How many petals around the rose

Question

How many petals around the rose

How many petals around the rose is a dice game you can play with your friends. Similar to "can I join the music box", there is a person that knows how the game works and the others have to discover the rule.

In this game, someone rolls some dice (usually two or more) and then people have to call "how many petals are around the rose".

Algorithm

If you want to play it by yourself, you can play it over at TIO. Just hide the header (which is where the rule is implemented) and try passing different arguments (from 1 to 6) in the function.

spoiler, what follows is the rule you are invited to find by yourself!

The "roses" here are the dice, and the "petals" are the black dots that are around a central dot. Because only the odd numbers have a central black dot, only numbers 1, 3, 5 matter for the petals. Those numbers have, respectively, 0, 2, 4 dots around the central dot, the "petals".

Input

Your input will be a non-empty list (or equivalent) of integers in the range [1, 6].

Output

The number of petals around the rose.

Test cases

Reference implementation in Python, that also generated the test cases.

1, 1 -> 0
1, 2 -> 0
1, 3 -> 2
1, 4 -> 0
1, 5 -> 4
1, 6 -> 0
2, 1 -> 0
2, 2 -> 0
2, 3 -> 2
2, 4 -> 0
2, 5 -> 4
2, 6 -> 0
3, 1 -> 2
3, 2 -> 2
3, 3 -> 4
3, 4 -> 2
3, 5 -> 6
3, 6 -> 2
4, 1 -> 0
4, 2 -> 0
4, 3 -> 2
4, 4 -> 0
4, 5 -> 4
4, 6 -> 0
5, 1 -> 4
5, 2 -> 4
5, 3 -> 6
5, 4 -> 4
5, 5 -> 8
5, 6 -> 4
6, 1 -> 0
6, 2 -> 0
6, 3 -> 2
6, 4 -> 0
6, 5 -> 4
6, 6 -> 0
3, 1, 5 -> 6
4, 5, 2 -> 4
4, 3, 5 -> 6
1, 4, 4 -> 0
5, 5, 2 -> 8
4, 1, 1 -> 0
3, 4, 1 -> 2
4, 3, 5 -> 6
4, 4, 5 -> 4
4, 2, 1 -> 0
3, 5, 5, 2 -> 10
6, 1, 4, 6, 3 -> 2
3, 2, 2, 1, 2, 3 -> 4
3, 6, 1, 2, 5, 2, 5 -> 10

---

This is code-golf so shortest submission in bytes, wins! If you liked this challenge, consider upvoting it... And happy golfing!

Answer 1

Python 3, 30 bytes

lambda l:sum(n**3&6for n in l)

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(test cases shamelessly borrowed from xnor's answer)

How?

Given \$1\le n\le6\$, the number of petals can be computed with:

$$p=n^{2k+1} \operatorname{and}6,\:k\in\mathbb{N}^*$$

where \$\operatorname{and}\$ is a bitwise operator.

This can also be written as:

$$p=2\times\left\lfloor\frac{n^{2k+1}\bmod 8}{2}\right\rfloor$$

And is based on the fact that, for any \$k\ge1\$:

$$n^{2k+1}\bmod 8=\cases{ n&\text{if $n$ is odd ($1$, $3$ or $5$)}\\ 0&\text{if $n$ is even ($2$, $4$ or $6$)} }$$

More specifically, choosing \$k=1\$:

$$p=n^3 \operatorname{and}6$$

As Python code, the resulting expression is just as long as the nice n%-2%n found by xnor. But because it ends with a digit, we can get rid of the space just before the for, saving a byte.

 n | n**3 | as binary  | AND 6
---+------+------------+-------
 1 |    1 | 00000 00 1 |   0
 2 |    8 | 00001 00 0 |   0
 3 |   27 | 00011 01 1 |   2
 4 |   64 | 01000 00 0 |   0
 5 |  125 | 01111 10 1 |   4
 6 |  216 | 11011 00 0 |   0

Answer 2

Retina 0.8.2, 11 bytes

5
33
3
33
3

Try it online! Link includes test cases. Input can be in almost any format really as only the 5s and 3s count. Explanation:

5
33

A 5 has as many petals as two 3s.

3
33

Speaking of 3s, they have two petals, so make them count twice.

3

Count the petals.

Answer 3

Python, 31 bytes

lambda l:sum(n%-2%n for n in l)

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The core expression n%-2%n evaluates to zero for even n and to n-1 for odd n.

       | n%-2   n%-2%n
-------+--------------
n even |    0        0  
n odd  |   -1      n-1

Answer 4

Jelly,  7  4 bytes

-3 using Giuseppe's method!

Ḃ×’S

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How?

We know that 3's and 5's "score", that they score one less than their pips, and that \$1-1=0\$ so:

        throw   1  2  3  4  5  6
  x=throw - 1   0  1  2  3  4  5
  y=throw % 2   1  0  1  0  1  0
score = x * y   0  0  2  0  4  0

So:

Ḃ×’S - Link: throws, list of integers in [1,6]   e.g. [1,2,3,4,5,6]
Ḃ    - (throws) % 2                                   [1,0,1,0,1,0]
  ’  - (throws) - 1                                   [0,1,2,3,4,5]
 ×   - multiply                                       [0,0,2,0,4,0]
   S - sum                                            6

Answer 5

R, 26 23 bytes

function(l)l%%2%*%(l-1)

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-3 thanks to Robin Ryder!

Returns a 1x1 matrix with the result.

---

Old answer, since there are a handful of explicit ports:

function(l)sum((l-1)*l%%2)

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Answer 6

Perl 6, 17 16 bytes

{sum $_>>³X+&6}

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Uses Arnauld's formula of \$n^3\&6\$. If HyperWhatevers worked, then something like (**³+&6).sum should be possible for 13 bytes.

My old regex based solution:

{sum m:g/3|5/X-1}

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Match 3s and 5s from the input, subtract one from each and sum them together.

Answer 7

Python, 32 bytes

lambda d:sum(r%2*~-r for r in d)

Port of Giuseppe's R answer.

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Answer 8

J, 9 bytes

1#.2&|*<:

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• 1#. Sum of applying the following to each element...
• 2&| Remainder when divided by 2
• * Times...
• <: Number decremented by 1

Answer 9

Japt -mx, 3 bytes

³&6

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Saved 2 bytes using Arnauld's formula so be sure to +1 him if you're +1ing this.

Original, 5 bytes

fu xÉ

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Answer 10

AWK, ^{29 28} 26 bytes

/3|5/{x+=$0-1}END{print+x}

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-1 byte thanks to Jo King

-2 bytes thanks to user41805

+x is needed so that inputs with no odd numbers still return 0.

Answer 11

PHP, 55 53 51 bytes

fn($a)=>array_sum(array_map(fn($v)=>$v%2*~-$v,$a));

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-2 bytes thanks to @RGS

Answer 12

Haskell, 23 bytes

f x=sum[n-1|n<-x,odd n]

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Answer 13

Python 3, ^{40 33} 32 bytes

Saved 7 bytes thanks to RGS!!!

lambda d:sum(i%2*~-i for i in d)

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This turns out to be a total rip off of Jonathan Allan's Python answer so update him instead.

Answer 14

Wolfram Language (Mathematica), 25 bytes

Tr[(Mod[#,2]#//.0->1)-1]&

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Answer 15

C# (Visual C# Interactive Compiler), 20 bytes

a=>a.Sum(x=>x--%2*x)

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Answer 16

C (gcc), ⁵⁰ 48 bytes

Saved 2 bytes thanks to xibu!!!

s;f(l,p)int*p;{for(s=0;l--;)s+=*p%2*~-*p++;l=s;}

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Inputs an integer array pointer preceded by its length.

Answer 17

Java (JDK), 47 45 bytes

r->{int s=0;for(int i:r)s+=i%2*~-i;return s;}

First time code golfing, hope I posted this right.

-2 Bytes by removing unneeded curly brackets

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Answer 18

TI-BASIC, 13 12 bytes

I found some 12 byte solutions

Ans-1:sum(Ansnot(fPart(Ans/2

The previous is the same byte count as below due to the line separator costing a byte

sum((Ans-1)2fPart(Ans/2

The above solutions use NfPart(Ans/N as a modulus operation to judge even/odd, giving us a list of dice that can have petals. Multiplying that by the original list -1 restores the petal counts to the list which can then be summed

2sum(Ans=3)+4sum(Ans=5

The original 13 byte solution simply compared the list once against 3 and separately against 5 before summing those results independently to get the petal count

All the above solutions take input as a list in Ans

Answer 19

Brain-Flak, 66 60 44 42 bytes

Two bytes saved by Jo King

({(({}[()])<>)<>{<({}[()])><>([{}])<>}{}})

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This challenge seems to ellude a straight forward solution in Brain-flak. It was quite fun to golf.

Answer 20

Julia 1.0, 23 16 bytes

l->sum(l.^3 .&6)

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-7 Bytes thanks to Maria Miller

Answer 21

05AB1E, 5 4 bytes

ÉÏ<O

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-7 byes thanks to Expired Data

-1 byte thanks to Grimmy

As promised, someone who knows 05AB1E better than I came and golfed it somehow.

11 bytes

ε2%}Iε1-}*O

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I'm sure someone who knows 05AB1E better than I do will come along and golf this somehow. Really simple:

ε2%}   # Map the code 2% (mod 2) to each element of the implicit input
Iε1-}  # Map the code 1- (sub 1) to each element of the explicit input
*O     # Multiply the two lists and sum

Answer 22

Perl 5 -pl, 15 bytes

$\+=$_%2*$_&6}{

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Takes the input list as one entry per line.

Answer 23

Roj, 134 92 88 86 65 bytes

Considerably easier than the previous challenge. Takes numbers in separate lines and end input with the input 0. Outputs via return value. (Dramatically saved 42 bytes by using dynamic input)

O=0;i=1;while i do readint i;if i==3 or i==5 do O=O+i-1 end end;O

Answer 24

APL (Dyalog Unicode), 8 bytes^SBCS

+/⊢|¯2|⊢

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A port of xnor's double modulo trick.

How it works

+/⊢|¯2|⊢
    ¯2|⊢  ⍝ Input modulo -2; 0 if even, -1 if odd
  ⊢|      ⍝ That modulo input; 0 if even, n-1 if odd
+/        ⍝ Sum

---

APL (Dyalog Unicode), 9 bytes^SBCS

2∘|+.×-∘1

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How it works

2∘|+.×-∘1
      -∘1  ⍝ Decrement input
   +.×     ⍝ Dot product with
2∘|        ⍝ Input modulo 2

Answer 25

K (oK), 13 bytes

{+/(2!x)*x-1}

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Essentially same as Jonah's J solution

Answer 26

W, 5 bytes

►S≡5u

Uncompressed:

(S2m*J

(      % Decrement input
 S     % Swap a copy of the input up
  2m   % Modulo the value by 2
    *  % Multiply them
     J % Find their sum
```

Answer 27

Icon, 50 bytes

procedure f(a)
n:=0&n+:=!a-1=(2|4)&\z
return n
end

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Explanation:

procedure f(a)         # the argument a is the list 
    n:=0 &             # sets the sum to 0 and  
    n+:=!a-1=(2|4) &   # add to n (n+:=) each element of a decreased by one (!a-1),
                       # if it now equals 2 or 4 (=(2|4)) and
    \z                 # loop (in fact checks if a variable z exists - since it 
                       # doesn't, backtracks to get the next element of a, if any)
    return n           # returns the sum
end

Answer 28

Actually, 8 bytes

A port using Arnauld's formula.

⌠3ⁿ6&⌡MΣ

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Answer 29

Excel, 34 bytes

=SUMIF(A:A,3)/3*2+SUMIF(A:A,5)/5*4

SUMIF does much of the work for us.

Feel like there should be a solution using SUMPRODUCT, but have not found it yet.

Answer 30

Burlesque, 12 bytes

psf{2.%}?d++

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ps     # Parse input to array
f{2.%} # Filter for mod2 == 1
?d     # Decrement each
++     # Sum