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I want to write a function that can automatically analyze any square matrices with periodic band diagonals and give continuations of them. But I can't figure out an elegant way to do this.

Suppose we have written such a function called continuation, and suppose we have an example square matrix mat of order 8 as

$\small \mathtt{mat}=\begin{pmatrix} y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 & 0 & 0 \\ 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 & 0 & 0 \\ 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 \\ \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 \\ 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} \\ 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 \\ 0 & 0 & 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} \\ 0 & 0 & 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y \\ \end{pmatrix}$

then continuation[mat][n] can give a square matrix that is a continuation of original mat with order n. For example:

continuation[mat][9]

should give a continuation of mat of order 9 as following

$\small \begin{pmatrix} y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 & 0 & 0 & 0 \\ 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 & 0 & 0 & 0 \\ 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 & 0 \\ \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 & 0 \\ 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 \\ 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 \\ 0 & 0 & 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 \\ 0 & 0 & 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0.2 & 0 & y+0.2 \\ \end{pmatrix}$

How to write such a continuation function which can automatically analyze square matrices with periodic diagonal band and give general continuation version?

The list representation of mat is here:

mat = {{0.2 + y, 2 t1, 0.2, t1, 0, 0, 0, 0},
       {2 t1, 0.2 - y, 0, 0.2, 0, 0, 0, 0},
       {0.2, 0, 0.2 + y, 2 t1, 0.2, t1, 0, 0},
       {t1, 0.2, 2 t1, 0.2 - y, 0, 0.2, 0, 0},
       {0, 0, 0.2, 0, 0.2 + y, 2 t1, 0.2, t1},
       {0, 0, t1, 0.2, 2 t1, 0.2 - y, 0, 0.2},
       {0, 0, 0, 0, 0.2, 0, 0.2 + y, 2 t1},
       {0, 0, 0, 0, t1, 0.2, 2 t1, 0.2 - y}}
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  • $\begingroup$ This is quite a complex task, unless you specify what exact type of continuations you mean... You say periodic, so I would guess start by writing a peroid-finding function for lists... $\endgroup$
    – Per Alexandersson
    Commented Dec 15, 2014 at 11:47

3 Answers 3

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Using the function FindTransientRepeat[], we can do the following:

mat = SparseArray[mat];
m = 10; {lo, up} = MinMax[mat["NonzeroPositions"].{-1, 1}];
sp = SparseArray[Table[With[{d = Diagonal[mat, k], p = Abs[k]}, 
                            Band[If[k >= 0, Identity, Reverse][{1, p + 1}]] ->
                            (PadRight[#1, m - p, #2] & @@ FindTransientRepeat[d, 2])],
                       {k, lo, up}]];

MatrixForm[sp]

$$\tiny\begin{pmatrix} y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 & 0 & 0 \\ \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 \\ 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} \\ 0 & 0 & 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} \\ 0 & 0 & 0 & 0 & 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y \end{pmatrix}$$

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  • $\begingroup$ Thank you so much for this new FindTransientRepeat[] function +1 $\endgroup$
    – matheorem
    Commented Aug 15, 2017 at 7:59
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Since no one gave a satisfactory answer, I figured out a way myself which combines Band, Diagonal and SparseArray.

  1. The Diagonal[m,k]gives the elements on the $k^{th}$ diagonal of m, illustrated as below:

    enter image description here

  2. Band in SparseArray can repeat the values cyclically, for example:

    SparseArray[Band[{1, 1}, {5, 5}] -> {x, y, z}, {5, 5}] // MatrixForm

    gives

    $\begin{pmatrix} x & 0 & 0 & 0 & 0 \\ 0 & y & 0 & 0 & 0 \\ 0 & 0 & z & 0 & 0 \\ 0 & 0 & 0 & x & 0 \\ 0 & 0 & 0 & 0 & y \\ \end{pmatrix}$

  3. Tally can find the cyclic sequence, since Tally tallies the elements in list, listing all distinct elements together with their multiplicities. for example:

    Tally[Diagonal[mat]]

    gives

    {{0.2 + y, 4}, {0.2 - y, 4}}

    so 

    Tally[Diagonal[mat]][[;;,1]]

    give the cyclic sequence {0.2 + y, 0.2 - y}

So combine these 3 features we got a version of the code of continuation as below:

Clear[continuation];
continuation[mat_, order_] := 
 Module[{DiagtoBand, orderofmat = Length@mat},

  (*DiagtoBand is used for changing diagonal index to Band parameter*)  

  DiagtoBand[i_] := 
   If[Positive[
     i], {{1, i + 1}, {order - i, order}}, {{1 - i, 1}, {order, 
      order + i}}];

  (*if the order required is less then the order of mat, 
  then we just simply ArrayPad it. On the other hand, we continue it. *)

  If[order <= orderofmat,
   ArrayPad[mat, {0, order - orderofmat}], 
   MatrixForm@
    Normal@SparseArray[
      Table[(Band @@ DiagtoBand[i]) -> 
        Tally[Diagonal[mat, i]][[;; , 1]], {i, -orderofmat + 1, 
        orderofmat - 1}], {order, order}]]]

The continuation function is now universal. 

continuation[mat,9]

gives

$\small \begin{pmatrix} y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 & 0 & 0 & 0 \\ 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 & 0 & 0 & 0 \\ 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 & 0 & 0 \\ \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 & 0 & 0 \\ 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 & \text{t1} & 0 \\ 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 & 0.2 & 0 \\ 0 & 0 & 0 & 0 & 0.2 & 0 & y+0.2 & 2 \text{t1} & 0.2 \\ 0 & 0 & 0 & 0 & \text{t1} & 0.2 & 2 \text{t1} & 0.2\, -y & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0.2 & 0 & y+0.2 \\ \end{pmatrix}$

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  • $\begingroup$ You might be able to do better with FindLinearRecurrence than with the Tally approach. In any case, glad to see you've come up with something you can use. $\endgroup$
    – bill s
    Commented Dec 16, 2014 at 4:09
  • $\begingroup$ @bills Thank you for suggesting FindLinearRecurrence. But I found that FindLinearRecurrence needs at least 5 elements in a list. So I think I just stick with Tally approach which is easy and understandable. $\endgroup$
    – matheorem
    Commented Dec 16, 2014 at 6:08
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The functions Band and SparseArray may be what you are looking for:

n = {9, 9}; 
Normal[SparseArray[{Band[{1, 1}] -> y + 0.2, Band[{1, 3}] -> 0.2, 
    Band[{1, 2}, n] -> {2 t1, 0}, Band[{3, 1}, n] -> {t1, 0}},  n]] // MatrixForm

Just increase the dimensions in the variable n and the specified pattern(s) continue.

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  • $\begingroup$ Thank you! I know band function. But you misunderstood me. I want a function continuation which can automatically analyze any square matrix mat we feed to it, and give general version continuation[mat]? $\endgroup$
    – matheorem
    Commented Dec 15, 2014 at 6:19
  • $\begingroup$ What you gave is a manual version. $\endgroup$
    – matheorem
    Commented Dec 15, 2014 at 6:27
  • $\begingroup$ So -- you want a function that's like FindSequenceFunction but works on matrices as well? $\endgroup$
    – bill s
    Commented Dec 15, 2014 at 14:29

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