Woo Hoo Math! Triads.

3 X 3 X 4 = 36 - That's if we use all strings. Maybe... :lol:

That would only be three strings. For that G triad, it would be 3 x 4 X 4 X 4 possible chord tone combinations because every string has 4 G major chord tones except the first string which has only 3.

But the time consuming part would be subtracting those combinations that aren't true triads, such as DGDG (0033) which has no third so doesn't count. I don't know of any quick way to do that.

Pick any 1 string and on that string pick any one note to be the triad base. It appears in 1 place over frets 0-11 so there is one way to play it:

1 x ...


now, we can play the 3rd on any of the 3 remaining strings, but again it appears only once over each string, so we have 3 choices:

1 x 3 . . .


Lastly we have a fifth to play over the remaining 2 strings, appearing once on each string, so giving us 2 choices:

1 x 3 x 2


So for any one triad, there are 6 ways to play it over a 4 strings between frets 1 and 12 without doubling notes. If you allow for a note to be doubled, then you have the choice of doubling the 1, 3 or 5 or no notes, so on the last string you ahve 4 choices:

1 x 3 x 2 x 4


Means that there are 24 ways to play any one triad with one note doubled or not over 4 strings.

I think the quick way is a different logic.

There must be three distinct notes in the triad, and the 'extra' string must double one of the three triad tones. That gives just three spellings:

1351
1353
1355

Now consider that the notes can appear in any sequence - so any one of the four strings can be the root, any of the remaining three can be the third, either of the remaining two will be the fifth, and the last string will be dictated by the note that's doubled. You can express that as:

4x3x2x1 = 24 options

times the three possible spellings is 72 voicings, max. It'll end up being less for any given triad, because some voicings won't be practical to reach.

Note -- you ignore spellings when you come up with your order, so inversions are already counted when you try and account for them again. 24 is right, 72 is one multiplier too many :)

Nope, my math is sound.... however, you bring up a good point; you can't distinguish between the triad note and its doubling; that halves the possibilities, so there are only 36, not 72 - but definately more than 24!

Take a C triad - there are three possible doublings:

CCEG
CEEG
CEGG

Now apply the variations (I'll just do the first spelling):

CCEG
CCGE
CECG
CEGC
CGCE
CGEC
ECCG
ECGC
EGCC
GCCE
GCEC
GECC

That's twelve choices, and the process can be repeated for each of the other two spellings with identical results.

If you're able to distinguish the 'double', each of these breaks down into two choices: CCEG vs. CCEG, etc - that's why there's mathematically 72 :)

NEZTOK, the reason you're getting so many is that you're figuring out every possible voicing that contains any triad tones, whether or not it's a triad...

in other words, one of your 81 possibilities would be CCCC, another EEEE, still others CCCE or CCEE - most won't really be triads.

Now that Neztok's changed the rules to exclude the 12th fret, :twisted: I agree with his results.

3x3x3x3 = 81 possible chord tone combinations. 45 are NOT true triads leaving 36 ways to play the same triad (assuming fingers that are 10 frets long :lol: )

Anyone up for 5 strings? :lol: :lol: :lol:

5 strings would be 243 possible chord tone combinations - but after that I'm stuck because I don't know how to calculate how many of those are non-triads. My brain switched off half an hour ago. :lol:

I'll take a stab at it....

First, the spellings - you can have two unique tones and a triple (R3555), or one unique tone and two doubles (R3355)

For two unique tones, the first one can lie on any of the five strings; the second can lie on any of the remaining four strings, and all the other strings are then fixed on the third tone. So that's 5x4 possibilities, or 20. But we have to multiply that by the number of ways we can choose two unique chord tones, which is three: R3, R5, 35. So we start with 60 possibilities there.

If there is only one unique tone, it can lie on any of the five strings, and be any of the three tones... 15 ways to do that. We then have to figure out how many ways we can uniquely arrange remaining tone 1 or tone 2:

1122
1212
1221
2112
2121
2211

six ways. Times the 15 = 90. Plus the 60 with two unique tones, and I get 150 total.

Yup i figured out what I did wrong -- don't have to start with the root -- can start with any of the three tones.

Let's try it across 6 strings and 24 frets!

So the first of three tones can be on any string in 2 octaves, so that's 6 choices

times

the second tone can be on any of the remaining 5 strings in 2 octaves, so that's 10 choices

times

the third tone can be on any of the remaining 4 strings in 2 octaves so that's 8 choices

times

the 3 remaining strings can each have any of the 3 tones in 2 octaves, or no note at all, so each of those strings has 7 choices.

so:

3 x 10 x 8 x 7 x 7 x 7 = 82,320 combinations

unless my math is wrong, which it probably is :)

Seems like a lot!

In fact, I think it seems like too many - because you've got any tone on any string to start with... so lots of the permutations will be exact duplicates of each other (CEGCEG is just one choice, but you'll have it showing up in about 30 of your possibilities)

I'll stick to the unique tone logic I came up with for five strings... but thinking through this, I have to refine it a little bit.

We can have three basic solutions:

1) two unique tones, one quadrupled
2) one unique tone, one doubled, and one tripled
3) three doubled tones

Two unique tones
Unique tone 1 will lie on any of the 6 strings, and unique tone 2 will lie on any of the remaining 5; that's 6x5=30 places to put 'em. In each of the 30 cases, the other 4 strings are mandated, so 30 choices for location. We've got the same 3 possible combinations of two unique voices (R3, R5, 35), so 30x3=90.

But wait - two octaves! That means a binary choice - for any of the 90 possibilities, each string can be in octave 1 or octave 2; there are 6^2 ways to arrange the octaves for each arrangement of letter names, so we have 90x36=3,240 options.

one unique, one doubled
The unique tone will be any of the three triad tones, and can be on any of the six strings; that's 18. We've now got five strings left, and the doubled note can be on strings 1-2, 1-3, 1-4, 1-5, 2-3, 2-4, 2-5, 3-4, 3-5, or 4-5... 10 choices, with the doubled tone being either of the two tones that aren't the unique one; that's 20. 20x18=360, and multiply by 36 for the binary octaves = 12,960

three doubled
The first string gets any one of three tones, and one of the other five strings must match it. 3x5=15 ways to work that out.

Now we've got two tones left, one of which will be on the next available string (if the first set ends up on strings 1 AND 2, the next avaible is string 3, otherwise it's string 2) - and the matching tone is on one of the other three unassigned strings. So that's 2x3=6, times the 15 ways for the first set = 90. [EDIT: re-reading this, I wasn't very clear in expressing myself... our first 'unassigned' string must be one of the two remaining chord tones, and the matching tone will be on one of the three other 'unassigned' strings - that's how I get the 2x3 here]

The last two strings are the remaining tone, just one choice.

90x36 for the binary octaves comes to 3,240.

So we've got 3240+12960+3240=19,440.

Oh yeah - all of these assume a dictated triad (i.e. how many ways can you do A major?). If we want to find ANY triad, we have 12 chromatic steps, and 4 triad types... 19,440x48=933,120.

Play one a second around the clock, and you have just enough time left this year to play them all :)

Come on, everybody sing, "tooooo much [clap clap] time on my hands....." :lol:

NEZTOK, I sure as Hades hope you are going to get credit for this exercise in some math or statistics class or something. Maybe you said so already.... I admit, I did NOT read every word of every post above.

You guys are just sick..... :twisted: :lol:

Margaret

*cough* fret-span *cough* :twisted: