Vertical movement on the fretboard is much simpler than horizontal movement. You might have already realized, but to make it formal, given that your finger is at some fret on the fretboard and has some ai x (where x is one of 0 1 2 3 4 5 6 7 8 9 10 11), the note produced when you move up the fretboard by n many frets has the ai x + n. (Note that when we say up, that means movement towards the center of the instrument meaning that the pitch goes up)
Concretely let's set our anchor as 0*, then the 3rd fret on string 9* is a 0, and when we go up to 7 frets, then the new ai is 0 + 7 which is just 7.
Similarly when we have an ai of x and we move down by n steps our new ai is x - n. An example for this one will be 7* minor with a 10 on string 2* at fret 3. If we release our finger from the string, we will have gone down by 3 frets and therefore our new ai is 7.
Horizontal movement is more complicated. But by remembering our tuning we can figure everything out.
To start things off let's start with a 0 on the thickest string without worrying about which fret it is at or what the anchor note is. We are aware that moving one string up (where up represents the direction towards the thinner strings, where the pitch goes up) increases our current ai by 5, this is true because of the way we have tuned our instrument.
So the thickest string at some fret is 0, at the same fret on the next string up we have 5, then on the next we would have 10, after that we would get 3, since 10 + 5 = 15 = 12 + 3 (and our system is cyclic by 12 steps). Continuing on we would have 8 and then 1:
| 0 | 5 | 10 | 3 | 8 | 1 |
We could do the same procedure but this time going down the strings, so we have a 0 on our thinnest string at some fret and so moving backwards we would subtract 5 from our current ai, to get 7 (5 steps before 0 is 7, since 7 + 5 = 0), again would be 2, then 9, then 4 then finally 11. To give us
| 11 | 4 | 9 | 2 | 7 | 0 |
Going back to our first example, which starts with an ai of 0, we can generalize that even further, considering that our first ai was actually x (where we know x represnts one of the possible ai's) then our horizontal slice would be:
| x + 0 | x + 5 | x + 10 | x + 3 | x + 8 | x + 1 |
So, for example, if x = 3, then the horizontal slice would become
| 3 | 8 | 1 | 6 | 11 | 4 |
Earlier on we discussed that we would never use negative numbers to represent notes or ai's, the reason behind this was that we only want a single way to notate a note or an ai. But we don't have any such restriction when we are performing calculations of computations. And not having this restriction turns out to be quite useful.
To understand why this might be useful, consider an ai of 10, if we wanted to know the ai produced by 10 + 10, then one method might be to do 10 + 10 = 20 = 12 + 8 = 8, so the ai is 8. But an easier understanding would be that going up 10 steps is the same as going down 2 (to see why start at 0 go back 2, you get 11 and then 10), so instead of having to figure out 10 + 10, we can just compute 10 - 2 = 8 which is much quicker. In this type of situation we can call -2 and 10 inverses of each other by 12.
From the above discussion we can derive each of the inverses:
| ai | that ai's inverse | smallest of the two |
|---|---|---|
| 0 | 0 | 0 |
| 1 | -11 | 1 |
| 2 | -10 | 2 |
| 3 | -9 | 3 |
| 4 | -8 | 4 |
| 5 | -7 | 5 |
| 6 | -6 | 6 |
| 7 | -5 | -5 |
| 8 | -4 | -4 |
| 9 | -3 |
-3 |
| 10 | -2 | -2 |
| 11 | -1 | -1 |
The above table will always allow us to minimize computations whenever possible as we will only have to ever consider numbers from the range -6 to 6 when adding or subtracting.
If you recall from before, we determined that given some ai x, that the horizontal movement would produce the following
| x + 0 | x + 5 | x + 10 | x + 3 | x + 8 | x + 1 |
But we know now that we can simplify some of the calculations, so using the inverse table, we get
| x + 0 | x + 5 | x - 2 | x + 3 | x - 4 | x + 1 |
We can also do the same with the reverse table, which originally looked like this:
| x + 11 | x + 4 | x + 9 | x + 2 | x + 7 | x + 0 |
Which gets simplified to
| x - 1 | x + 4 | x - 3 | x + 2 | x - 5 | x + 0 |
No worries if the above isn't perfectly clear straight away, we'll make it clear with many examples.
Consider an ai of 2, on some fret, what would go in the question mark?
| 2 | X | ? | X | X | X |
We can start with:
| x + 0 | x + 5 | x - 2 | x + 3 | x - 4 | x + 1 |
If we were to replace x by 2, then we can see that the ? = 0
Consider an ai of 4, on some fret, what would go in the question mark?
| X | X | ? | X | X | 4 |
We can start with the other table:
| x - 1 | x + 4 | x - 3 | x + 2 | x - 5 | x + 0 |
If we were to replace x by 4, then we can see that the ? = 1
Consider an ai of 8, on some fret, what would go in the question mark?
| X | ? | X | 8 | X | X |
We notice that none of our current tables have the x + 0 at the right location, so replacing x by 8 isn't going to work like it did previously. Instead we just have to understand the way the tables have come about anyways.
When we derived the table :
| x - 1 | x + 4 | x - 3 | x + 2 | x - 5 | x + 0 |
We can see that going two strings down made us subtract 5 twice which was the same as -10 which is the same as +2. This fact holds true no matter which string you start on, so in this case moving two strings left from the 8, gives us 8 + 2 = 10.
From the work done with the examples, we were able to figure out that there is a more general way to understand our movement tables, originally we had seen that the following table works
| x - 1 | x + 4 | x - 3 | x + 2 | x - 5 | x + 0 |
But for example, the fact that when you go two strings left makes you go up by 2, isn't just a property that is true for the thinnest string, it holds true for any string since moving left by two reduces your current number by 10, or rather adds 2. So in a sense every one of the results gotten from the two tables earlier hold in the same manner so we can combine them to create a horizontal movement overlay (note that the numbers do not represent ai's, but distances that need to be added to get to a new ai).
| 6 | -1 | 4 |
-3 | 2 | -5 | 0 | 5 | -2 | 3 | -4 | 1 | -6 |
To see how to use this overlay we can use the following example where our goal is to find out what the ? is.
| X | 3 | X | X | X | ? |
We can then see
| X | 3 | X | X | X | ? | |||||||
| 6 | -1 | 4 |
-3 | 2 | -5 | 0 | 5 | -2 | 3 | -4 | 1 | -6 |
That we just have to subtract 4 from our current ai which gives us 11.
So by placing it so that it's 0 aligns with the current ai we will be able to deduce the other note. In reality we won't ever need all the information in the overlay though we just need to be able to recall the horizontal distance and the change associated with that. So the overlay in another form can look like
| distance from horizontal anchor point | <- (direction) | -> (direction) |
|---|---|---|
| 1 | -5 | 5 |
| 2 | +2 |
-2 |
| 3 | -3 | 3 |
| 4 | +4 | -4 |
| 5 | -1 | +1 |
Due to it's mirrored form, this reduces the amount of information you need to know by half, as by moving in opposite directions inverts the sign.
So finally coming back to the question we answered previously
| X | 3 | X | X | X | ? |
We can alternately just see that the ? is distance 4 away in the -> direction, and thus we must subtract 4, from the above table giving us 11 even quicker.
The table we have just created defines the horizontal movement across the fretboard and should be trained so that you become quick at using it. In order to do that you should use this training app
Instead of memorizing more information, we can simply break diagonal movement into it's horizontal and vertical components and over time we will indirectly memorize it. So for example, given
| X | X | X | X | 1 | X |
| X | X | X | X | X | X |
| ? | X | X | X | X | X |
We can determine that the ? is 4 left so from our movement table, we can see that adds 4 to our current ai, and then we are also 2 up in the vertical direction, so therefore we get 1 + 4 + 2 = 5 + 2 = 7 as the ?
In this manner we will be able to figure any type of movement across the fretboard.