Visible Dice Faces

Question

A traditional Western die is a cube, on which the integers 1 to 6 are marked on the faces. Pairs that add to 7 are placed on opposite faces.

As it is cube, we can only see between 1 and 3 faces (inclusive)¹ at any given time. Opposite faces can never be seen at the same time.

Your task is to write a program or function which, given a list of integers representing sides on a die, determines if it is possible to see these faces at the same time.

¹_{Okay, maybe you can see 4 or 5 faces with a pair of eyes, but for the purpose of this challenge we observing the die from a single point.}

Rules:

Your submission may assume the input list:
- Is non-empty.
- Only contains values which satisfy 1 ≤ n ≤ 6.
- Contains no duplicate elements.
You may not assume that the input is sorted.
Your submission should output a truthy / falsy value: truthy is the faces can be seen at the same time, falsy otherwise.
This is code-golf, so the shortest answer (in bytes) wins!
Standard loopholes are forbidden by default.

Test Cases

Truthy:

[6]                 (One face)
[6, 2]              (Share a side)
[1, 3]              (Share a side)
[2, 1, 3]           (Share a vertex)
[3, 2, 6]           (Share a vertex)

Falsy:

[1, 6]              (1 and 6 are opposite)
[5, 4, 2]           (2 and 5 are opposite)
[3, 1, 4]           (3 and 4 are opposite)
[5, 4, 6, 2]        (Cannot see 4 faces)
[1, 2, 3, 4, 5, 6]  (Cannot see 6 faces)

It seems that the last two falsey cases are superfluous as any list longer than 3 will contain opposing values, no? — Weckar E., Dec 6 '17 at 12:22
@WeckarE yes, obviously - if you have a look at the answers they all exploit this. It was just an easier explanation to write. — FlipTack, Dec 6 '17 at 15:09
@FlipTack Actually you don't have to check the length at all, every list longer than 3 elements has at least one pair of opposite sides. — Erik the Outgolfer, Dec 7 '17 at 18:12
You can still see up to 5 faces from a single point if you bent the light waves with something heavy like a black hole — Ferrybig, Dec 14 '17 at 6:46

Chas Brown · Accepted Answer · 2017-12-04 20:53:19Z

15

Python 2, 35 bytes

lambda a:any(7-i in a for i in a)<1

Try it online!

edited Dec 4 '17 at 20:53

answered Dec 4 '17 at 20:46

Chas Brown

8,9591 gold badge7 silver badges32 bronze badges

5

\$\begingroup\$ -1 byte in Python 3. \$\endgroup\$ – notjagan Dec 4 '17 at 22:53

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Arnauld · Accepted Answer · 2017-12-06 00:01:07Z

JavaScript (ES6), 38 34 30 29 28 bytes

Takes input as any number of separate parameters. Returns 0 or 1.

(a,b,c,d)=>!(d|(a^b^c)%7)^!c

Test cases

let f =

(a,b,c,d)=>!(d|(a^b^c)%7)^!c

console.log('[Truthy]')
console.log(f(6      ))
console.log(f(6, 2   ))
console.log(f(1, 3   ))
console.log(f(2, 1, 3))
console.log(f(3, 2, 6))

console.log('[Falsy]')
console.log(f(1, 6            ))
console.log(f(5, 4, 2         ))
console.log(f(3, 1, 4         ))
console.log(f(5, 4, 6, 2      ))
console.log(f(1, 2, 3, 4, 5, 6))

How?

Below are simplified versions of the main expression according to the number of parameters provided, undefined variables being coerced to either 0 or false:

# of param. | simplified expression        | comment
------------+------------------------------+---------------------------------------------
     1      | !(a % 7) ^ 1                 | always true
     2      | !((a ^ b) % 7) ^ 1           | false for (1,6), (2,5) and (3,4)
     3      | !((a ^ b ^ c) % 7)           | see the table below
     4+     | !(d | (a ^ b ^ c) % 7)       | always false

NB: The order of (a,b,c) doesn't matter because they're always XOR'd together.

The trickiest case is the 3rd one. Here is a table showing all possible combinations:

a | b | c | a^b^c | %7 | =0? | faces that sum to 7
--+---+---+-------+----+-----+--------------------
1 | 2 | 3 |   0   |  0 | Yes | none
1 | 2 | 4 |   7   |  0 | Yes | none
1 | 2 | 5 |   6   |  6 | No  | 2 + 5
1 | 2 | 6 |   5   |  5 | No  | 1 + 6
1 | 3 | 4 |   6   |  6 | No  | 3 + 4
1 | 3 | 5 |   7   |  0 | Yes | none
1 | 3 | 6 |   4   |  4 | No  | 1 + 6
1 | 4 | 5 |   0   |  0 | Yes | none
1 | 4 | 6 |   3   |  3 | No  | 1 + 6
1 | 5 | 6 |   2   |  2 | No  | 1 + 6
2 | 3 | 4 |   5   |  5 | No  | 3 + 4
2 | 3 | 5 |   4   |  4 | No  | 2 + 5
2 | 3 | 6 |   7   |  0 | Yes | none
2 | 4 | 5 |   3   |  3 | No  | 2 + 5
2 | 4 | 6 |   0   |  0 | Yes | none
2 | 5 | 6 |   1   |  1 | No  | 2 + 5
3 | 4 | 5 |   2   |  2 | No  | 3 + 4
3 | 4 | 6 |   1   |  1 | No  | 3 + 4
3 | 5 | 6 |   0   |  0 | Yes | none
4 | 5 | 6 |   7   |  0 | Yes | none

Alt. version #1, 32 bytes

Takes input as an array. Returns a boolean.

a=>a.every(x=>a.every(y=>x+y-7))

Test cases

let f =

a=>a.every(x=>a.every(y=>x+y-7))

console.log('[Truthy]')
console.log(f([6]      ))
console.log(f([6, 2]   ))
console.log(f([1, 3]   ))
console.log(f([2, 1, 3]))
console.log(f([3, 2, 6]))

console.log('[Falsy]')
console.log(f([1, 6]            ))
console.log(f([5, 4, 2]         ))
console.log(f([3, 1, 4]         ))
console.log(f([5, 4, 6, 2]      ))
console.log(f([1, 2, 3, 4, 5, 6]))

Alt. version #2, Chrome/Firefox, 34 bytes

This one abuses the sort methods of Chrome and Firefox. It doesn't work with Edge.

Takes input as an array. Returns 0 or 1.

a=>a.sort((a,b)=>k&=a+b!=7,k=1)&&k

Test cases

let f =

a=>a.sort((a,b)=>k&=a+b!=7,k=1)&&k

console.log('[Truthy]')
console.log(f([6]      ))
console.log(f([6, 2]   ))
console.log(f([1, 3]   ))
console.log(f([2, 1, 3]))
console.log(f([3, 2, 6]))

console.log('[Falsy]')
console.log(f([1, 6]            ))
console.log(f([5, 4, 2]         ))
console.log(f([3, 1, 4]         ))
console.log(f([5, 4, 6, 2]      ))
console.log(f([1, 2, 3, 4, 5, 6]))

totallyhuman · Accepted Answer · 2017-12-05 20:43:39Z

Haskell, 24 bytes

-3 bytes thanks to H.PWiz.

f l=all(/=7)$(+)<$>l<*>l

Try it online!

Explanation

f l=all(/=7)$(+)<$>l<*>l

f l=                      -- make a function f that takes a single argument l
             (+)<$>l<*>l  -- take the sum of each pair in the cartesian product...
    all(/=7)$             -- ...and check if they're all inequal to 7

Uriel · Accepted Answer · 2017-12-04 21:02:23Z

APL (Dyalog), 7 bytes

~7∊∘.+⍨

Try it online!

∘.+⍨ - addition table (every element with every element)

7∊ - 7 exists?

~ - negate

APL (Dyalog), 7 bytes

⍬≡⊢∩7-⊢

Try it online!

7-⊢ - subtract each element from 7

⊢∩ - intersect with the original array

⍬≡ - empty?

Giuseppe · Accepted Answer · 2017-12-05 20:35:27Z

5

R, 27 bytes

thanks to Gregor for fixing a bug

function(d)!any((7-d)%in%d)

Try it online!

Port of Chas Brown's answer. Having vectorized operations helps make this much shorter in R.

edited Dec 5 '17 at 20:35

answered Dec 4 '17 at 21:24

Giuseppe

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\$\begingroup\$ I think you need some parens around (7-d), other wise d%in%d takes precedence. \$\endgroup\$ – Gregor Thomas Dec 5 '17 at 20:32
\$\begingroup\$ @Gregor you're absolutely right. \$\endgroup\$ – Giuseppe Dec 5 '17 at 20:35

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J42161217 · Accepted Answer · 2017-12-05 09:13:15Z

4

Mathematica, 20 bytes

xFreeQ[#+x&/@x,7]

The  is \[Function]

-12 bytes from Martin Ender
-7 bytes from Misha Lavrov

Try it online!

edited Dec 5 '17 at 9:13

answered Dec 4 '17 at 20:40

J42161217

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H.PWiz · Accepted Answer · 2017-12-04 20:52:19Z

3

Haskell, 26 bytes

f x=null[1|7<-(+)<$>x<*>x]

Try it online!

answered Dec 4 '17 at 20:52

H.PWiz

11k2 gold badges14 silver badges53 bronze badges

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Erik the Outgolfer · Accepted Answer · 2017-12-04 21:03:11Z

3

Pyth, 5 bytes

!@-L7

Try it here.

answered Dec 4 '17 at 21:03

Erik the Outgolfer

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\$\begingroup\$ I could swear I tried this two minutes ago and it didn’t work... Huh I wonder what went wrong \$\endgroup\$ – Mr. Xcoder Dec 4 '17 at 21:04
\$\begingroup\$ First thing I thought of? The Sandlot. \$\endgroup\$ – maxathousand Dec 4 '17 at 22:23

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Mego · Accepted Answer · 2017-12-04 21:10:11Z

3

Actually, 8 bytes

;∙♂Σ7@cY

Try it online! (runs all the test cases)

Explanation:

;∙♂Σ7@cY
;∙        Cartesian product with self
  ♂Σ      sum all pairs
    7@c   count 7s
       Y  logical negate

answered Dec 4 '17 at 21:10

Mego

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Martin Ender · Accepted Answer · 2017-12-04 21:49:43Z

3

Husk, 5 bytes

Ëo≠7+

Try it online!

Explanation

Ëo     Check that the following function gives a truthy value for all pairs 
       from the input.
    +    Their sum...
  ≠7     ...is not equal to 7.

answered Dec 4 '17 at 21:49

Martin Ender

185k61 gold badges410 silver badges912 bronze badges

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Neil · Accepted Answer · 2017-12-04 23:46:16Z

Retina, 21 20 bytes

O`.
M`1.*6|2.*5|34
0

Try it online! Link includes test cases. Edit: Saved 1 byte thanks to @MartinEnder. Explanation:

O`.

Sort the input.

M`1.*6|2.*5|34

Check for a pair of opposite sides (3 and 4 sort next to each other). This returns 1 for an invalid die or 0 for a valid one.

Logically negate the result.

Erik the Outgolfer · Accepted Answer · 2017-12-04 20:55:54Z

2

Jelly, 5 bytes

7_f⁸Ṇ

Try it online!

Port of Chas Brown's answer.

Explanation

7_f⁸Ṇ
7_    Subtract each element from 7
  f⁸  Filter with the original list
    Ṇ Check if empty

edited Dec 4 '17 at 20:55

answered Dec 4 '17 at 20:39

Erik the Outgolfer

38.1k4 gold badges31 silver badges113 bronze badges

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Martin Ender · Accepted Answer · 2017-12-04 22:53:51Z

Alice, 18 bytes

/..y1nr@ 
\iReA6o/

Try it online!

Prints Jabberwocky for valid inputs and nothing otherwise.

Explanation

Unfolding the zigzag control flow, the program is really just:

i.e16r.RyAno

i.  Read all input and duplicate it.
e16 Push "16".
r   Range expansion to get "123456".
.R  Duplicate and reverse.
y   Transliterate, replaces each face with its opposite.
A   Intersection with input.
n   Logical NOT, turns empty strings into "Jabberwocky"
    and everything else into an empty string.

Mr. Xcoder · Accepted Answer · 2017-12-05 09:06:57Z

2

Pyth, 5 bytes

-7sM*

Test suite.

isaacg saved a byte!

edited Dec 5 '17 at 9:06

answered Dec 4 '17 at 20:53

Mr. Xcoder

39.8k8 gold badges67 silver badges206 bronze badges

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G B · Accepted Answer · 2017-12-05 11:27:32Z

2

Ruby, 36 31 24 23 bytes

->l{l-l.map{|x|7-x}==l}

Try it online!

It was so simple, I was looking for the solution to the wrong problem all the time.

edited Dec 5 '17 at 11:27

answered Dec 4 '17 at 21:05

G B

11.1k1 gold badge14 silver badges35 bronze badges

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Emigna · Accepted Answer · 2017-12-04 21:24:02Z

05AB1E, 5 bytes

7αå_P

Try it online!

Explanation

7α      # absolute difference between 7 an each in input list
  å     # check each element if it exist in input
   _    # logical negate
    P   # product of list

One of several ways to do this in 5 bytes in 05AB1E

Xcali · Accepted Answer · 2017-12-04 21:30:35Z

1

Perl 5, 51 + 1 (`-a`) = 52 bytes

say$F[0]+($t=$F[1])-7&&$F[0]+$t-7&&$t+$F[2]-7&&@F<4

Try it online!

answered Dec 4 '17 at 21:30

Xcali

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\$\begingroup\$ 3 1 4 gives true, what about say 1>grep{$x=$_;grep$_+$x==7,@F}@F \$\endgroup\$ – Nahuel Fouilleul Dec 5 '17 at 16:14
\$\begingroup\$ 25 + 1 (-p) inspired from sed solution \$\endgroup\$ – Nahuel Fouilleul Dec 5 '17 at 16:51

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Martin Ender · Accepted Answer · 2017-12-05 09:48:14Z

Retina, 20 bytes

T`_654`d
M`(.).*\1
0

Try it online!

An alternative to Neil's approach.

Explanation

T`_654`d

Turn 6,5, 4 into 1, 2, 3, respectively.

M`(.).*\1

Try to find repeated characters and count the number of matches.

Make sure the result was zero (effectively a logical negation).

Olivier Grégoire · Accepted Answer · 2017-12-05 11:00:36Z

1

Java (OpenJDK 8), 43 bytes

l->l.stream().noneMatch(x->l.contains(7-x))

Try it online!

answered Dec 5 '17 at 11:00

Olivier Grégoire

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Jordan · Accepted Answer · 2017-12-05 15:25:02Z

1

GNU sed, 37 22 + 1 = 23 bytes

+1 byte for -r flag. Takes input as digits; prints the input for truthy and 0 for falsy.

-10 bytes thanks to @MartinEnder.

y/123/654/
/(.).*\1/c0

Try it online!

Explanation

Similar to @MartinEnder’s Alice answer.

y/123/654/   # Transliterate 1-3 to 6-4
/(.).*\1/c0  # If a digit appears twice, replace the pattern space with 0

edited Dec 5 '17 at 15:25

answered Dec 5 '17 at 5:31

Jordan

5,00114 silver badges27 bronze badges

\$\begingroup\$ It doesn't exactly print the input for truthy, it prints the input with 1-3 replaced by 6-4. \$\endgroup\$ – Andrew Dec 5 '17 at 19:05

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Sean · Accepted Answer · 2017-12-05 21:41:12Z

Perl 6, 18 bytes

!(1&6|2&5|3&4∈*)

Try it online!

1 & 6 | 2 & 5 | 3 & 4 is a junction consisting of the numbers 1 and 6, OR the numbers 2 and 5, OR the numbers 3 and 4. This junction is an element of (∈) the input list * if it contains 1 and 6, or 2 and 5, or 3 and 4. That result is then negated (!) to get the required boolean value.

Draconis · Accepted Answer · 2017-12-04 20:54:05Z

0

Haskell, 46 41 37 bytes

f z=all ((/=7).sum) [[x,y]|x<-z,y<-z]

Takes the Cartesian product of the list with itself, then checks if all of the resulting lists do not sum to 7. (If any of them summed to 7, opposite faces would be visible, and "all do not" is shorter than "not any do".)

edited Dec 4 '17 at 20:54

answered Dec 4 '17 at 20:46

Draconis

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Galen Ivanov · Accepted Answer · 2017-12-05 09:16:28Z

0

J, 12 bytes

7-.@e.[:,+/~

This is pretty much the same as the first APL solution by Uriel

Try it online!

answered Dec 5 '17 at 9:16

Galen Ivanov

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ElPedro · Accepted Answer · 2017-12-05 09:54:52Z

IBM/Lotus Notes Formula, 7 bytes

!7-i*=i

Takes input from a multi-value number field i.

Recursively subtracts each value of i from 7 and checks to see if it is in the original list. ! changes the 1 if it is to a 0 (faces cannot all be seen).

Test Cases (no TIO available for Notes Formula)

Οurous · Accepted Answer · 2017-12-06 02:46:41Z

0

Clean, 49 bytes

import StdEnv
f l=and[(a+b)<>7\\(a,b)<-diag2 l l]

Try it online!

edited Dec 6 '17 at 2:46

answered Dec 4 '17 at 21:33

Οurous

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1

\$\begingroup\$ @StephenLeppik I don't think this answer is necessarily using the file name to store information. \$\endgroup\$ – Steadybox Dec 5 '17 at 1:02
\$\begingroup\$ @StephenLeppik Oh fair point there, the import is required but the function doesn't need to be declared in a specific module. Thanks. \$\endgroup\$ – Οurous Dec 5 '17 at 3:27

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Herman L · Accepted Answer · 2017-12-08 20:04:54Z

Swift, 46 bytes

f(a:Int)->[Int]{!a.contains{a.contains(7-$0)}}

Takes input as [Int] (integer array) and returns a Bool (boolean).

Short Explanation (ungolfed)

f(a:Int)->[Int]{
    !a.contains (where: {n in    // The input doesn't have an element where
        a.contains(7 - n)        //    the opposite side is in the input
    })
}

Carcigenicate · Accepted Answer · 2018-01-01 20:24:30Z

Clojure, 89 80 72 bytes

-9 bytes because I realized use of reduced is unnecessary

-8 bytes by changing from using reduce to using some

#(if(< 0(count %)4)(not(some(fn[[a b]](=(+ a b)7))(for[a % b %][a b]))))

I tried writing this without looking at any of the tricks other answers use. I may be able to improve on this later.

Returns true for true cases, false if any sides sum to 7, and nil if the number of sides is illegal.

(defn can-see? [dice-sides]
  ; Ensure its a valid length
  (if (< 0 (count dice-sides) 4)

    ; Then check if any of the permutations sum to 7
    (not (some (fn [[a b]]
                 (= (+ a b) 7))

               ; Generate all permutations
               (for [a dice-sides, b dice-sides] [a b])))))

Visible Dice Faces

Rules:

Test Cases

27 Answers 27

Python 2, 35 bytes

JavaScript (ES6), 38 34 30 29 28 bytes

Test cases

How?

Alt. version #1, 32 bytes

Test cases

Alt. version #2, Chrome/Firefox, 34 bytes

Test cases

Haskell, 24 bytes

Explanation

APL (Dyalog), 7 bytes

APL (Dyalog), 7 bytes

R, 27 bytes

Mathematica, 20 bytes

Haskell, 26 bytes

Pyth, 5 bytes

Actually, 8 bytes

Husk, 5 bytes

Explanation

Retina, 21 20 bytes

Jelly, 5 bytes

Explanation

Alice, 18 bytes

Explanation

Pyth, 5 bytes

Ruby, 36 31 24 23 bytes

05AB1E, 5 bytes

Perl 5, 51 + 1 (-a) = 52 bytes

Retina, 20 bytes

Explanation

Java (OpenJDK 8), 43 bytes

GNU sed, 37 22 + 1 = 23 bytes

Explanation

Perl 6, 18 bytes

Haskell, 46 41 37 bytes

J, 12 bytes

IBM/Lotus Notes Formula, 7 bytes

Clean, 49 bytes

Swift, 46 bytes

Short Explanation (ungolfed)

Clojure, 89 80 72 bytes

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