Results 1 to 6 of 6

Math Help - Permute number of on/off possibilities for a 6x6 buss matrix

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    3

    Permute number of on/off possibilities for a 6x6 buss matrix

    I'm an electroacoustic musician using a single microphone input into a laptop Logic Pro audio mixing environment with 6 auxiliary channels. Each auxiliary has unique fx/eq and I can send its audio to any of the other 5 auxiliary channels via their respective input busses:

    Auxiliary 1 -> Buss 2 3 4 5 6
    Auxiliary 2 -> Buss 1 3 4 5 6
    Auxiliary 3 -> Buss 1 2 4 5 6
    Auxiliary 4 -> Buss 1 2 3 5 6
    Auxiliary 5 -> Buss 1 2 3 4 6
    Auxiliary 6 -> Buss 1 2 3 4 5

    I'd like to calculate the number of possible bussing combinations based on whether a buss is either on or off. Not sure if this is a basic or advanced calculation as my 2nd year university calculus and stats days are far too long ago to remember the details.

    Any takers?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by zeug View Post
    I'm an electroacoustic musician using a single microphone input into a laptop Logic Pro audio mixing environment with 6 auxiliary channels. Each auxiliary has unique fx/eq and I can send its audio to any of the other 5 auxiliary channels via their respective input busses:

    Auxiliary 1 -> Buss 2 3 4 5 6
    Auxiliary 2 -> Buss 1 3 4 5 6
    Auxiliary 3 -> Buss 1 2 4 5 6
    Auxiliary 4 -> Buss 1 2 3 5 6
    Auxiliary 5 -> Buss 1 2 3 4 6
    Auxiliary 6 -> Buss 1 2 3 4 5

    I'd like to calculate the number of possible bussing combinations based on whether a buss is either on or off. Not sure if this is a basic or advanced calculation as my 2nd year university calculus and stats days are far too long ago to remember the details.

    Any takers?
    If I understand your question correctly (no guarantee of that), you have 6 switches, each of which can be put in any one of 5 positions independent of the other switch positions, so the total number of possibilities is 5^6.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    3
    Each of the 6 'switches' (aux channels) has 5 switches (busses) any one of which can be on or off (2 positions).

    For instance aux 1 can have all of its 5 busses to the other 5 auxiliaries (2-6) open (sending audio) or closed, or any combination between. Likewise for aux 2 and so on.

    So... does that mean aux 1 has 2^5 or 32 possible switching positions and thus all six auxiliaries together would have 32^6 or about a billion possible combinations?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    EDIT.
    Reading the above reply the number is 32^6=2^{30}=1073741824.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by zeug View Post
    Each of the 6 'switches' (aux channels) has 5 switches (busses) any one of which can be on or off (2 positions).

    For instance aux 1 can have all of its 5 busses to the other 5 auxiliaries (2-6) open (sending audio) or closed, or any combination between. Likewise for aux 2 and so on.

    So... does that mean aux 1 has 2^5 or 32 possible switching positions and thus all six auxiliaries together would have 32^6 or about a billion possible combinations?
    Aha, that's not what I understood before.

    So now, if I understand you rightly, you have an array of 5 x 6 = 30 on/off switches, each of which can be toggled independently of all the others. If this is the case, then there are 2^{30} possible configurations.

    [Edit] Beaten to the punch by Plato!
    Last edited by awkward; April 25th 2010 at 02:08 PM. Reason: Beaten to the punch
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2010
    Posts
    3
    Cool, now 2^{30} aux bussing possibilities is a nice bit of complexity in compositional structure. These auxiliary channels are software analogues of the channel strips on an audio mixer, they have volume faders, pan, EQ and so on:

    .

    I use them along with a single audio channel strip in live performance to create different FX loops on the fly within the 6 auxiliary channels. Their aux busses are the audio connections between them and the Audio Channel is the originating audio source.

    I'd like to complicate things a bit now by introducing another concept - aux channel output can be either on (its buss is receiving audio) or off (no audio bussed to that channel). The original audio input comes from the Audio Channel with 6 busses, each of which can send audio to its respective aux channel 1-6. This audio channel strip then has 2^6 = 64 possible bussing combinations sending audio out to the auxiliaries.

    Two simplest case examples would be audio channel busses all off thus sending no audio to any aux (aux outputs 1-6 = off)... or all busses on to each aux (aux outputs 1-6 = on) ... and 62 other possible variations in between.

    Audio chnl -> busses 1-6 -> aux 1-6 -> output 1-6

    1. Firstly, how many total bussing combinations would there be if I add the 6 buss audio channel to the 2^{30} aux bussing array?
    2. And of this total how many are useful (aux output = on)?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Number of possibilities?
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: June 9th 2011, 01:52 PM
  2. Replies: 3
    Last Post: March 17th 2009, 11:10 AM
  3. number of possibilities
    Posted in the Statistics Forum
    Replies: 2
    Last Post: November 20th 2008, 12:09 AM
  4. number of possibilities
    Posted in the Statistics Forum
    Replies: 2
    Last Post: November 9th 2008, 11:17 AM
  5. number of possibilities
    Posted in the Statistics Forum
    Replies: 6
    Last Post: April 22nd 2008, 11:52 AM

Search Tags


/mathhelpforum @mathhelpforum