## Binary self-rotation

13

Given a binary 3D array, for each layer, cyclically rotate up each of its columns as many steps as indicated by the binary encoding of the columns of the layer above it, and then cyclically rotate left each of its rows as many steps as indicated by the binary encoding of the rows of the layer below it.

There will always be at least three layers. The top layer's columns and the bottom layer's rows should not be rotated.

## Walk-through

[[[1,0,1],
[1,0,0]],

[[1,0,1],
[0,1,1]],

[[0,1,1],
[1,1,1]],

[[1,1,0],
[1,1,1]]]


The first step is evaluating the numbers encoded in binary by the columns and rows of each layer:

     3 0 2
5 [[[1,0,1],
4   [1,0,0]],

2 1 3
5  [[1,0,1],
3   [0,1,1]],

1 3 3
3  [[0,1,1],
7   [1,1,1]],

3 3 1
6  [[1,1,0],
7   [1,1,1]]]


The first layer, [[1,0,1],[1,0,0]] will not have its columns rotated, but its rows will be cyclically rotated left 5 steps and 3 step respectively, thus becoming [[1,1,0],[1,0,0]].
The second layer, [[1,0,1],[0,1,1]], will have its columns cyclically rotated up 3, 0, and 2 steps respectively, giving [[0,0,1],[1,1,1]], and then the rows are cyclically rotated left 3 and 7 steps respectively, with no visible change.
The third layer, [[0,1,1],[1,1,1]] rotated up 2, 1, and 3 steps stays the same, and neither does rotating left 6 and 7 steps do anything.
Finally, the fourth layer, [[1,1,0],[1,1,1]] rotated up 1, 3, and 3 steps is [[1,1,1],[1,1,0]], but its rows are not rotated afterwards, as it is the last layer.
Putting all the layers together again, gives us the binary self-rotated 3D array:

[[[1,1,0],
[1,0,0]],

[[0,0,1],
[1,1,1]],

[[0,1,1],
[1,1,1]],

[[1,1,1],
[1,1,0]]]


Example cases:

[[[1,0,1],[1,0,0]],[[1,0,1],[0,1,1]],[[0,1,1],[1,1,1]],[[1,1,0],[1,1,1]]] gives
[[[1,1,0],[1,0,0]],[[0,0,1],[1,1,1]],[[0,1,1],[1,1,1]],[[1,1,1],[1,1,0]]]

[[[1]],[[1]],[[0]]] gives
[[[1]],[[1]],[[0]]]

[[[1,0,1],[1,0,1],[1,0,1]],[[0,0,1],[0,0,1],[0,0,1]],[[1,0,0],[1,0,1],[0,0,1]]] gives
[[[0,1,1],[0,1,1],[0,1,1]],[[0,1,0],[1,0,0],[0,1,0]],[[1,0,1],[1,0,1],[0,0,0]]]

3

# Jelly,  18  17 bytes

ṙ""Ḅ}
Z€çŻṖ$$Z€çḊ  Try it online! ### How? ṙ""Ḅ} - Link 1, rotation helper: 3d matrix to rotate, 3d matrix of rotation instructions } - use the right argument for: Ḅ - un-binary (vectorises) - get the rotation amounts as a 2d matrix " - zip with: " - zip with: ṙ - rotate (the current row) left by (the current amount) Z€çŻṖ Z€çḊ - Main Link: 3d matrix, M Z€ - transpose €ach (layer of M)  - last two links as a monad:  - last two links as a monad: Ż - prepend a zero Ṗ - pop (i.e. remove the tail) ç - call the last Link as a dyad (i.e. f(Z€ result, ŻṖ result) ) Z€ - transpose €ach (layer of that) Ḋ - dequeue (i.e. remove the head layer of M) ç - call the last Link as a dyad (i.e. f(Z€çŻṖ$$Z€ result, Ḋ result) )


Note:  (or possibly $$...$$?) seems to mess up the code-block formatting (but only once posted, not in the preview), so I added a space to make my life easier.

3

# Python 2, 220211209185176174164161 159 bytes

lambda m:map(R,z(map(R,z(m,['']+[z(*l)for l in m])),m[1:]+['']))
R=lambda(l,L):map(lambda r,i:r[i:]+r[:i or 0],z(*l),[int(b[1::3],2)%len(b)for b in L])
z=zip


Try it online!

-2 bytes, thanks to Jonathan Allan

Since you handle None during the slicing for the rotation I believe both of the ['0'] can become [[]]. – Jonathan Allan – 2019-01-22T17:29:22.617

@JonathanAllan Thanks :) – TFeld – 2019-01-23T08:01:35.377

2

# APL+WIN, 53 39 bytes

Many thanks to Adám for saving 14 bytes

(1 0↓⍉2⊥⍉m⍪0)⌽(¯1 0↓2⊥2 1 3⍉0⍪m)⊖[2]m←⎕


Try it online! Courtesy of Dyalog Classic

Prompts for input of a 3d array of the form:

4 2 3⍴1 0 1 1 0 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 1


which yields:

1 0 1
1 0 0

1 0 1
0 1 1

0 1 1
1 1 1

1 1 0
1 1 1


Explanation:

m←⎕ Prompt for input

(¯1 0↓2⊥2 1 3⍉0⍪m) Calculate column rotations

(1 0↓⍉2⊥⍉m⍪0) Calculate row rotations

(...)⌽(...)⊖[2]m Apply column and row rotation and output resulting 3d array:

1 1 0
1 0 0

0 0 1
1 1 1

0 1 1
1 1 1

1 1 1
1 1 0


Instead of enclosing and using ¨, just process the entire array at once. Try it online!

@Adám Many thanks. I do not know why I over thought this one and went the nested route :( Getting old? – Graham – 2019-01-22T18:44:07.433

2

# R, 226216 205 bytes

-21 bytes thanks to digEmAll

function(a,L=for){d=dim(b<-a)
r=function(a,n,l=sum(a|1))a[(1:l+sum(n*2^(sum(n|1):1-1))-1)%%l+1]
L(i,I<-2:d[3],L(j,1:d,b[j,,i]<-r(b[j,,i],a[j,,i-1])))
L(i,I-1,L(k,1:d[2],b[,k,i]<-r(b[,k,i],a[,k,i+1])))
b}


Try it online!

1

# 05AB1E, 41 39 bytes

εNĀiø¹N<èøJC‚øε._}ø}N¹g<Êi¹N>èJC‚øε._


This feels way too long.. Can definitely be golfed some more.

Explanation:

ε                    # Map each layer in the (implicit) input to:
# (N is the layer-index of this map)
NĀi                 #  If it is not the first layer:
ø                #   Zip/transpose the current layer; swapping rows/columns
¹N<è             #   Get the N-1'th layer of the input
ø            #   Zip/transpose; swapping rows/columns
J           #   Join all inner lists (the columns) together
C          #   And convert it from binary to integer
‚         #   Pair it with the current layer's columns we're mapping
ø        #   Zip/transpose; to pair each integer with a layer's columns
ε   }   #   Map over these pairs:
      #    Push both values of the pair separately to the stack
._    #    Rotate the column the integer amount of times
ø                #   Zip/transpose the rows/columns of the current layer back
}                 #  Close the if-statement
N¹g<Êi              #  If this is not the last layer (layer-index-1 != amount_of_layers):
¹N>è          #   Get the N+1'th layer of the input
J         #   Join all inner lists (the rows) together
C        #   And convert it from binary to integer
‚       #   Pair it with the current layer's rows we're mapping
ø      #   Zip/transpose; to pair each integer with a layer's rows
ε     #   Map over these pairs:
#    Push both values of the pair separately to the stack
._  #    Rotate the row the integer amount of times
# (implicitly output the result after the layer-mapping is done)


0

# Wolfram Language (Mathematica), 138131125 123 bytes

t=Map@Thread

• Map[Thread] is equivalent to Transpose[a, {1,3,2}], which transposes the columns and rows.
• Fold[#+##&] is shorter than IntegerDigits[#,2] for converting from binary.