A polyomino with \$n\$ cells is a shape consisting of \$n\$ equal squares connected edge to edge.
No free polyomino is the rotation, translation or reflection (or a combination of these transformations, i.e. an isometry) of another (they're equal).
The sequence of free polyominoes with \$n\$ cells is as follows:

1, 1, 1, 2, 5, 12, 35, 108, 369, 1285, ...

Polyominoes image


  • \$n=0\$ (1) may be excluded
  • This is A000105
  • Standard rules: you may output the \$n^{th}\$ term, all terms up to the \$n\$ terms or the entire sequence
  • This is , so the shortest answer wins
  • 1
    \$\begingroup\$ Related (one-sided instead of free) \$\endgroup\$
    – Arnauld
    Apr 5, 2023 at 10:37

5 Answers 5


JavaScript (ES10), 259 bytes

For now, this is just a slightly modified version of my answer to this other challenge so that reflections are discarded. Supports \$n=0\$.


Try it online!


05AB1E, 73 bytes

Should have been 1 byte less by replacing })}€ê with }){}, but the many nested maps/loops cause a bug in 05AB1E here (})ê} and })}€{ also doesn't work strangely enough..)


Basically the exact same approach as my answer for the related challenge, but with a single added Duplicate (in step 4) to also remove reflections in addition to the rotations.

Extremely slow brute-force, so is only able to output up to \$a(4)\$ on TIO.

Try it online or verify the first few results.

Also mostly a copy-paste from my answer of the related challenge, except for step 4.

Step 1: Create all possible \$n^2\$-sized lists using 0s and 1s, consisting of \$n\$ amount of 1s:

1Ý          # Push pair [0,1]
  In        # Push the squared input
    ã       # Cartesian power
ʒ           # Filter this list of lists by:
   i        #  If
 O          #  the sum of the current list
  Q         #  is equal to the (implicit) input-integer:
            #   Continue with the check in step 2 below
            #  (implicit else: implicitly use the implicit input for the filter;
            #   this is only truthy for edge case n=1, which fails step 2 due to the `ü3`)

Try just this first step online (without trailing i).

Step 2: Filter it further to only keep single polynominos, using a flood-fill approach:

 y          #   Push the current list again
  ƶ         #   Multiply each value by its 1-based index
   Iô       #   Convert the list to an input-by-input block
     ©      #   Store this block in variable `®` (without popping)
 Δ          #   Loop until the result no longer changes to flood-fill the matrix:
  2Fø0δ.ø}  #    Add a border of 0s around the matrix:
  2F     }  #     Loop 2 times:
    ø       #      Zip/transpose; swapping rows/columns
      δ     #      Map over each row:
     0 .ø   #       Add a leading/trailing 0
  2Fø€ü3}   #    Convert it into overlapping 3x3 blocks: 
  2F    }   #     Loop 2 times again:
    ø       #      Zip/transpose; swapping rows/columns
     €      #      Map over each inner list:
      ü3    #       Convert it to a list of overlapping triplets
  ®Ā        #    Push matrix `®` and convert all its positive values back to 1s
    *       #    Multiply each 3x3 block by this matrix of 0s/1s (so 0s will remain 0s)
  εεÅsyøÅsM #    Get the largest value from the horizontal/vertical cross of each 3x3 block:
  εε        #     Nested map over each 3x3 block:
    Ås      #       Pop and push its middle row
      y     #       Push the 3x3 block again
       ø    #       Zip/transpose; swapping rows/columns
        Ås  #       Pop and push its middle rows as well (the middle column)
          M #       Push the flattened maximum of the entire (scoped) stack,
            #       which is the flattened maximum of the cross of the current 3x3 block
  }}        #     Close the nested map
 }˜         #   After the flood-fill loop: flatten the block to a list
   Ù        #   Uniquify its values
    g       #   Pop and push its length
     <      #   Decrease it by 1 to account for the 0s
            #   (only 1 is truthy in 05AB1E, so only single islands remain)
]           # Close both the if-statement and filter

Try just the first two steps online.

Step 3: Convert all valid lists to matrices, and slash off any rows/columns of 0s to have the actual polynominos:

ε           # Map over each inner list
 Iô         #  Convert it to an n-by-n block
   2F       #  Inner loop 2 times:
     ø      #   Zip/transpose; swapping rows/columns
      ʒ     #   Filter this list of rows by:
       à    #    Get the maximum of the row (so if it only contains 0s, it'll be removed)
      }     #   Close the filter
    }       #  Close the inner loop

Try just the first three steps online.

Step 4: Remove all duplicated rotations and reflections, by first converting each polynomino to a quartet of its four sorted rotations, then get the reflection of each rotation, and then uniquify that list of octets.

 4F         #  Inner loop 4 times:
   D        #   Duplicate the current polynomino-matrix
    í       #   Reverse each inner row to reflect it
     D      #   Duplicate this new reflected polynomino-matrix again
      ø     #   Zip/transpose the matrix; swapping rows/columns
  })        #  After the loop: wrap the eight rotations + reflections on the stack into a list

            # Explanation if the 05AB1E bug mentioned at the top wasn't present:
    {       #  Sort the octet of rotations + reflections
}Ù          # After the map: uniquify the list of polynomino-rotations/reflections

            # Actual explanation with bug:
}€          # After the map: open a new map:
  ê         #  Sort and uniquify each octet
   Ù        # After the map: uniquify the list of distinct polynomino-orientations/reflections

Try just the first four steps online.

Step 5: Get the amount of unique polynominos left, and output it as result:

g           # Pop and push the length
            # (which is output implicitly as result)

Python3, 1374 bytes*:

def U(b):
 while q:
  if(T:=[i[::-1]for i in b])not in s:q+=T,;s+=T,
  if(T:=[*map(list,zip(*[i[::-1]for i in zip(*b)]))])not in s:q+=T,;s+=T,
  if(T:=[*map(list,zip(*b))])not in s:q+=T,;s+=T,
 return s
Z=lambda B:[[j[0]if j else j for j in i]for i in B]
def S(b):
 b=[[*i]for i in b if any(i)]
 if 1-any([*zip(*b)][0]):b=[i[1:]for i in b]
 if 1-any([*zip(*b)][-1]):b=[i[:-1]for i in b]
 return b
C=lambda b:[(x,y)for x,t in E(b)for y,j in E(t)if j]
def M(b):
 while q:
  for j,k in P:
   if c in t:t.remove(c);q+=c,
V=lambda b,x,y:x*y==0 or len(b)==x or len(b[0])==y
def f(n):
 b=[[[1,0]for _ in R(w)]for _ in R(n//(w or 1))]+[[[1,0]for _ in R(n%(w or 1))]]*(n%(w or 1)>0)
 B=[i+[0]*(max(map(len,b))-len(i))for i in b]
 while q:
  for x,y in c:
   if 0==b[x][y][1]:
    for X,Y in c:
      for j,k in P:
       if(J,K)not in c:
        if K==len(B[0]):B=[i+[0]for i in B]
        if K<0:B=[[0]+i for i in B];K=0
        if J==len(B):B=B+[[0 for _ in B[0]]]
        if J<0:B=[[0 for _ in B[0]]]+B;J=0
        if M(S(Z(B)))and all(S(Q)not in s for Q in U(S(Z(B)))):q+=B,;s+=S(Z(B)),
 return len(s)

Try it online!

* Rather long in bytes, but computes solutions up to and including \$n = 9\$ in under 40 seconds on TIO.

  • \$\begingroup\$ 0==any(...) -> 1-any(...) (in function S) \$\endgroup\$
    – The Thonnu
    Apr 6, 2023 at 15:26
  • \$\begingroup\$ q+=[T] -> q+=T, (you have 9 similar occurences where that can be changed) \$\endgroup\$
    – The Thonnu
    Apr 6, 2023 at 15:27
  • 1
    \$\begingroup\$ @TheThonnu Thank you, updated. \$\endgroup\$
    – Ajax1234
    Apr 6, 2023 at 15:53

R, 205 199 bytes

f=\(n,o={},v=0,u=0,`~`=c,`?`=sort,d=-1~1i~1~-1i){for(j in u)F=F+`if`(length(p<-?j~o)<n,f(n,p,v,setdiff((u=u[-1])~j+d,v<-j~v)),2^(-all(apply(Conj(p)%o%d,2,a<-\(x)any(diff(p-?x))))-a(-p)-a(p*1i))/n);F}

Attempt This Online!

A variation of my answer to the related challenge adapted to account for reflections.

Specifically, since the polyominoes \$p\$ are stored as complex numbers, reflections across both axes and diagonals can be represented by their complex conjugates \$Conj(p)\$ multiplied by directional vectors in \$d\$.


Wolfram Language(Mathematica), 532 496 366 bytes

saved so many bytes thanks to the comment of @Aiden Chow

Modified from the Mathematica code provided by A000105. Really naive. I don't know what I'm doing.


Try it online!

The original Mathematica code looks like

(* In this program by Jaime Rangel-Mondragón, polyominoes are represented as a list of Gaussian integers. *)
polyominoQ[p_List] := And @@ ((IntegerQ[Re[#]] && IntegerQ[Im[#]])& /@ p);
rot[p_?polyominoQ] := I*p;
ref[p_?polyominoQ] := (# - 2 Re[#])& /@ p;
cyclic[p_] := Module[{i = p, ans = {p}}, While[(i = rot[i]) != p, AppendTo[ans, i]]; ans];
dihedral[p_?polyominoQ] := Flatten[{#, ref[#]}& /@ cyclic[p], 1];
canonical[p_?polyominoQ] := Union[(# - (Min[Re[p]] + Min[Im[p]]*I))& /@ p];
allPieces[p_] := Union[canonical /@ dihedral[p]];
polyominoes[1] = {{0}};
polyominoes[n_] := polyominoes[n] = Module[{f, fig, ans = {}}, fig = ((f = #1; ({f, #1 + 1, f, #1 + I, f, #1 - 1, f, #1 - I}&) /@ f)&) /@ polyominoes[n - 1]; fig = Partition[Flatten[fig], n]; f = Select[Union[canonical /@ fig], Length[#1] == n &]; While[f != {}, ans = {ans, First[f]}; f = Complement[f, allPieces[First[f]]]]; Partition[Flatten[ans], n]];
a[n_] := a[n] = Length[ polyominoes[n]];
Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 1, 12}] (* Jean-François Alcover, Mar 24 2015, after Jaime Rangel-Mondragón *)
  • \$\begingroup\$ Changing all the two letter variables to one letter gives 513 bytes \$\endgroup\$
    – Aiden Chow
    Apr 18, 2023 at 4:16
  • \$\begingroup\$ With the semicolons, it looks like you don't actually need the newlines at all, giving 496 bytes \$\endgroup\$
    – Aiden Chow
    Apr 18, 2023 at 4:21
  • \$\begingroup\$ A bunch of golfs here and there: 366 bytes \$\endgroup\$
    – Aiden Chow
    Apr 18, 2023 at 6:36
  • \$\begingroup\$ The only thing I couldn't figure out is how to remove the z function (I have removed all the other functions except for Q and F, as they were unnecessary), maybe someone with more experience with Mathematica could help with that. It seems to boil down to creating an anonymous function, but I don't know much Mathematica so I'm unsure of how to do that. \$\endgroup\$
    – Aiden Chow
    Apr 18, 2023 at 6:37

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