# Scala algorithm: Compute modulo of an exponent without exponentiation

Published

## Algorithm goal

Calculate $$(b^e) \% m$$ without using exponentiation. This could be useful where the exponent is very big, potentially overflowing.

## Test cases in Scala

assert(exponentModulo(base = 2, exponent = 3, modulo = 3) == 2)
assert(exponentModulo(base = 4, exponent = 3, modulo = 3) == 1)
assert(
exponentModulo(base = 5, exponent = 10, modulo = 7) ==
exponentModuloRaw(base = 5, exponent = 10, modulo = 7)
)

## Algorithm in Scala

7 lines of Scala (compatible versions 2.13 & 3.0), showing how concise Scala can be!

## Explanation

Note that $$(a*b) \% m = ((a \% m) * (b \% m)) \% m$$. Proof: if $$a = xm + y$$ and $$b = wm + z$$, then , $$a * b = xwm^2 + (w + x)m + yz$$ and $$(a * b) \% m = yz \% m$$; $$((a \% m) * (b \% m)) \% m = (y * z) \% m = yz \% m$$.

Then, $$(b^e) \% m = (((b ^ {e - 1}) \% m) * (b \% m)) \% m$$. (this is Â© from www.scala-algorithms.com)

## Scala concepts & Hints

1. ### foldLeft and foldRight

A 'fold' allows you to perform the equivalent of a for-loop, but with a lot less code.

def foldMutable[I, O](initialState: O)(items: List[I])(f: (O, I) => O): O =
items.foldLeft(initialState)(f)

2. ### Pattern Matching

Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.

assert("Hello World".collect {
case character if Character.isUpperCase(character) => character.toLower
} == "hw")

3. ### Range

The (1 to n) syntax produces a "Range" which is a representation of a sequence of numbers.

assert((1 to 5).toString == "Range 1 to 5")

assert((1 to 5).reverse.toString() == "Range 5 to 1 by -1")

assert((1 to 5).toList == List(1, 2, 3, 4, 5))

4. ### Stack Safety

Stack safety is present where a function cannot crash due to overflowing the limit of number of recursive calls.

This function will work for n = 5, but will not work for n = 2000 (crash with java.lang.StackOverflowError) - however there is a way to fix it :-)

In Scala Algorithms, we try to write the algorithms in a stack-safe way, where possible, so that when you use the algorithms, they will not crash on large inputs. However, stack-safe implementations are often more complex, and in some cases, overly complex, for the task at hand.

def sum(from: Int, until: Int): Int =
if (from == until) until else from + sum(from + 1, until)

def thisWillSucceed: Int = sum(1, 5)

def thisWillFail: Int = sum(1, 300)

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