Yes, if you look at the for loops that I provided, they do just that.

Hence why in my examples I re-use variables within the loop, that have just been used to start the loop (e.g. $e).

Going though the loops:

$a) Starts the main loop, but is then used as a countdown to zero, to limit the number of subtractions (done by the second loop).

$b) The divisor, needs to always be available, so the subtraction can be performed on each loop.

$c) Monitors the number of times that the main loop has been run… as this provides the answer to $d, when the second loop cannot be run any more ($a <= 0).

$d) Is first used to perform the subtraction (the temporary variable), but after its done that, it stores the result for when the main loop finishes… I suspect out of all the variables, this is the one to remove.

$e) I used first to calculate ($b – $a) without changing $a (as the loop runs many times – to create the fake "if" condition), and when its completed, $a is updated with the new value from $e.

Your right about it sounding like a hack, but that does seem to be the only way to get this effect.

I like your examples, but they do use more than the 4 variable target (e.g. in the division, there are 4 variables in that function, but more are used by calling the other functions).

Which is why I removed the use of functions (something that isn't in the list of allowed constructs), and tried to reduce the number of variables even further (getting it down to 5).

Oh, and as to having only "two" loops in the division… if you look at the code flow, there is actually 5 (2 in main body, 2 from the SUB's, and 1 from the DEC)… mine on the other hand only uses 4 loops.

Remember that "loop", as defined in the problem statement, allows one to redefine its value without changing the number of iterations. Thus in something like "loop(v1) { v1 = 0}", it would still loop 5 times if v1 was set to 5 before hitting the loop. My macro of dec works because v1 is always one behind the value of temp.

I like the DIV with only two loops but I feel like the use of c to reduce the looped increment to a single increment is a hack.

@ronen cohen:

Aww, that's cheating… not only does it use 5 variables (a, b, c, k and m), it also goes out to a function (no if statements or anything like that) which presumably uses more variables… and you have an "m = 1", and a "subtract(a, 1)"… where you can only set a variable to 0.

And later on, I don't understand why you have a c++, when it doesn't seem to be used… did you copy the right version over?

So, the question remains, can it be done in less than or equal to 4 variables?

@kurt zoglmann:

I've already got stuck on your "dec" function (copy below)… as temp = 0, how would the loop(temp) run to increment v1?

I am presuming that the inc() function just increments the value by 1 (e.g. v1++).

macro dec (v1, temp) {
temp = 0
loop(v1) {
v1 = 0
loop(temp) {
inc(v1)
}
inc(temp)
}
}

this is pseudo code.

//decided to make it more challenging by not allowing post increment

// when v1 is 0 returns 0

macro dec (v1, temp) {

temp = 0

loop(v1) {

v1 = 0

loop(temp) {

inc(v1)

}

inc(temp)

}

}

//equivalent to x = x – y or x-= y

//when y > x in x – y, then x = 0

macro subtract-equal(v1, v2, temp) {

loop(v2) {

dec(v1,temp)

}

}

//equivalent to x = x / y or x /= y

//throws away remainder

//infinite loop when given 0 for v2 and v1 >= 1

//when 0 / 0 it equals 0 instead of error or infinite loop

macro divide-equal(v1, v2, temp, div-temp) {

div-temp = 0

loop(v1) {

subtract-equal(v1,v2,temp)

loop(v1) {

inc(div-temp)

}

dec(v1,temp)

loop(v1) {

dec(div-temp)

}

inc(v1)

}

v1 = div-temp

}

hopefully i didn't make a mistake. i didn't test this except by hand.

a = 10

b = 2

c = 0

loop (a)

{

k = b

loop (a)

{

m = 1

loop (k)

{

loop (m)

{

c++

m = 0

}

a = subtract(a, 1)

k = 0

}

}

}

// outputs 5

print c

a = 10

b = 2

c = 0

loop (a)

{

loop(a)

{

a = subtract(a, b)

c++;

}

}

// outputs 5

print c