Surface of the 3x3x3 cube as a graphFind the Chromatic NumberEdge Elimination NumberUndirect a GraphIs My Graph Planar?Strongly Connected ComponentsFind a dual graphFind a set of maximal matching edgesCalculate TreewidthDetermine if a Graph is ToroidalConstruct a line graph / conjugate graph
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Surface of the 3x3x3 cube as a graph
Find the Chromatic NumberEdge Elimination NumberUndirect a GraphIs My Graph Planar?Strongly Connected ComponentsFind a dual graphFind a set of maximal matching edgesCalculate TreewidthDetermine if a Graph is ToroidalConstruct a line graph / conjugate graph
$begingroup$
Your task is to generate a graph with 54 vertices, each corresponds to a facet on a Rubik's cube. There is an edge between two vertices iff the corresponding facets share a side.
Rules
- You may choose to output an adjacency list, adjacency matrix, edge list, or any reasonable format to represent a graph in an algorithm. (A visual graph readable by a human is generally not a reasonable format in an algorithm in most cases.)
- You may make either every vertex adjacent to itself, or none adjacent to itself.
- You may either include both directions for each edge (count one or two times for self-loops), or output exactly one time for each edge, but not mix the ways.
- You may renumber the vertices, skip some numbers, or even use non-number labels for the vertices in any way you want. You should also post the numbering if it isn't obvious, so others could check your answer in easier ways.
- This is code-golf. Shortest code in bytes wins.
Example output
This is the numbering of vertices used in the example:
0 1 2
3 4 5
6 7 8
9 10 11 18 19 20 27 28 29 36 37 38
12 13 14 21 22 23 30 31 32 39 40 41
15 16 17 24 25 26 33 34 35 42 43 44
45 46 47
48 49 50
51 52 53
Output as an adjacency list (vertex number before each list is optional):
0 [1 3 9 38]
1 [2 4 0 37]
2 [29 5 1 36]
3 [4 6 10 0]
4 [5 7 3 1]
5 [28 8 4 2]
6 [7 18 11 3]
7 [8 19 6 4]
8 [27 20 7 5]
9 [10 12 38 0]
10 [11 13 9 3]
11 [18 14 10 6]
12 [13 15 41 9]
13 [14 16 12 10]
14 [21 17 13 11]
15 [16 51 44 12]
16 [17 48 15 13]
17 [24 45 16 14]
18 [19 21 11 6]
19 [20 22 18 7]
20 [27 23 19 8]
21 [22 24 14 18]
22 [23 25 21 19]
23 [30 26 22 20]
24 [25 45 17 21]
25 [26 46 24 22]
26 [33 47 25 23]
27 [28 30 20 8]
28 [29 31 27 5]
29 [36 32 28 2]
30 [31 33 23 27]
31 [32 34 30 28]
32 [39 35 31 29]
33 [34 47 26 30]
34 [35 50 33 31]
35 [42 53 34 32]
36 [37 39 29 2]
37 [38 40 36 1]
38 [9 41 37 0]
39 [40 42 32 36]
40 [41 43 39 37]
41 [12 44 40 38]
42 [43 53 35 39]
43 [44 52 42 40]
44 [15 51 43 41]
45 [46 48 17 24]
46 [47 49 45 25]
47 [33 50 46 26]
48 [49 51 16 45]
49 [50 52 48 46]
50 [34 53 49 47]
51 [52 44 15 48]
52 [53 43 51 49]
53 [35 42 52 50]
code-golf kolmogorov-complexity graph-theory rubiks-cube
asked May 19 at 20:25
jimmy23013jimmy23013
30k561132
$endgroup$
add a comment |
$begingroup$
Your task is to generate a graph with 54 vertices, each corresponds to a facet on a Rubik's cube. There is an edge between two vertices iff the corresponding facets share a side.
Rules
- You may choose to output an adjacency list, adjacency matrix, edge list, or any reasonable format to represent a graph in an algorithm. (A visual graph readable by a human is generally not a reasonable format in an algorithm in most cases.)
- You may make either every vertex adjacent to itself, or none adjacent to itself.
- You may either include both directions for each edge (count one or two times for self-loops), or output exactly one time for each edge, but not mix the ways.
- You may renumber the vertices, skip some numbers, or even use non-number labels for the vertices in any way you want. You should also post the numbering if it isn't obvious, so others could check your answer in easier ways.
- This is code-golf. Shortest code in bytes wins.
Example output
This is the numbering of vertices used in the example:
0 1 2
3 4 5
6 7 8
9 10 11 18 19 20 27 28 29 36 37 38
12 13 14 21 22 23 30 31 32 39 40 41
15 16 17 24 25 26 33 34 35 42 43 44
45 46 47
48 49 50
51 52 53
Output as an adjacency list (vertex number before each list is optional):
0 [1 3 9 38]
1 [2 4 0 37]
2 [29 5 1 36]
3 [4 6 10 0]
4 [5 7 3 1]
5 [28 8 4 2]
6 [7 18 11 3]
7 [8 19 6 4]
8 [27 20 7 5]
9 [10 12 38 0]
10 [11 13 9 3]
11 [18 14 10 6]
12 [13 15 41 9]
13 [14 16 12 10]
14 [21 17 13 11]
15 [16 51 44 12]
16 [17 48 15 13]
17 [24 45 16 14]
18 [19 21 11 6]
19 [20 22 18 7]
20 [27 23 19 8]
21 [22 24 14 18]
22 [23 25 21 19]
23 [30 26 22 20]
24 [25 45 17 21]
25 [26 46 24 22]
26 [33 47 25 23]
27 [28 30 20 8]
28 [29 31 27 5]
29 [36 32 28 2]
30 [31 33 23 27]
31 [32 34 30 28]
32 [39 35 31 29]
33 [34 47 26 30]
34 [35 50 33 31]
35 [42 53 34 32]
36 [37 39 29 2]
37 [38 40 36 1]
38 [9 41 37 0]
39 [40 42 32 36]
40 [41 43 39 37]
41 [12 44 40 38]
42 [43 53 35 39]
43 [44 52 42 40]
44 [15 51 43 41]
45 [46 48 17 24]
46 [47 49 45 25]
47 [33 50 46 26]
48 [49 51 16 45]
49 [50 52 48 46]
50 [34 53 49 47]
51 [52 44 15 48]
52 [53 43 51 49]
53 [35 42 52 50]
code-golf kolmogorov-complexity graph-theory rubiks-cube
asked May 19 at 20:25
jimmy23013jimmy23013
30k561132
$endgroup$
$begingroup$
@DavidC 45 was also wrong. Thanks. Fixed.
$endgroup$
– jimmy23013
May 20 at 16:28
add a comment |
$begingroup$
Your task is to generate a graph with 54 vertices, each corresponds to a facet on a Rubik's cube. There is an edge between two vertices iff the corresponding facets share a side.
Rules
- You may choose to output an adjacency list, adjacency matrix, edge list, or any reasonable format to represent a graph in an algorithm. (A visual graph readable by a human is generally not a reasonable format in an algorithm in most cases.)
- You may make either every vertex adjacent to itself, or none adjacent to itself.
- You may either include both directions for each edge (count one or two times for self-loops), or output exactly one time for each edge, but not mix the ways.
- You may renumber the vertices, skip some numbers, or even use non-number labels for the vertices in any way you want. You should also post the numbering if it isn't obvious, so others could check your answer in easier ways.
- This is code-golf. Shortest code in bytes wins.
Example output
This is the numbering of vertices used in the example:
0 1 2
3 4 5
6 7 8
9 10 11 18 19 20 27 28 29 36 37 38
12 13 14 21 22 23 30 31 32 39 40 41
15 16 17 24 25 26 33 34 35 42 43 44
45 46 47
48 49 50
51 52 53
Output as an adjacency list (vertex number before each list is optional):
0 [1 3 9 38]
1 [2 4 0 37]
2 [29 5 1 36]
3 [4 6 10 0]
4 [5 7 3 1]
5 [28 8 4 2]
6 [7 18 11 3]
7 [8 19 6 4]
8 [27 20 7 5]
9 [10 12 38 0]
10 [11 13 9 3]
11 [18 14 10 6]
12 [13 15 41 9]
13 [14 16 12 10]
14 [21 17 13 11]
15 [16 51 44 12]
16 [17 48 15 13]
17 [24 45 16 14]
18 [19 21 11 6]
19 [20 22 18 7]
20 [27 23 19 8]
21 [22 24 14 18]
22 [23 25 21 19]
23 [30 26 22 20]
24 [25 45 17 21]
25 [26 46 24 22]
26 [33 47 25 23]
27 [28 30 20 8]
28 [29 31 27 5]
29 [36 32 28 2]
30 [31 33 23 27]
31 [32 34 30 28]
32 [39 35 31 29]
33 [34 47 26 30]
34 [35 50 33 31]
35 [42 53 34 32]
36 [37 39 29 2]
37 [38 40 36 1]
38 [9 41 37 0]
39 [40 42 32 36]
40 [41 43 39 37]
41 [12 44 40 38]
42 [43 53 35 39]
43 [44 52 42 40]
44 [15 51 43 41]
45 [46 48 17 24]
46 [47 49 45 25]
47 [33 50 46 26]
48 [49 51 16 45]
49 [50 52 48 46]
50 [34 53 49 47]
51 [52 44 15 48]
52 [53 43 51 49]
53 [35 42 52 50]
code-golf kolmogorov-complexity graph-theory rubiks-cube
asked May 19 at 20:25
jimmy23013jimmy23013
30k561132
$endgroup$
Your task is to generate a graph with 54 vertices, each corresponds to a facet on a Rubik's cube. There is an edge between two vertices iff the corresponding facets share a side.
Rules
- You may choose to output an adjacency list, adjacency matrix, edge list, or any reasonable format to represent a graph in an algorithm. (A visual graph readable by a human is generally not a reasonable format in an algorithm in most cases.)
- You may make either every vertex adjacent to itself, or none adjacent to itself.
- You may either include both directions for each edge (count one or two times for self-loops), or output exactly one time for each edge, but not mix the ways.
- You may renumber the vertices, skip some numbers, or even use non-number labels for the vertices in any way you want. You should also post the numbering if it isn't obvious, so others could check your answer in easier ways.
- This is code-golf. Shortest code in bytes wins.
Example output
This is the numbering of vertices used in the example:
0 1 2
3 4 5
6 7 8
9 10 11 18 19 20 27 28 29 36 37 38
12 13 14 21 22 23 30 31 32 39 40 41
15 16 17 24 25 26 33 34 35 42 43 44
45 46 47
48 49 50
51 52 53
Output as an adjacency list (vertex number before each list is optional):
0 [1 3 9 38]
1 [2 4 0 37]
2 [29 5 1 36]
3 [4 6 10 0]
4 [5 7 3 1]
5 [28 8 4 2]
6 [7 18 11 3]
7 [8 19 6 4]
8 [27 20 7 5]
9 [10 12 38 0]
10 [11 13 9 3]
11 [18 14 10 6]
12 [13 15 41 9]
13 [14 16 12 10]
14 [21 17 13 11]
15 [16 51 44 12]
16 [17 48 15 13]
17 [24 45 16 14]
18 [19 21 11 6]
19 [20 22 18 7]
20 [27 23 19 8]
21 [22 24 14 18]
22 [23 25 21 19]
23 [30 26 22 20]
24 [25 45 17 21]
25 [26 46 24 22]
26 [33 47 25 23]
27 [28 30 20 8]
28 [29 31 27 5]
29 [36 32 28 2]
30 [31 33 23 27]
31 [32 34 30 28]
32 [39 35 31 29]
33 [34 47 26 30]
34 [35 50 33 31]
35 [42 53 34 32]
36 [37 39 29 2]
37 [38 40 36 1]
38 [9 41 37 0]
39 [40 42 32 36]
40 [41 43 39 37]
41 [12 44 40 38]
42 [43 53 35 39]
43 [44 52 42 40]
44 [15 51 43 41]
45 [46 48 17 24]
46 [47 49 45 25]
47 [33 50 46 26]
48 [49 51 16 45]
49 [50 52 48 46]
50 [34 53 49 47]
51 [52 44 15 48]
52 [53 43 51 49]
53 [35 42 52 50]
code-golf kolmogorov-complexity graph-theory rubiks-cube
code-golf kolmogorov-complexity graph-theory rubiks-cube
asked May 19 at 20:25
jimmy23013jimmy23013
30k561132
asked May 19 at 20:25
jimmy23013jimmy23013
30k561132
edited May 20 at 16:27
jimmy23013
asked May 19 at 20:25
jimmy23013jimmy23013
30k561132
asked May 19 at 20:25
jimmy23013jimmy23013
30k561132
asked May 19 at 20:25
jimmy23013jimmy23013
30k561132
30k561132
$begingroup$
@DavidC 45 was also wrong. Thanks. Fixed.
$endgroup$
– jimmy23013
May 20 at 16:28
add a comment |
$begingroup$
@DavidC 45 was also wrong. Thanks. Fixed.
$endgroup$
– jimmy23013
May 20 at 16:28
$begingroup$
@DavidC 45 was also wrong. Thanks. Fixed.
$endgroup$
– jimmy23013
May 20 at 16:28
$begingroup$
@DavidC 45 was also wrong. Thanks. Fixed.
$endgroup$
– jimmy23013
May 20 at 16:28
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
APL (Dyalog Classic), 34 30 bytes
-4 thanks to jimmy23013
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
Try it online!
outputs an adjacency matrix with each vertex adjacent to itself
⍳3 3 generate an array of (0 0)(0 1)(0 2)(1 0)(1 1)(1 2)(2 0)(2 1)(2 2)
○ multiply all by π
7 ¯1∘., prepend 7 or -1 in all possible ways
(⍳3)∘.⌽ rotate coord triples by 0 1 2 steps in all possible ways
+/¨|∘.-⍨, compute manhattan distance between each pair
4≥ it must be no greater than 4 for neighbouring facets
answered May 19 at 21:32
ngnngn
7,57612661
$endgroup$
$begingroup$
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
$endgroup$
– jimmy23013
May 19 at 23:41
$begingroup$
@jimmy23013 using π is very nice :) thank you!
$endgroup$
– ngn
May 19 at 23:48
add a comment |
$begingroup$
Ruby, 79 bytes
54.timesi
Try it online!
Prints a representation of a unidirectional graph, as a list of the vertices to the right of and below each vertex as shown in the map below.
0 1 2 3 4 5
6 7 8 9 10 11
12 13 14 15 16 17
18 19 20 21 22 23
24 25 26 27 28 29
30 31 32 33 34 35
36 37 38 39 40 41
42 43 44 45 46 47
48 49 50 51 52 53
answered May 20 at 4:25
Level River StLevel River St
20.6k32683
$endgroup$
add a comment |
$begingroup$
Python 2.7, 145
def p(n):l=(3-n%2*6,n/6%3*2-2,n/18*2-2);k=n/2%3;return l[k:]+l[:k]
r=range(54)
x=[[sum((x-y)**2for x,y in zip(p(i),p(j)))<5for i in r]for j in r]
Try it online!
Defines an adjacency matrix x as a list of lists of boolean values. Facets count as being adjacent to themselves.
p(n) computes the coordinates of the center of the nth facet of a 3x3x3 cube whose facets are 2 units across. Adjacency is determined by testing if 2 facets have a square distance under 5 (adjacent facets have square distance at most 4, non-adjacent facets have square distance at least 6).
answered May 19 at 23:18
cardboard_boxcardboard_box
4,1751431
$endgroup$
add a comment |
$begingroup$
Charcoal, 48 bytes
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧
Generate all sets of 3-dimensional coordinates in the range [0..6] for each dimension.
≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υ
Keep only those coordinates that are centres of 2x2 squares on one of the faces x=0, y=0, z=0, x=6, y=6, z=6.
ＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
For each coordinate, print the indices of those coordinates whose taxicab distance is 2.
The vertices are numbered as follows:
33 34 35
21 22 23
9 10 11
36 24 12 0 1 2 13 25 37 47 46 45
38 26 14 3 4 5 15 27 39 50 49 48
40 28 16 6 7 8 17 29 41 53 52 51
18 19 20
30 31 32
42 43 44
answered May 19 at 23:45
NeilNeil
84.6k845182
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
APL (Dyalog Classic), 34 30 bytes
-4 thanks to jimmy23013
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
Try it online!
outputs an adjacency matrix with each vertex adjacent to itself
⍳3 3 generate an array of (0 0)(0 1)(0 2)(1 0)(1 1)(1 2)(2 0)(2 1)(2 2)
○ multiply all by π
7 ¯1∘., prepend 7 or -1 in all possible ways
(⍳3)∘.⌽ rotate coord triples by 0 1 2 steps in all possible ways
+/¨|∘.-⍨, compute manhattan distance between each pair
4≥ it must be no greater than 4 for neighbouring facets
answered May 19 at 21:32
ngnngn
7,57612661
$endgroup$
$begingroup$
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
$endgroup$
– jimmy23013
May 19 at 23:41
$begingroup$
@jimmy23013 using π is very nice :) thank you!
$endgroup$
– ngn
May 19 at 23:48
add a comment |
$begingroup$
APL (Dyalog Classic), 34 30 bytes
-4 thanks to jimmy23013
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
Try it online!
outputs an adjacency matrix with each vertex adjacent to itself
⍳3 3 generate an array of (0 0)(0 1)(0 2)(1 0)(1 1)(1 2)(2 0)(2 1)(2 2)
○ multiply all by π
7 ¯1∘., prepend 7 or -1 in all possible ways
(⍳3)∘.⌽ rotate coord triples by 0 1 2 steps in all possible ways
+/¨|∘.-⍨, compute manhattan distance between each pair
4≥ it must be no greater than 4 for neighbouring facets
answered May 19 at 21:32
ngnngn
7,57612661
$endgroup$
$begingroup$
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
$endgroup$
– jimmy23013
May 19 at 23:41
$begingroup$
@jimmy23013 using π is very nice :) thank you!
$endgroup$
– ngn
May 19 at 23:48
add a comment |
$begingroup$
APL (Dyalog Classic), 34 30 bytes
-4 thanks to jimmy23013
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
Try it online!
outputs an adjacency matrix with each vertex adjacent to itself
⍳3 3 generate an array of (0 0)(0 1)(0 2)(1 0)(1 1)(1 2)(2 0)(2 1)(2 2)
○ multiply all by π
7 ¯1∘., prepend 7 or -1 in all possible ways
(⍳3)∘.⌽ rotate coord triples by 0 1 2 steps in all possible ways
+/¨|∘.-⍨, compute manhattan distance between each pair
4≥ it must be no greater than 4 for neighbouring facets
answered May 19 at 21:32
ngnngn
7,57612661
$endgroup$
APL (Dyalog Classic), 34 30 bytes
-4 thanks to jimmy23013
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
Try it online!
outputs an adjacency matrix with each vertex adjacent to itself
⍳3 3 generate an array of (0 0)(0 1)(0 2)(1 0)(1 1)(1 2)(2 0)(2 1)(2 2)
○ multiply all by π
7 ¯1∘., prepend 7 or -1 in all possible ways
(⍳3)∘.⌽ rotate coord triples by 0 1 2 steps in all possible ways
+/¨|∘.-⍨, compute manhattan distance between each pair
4≥ it must be no greater than 4 for neighbouring facets
answered May 19 at 21:32
ngnngn
7,57612661
edited May 19 at 23:51
answered May 19 at 21:32
ngnngn
7,57612661
answered May 19 at 21:32
ngnngn
7,57612661
answered May 19 at 21:32
ngnngn
7,57612661
7,57612661
$begingroup$
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
$endgroup$
– jimmy23013
May 19 at 23:41
$begingroup$
@jimmy23013 using π is very nice :) thank you!
$endgroup$
– ngn
May 19 at 23:48
add a comment |
$begingroup$
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3
$endgroup$
– jimmy23013
May 19 at 23:41
$begingroup$
@jimmy23013 using π is very nice :) thank you!
$endgroup$
– ngn
May 19 at 23:48
$begingroup$
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3$endgroup$
– jimmy23013
May 19 at 23:41
$begingroup$
4≥+/¨|∘.-⍨,(⍳3)∘.⌽7 ¯1∘.,○⍳3 3$endgroup$
– jimmy23013
May 19 at 23:41
$begingroup$
@jimmy23013 using π is very nice :) thank you!
$endgroup$
– ngn
May 19 at 23:48
$begingroup$
@jimmy23013 using π is very nice :) thank you!
$endgroup$
– ngn
May 19 at 23:48
add a comment |
$begingroup$
Ruby, 79 bytes
54.timesi
Try it online!
Prints a representation of a unidirectional graph, as a list of the vertices to the right of and below each vertex as shown in the map below.
0 1 2 3 4 5
6 7 8 9 10 11
12 13 14 15 16 17
18 19 20 21 22 23
24 25 26 27 28 29
30 31 32 33 34 35
36 37 38 39 40 41
42 43 44 45 46 47
48 49 50 51 52 53
answered May 20 at 4:25
Level River StLevel River St
20.6k32683
$endgroup$
add a comment |
$begingroup$
Ruby, 79 bytes
54.timesi
Try it online!
Prints a representation of a unidirectional graph, as a list of the vertices to the right of and below each vertex as shown in the map below.
0 1 2 3 4 5
6 7 8 9 10 11
12 13 14 15 16 17
18 19 20 21 22 23
24 25 26 27 28 29
30 31 32 33 34 35
36 37 38 39 40 41
42 43 44 45 46 47
48 49 50 51 52 53
answered May 20 at 4:25
Level River StLevel River St
20.6k32683
$endgroup$
add a comment |
$begingroup$
Ruby, 79 bytes
54.timesi
Try it online!
Prints a representation of a unidirectional graph, as a list of the vertices to the right of and below each vertex as shown in the map below.
0 1 2 3 4 5
6 7 8 9 10 11
12 13 14 15 16 17
18 19 20 21 22 23
24 25 26 27 28 29
30 31 32 33 34 35
36 37 38 39 40 41
42 43 44 45 46 47
48 49 50 51 52 53
answered May 20 at 4:25
Level River StLevel River St
20.6k32683
$endgroup$
Ruby, 79 bytes
54.timesi
Try it online!
Prints a representation of a unidirectional graph, as a list of the vertices to the right of and below each vertex as shown in the map below.
0 1 2 3 4 5
6 7 8 9 10 11
12 13 14 15 16 17
18 19 20 21 22 23
24 25 26 27 28 29
30 31 32 33 34 35
36 37 38 39 40 41
42 43 44 45 46 47
48 49 50 51 52 53
answered May 20 at 4:25
Level River StLevel River St
20.6k32683
edited May 20 at 4:32
answered May 20 at 4:25
Level River StLevel River St
20.6k32683
answered May 20 at 4:25
Level River StLevel River St
20.6k32683
answered May 20 at 4:25
Level River StLevel River St
20.6k32683
20.6k32683
add a comment |
add a comment |
$begingroup$
Python 2.7, 145
def p(n):l=(3-n%2*6,n/6%3*2-2,n/18*2-2);k=n/2%3;return l[k:]+l[:k]
r=range(54)
x=[[sum((x-y)**2for x,y in zip(p(i),p(j)))<5for i in r]for j in r]
Try it online!
Defines an adjacency matrix x as a list of lists of boolean values. Facets count as being adjacent to themselves.
p(n) computes the coordinates of the center of the nth facet of a 3x3x3 cube whose facets are 2 units across. Adjacency is determined by testing if 2 facets have a square distance under 5 (adjacent facets have square distance at most 4, non-adjacent facets have square distance at least 6).
answered May 19 at 23:18
cardboard_boxcardboard_box
4,1751431
$endgroup$
add a comment |
$begingroup$
Python 2.7, 145
def p(n):l=(3-n%2*6,n/6%3*2-2,n/18*2-2);k=n/2%3;return l[k:]+l[:k]
r=range(54)
x=[[sum((x-y)**2for x,y in zip(p(i),p(j)))<5for i in r]for j in r]
Try it online!
Defines an adjacency matrix x as a list of lists of boolean values. Facets count as being adjacent to themselves.
p(n) computes the coordinates of the center of the nth facet of a 3x3x3 cube whose facets are 2 units across. Adjacency is determined by testing if 2 facets have a square distance under 5 (adjacent facets have square distance at most 4, non-adjacent facets have square distance at least 6).
answered May 19 at 23:18
cardboard_boxcardboard_box
4,1751431
$endgroup$
add a comment |
$begingroup$
Python 2.7, 145
def p(n):l=(3-n%2*6,n/6%3*2-2,n/18*2-2);k=n/2%3;return l[k:]+l[:k]
r=range(54)
x=[[sum((x-y)**2for x,y in zip(p(i),p(j)))<5for i in r]for j in r]
Try it online!
Defines an adjacency matrix x as a list of lists of boolean values. Facets count as being adjacent to themselves.
p(n) computes the coordinates of the center of the nth facet of a 3x3x3 cube whose facets are 2 units across. Adjacency is determined by testing if 2 facets have a square distance under 5 (adjacent facets have square distance at most 4, non-adjacent facets have square distance at least 6).
answered May 19 at 23:18
cardboard_boxcardboard_box
4,1751431
$endgroup$
Python 2.7, 145
def p(n):l=(3-n%2*6,n/6%3*2-2,n/18*2-2);k=n/2%3;return l[k:]+l[:k]
r=range(54)
x=[[sum((x-y)**2for x,y in zip(p(i),p(j)))<5for i in r]for j in r]
Try it online!
Defines an adjacency matrix x as a list of lists of boolean values. Facets count as being adjacent to themselves.
p(n) computes the coordinates of the center of the nth facet of a 3x3x3 cube whose facets are 2 units across. Adjacency is determined by testing if 2 facets have a square distance under 5 (adjacent facets have square distance at most 4, non-adjacent facets have square distance at least 6).
answered May 19 at 23:18
cardboard_boxcardboard_box
4,1751431
answered May 19 at 23:18
cardboard_boxcardboard_box
4,1751431
answered May 19 at 23:18
cardboard_boxcardboard_box
4,1751431
answered May 19 at 23:18
cardboard_boxcardboard_box
4,1751431
4,1751431
add a comment |
add a comment |
$begingroup$
Charcoal, 48 bytes
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧
Generate all sets of 3-dimensional coordinates in the range [0..6] for each dimension.
≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υ
Keep only those coordinates that are centres of 2x2 squares on one of the faces x=0, y=0, z=0, x=6, y=6, z=6.
ＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
For each coordinate, print the indices of those coordinates whose taxicab distance is 2.
The vertices are numbered as follows:
33 34 35
21 22 23
9 10 11
36 24 12 0 1 2 13 25 37 47 46 45
38 26 14 3 4 5 15 27 39 50 49 48
40 28 16 6 7 8 17 29 41 53 52 51
18 19 20
30 31 32
42 43 44
answered May 19 at 23:45
NeilNeil
84.6k845182
$endgroup$
add a comment |
$begingroup$
Charcoal, 48 bytes
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧
Generate all sets of 3-dimensional coordinates in the range [0..6] for each dimension.
≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υ
Keep only those coordinates that are centres of 2x2 squares on one of the faces x=0, y=0, z=0, x=6, y=6, z=6.
ＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
For each coordinate, print the indices of those coordinates whose taxicab distance is 2.
The vertices are numbered as follows:
33 34 35
21 22 23
9 10 11
36 24 12 0 1 2 13 25 37 47 46 45
38 26 14 3 4 5 15 27 39 50 49 48
40 28 16 6 7 8 17 29 41 53 52 51
18 19 20
30 31 32
42 43 44
answered May 19 at 23:45
NeilNeil
84.6k845182
$endgroup$
add a comment |
$begingroup$
Charcoal, 48 bytes
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧
Generate all sets of 3-dimensional coordinates in the range [0..6] for each dimension.
≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υ
Keep only those coordinates that are centres of 2x2 squares on one of the faces x=0, y=0, z=0, x=6, y=6, z=6.
ＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
For each coordinate, print the indices of those coordinates whose taxicab distance is 2.
The vertices are numbered as follows:
33 34 35
21 22 23
9 10 11
36 24 12 0 1 2 13 25 37 47 46 45
38 26 14 3 4 5 15 27 39 50 49 48
40 28 16 6 7 8 17 29 41 53 52 51
18 19 20
30 31 32
42 43 44
answered May 19 at 23:45
NeilNeil
84.6k845182
$endgroup$
Charcoal, 48 bytes
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⁷Ｆ⁷Ｆ⁷⊞υ⟦ικλ⟧
Generate all sets of 3-dimensional coordinates in the range [0..6] for each dimension.
≔Φυ⁼Φ﹪ι⁶¬﹪λ²⟦⁰⟧υ
Keep only those coordinates that are centres of 2x2 squares on one of the faces x=0, y=0, z=0, x=6, y=6, z=6.
ＩＥυΦＬυ⁼²ΣＥ§υλ↔⁻ν§ιξ
For each coordinate, print the indices of those coordinates whose taxicab distance is 2.
The vertices are numbered as follows:
33 34 35
21 22 23
9 10 11
36 24 12 0 1 2 13 25 37 47 46 45
38 26 14 3 4 5 15 27 39 50 49 48
40 28 16 6 7 8 17 29 41 53 52 51
18 19 20
30 31 32
42 43 44
answered May 19 at 23:45
NeilNeil
84.6k845182
answered May 19 at 23:45
NeilNeil
84.6k845182
answered May 19 at 23:45
NeilNeil
84.6k845182
answered May 19 at 23:45
NeilNeil
84.6k845182
84.6k845182
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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@DavidC 45 was also wrong. Thanks. Fixed.
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– jimmy23013
May 20 at 16:28