Math help - Rock Band guitar ...

[Jeff Wilder]

How do i calculate the number of possible fret combinations when playing guitar in Rock Band? Parameters:

(1) All single frets can be played.
(2) Any two frets can be played as a chord.
(3) Three frets of any four adjacent frets can be played as a chord.
(4) There are no four-fret chords.

So if the frets are ABCDE ...

(1) 5 possibilities, each one of ABCDE.
(2) 10 possibilities, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.
(3) ? possibilities. E.g., ABD (but not ABE, because A to E is more than four frets).
(4) Eliminate ? possibilities, like ACDE, ABDE, and so on because there aren't any four-fret chords.

So how many combinations are there, and how would I figure that out? (I have never figured out permutations and combinations when restrictions are involved.)

 

[freyar]

I don't know the first thing about guitars, but I can do combinations! Wikipedia says most guitars have 6 strings, though...

Anyway, for the 5 string example:
1) 5 single frets
2) 10 = "5 choose 2" = 5*4/2 doubles
3) There are two groups of 4 adjacent frets (ABCD and BCDE), each of which have 4 possible combinations of triples played at once. Then you've double counted BCD, so there must be 7 total.
4) In this case, you're eliminating 5 4-fret non-chords

If you have N strings on your guitar
1) N singles
2) N choose 2 = N*(N-1)/2 doubles
3) N-3 groups of 4 adjacent frets with 4 chords each. However, for each of the first N-4 groups, you double count the 3-string chord that leaves off the first string of the group (since that is the same as the chord in the next group that leaves off the last string). In total you have 4*(N-3)-(N-4) = 3N-8. I think.
5) Eliminate N choose 4 = N(N-1)(N-2)(N-3)/24 nonexistent 4-string chords.

 

[Janx]

the OP is talking about the video game controller for Rock Band, not real guitars.

there are 5 fret buttons

1 button = 1^5 combinations
2 ubtton = 2^5 combinations
3 buttons = 3^5 combinations

resolve the "to the power of" bits, and then add them up

I'm assuming I expressed it correctly that 2 buttons out of five is 2^5 and not 5^2, which I'm pretty sure I'm right because 8^2 expresses 256 combinations which is a byte.

A real guitar typically has 6 strings and 22-24 frets. A true chord requires 3 notes, though a power chord is only 2 (note a true chord, it uses the root, and the fifth). A chord may use more notes than 3.

Thus, on a real guitar, for a "chord" you might hold down 2-6 notes and strum 2-6 strings. Not all chords use all the strings.

I also note that you called the buttons ABCDE, implying using the notes in order.

Assuming you were writing a game to play the guitar just as music:

If you were going to make each button be a chord, you might be better off thinking of them in terms of common chord progressions.

Thus, if the song was in the key of F, and the rock tune went in a I-IV-V-I progression, you'd want each button to represent that (not using all the buttons, you could make them be useful by programming them to be III and VI.

F =I (6th string, 1st fret)
G =II (6th string, 3rd fret)
A =III (5th string, open)
Ab=IV (5th string, 1st fret)
C =V (5th string, 3rd fret)
D =VI (4th string, open)
Eb=VII (4th string, 1st fret)
F = VIII (4th string, 3rd fret)

I play bass and guitar, so I had to learn where all the notes are. From there, learning the parts of a chord were needed, thus I learned where they are relative to the root note (in this example low F).

What they translates to is you'd want your notes on the buttons to be:
F, A, Ab, C, D

Play them as chords, and you could cover a ton of rock tunes.

 

[TDRandall]

Been literally decades since I had to figure out a problem like this one. I think the easiest way to attack and explain it is as "all possibilities - the undesirable ones". Presuming I understand the conditions correctly:

2^^5 (i.e. 32) is all possible combination with five buttons (each button can be either "down" or "up")
- 1 illegal five fret chord (all buttons down)
- 5 possible illegal four fret chords (one for each where only one button is not down)
- 3 three button illegal fret chords, where two of the three buttons down are the outside ones:

= 23 possible valid combinations

 

[Janx]

pretty sure TD's math is wrong.


Also, now that I pulled out the calculator and used the x^y function, I got the notation backwards.

It is 5^1 + 5^2 + 5^3 = 155 combinations

1^5 yields 1, which we clearly know is not the right answer, wheras as 5^2 =25 and 5^3=125 which are plausible, even if we're not math experts. Math by deduction is useful for those who aren't good at actual math.

 

[TDRandall]

You are ultimately correct that my final result is wrong, Janx. I also should have also removed the zero button possibility, so the final answer should have been 22.

But go ahead, show me either the 25 or 32 (depending on which way you want to read the ^ function) hand positions where you are touching the table with only two fingers. Now show me either the 125 or 243 three finger combinations.

Combinations and probabilities were notorious for throwing gut level estimates completely out the window, at least for me.

 

[freyar]

TD's right now (subtracting that 0-string possibility). Janx, what you're doing is appropriate if you play one note than the next: you're saying that playing A-B is different than playing B-A and also that you can play A-A. When you play the notes simultaneously, order doesn't count (and you can't play the same note twice at once). You don't want to use powers to do this counting, you want to use combinations, called "5 choose 2" for example and written something like 5C2. The formula is NCk = N!/(N-k)! k!, where ! is the factorial N!=N(N-1)(N-2)... 1. The combination means that you take one key to play, then there are N-1 other keys left to play at the same time, etc. If you're playing k keys, that gives you N!/(N-k)!. But you shouldn't count playing A first then B separately from B then A, so you've overcounted by what turns out to be k!.

 

[Janx]

ah, i see now the error of my math. Freyar's explanation nailed it, by pointing out the actual difference.

As a manually tedious exercise, here's the combinations:

1 finger: 5
a,b,c,d,e

2 fingers: 10
ab,ac,ad,ae
bc,bd,be
cd,ce
de

3 fingers: 9
abc,abd,abe,acd,ace
bcd,bce,bde
cde

5+10+9=24 combinations

I did it manually. In theory, for any fingering, the letters are not repeated in any other fingering (either as exact copy, or by re-arrangement, but same letters where abc=cba=bac)

I suspect it is correct, if I made a mistake, it is more likely that I missed a combination, than included a repeat.

For a smaller problem, sometimes there's merit in doing it the hardware to double check the math, and to appreciate why the math works.

In this case, I suspect all the math is wrong, but in TD's case, being off by 2 isn't that bad.

In reading the OP's fine print, he wants to eliminate AE, because RB/GH has no large span chords. Thus subtracting those, would bring us closer in line to TD, where he probably did that subtraction.

Showing one's work helps...

 

[TDRandall]

Yes thank you Freyar for figuring it out. With the numbers he was throwing around I knew Janx was working along some plan, but I wasn't seeing it.

Unfortunately, being at least the part time bean-counter here in the office it's going to dig at me until we have matching numbers.

Looking closer I think the abe and ace combos shouldn't have been included in the explicit list, as the five key span for a 3 finger chord was specifically excluded by the parameters. So we could meet at 22.

I don't know the game so don't know about the AE two button chord. Parameter #2 seems to explicitly allow it, but parameter #3 implicitly disallows it - although that line probably addresses only to three button chords.

I don't know, I don't play it. Sounds like Janx has and knows the AE chord is never used, so after using my "phone a friend" help line I will lock in my final answer for Regis at ....

21 !!!!!

Ahhhhh... I can sleep soundly tonight knowing every paper clip has finally been accounted for.

 

[Janx]

At this point, TD/freyar's formula is right for figuring out the actual combinations with math, as is my manual method.

that yields a base of 24, and then subtract whatever "undesired" exceptions like AE, ABE, and ACE for spanning all 5 frets

Thus, as TD says, it would be 21 remaining valid fingerings.

I enjoy RB/GH, but I don't spend a lot of time at it. I could spend oodles of time getting good at it, or spend oodles of time getting good at real guitar...

as a side note, I use the term "fingering" because that's what you call where to place your fingers to do a specific chord. As in "show me the fingering for the Fm7 chord"

good times with math and guitars...

Now that we've all done JeffWilder's work, would he mind explaining why he wants to know? Or be known as a thankless and random question asker?