 23, making your score 27. Nice find, though. – algorithmshark Sep 2 '14 at 15:29