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How can I construct this matrix by MMA?

$\left(\begin{array}{cccccc}1 & 2 & 3 & \cdots & n-1 & n \\ n & 1 & 2 & \cdots & n-2 & n-1 \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 2 & 3 & 4 & \cdots & n & 1\end{array}\right)$

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4
  • 3
    $\begingroup$ Tenfold challenge: completed ✓ $\endgroup$
    – bmf
    Apr 19, 2022 at 22:47
  • $\begingroup$ @bmf It's amazing! The answers are all great. I gave the ✓ to the one with the highest vote. $\endgroup$
    – lotus2019
    Apr 20, 2022 at 2:17
  • $\begingroup$ Good choice. The accept should be either the answer by Roman or kglr. Quite fun reading through all the alternatives. Nice question! $\endgroup$
    – bmf
    Apr 20, 2022 at 2:21
  • $\begingroup$ Cool question: I wrote it down, wil think about this over the night. $\endgroup$
    – OCP30pt1c1l1l43-X1z17
    Apr 22, 2022 at 13:33

11 Answers 11

Reset to default
22
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a[n_Integer?Positive] := Array[Mod[#2 - #1, n] + 1 &, {n, n}]

a[6] // MatrixForm

$$ \left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 1 & 2 & 3 & 4 & 5 \\ 5 & 6 & 1 & 2 & 3 & 4 \\ 4 & 5 & 6 & 1 & 2 & 3 \\ 3 & 4 & 5 & 6 & 1 & 2 \\ 2 & 3 & 4 & 5 & 6 & 1 \\ \end{array} \right) $$

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16
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Here is an approach (pulling out the length as a sort of parameter--set len to whatever you want):

With[
  {len=5},
  NestList[RotateRight,Range[len],len-1]]
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16
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tm[n_] := ToeplitzMatrix[RotateRight @ Reverse @ Range @ n, Range @ n] 

TeXForm @ tm[5]

$$\left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 2 & 3 & 4 \\ 4 & 5 & 1 & 2 & 3 \\ 3 & 4 & 5 & 1 & 2 \\ 2 & 3 & 4 & 5 & 1 \\ \end{array} \right)$$

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8
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You can use SparseArray and Band

sa[n_] :=
 Normal@SparseArray[
   Flatten[
    {Band[{1, 1}] -> 1,
     Table[Band[{1, i}] -> i, {i, 2, n}], 
     Diagonal[
      Table[Table[Band[{k, 1}] -> n - x, {k, 2, n}], {x, 0, 
        n - 2}]]},
    1
    ],
   {n, n}
   ]

Grid@Partition[MatrixForm /@ Table[sa[xx], {xx, 2, 13}], 3]

res

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7
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rrmat[n_] := RotateRight[Range[n], #] & /@ Range[0, n - 1]

Test:

MatrixForm /@ (rrmat /@ Range[2, 6])

$$\left\{\left( \begin{array}{cc} 1 & 2 \\ 2 & 1 \\ \end{array} \right),\left( \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \\ \end{array} \right),\left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \\ 3 & 4 & 1 & 2 \\ 2 & 3 & 4 & 1 \\ \end{array} \right),\left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 2 & 3 & 4 \\ 4 & 5 & 1 & 2 & 3 \\ 3 & 4 & 5 & 1 & 2 \\ 2 & 3 & 4 & 5 & 1 \\ \end{array} \right),\left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 1 & 2 & 3 & 4 & 5 \\ 5 & 6 & 1 & 2 & 3 & 4 \\ 4 & 5 & 6 & 1 & 2 & 3 \\ 3 & 4 & 5 & 6 & 1 & 2 \\ 2 & 3 & 4 & 5 & 6 & 1 \\ \end{array} \right)\right\}$$

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7
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Another use of FoldList

len = 6; 
FoldList[RotateRight, Range[len], ConstantArray[1, len - 1]]
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3
  • $\begingroup$ (+1) and just a related minor comment: @N.J.Evans already used FoldList. $\endgroup$
    – bmf
    Apr 19, 2022 at 19:30
  • $\begingroup$ I think/hope the answer is different enough though... $\endgroup$
    – bill s
    Apr 19, 2022 at 19:31
  • $\begingroup$ totally agree with you on this :-) $\endgroup$
    – bmf
    Apr 19, 2022 at 19:32
7
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Another approach using Fold:

With[{len = 6}, Map[Fold[RotateRight, Array[# - 1 &, len, 2], {#}] &, RotateRight@Range[len]]]

Another approach using Nest:

With[{len = 6}, Map[Nest[RotateRight, Array[# - 1 &, len, 2], #] &, RotateRight@Range[len]]]

enter image description here

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6
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Since @bmf challenged us to get to 10:

With[
  {len = 5},
  MapIndexed[RotateRight[#1, #2 - 1] &, ConstantArray[Range[len], len]]]
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1
  • 2
    $\begingroup$ Also a (+1) as the other answer. I think that this 10-fold answers challenge will become a theme for any list and/or array related question. All credits to @Nasser $\endgroup$
    – bmf
    Apr 19, 2022 at 16:27
6
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Is this sufficiently different?

With[
  {len=5},
  NestList[Mod[#-1,len,1]&,Range[len],len-1]]
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1
  • $\begingroup$ For me it's sufficient $\endgroup$
    – bmf
    Apr 19, 2022 at 18:40
6
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Another silly option:

rotmat[n_] := Module[
  {r = Range[n]},
  ArrayReshape[
   FoldList[{r[[-#2[[1]] ;;]], r[[;; -#2[[2]]]]} &, r, 
    Partition[r, 2, 1]], {n, n}]
  ]
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1
  • $\begingroup$ Nicely done and we are one step closer :-) $\endgroup$
    – bmf
    Apr 19, 2022 at 19:15
6
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Since @kglr used the ToeplitzMatrix, I thought it'd be a good idea to use the HankelMatrix. Well, it was not but I managed to get the following

idm[n_] := IdentityMatrix[n]
hm[n_] := LowerTriangularize[Reverse@HankelMatrix[Range@n] + 1, -1]
diag[n_] := 
 Diagonal[Map[Sort, 
     UpperTriangularize[Transpose@Reverse@HankelMatrix[n]], 1], #] & /@
   Range[n - 1]
aux[n_] := Table[diag[n][[xx]] - (xx - 1), {xx, 1, n - 1}]
last[n_] := PadRight[ArrayPad[PadLeft[aux[n]], {0, {1, 0}}], {n, n}]
res[n_] := idm[n] + hm[n] + last[n]

We can check

Grid@Partition[MatrixForm /@ Table[res[i], {i, 2, 13}], 3]

matgrid

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