Matrix rotation sort

Question

Lets define a non-empty, unsorted and finite matrix with unique numbers as follow: $$N = \begin{Bmatrix} 4&5&7\\1&3&6 \end{Bmatrix}$$

Lets define 4 matrix moves as:

• ↑* (up): Moves a column up
• ↓* (down): Moves a column down
• →* (right): Moves a row to the right
• ←* (left): Moves a row to the left

The asterisk(*) represents the column/row that is affected by the move (It can be 0-indexed or 1-indexed. Up to you. Please state which one in your answer).

---

The challenge is, using above moves, sort the matrix in a ascendant order (being the top left corner the lowest and the bottom right corner the highest).

Example

Input: $$N=\begin{Bmatrix}4&2&3\\1&5&6 \end{Bmatrix}$$ Possible Output: ↑0 or ↓0. (Notice any of those moves can sort the matrix so both answer are correct)

---

Input: $$N=\begin{Bmatrix}2&3&1\\4&5&6 \end{Bmatrix}$$ Possible Output: →0

---

Input (Example test case): $$N = \begin{Bmatrix} 4&5&7\\1&3&6 \end{Bmatrix}$$ Possible Output: ↑0↑1←1↑2

---

Input: $$N = \begin{Bmatrix} 5&9&6\\ 8&2&4\\ 1&7&3 \end{Bmatrix}$$ Possible Output: ↑0↑2→0→2↑0→2↑1↑2←1

---

Input: $$N = \begin{Bmatrix} 1 & 27 & 28 & 29 & 6 \\10 & 2 & 3 & 4 & 5 \\17 & 7 & 8 & 13 & 9 \\15 & 11 & 12 & 18 & 14 \\26 & 16 & 21 & 19 & 20 \\30 & 22 & 23 & 24 & 25 \end{Bmatrix}$$ Possible Output: ↑2↑1←3→0←3↓0←0←2→3↑3↑4

---

Input: $$N = \begin{Bmatrix} 1 \end{Bmatrix} $$ Output: or any move

---

Input: $$N = \begin{Bmatrix} 1&2\\3&4 \end{Bmatrix} $$ Output:

---

Notes

• There can be different correct outputs (there don't need to be necessarily the same as the test cases or the shortest one)
• You can assume it will be always a way to order the matrix
• Edges connects (like pacman :v)
• There wont be a matrix with more than 9 columns or/and rows
• Assume matrix contains only positive non-zero unique integers
• You can use any 4 distinct values other than numbers to represent the moves (in case of that, please state that in your answer)
• Column/row can be 0 or 1 indexed
• Winning criteria code-golf

---

Extra test cases are always welcome

Answer 1

JavaScript (ES6),  226  219 bytes

Brute force search, using right ("R") and down ("D") moves.

Returns either a string describing the moves, or an empty array if the input matrix is already sorted. Columns and rows in the output are 0-indexed.

f=(m,M=2)=>(g=(s,m)=>m[S='some'](p=r=>r[S](x=>p>(p=x)))?!s[M]&&m[0][S]((_,x,a)=>g(s+'D'+x,m.map(([...r],y)=>(r[x]=(m[y+1]||a)[x])&&r)))|m[S]((_,y)=>g(s+'R'+y,m.map(([...r])=>y--?r:[r.pop(),...r]))):o=s)([],m)?o:f(m,M+2)

Try it online!

Commented

f =                              // f = main recursive function taking:
(m, M = 2) => (                  //   m[] = input matrix; M = maximum length of the solution
  g =                            // g = recursive solver taking:
  (s, m) =>                      //   s = solution, m[] = current matrix
    m[S = 'some'](p =            // we first test whether m[] is sorted
      r =>                       // by iterating on each row
        r[S](x =>                // and each column
          p > (p = x)            // and comparing each cell x with the previous cell p
        )                        //
    ) ?                          // if the matrix is not sorted:
      !s[M] &&                   //   if we haven't reached the maximum length:
      m[0][S]((_, x, a) =>       //     try all 'down' moves:
        g(                       //       do a recursive call:
          s + 'D' + x,           //         append the move to s
          m.map(([...r], y) =>   //         for each row r[] at position y:
            (r[x] =              //           rotate the column x by replacing r[x] with
              (m[y + 1] || a)[x] //           m[y + 1][x] or a[x] for the last row (a = m[0])
            ) && r               //           yield the updated row
      ))) |                      //
      m[S]((_, y) =>             //     try all 'right' moves:
        g(                       //       do a recursive call:
          s + 'R' + y,           //         append the move to s
          m.map(([...r]) =>      //         for each row:
            y-- ?                //           if this is not the row we're looking for:
              r                  //             leave it unchanged
            :                    //           else:
              [r.pop(), ...r]    //             rotate it to the right
      )))                        //
    :                            // else (the matrix is sorted):
      o = s                      //   store the solution in o
)([], m) ?                       // initial call to g(); if we have a solution:
  o                              //   return it
:                                // else:
  f(m, M + 2)                    //   try again with a larger maximum length

Answer 2

Python 2, 296 277 245 Python 3, 200 194 bytes

from numpy import*
def f(p):
 s='';u=[]
 while any(ediff1d(p)<0):u+=[(copy(p),s+f'v{v}',f':,{v}')for v in r_[:shape(p)[1]]]+[(p,s+'>0',0)];p,s,i=u.pop(0);exec(f'p[{i}]=roll(p[{i}],1)')
 return s

Try it online!

-19: unicode arrows weren't required...
-32: slightly reworked, but much slower performance on average.
-45: took some inspiration from @Arnauld's answer. Switched to Python 3 for f'' (-4 bytes)
-6: range( )→r_[: ], diff(ravel( ))→ediff1d( )

---

Exhaustively searches combinations of all possible ↓ moves and →0. Times out on the third test case.

Since →n is equivalent to

↓0↓1...↓(c-1) 	... repeated r-n times
→0
↓0↓1...↓(c-1)	... repeated n times

where r and c are the numbers of rows and columns, these moves are sufficient to find every solution.

---

from numpy import*
def f(p):
    s=''                                    #s: sequence of moves, as string
    u=[]                                    #u: queue of states to check
    while any(ediff1d(p)<0):                #while p is not sorted
        u+=[(copy(p),s+f'v{v}',f':,{v}')    #add p,↓v to queue
            for v in r_[:shape(p)[1]]]      # for all 0<=v<#columns
        u+=[(p,s+'>0',0)]                   #add p,→0
        p,s,i=u.pop(0)                      #get the first item of queue
        exec(f'p[{i}]=roll(p[{i}],1)')      #transform it
    return s                                #return the moves taken

---

>v correspond respectively to →↓. (others undefined)

Answer 3

Jelly, 35 bytes

ṙ€LXȮƊ¦1
ÇZÇZƊ⁾ULXȮOịØ.¤?F⁻Ṣ$$¿,“”Ṫ

Try it online!

Full program. Outputs moves to STDOUT using L for left and R for right. Keeps trying random moves until the matrix is sorted, so not very efficient in terms of speed or algorithmic complexity.