How much is my dice matrix worth?

Question

Input

A non-empty binary matrix consisting of 3x3 sub-matrices put side by side.

Task

Your task is to identify valid dice patterns (as described below) among the 3x3 sub-matrices. Each valid pattern is worth the value of the corresponding dice. Invalid patterns are worth 0.

Output

The sum of the valid dice values.

Dice patterns

$$\begin{align} &1:\pmatrix{\color{gray}0,\color{gray}0,\color{gray}0\\\color{gray}0,1,\color{gray}0\\\color{gray}0,\color{gray}0,\color{gray}0} &&2:\pmatrix{1,\color{gray}0,\color{gray}0\\\color{gray}0,\color{gray}0,\color{gray}0\\\color{gray}0,\color{gray}0,1}\text{or}\pmatrix{\color{gray}0,\color{gray}0,1\\\color{gray}0,\color{gray}0,\color{gray}0\\1,\color{gray}0,\color{gray}0}\\ &3:\pmatrix{1,\color{gray}0,\color{gray}0\\\color{gray}0,1,\color{gray}0\\\color{gray}0,\color{gray}0,1}\text{or}\pmatrix{\color{gray}0,\color{gray}0,1\\\color{gray}0,1,\color{gray}0\\1,\color{gray}0,\color{gray}0} &&4:\pmatrix{1,\color{gray}0,1\\\color{gray}0,\color{gray}0,\color{gray}0\\1,\color{gray}0,1}\\ &5:\pmatrix{1,\color{gray}0,1\\\color{gray}0,1,\color{gray}0\\1,\color{gray}0,1} &&6:\pmatrix{1,\color{gray}0,1\\1,\color{gray}0,1\\1,\color{gray}0,1}\text{or}\pmatrix{1,1,1\\\color{gray}0,\color{gray}0,\color{gray}0\\1,1,1} \end{align}$$

Example

The expected output for the following matrix is 14 because it contains the dice 5, 6 and 3, followed by an invalid pattern (from left to right and from top to bottom).

$$\pmatrix{1,0,1,1,1,1\\ 0,1,0,0,0,0\\ 1,0,1,1,1,1\\ 1,0,0,0,0,0\\ 0,1,0,0,1,0\\ 0,0,1,0,1,0}$$

Rules

• Both the width and the height of the matrix are guaranteed to be multiples of 3.
• You must ignore sub-matrices that are not properly aligned on the grid (see the 3rd test case). More formally and assuming 0-indexing: the coordinates of the top-left cell of each sub-matrix to be considered are of the form \$(3x, 3y)\$.
• This is code-golf.

Test cases

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

// 2
[ [ 0,0,1 ],
  [ 0,0,0 ],
  [ 1,0,0 ] ]

// 0 (0 + 0)
[ [ 0,0,1,0,1,0 ],
  [ 0,0,0,1,0,0 ],
  [ 0,0,1,0,1,0 ] ]

// 9 (3 + 3 + 3)
[ [ 1,0,0,0,0,1,1,0,0 ],
  [ 0,1,0,0,1,0,0,1,0 ],
  [ 0,0,1,1,0,0,0,0,1 ] ]

// 6 (6 + 0)
[ [ 1,0,1 ],
  [ 1,0,1 ],
  [ 1,0,1 ],
  [ 1,0,1 ],
  [ 1,0,0 ],
  [ 1,0,1 ] ]

// 14 (5 + 6 + 3 + 0)
[ [ 1,0,1,1,1,1 ],
  [ 0,1,0,0,0,0 ],
  [ 1,0,1,1,1,1 ],
  [ 1,0,0,0,0,0 ],
  [ 0,1,0,0,1,0 ],
  [ 0,0,1,0,1,0 ] ]

// 16 (1 + 2 + 3 + 4 + 0 + 6)
[ [ 0,0,0,1,0,0,1,0,0 ],
  [ 0,1,0,0,0,0,0,1,0 ],
  [ 0,0,0,0,0,1,0,0,1 ],
  [ 1,0,1,1,1,1,1,0,1 ],
  [ 0,0,0,1,0,1,1,0,1 ],
  [ 1,0,1,1,1,1,1,0,1 ] ]

Answer 1

Python 3, 195 189 bytes

-6 bytes thanks to @Jo King

lambda m:sum({16:1,257:2,68:2,273:3,84:3,325:4,341:5,455:6,365:6}.get(int(''.join(str(e)for c in m[3*i:][:3]for e in c[3*j:][:3]),2),0)for i in range(len(m)//3)for j in range(len(m[0])//3))

Try it online! (189) Try it online! (195)

Human readable version:

# 3x3 part matrix to dice, beginning at coordinates 3*i, 3*j
def single_matrix_to_dice(matrix, i, j):
    # Example: matrix = [[0, 0, 0], [0, 1, 0], [0, 0, 0]], i=0, j=0 (result is 1)

    matrix_string = ''.join(
        str(e) for column in matrix[3*i:3*i+3] 
        for entry in column[3*j:3*j+3]
    ) # Slicing the matrix so that only the valid entries remain, here '000010000'

    # Interpreting the matrix string as binary number, here 16
    binary_number = int(matrix_string,2)

    # binary representations of all valid dice rolls
    dct = {16:1,257:2,68:2,273:3,84:3,325:4,341:5,455:6,365:6}

    return dct.get(binary_number, 0)

def f(matrix):
    return sum(
        single_matrix_to_dice(matrix, i, j) for i in range(len(m)//3) 
        for j in range(len(m[0])//3))
    ) # len(m)/3 would generate a float, so len(m)//3 is used

Answer 2

R, 134 bytes

function(m,d=dim(m)/3-1){for(a in 0:d)for(b in 0:d[2])F=F+sum(y<-m[1:3+a*3,1:3+b*3])*sum(y*2^(8:0))%in%utf8ToInt("āDđTŅŕLJŭ");F}

Try it online!

I noticed I had the same idea of @Heteira

History :

• 171 : -10 bytes thanks to @JayCe !
• 161 : -3 bytes thanks to @Giuseppe !
• 158 : -13 bytes saved !
• 145 : -2 bytes thanks to @Giuseppe!
• 143 : -6 bytes saved !
• 137 : -3 bytes thanks to @JayCe!

Answer 3

Perl 6, 113 105 97 94 bytes

{sum (|@_[*;^3+3*$_]for ^@_[0]).rotor(9).map:{"@āđŅŕLJ@@DT@@ŭ".ords.first(:2[$_],:k)%7}}

Try it online!

Splits up the matrix into sub matrices of 3x3, converts the nine 1s and 0s to base 2 and then indexes it into a list of integers for the value.

Explanation:

{  #Start anonymous code block
  sum   # Sum of all
     (|@_[*;^3+3*$_]   # Get the n*3 to n*3+3th elements in every sub-list
           for ^@_[0]) # For n in the range 0 to width (divide by 3 to avoid warnings)
     .rotor(9)  # Split this list into groups of 9 (split the dice up)
     .map:{     # And map each die to 
        "@āđŅŕLJ@@DT@@ŭ".ords  # In the list of integers
           .first(      # The first appearance of 
               :2[$_],  # The dice converted from a list of 0s and 1s to base 2
                 :k     # Returning the index
             )%7        # And modulo by 7 to get the alternate versions of 2, 3 and 6
          }
}

Answer 4

Jelly,  29 28 bytes

-1 thanks to Mr. Xcoder (use Ṁ to replace ṢṪ)

s€3ZẎs3µZU,ƊṀṙ1FḄ“°€⁼-Ḍ?‘i)S

A monadic link.

Try it online! Or run the tests.

How?

s€3ZẎs3µZU,ƊṀṙ1FḄ“°€⁼-Ḍ?‘i)S - Link: list of lists of 1s and 0s
s€3                          - split each into threes
   Z                         - transpose
    Ẏ                        - tighten
     s3                      - split into threes -> the sub-matrices in column-major order
       µ                  )  - for each sub-matrix, say D:
           Ɗ                 -   last three links as a monad:
        Z                    -     transpose D
         U                   -     reverse each -> D rotated a quarter turn clockwise
          ,                  -     pair with D
            Ṁ                -   get the maximum of the two orientations
             ṙ1              -   rotate left by one (to ensure FḄ will yield integers <256 for all non-zero valued D)
               F             -   flatten
                Ḅ            -   convert from binary
                         i   -   first 1-based index in (0 if not found):
                 “°€⁼-Ḍ?‘    -     code-page indices list = [128,12,140,45,173,63]
                           S - sum

For example when a sub-matrix is:

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

Then ZU,Ɗ yields:

[[[1, 1, 1],
  [0, 0, 0],
  [1, 1, 1]],   ...which has maximum (Ṁ):    ...and after ṙ1:
 [[1, 0, 1],                   [[1, 1, 1],         [[0, 0, 0],
  [1, 0, 1],                    [0, 0, 0],          [1, 1, 1],
  [1, 0, 1]]]                   [1, 1, 1]]          [1, 1, 1]]

...which flattens to [0, 0, 0, 1, 1, 1, 1, 1, 1], which, converting from binary, is 63 which is the sixth entry in the code-page index list “°€⁼-Ḍ?‘ (? being byte 3F in Jelly's code-page)

Answer 5

Japt -x, 36 bytes

Now it's getting interesting. I'm sure can be golfed down even more

ò3 ®®ò3Ãy f@"0ıtŵġdťƍǧ"øºXm¬¬Í+H)d
c

Try it online!

Answer 6

Retina 0.8.2, 90 bytes

+`(...)(.+¶)(...)(.+¶)(...)
$1¶$3¶$5¶$2$4
¶

M!`.{9}
G`111000111|(101){3}|(.)0(.0).0\3\2
1

Try it online! Explanation:

+`(...)(.+¶)(...)(.+¶)(...)
$1¶$3¶$5¶$2$4

Repeatedly remove \$3\times 3\$ blocks from each \$3\times n\$ row until all rows have 3 columns.

¶

M!`.{9}

Join all the blocks together and then split back up into rows of 9 columns.

G`111000111|(101){3}|(.)0(.0).0\3\2

Only keep valid dice patterns (two patterns for 6, then one matches any number from 0 to 5, although 0 of course will not contribute to the count below.)

1

Count the pips on the valid dice.

Answer 7

Ruby, 151 bytes

->m{m.each_slice(3).flat_map{|r|r.transpose.each_slice(3).map{|d|" \x10āđŅŕLJ  DT  ŭ".chars.map(&:ord).index(d.flatten.join.to_i 2)&.%7}-[p]}.sum}

Try it online!

A lambda accepting a 2d array of ints (or strings, I guess). Takes inspiration from Jo King's answer. I feel like slicing the dice out of the input matrix took a lot of space, so I may well be outgolfed. Fortunately, dealing with nils only cost me a handful of bytes.

Ungolfed:

->m{
  m.each_slice(3).flat_map{|r|             # Split into groups of 3 rows
    r.transpose.each_slice(3).map{|d|      # Split into groups of 3 columns
      " \x10āđŅŕLJ  DT  ŭ".chars.map(&:ord) # [0,16,257,273,325,341,455,0,0,68,84,0,0,365]
        .index(                            # Find in that array
          d.flatten.join.to_i 2            #   the die flattened into a bitstring (nil if not found)
        )&.%7                              # Safe-modulo 7 (leaves nils as nil)
    }-[p]                                  # Remove nils
  }.sum                                    # Add 'em up
}

Answer 8

Clojure, 197 bytes

#(apply +(for[R[range]i(R 0(count %)3)j(R 0(count(% 0))3)](case(apply +(map *(iterate(partial * 2)1)(for[x(R 3)y(R 3)]((%(+ i x))(+ j y)))))16 1 257 2 68 2 273 3 84 3 325 4 3 4 1 5 455 6 365 6 0)))

I should have come up with something smarter.

Answer 9

Python 2, 159 bytes

f=lambda a:a>[]and sum(u'ȀāDđTŅȀŕȀLJŭ'.find(unichr(int(J(J(map(str,r[i:i+3]))for r in a[:3]),2)))/2+1for i in range(0,len(a[0]),3))+f(a[3:])
J=''.join

Try it online!

Hat tip to Jonathan Frech for the unicode encoding approach.