power and modulo on the fly for big numbers - scala

Question

I raise some basis b to the power p and take the modulo m of that.
Let's assume b=55170 or 55172 and m=3043839241 (which happens to be the square of 55171). The linux-calculator bc gives the results (we need this for control):
echo "p=5606;b=55171;m=b*b;((b-1)^p)%m;((b+1)^p)%m" | bc
2734550616
309288627
Now calculating 55170^5606 gives a somewhat large number, but since I have to do a modulooperation, I can circumvent the usage of BigInt, I thought, because of:
(a*b) % c == ((a%c) * (b%c))%c i.e.
(9*7) % 5 == ((9%5) * (7%5))%5 =>
63 % 5 == (4 * 2) %5 =>
3 == 8 % 5
... and a^d = a^(b+c) = a^b * a^c, therefore I can divide b+c by 2, which gives, for even or odd ds d/2 and d-(d/2), so for 8^5 I can calculate 8^2 * 8^3.
So my (defective) method, which always cut's off the divisor on the fly looks like that:
def powMod (b: Long, pot: Int, mod: Long) : Long = {
if (pot == 1) b % mod else {
val pot2 = pot/2
val pm1 = powMod (b, pot2, mod)
val pm2 = powMod (b, pot-pot2, mod)
(pm1 * pm2) % mod
}
}
and feeded with some values,
powMod (55170, 5606, 3043839241L)
res2: Long = 1885539617
powMod (55172, 5606, 3043839241L)
res4: Long = 309288627
As we can see, the second result is exactly the same as the one above, but the first one looks quiet different. I'm doing a lot of such calculations, and they seem to be accurate as long as they stay in the range of Int, but I can't see any error. Using a BigInt works as well, but is way too slow:
def calc2 (n: Int, pri: Long) = {
val p: BigInt = pri
val p3 = p * p
val p1 = (p-1).pow (n) % (p3)
val p2 = (p+1).pow (n) % (p3)
print ("p1: " + p1 + " p2: " + p2)
}
calc2 (5606, 55171)
p1: 2734550616 p2: 309288627
(same result as with bc) Can somebody see the error in powMod?

Answer 1

I think the answer is here:
scala> math.sqrt(Long.MaxValue).toLong < 3043839241L
res9: Boolean = true
That means you can have a long overflow even for numbers which are less than that particular module value. Let's try to catch it:
scala> def powMod (b: Long, pot: Int, mod: Long) : Long = {
| if (pot == 1) b % mod else {
| val pot2 = pot/2
| val pm1 = powMod (b, pot2, mod)
| val pm2 = powMod (b, pot-pot2, mod)
| val partial = ((pm1 % mod) * (pm2 % mod)).ensuring(res =>
| res > pm1 % mod && res > pm2 % mod, "Long overflow multiplying "+pm1+" by "+pm2)
| partial % mod
| }
| }
powMod: (b: Long,pot: Int,mod: Long)Long
scala> powMod (55170, 5606, 3043839241L)
java.lang.AssertionError: assertion failed: Long overflow multiplying 3042625480 by 3042625480
There you have it.

Answer 2

Not familiar with Scala, but...
def powMod (b: Long, pot: Int, mod: Long) : Long = {
if (pot == 1) b % mod else {
val pot2 = pot/2
val pm1 = powMod (b, pot, mod)
val pm2 = powMod (b, pot-pot2, mod)
(pm1 * pm2) % mod
}
}
Did you mean
val pm1 = powMod (b, pot2, mod)
Notice the pot2 instead of pot.
Strangely, it seems that this should loop forever/overflow the stack, but who knows what Scala is doing.

Answer 3

Ok fellows, it took me some time, and finally destroyed a long but never proven assumption, which was, that if you multiply two 64-bit-positive integral values (aka: Longs, and practically only 63-bit, after all), you could overrun, and get negative values, but not get an overrun to reach positive (but wrong) values again.
So I had tried to put a guard into my code, to calculate my value with BigInt, it too big, but the guard was insufficient, because I tested for res < 0. res < pm1 && res < pm2 isn't sufficient too.
To increase the speed I used a mutable.HashMap, and now the code looks like this:
val MVL : Long = Integer.MAX_VALUE
var modPow = new scala.collection.mutable.HashMap [(Long, Int, Long), Long ] ()
def powMod (b: Long, pot: Int, mod: Long) : Long = {
if (pot == 1) b % mod else modPow.getOrElseUpdate ((b, pot, mod), {
val pot2= pot/2
val pm1 = powMod (b, pot2, mod)
val pm2 = powMod (b, pot-pot2, mod)
val res = (pm1 * pm2)
// avoid Long-overrun
if (pm1 < MVL && pm2 < MVL)
res % mod else {
val f1: BigInt = pm1
val f2: BigInt = pm2
val erg = (f1 * f2) % mod
erg.longValue
}
})
}
You might ask yourself, whether the Long-declared MVL is really needed, or whether a
if (pm1 < Integer.MAX_VALUE && ...
would have worked too. No. It wouldn't. :) Another trap to avoid. :)
Finally it is pretty fast and correct and I learned two lessons about overruns and MAX_VALUE - comparision.