# 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

### foldLeft and foldRight

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

### Pattern Matching

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

### Range

The

`(1 to n)`

syntax produces a "Range" which is a representation of a sequence of numbers.### 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.