# Chapter 2: Splitting Bars
### Recursive problem #2: Split
Imagine a bar of solid gold.
You want to find out how much it weighs, but you're held back by one thing:
**You only know that a square of gold weighs 5 pounds.**
Your gold bar is larger than a square.
As we did before, let's go ahead and define a function. The function **split** has one parameter: **gold\_bar**, the bar of gold.
**split** aims to return how many pounds the gold bar weighs.
```python
def split(gold_bar):
#this function should return how much "gold_bar" weighs!
```
If our problem is weighing the gold bar, let's think about how we can make the problem smaller! True to the function's name, why don't we try **splitting** the gold bar...
In this case, we can **split** the gold bar into:
```python
square_of_gold + (rest_of_bar)
```
The square of gold is a known quantity: we know it weighs 5 pounds. The rest of the bar is an unsolved problem. However, it's smaller than our original problem! It's important to realize that when added together, these two **smaller** problems are **equivalent** to our original problem, but they're easier for us to solve.
We can take the rest of the bar and **split** it again, and repeat this process. What does this look like in code?
```python
def split(gold_bar):
return 5 + split(rest_of_bar)
```
We keep splitting the rest of the bar in order to make our problem **smaller**.
```python
split(gold_bar) -> split(rest_of_bar) -> split(rest of rest_of_bar) -> ...
```
Again, because we are making the problem smaller and smaller; this line is the **recursive call**.
For now, let's continue splitting.
```python
square_of_gold + split(rest_of_bar)
square_of_gold + (square_of_gold + split(rest_of_bar))
```
And split the rest of the bar again....
We are left with
```python
square_of_gold + ( square_of_gold + (square_of_gold + split(rest_of_bar))
```
but the rest of the bar is a square! That's a known quantity! This is **equivalent** to
```python
square_of_gold + ( square_of_gold + (square_of_gold + square_of_gold))
```
Which is 5 + 5 + 5 + 5 = 20 pounds!
By now you should have realized that our **base case** is when we have split the gold bar enough times so that it is a square, a known quantity. In code:
```python
def split(gold_bar):
if gold_bar == square_of_gold:
return 5
return 5 + split(rest_of_bar)
```
Now, every time we split, we see if the **gold\_bar** we're splitting is a square; if it is, we **return** it, and we're finished! Otherwise, we return square of gold + **split**(rest of bar), knowing that **split** will make our problem smaller.
If you're still confused or not completely sold, here's a quick recap to show you what's happening:
```python
split(gold_bar) -> square + split(rest_of_bar) #4 squares = 20 pounds
V
(square + split(rest_of_previous_bar)) #3 squares
V
(square + split(rest_of_previous_bar)) #2 squares
V
square
```
### Great, now what?
So, what was the point of this example? The goal of this example was to introduce the idea of **split**ting a problem into two smaller ones; In our example, we **split** the problem of weighing a gold bar into the problems of:
1. Weighing a square of gold
2. Weighing the rest of the bar
The first problem we could solve easily, and the second could be **split** further. In order to generalize this kind of thinking, consider this **split**ting strategy as splitting a problem into:
1. A solved or easily solvable problem
2. A problem that can be further **split**
I'll give concrete examples in the next chapter, to show you how we can use this approach in real code.
---
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