Gambling with an Alien

Question

You met a 4-th dimensional being who challenged you to a game of dice. The rules are simple: each player rolls 3 6-sided dice and takes the sum of each combination of 2 dice. The player with the highest sum wins. If the first-highest sum is a tie, consider the second-highest sum, and so on.

Your opponent's dice look normal, but you think they might have more than 6 sides! Your goal is to find out if the alien rolled a number higher than 6.

Your input is a set of 3 integers, separated by at least one character (space or otherwise). Input must not be hard coded.

// All valid inputs
6 7 9
6, 7, 9
(6, 7, 9)

Your output is a string representing if the alien cheated. Any output is allowed, as long as two distinct values exist for cheating/valid. Appended whitespace is permitted.

"😀" // (U+1F600) Valid set of dice
"😠" // (U+1F620) Invalid set of dice

Examples:

<-: 8 11 9
->: 😀       // Dice can be found to be 3, 5, and 6, which are all on 6-side die

<-: 9 11 6
->: 😠       // Dice can be found to be 2, 7, and 4, which means the alien cheated!

Assumptions:

Input will be 3 integers, 2-12
Input numbers can be in any order
Die values will be positive integers
Alien only cheated if they rolled greater than 6

Write a program or function that takes inputs, returns or prints outputs.

Fewest bytes wins!

"Any stdout is allowed, as long as two distinct values exist for cheating/valid". Does this mean you can choose your own output string to stdout, as long as the output for chating/valid is distinct? For example, you can output c for cheating and v for valid? — xiver77, Commented Mar 3, 2022 at 14:57
@badatgolf oh thanks for the clarification. I got caught up in the format. Yes, that would be invalid, per the clause "Die values will be positive integers" — Fruity Medley, Commented Mar 3, 2022 at 15:38
I recommend allowing that the implementation can just be a function returning a truthy/falsy value. As you can see, 2 answers are already just a function that doesn't write to stdout. That's the kind of the norm for this kind of challenges in this site. — xiver77, Commented Mar 3, 2022 at 15:42
@badatgolf That shouldn't invalidate answers. The clause is meant to indicate that an input will always be legal. The answers aren't meant to check for validity. — Fruity Medley, Commented Mar 3, 2022 at 15:48

Dominic van Essen · Accepted Answer · 2022-03-03 15:34:57Z

13

Husk, 8 bytes

¬>6-▼¹½Σ

Try it online!

Outputs 1 for normal dice, 0 if the alien cheated.

Checks whether half (½) of the sum (Σ) minus (-) the smallest (▼) element is not (¬) greater than 6 (>6).

edited Mar 3, 2022 at 15:34

answered Mar 3, 2022 at 15:29

Dominic van Essen

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Dominic van Essen · Accepted Answer · 2022-03-03 15:45:15Z

7

R, 28 bytes

function(x)sum(x/2)-min(x)>6

Try it online!

Same approach as my Husk answer.

edited Mar 3, 2022 at 15:45

answered Mar 3, 2022 at 15:32

Dominic van Essen

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AZTECCO · Accepted Answer · 2022-03-03 21:36:51Z

7

Haskell, 20 bytes

f l=any(<sum l/2-6)l

Try it online!

Inspired by @DominicVanEssen R answer.
Tells if it's a cheater (the alien.. not Dominic).

answered Mar 3, 2022 at 21:36

AZTECCO

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l4m2 · Accepted Answer · 2022-03-03 15:02:18Z

5

JavaScript (Node.js), 36 bytes

f=(a,b,c)=>a>b|a>c?f(b,c,a):b+c-a<13

Try it online!

Like the hypot one

answered Mar 3, 2022 at 15:02

l4m2

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3

\$\begingroup\$ "Like the hypot one" what are you referring to? \$\endgroup\$
– Jonah
Commented Mar 3, 2022 at 15:35

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Arnauld · Accepted Answer · 2022-03-03 15:49:52Z

5

JavaScript (ES6), 35 bytes

Expects ([a,b,c]). Returns true for cheating or false for valid.

A=>A.some(x=>eval(A.join`+`)/2>x+6)

Try it online!

answered Mar 3, 2022 at 15:49

Arnauld

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badatgolf · Accepted Answer · 2022-03-04 01:53:44Z

5

C (gcc) (-O0), 73 bytes

#define g(a,b) if(a>b)a^=b^=a^=b;
f(a,b,c){g(a,b)g(b,c)g(a,b)a=b+c-a<13;}

Try it online!

Not the prettiest solution

C (gcc) (-O0), 38 bytes, shamelessly copy this beautiful answer.

f(a,b,c){a=a>b|a>c?f(b,c,a):b+c-a<13;}

Try it online!

-2 bytes thanks to AZTECCO

edited Mar 4, 2022 at 1:53

answered Mar 3, 2022 at 15:32

badatgolf

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2

\$\begingroup\$ For the 40Byte answer you can do a=.. and remove s \$\endgroup\$
– AZTECCO
Commented Mar 3, 2022 at 20:01

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Kevin Cruijssen · Accepted Answer · 2022-03-03 18:01:24Z

4

05AB1E, 7 6 bytes

O;αà6›

Inspired by @DominicVanEssen's R answer.
Takes a list [a,b,c] as input, and outputs 1/0 for cheating/valid respectively.

Try it online or verify all test cases.

Explanation:

O       # Sum the three values in the (implicit) input-list
        #  e.g. [8,11,9] → 28
        #  e.g. [9,11,6] → 26
 ;      # Halve it
        #  → 14
        #  → 13
  α     # Take the absolute difference of this sum with each value in the
        # (implicit) input-list
        #  → [6,3,5]
        #  → [4,2,7]
   à    # Pop and push its maximum
        #  → 6
        #  → 7
    6›  # Check if this is larger than 6
        #  → 0
        #  → 1
        # (after which this is output implicitly as result)

edited Mar 3, 2022 at 18:01

answered Mar 3, 2022 at 16:19

Kevin Cruijssen

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\$\begingroup\$ Code and byte count do not match \$\endgroup\$
– Seggan
Commented Mar 3, 2022 at 16:21
\$\begingroup\$ Also the TIO link \$\endgroup\$
– Seggan
Commented Mar 3, 2022 at 16:27
\$\begingroup\$ @Seggan Oops, you're completely right. I made a mistake with the 6-byter and forgot to change the byte-count and single TIO when I fixed it. Should be fixed now. \$\endgroup\$
– Kevin Cruijssen
Commented Mar 3, 2022 at 16:27

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Sundar R · Accepted Answer · 2022-03-10 12:17:34Z

MATL, 10 bytes

2/sGX<-IE>

Try it out / Test three cases

Port of @Dominic van Essen's answers. 1 = You cheating alien! 0 = Never had a moment of doubt.

4-sGX<E-F> Try it online! is another 10 byter.

MATL, 11 bytes

.5IXy-iY*7<

Try it out / Test three cases

Outputs truthy (all 1s) for no cheating, falsy (some 0s) for cheating.

For given input [a, b, c], the system of linear equations is : \$ d_2 + d_3 = a \$ , \$ d_1 + d_3 = b \$ , \$ d_1 + d_2 = c \$

Or in matrix form: \$ \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \$

The solution - the set of dice values - is the inverse of the matrix on the left, multiplied by the input. This inverse is equal to \$ 0.5 - I_3 \$ (where \$I_3\$ is the 3x3 identity matrix).

.5IXy- computes this inverse, iY* multiplies that by the input, 7< checks if all of the resulting dice values are less than 7 (and hence valid).

I love the matrix one! I think you can use 6+G2/s<a or similar for 8 (maybe 7, though, I know there's a multiply by 2 command but I never remember what it is) porting my R answer. — Giuseppe, Commented Mar 10, 2022 at 13:00

Neil · Accepted Answer · 2022-03-03 21:29:52Z

3

Retina 0.8.2, 34 bytes

\d+
$*
O`1+
^
12$*
(1+)(1+),\1,\2$

Try it online! Link includes test cases. Outputs 1 for normal dice, 0 if the alien cheated. Explanation:

\d+
$*

Convert to unary.

O`1+

Sort into order.

^
12$*

Add 12 to the smallest value.

(1+)(1+),\1,\2$

The sum should then equal or exceed the sum of the other two values.

answered Mar 3, 2022 at 21:29

Neil

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Seggan · Accepted Answer · 2022-03-03 17:09:20Z

2

Vyxal, 7 bytes

∑½?g-6≤

Try it Online!

Port of @Dominic van Essen's Husk answer. 1 if the dice are normal, 0 if the alien used tesseracts for its dice.

∑½?g-6≤ # Takes list as input
∑       # Sum of the list
 ½      # Halved..
    -   # Minus...
  ?g    # Smallest element of the input
     6≤ # Is less than or equal to 6?

edited Mar 3, 2022 at 17:09

answered Mar 3, 2022 at 16:33

Seggan

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1

\$\begingroup\$ Oops, messed up my words. Thanks for noticing that. \$\endgroup\$
– Seggan
Commented Mar 3, 2022 at 17:09

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sinvec · Accepted Answer · 2022-03-03 23:51:02Z

2

J, ²⁰ 11 bytes

12<+/-2*<./

-9 bytes thanks @Bubbler

Try it online!

edited Mar 3, 2022 at 23:51

answered Mar 3, 2022 at 22:40

sinvec

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\$\begingroup\$ You can use a noun in the left side of a fork (so 6 works instead of 6:), you don't need abs if you reverse the direction of subtraction, you can double everything else instead of halving the sum, and you can change "max is greater than 6" 6<>./@(...) to "some number is greater than 6" 1 e.6<. These changes give 12 bytes. \$\endgroup\$
– Bubbler
Commented Mar 3, 2022 at 23:38
\$\begingroup\$ Actually this is shorter. \$\endgroup\$
– Bubbler
Commented Mar 3, 2022 at 23:43

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Giuseppe · Accepted Answer · 2022-03-04 16:05:21Z

2

R, 28 bytes

function(x)any(sum(x)/2>6+x)

Try it online!

Alternate and equally-golfy solution porting this python answer -- the naive port would be something like sum(x)-2*x>12 which is a byte longer.

answered Mar 4, 2022 at 16:05

Giuseppe

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Sundar R · Accepted Answer · 2022-03-05 09:46:56Z

2

Julia 1.0, 23 bytes

x->sum(x/2)-min(x...)>6

Try it online!

Port of @Dominic van Essen's answers. Output is the answer to the question "Did the alien cheat?"

Julia 1.2 or above, with `using LinearAlgebra`, 23 bytes

x->any((.5 .-I(3))x.>6)

(doesn't work on TIO since it only has Julia 1.0)

18 bytes without the any call, if output can be all 0s for no cheating, at least one 1 for cheating alien.

Find the dice values by solving the system of linear equations via matrices, and check if any are greater than 6. (More explanation in my MATL answer).

edited Mar 5, 2022 at 9:46

answered Mar 5, 2022 at 4:38

Sundar R

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\$\begingroup\$ 22 bytes \$\endgroup\$
– Giuseppe
Commented Mar 10, 2022 at 13:01

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Lecdi · Accepted Answer · 2022-03-05 12:32:25Z

2

Python 3, 27 26 bytes

Code

lambda v:sum(v)/2-min(v)>6

Outputs True for cheated and False for valid.

Try it online!

Explanation

Let \$a\$, \$b\$ and \$c\$ be the three dice rolls, where \$a\$ is the highest roll. Therefore, if \$a > 6\$, the alien cheated, and otherwise they did not.

The sum of the numbers is \$2a + 2b + 2c\$. The minimum of the numbers is \$b + c\$. Therefore, sum(v)/2-min(v) is \$(2a + 2b + 2c)/2 - (b + c)\$ which is \$a\$. If this is greater than \$6\$, the alien cheated.

edited Mar 5, 2022 at 12:32

answered Mar 5, 2022 at 12:27

Lecdi

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Saphereye · Accepted Answer · 2022-03-06 05:08:40Z

2

Python 3.8 (pre-release),38 28 27 bytes

Latest Answer

lambda *a:sum(a)/2-min(a)>6

Try it online!

Old Answer

lambda x,y,z:max(x-y+z,x+y-z,y+z-x)>12

Try it online!

Output : Gives True when cheated and False when not.

Thank you Aetol for the suggestion of using a+b+c-2*min(a,b,c)>12 instead of max(x-y+z,x+y-z,y+z-x)>12

This R answer gave the nice sum(a)/2-min(a)>6 Idea

edited Mar 6, 2022 at 5:08

answered Mar 4, 2022 at 12:46

Saphereye

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1

\$\begingroup\$ Checking a+b+c-2*min(a,b,c)>12 is shorter by 4 characters \$\endgroup\$
– Aetol
Commented Mar 5, 2022 at 18:01

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sinvec · Accepted Answer · 2022-03-06 17:43:32Z

2

Pyth, ^{22 18} 11 bytes

L>-/sb2hSb6

Port of @Lecdi's Python 3 solution.

L>-/sb2hSb6
L             # lambda b:
    sb        # sum(b)
   /  2       # sum(b) / 2
       hSb    # min(b)
  -           # sum(b)/2 - min(b)
 >        6   # sum(b)/2 - min(b) > 6

Try it online!

edited Mar 6, 2022 at 17:43

answered Mar 6, 2022 at 17:08

sinvec

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Neil · Accepted Answer · 2022-03-03 21:34:58Z

1

Charcoal, 8 bytes

›Σθ⊗⁺⁶⌊θ

Try it online! Link is to verbose version of code. Takes input as an array and outputs - if the alien cheated, nothing if it didn't. Explanation: Inspired by @DominicvanEssen's answers.

  θ         Input array
 Σ          Summed
›           Is greater than
       θ    Input array
      ⌊     Minimum
    ⁺       Plus
     ⁶      Literal integer `6`
   ⊗        Doubled
            Implicitly print

answered Mar 3, 2022 at 21:34

Neil

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G B · Accepted Answer · 2022-03-04 09:03:12Z

1

Ruby, 21 bytes

->*s{s.sum/2-s.min<7}

Try it online!

answered Mar 4, 2022 at 9:03

G B

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Ginger · Accepted Answer · 2022-03-04 15:48:36Z

1

Etch, 47 bytes

s=:get;:split" ";:intify;;:out:sum s;/2-:min s;>6;

A straightforward point of this Husk answer.

answered Mar 4, 2022 at 15:48

Ginger

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Dominic van Essen · Accepted Answer · 2022-03-05 14:41:29Z

BQN, 9 bytes

¯12>-˜´∘∨

Try it at BQN online REPL

Outputs 0 if the dice are 6-sided, 1 if the alien is cheating.

We calculate the highest die roll using a single fold (or reduce in some languages) of a "minus" operation across the 3 sums-of-2-rolls in increasing order.
Consider initial die rolls of s, m and l, where l is the (possibly non-unique) largest, and s is the smallest. Then, the sums-of-2-rolls, in increasing order, are: s+m, s+l, m+l.
Folding "minus" across this yields (s+m minus s+l) minus m+l = m-l - (m+l) = -2l. So we just need to check whether the result is less than minus 12: if it is, then l was greater than 6 and the alien was cheating.

         ∨   # sort the input
        ∘    # and use that to
     -˜´     # fold 'subtracted from' from the right
¯12>         # and check whether it's less than minus 12

This comes out 1 byte shorter than the "subtract the lowest value from half the sum of the input" approach (12<+´-2×⌊´ = 10 bytes in BQN: try it)

138 Aspen · Accepted Answer · 2023-04-18 10:27:35Z

1

Scala, 35 bytes

Try it online!

def f(x:Seq[Int])=x.sum/2.0-x.min>6

answered Apr 18, 2023 at 10:27

138 Aspen

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Stack Exchange Network

Gambling with an Alien

21 Answers 21

Husk, 8 bytes

R, 28 bytes

Haskell, 20 bytes

JavaScript (Node.js), 36 bytes

JavaScript (ES6), 35 bytes

C (gcc) (-O0), 73 bytes

C (gcc) (-O0), 38 bytes, shamelessly copy this beautiful answer.

05AB1E, 7 6 bytes

MATL, 10 bytes

MATL, 11 bytes

Retina 0.8.2, 34 bytes

Vyxal, 7 bytes

J, ²⁰ 11 bytes

R, 28 bytes

Julia 1.0, 23 bytes

Julia 1.2 or above, with `using LinearAlgebra`, 23 bytes

Python 3, 27 26 bytes

Code

Explanation

Python 3.8 (pre-release),38 28 27 bytes

Pyth, ^{22 18} 11 bytes

Charcoal, 8 bytes

Ruby, 21 bytes

Etch, 47 bytes

BQN, 9 bytes

Scala, 35 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
decision-problem
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Gambling with an Alien

21 Answers 21

Husk, 8 bytes

R, 28 bytes

Haskell, 20 bytes

JavaScript (Node.js), 36 bytes

JavaScript (ES6), 35 bytes

C (gcc) (-O0), 73 bytes

C (gcc) (-O0), 38 bytes, shamelessly copy this beautiful answer.

05AB1E, 7 6 bytes

MATL, 10 bytes

MATL, 11 bytes

Retina 0.8.2, 34 bytes

Vyxal, 7 bytes

J, 20 11 bytes

R, 28 bytes

Julia 1.0, 23 bytes

Julia 1.2 or above, with using LinearAlgebra, 23 bytes

Python 3, 27 26 bytes

Code

Explanation

Python 3.8 (pre-release),38 28 27 bytes

Pyth, 22 18 11 bytes

Charcoal, 8 bytes

Ruby, 21 bytes

Etch, 47 bytes

BQN, 9 bytes

Scala, 35 bytes

Your Answer

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J, ²⁰ 11 bytes

Julia 1.2 or above, with `using LinearAlgebra`, 23 bytes

Pyth, ^{22 18} 11 bytes

Not the answer you're looking for? Browse other questions tagged
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decision-problem
or ask your own question.