# My triangle needs more nodes

Consider the standard equilateral triangle, with nodes labeled using barycentric coordinates:

We can turn this 3 node triangle into a 6 node triangle by adding a new line of 3 vertices (one more than was present on a side of the original 3 node triangle), remove any internal edges (but not internal nodes) and re-normalize the coordinates:

Repeating the process to go from a 6 node triangle to a 10 node triangle, add a line of 4 vertices (again, one more than was present on a side of the original 6 node triangle), remove any internal edges (but not internal nodes) and re-normalize the coordinates:

This process can be repeated indefinitely. The goal of this challenge is given an integer N representing how many times this process has been performed, output all the nodes for the associated triangle in barycentric coordinates.

# Input

Your program/function should take as input a single non-negative integer N representing how many times this process has been applied. Note that for N=0, you should output the original triangle with 3 nodes.

The input may come from any source (function parameter, stdio, etc.).

# Output

Your program/function should output all the nodes in normalized barycentric coordinates. The order of the nodes does not matter. A number can be specified as a fraction (fraction reduction not required) or a floating point number. You may also output "scaled" vectors to specify a node. For example, all 3 of the following outputs are equivalent and allowed:

0.5,0.5,0

1/2,2/4,0

[1,1,0]/2


If using floating point output, your output should be accurate to within 1%. The output may be to any sink desired (stdio, return value, return parameter, etc.). Note that even though the barycentric coordinates are uniquely determined by only 2 numbers per node, you should output all 3 numbers per node.

# Examples

Example cases are formatted as:

N
x0,y0,z0
x1,y1,z1
x2,y2,z2
...


where the first line is the input N, and all following lines form a node x,y,z which should be in the output exactly once. All numbers are given as approximate floating point numbers.

0
1,0,0
0,1,0
0,0,1

1
1,0,0
0,1,0
0,0,1
0.5,0,0.5
0.5,0.5,0
0,0.5,0.5

2
1,0,0
0,1,0
0,0,1
0.667,0,0.333
0.667,0.333,0
0.333,0,0.667
0.333,0.333,0.333
0.333,0.667,0
0,0.333,0.667
0,0.667,0.333

3
1,0,0
0.75,0,0.25
0.75,0.25,0
0.5,0,0.5
0.5,0.25,0.25
0.5,0.5,0
0.25,0,0.75
0.25,0.25,0.5
0.25,0.5,0.25
0.25,0.75,0
0,0,1
0,0.25,0.75
0,0.5,0.5
0,0.75,0.25
0,1,0


# Scoring

This is code golf; shortest code in bytes wins. Standard loopholes apply. You may use any built-ins desired.

• You say "If using floating point output". What alternatives are there? Fractions? If so, do they have to be reduced? How about scaled vectors like [1,2,3]/6? Commented Aug 23, 2016 at 9:54
• Yes, all of those alternatives are allowed. I'll update the problem statement. Commented Aug 23, 2016 at 9:55

## CJam (22 bytes)

{):X),3m*{:+X=},Xdff/}


This is an anonymous block (function) which takes N on the stack and leaves an array of arrays of doubles on the stack. Online demo

### Dissection

{         e# Define a block
):X     e# Let X=N+1 be the number of segments per edge
),3m*   e# Generate all triplets of integers in [0, X] (inclusive)
{:+X=}, e# Filter to those triplets which sum to X
Xdff/   e# Normalise
}


f n|m<-n+1=[map(/m)[x,y,m-x-y]|x<-[0..m],y<-[0..m-x]]


# Python 3, 87 bytes

This is actually supposed to be a comment to the solution by TheBikingViking but I don't have enough reputation for comments.

One can save a few bytes by only iterating over the variables i,j and using the fact that with the third one they add up to n+1.

def f(n):d=n+1;r=range(n+2);print([[i/d,j/d,(d-i-j)/d]for i in r for j in r if d>=i+j])


# 05AB1E, 10 bytes

ÌL<¤/3ãDOÏ


Explanation

ÌL<          # range(1,n+2)-1
¤/        # divide all by last element (n+1)
3ã      # cartesian product repeat (generate all possible triples)
DO    # make a copy and sum the triples
Ï   # keep triples with sum 1


Try it online

• Since ¤ consumes the array, why does / divide the array by that? Does it "remember" that last popped value and use it if needed? Commented Aug 23, 2016 at 13:40
• @LuisMendo: ¤ is one of the few commands that does not pop and consume from the stack. It pushes the last element of the list while leaving the list on the stack. Commented Aug 23, 2016 at 13:47
• Commented Aug 23, 2016 at 14:04
• @LuisMendo 05AB1E only outputs the top of the stack at the end of a program. Commented Aug 23, 2016 at 14:20
• Oh, of course! Thanks for the explanations Commented Aug 23, 2016 at 15:08

## Mathematica,  44  43 bytes

Select[Range[0,x=#+1]~Tuples~3/x,Tr@#==1&]&


This is an unnamed function taking a single integer argument. Output is a list of lists of exact (reduced) fractions.

Generates all 3-tuples of multiples of 1/(N+1) between 0 and 1, inclusive, and then selects those whose sum is 1 (as required by barycentric coordinates).

# MATL, 17 bytes

2+:qGQ/3Z^t!s1=Y)


Try it online!

### Explanation

The approach is the same as in other answers:

1. Generate the array [0, 1/(n+1), 2/(n+1), ..., 1], where n is the input;
2. Generate all 3-tuples with those values;
3. Keep only those whose sum is 1.

More specifically:

2+     % Take input and add 2: produces n+2
:q     % Range [0 1 ... n+1]
GQ/    % Divide by n+1 element-wise: gives [0, 1/(n+1), 2/(n+1)..., 1]
3Z^    % Cartesian power with exponent 3. Gives (n+1)^3 × 3 array. Each row is a 3-tuple
t      % Duplicate
!s     % Sum of each row
1=     % Logical index of entries that equal 1
Y)     % Use that index to select rows of the 2D array of 3-tuples


## Jellyfish, 37 33 bytes

Thanks to Zgarb for saving 4 bytes.

p
*%
# S

=E   S
/
1+r#>>i
3


Try it online!

Like my Mathematica and Peter's CJam answers, this generates a set of candidate tuples and then selects only those that sum to 1. I'm not entirely happy with the layout yet, and I wonder whether I can save some bytes with hooks or forks, but I'll have to look into that later.

# Perl 6: 50 40 bytes

{grep *.sum==1,[X] (0,1/($_+1)...1)xx 3}  Returns a sequence of 3-element lists of (exact) rational numbers. Explanation: • $_
Implicitly declared parameter of the lambda.
• 0, 1/(\$_ + 1) ... 1
Uses the sequence operator ... to construct the arithmetic sequence that corresponds to the possible coordinate values.
• [X] EXPR xx 3
Takes the Cartesian product of three copies of EXPR, i.e. generates all possible 3-tuples.
• grep *.sum == 1, EXPR
Filter tuples with a sum of 1.

# Ruby, 62

I'd be surprised if this can't be improved on:

->x{0.step(1,i=1.0/(x+1)){|a|0.step(1-a,i){|b|p [a,b,1-a-b]}}}


Taking the advice latent in the puzzle, this calculates the second node options based on the first, and the third node by subtracting the first two.

# Brachylog, 24 bytes

+:1f
yg:2j:eaL+?/g:Lz:*a


Try it online!

# Python 3, 106 bytes

def f(n):r=range(n+2);print([x for x in[[i/-~n,j/-~n,k/-~n]for i in r for j in r for k in r]if sum(x)==1])


A function they takes input via argument and prints a list of lists of floats to STDOUT.

Python is not good at Cartesian products...

How it works

def f(n):                         Function with input iteration number n
r=range(n+2)                      Define r as the range [0, n+1]
for i in r for j in r for k in r  Length 3 Cartesian product of r
[i/-~n,j/-~n,k/-~n]               Divide each element of each list in the product
by n+1
[x for x in ... if sum(x)==1]     Filter by summation to 1
print(...)                           Print to STDOUT


Try it on Ideone

# Actually, 15 bytes

This uses an algorithm similar to the one in TheBikingViking's Python answer. Golfing suggestions welcome. Try it online!

u;ur♀/3@∙Σ1=░


Ungolfed:

u;                Increment implicit input and duplicate.
ur              Range [0..n+1]
♀/            Divide everything in range by (input+1).
3@∙         Take the Cartesian product of 3 copies of [range divided by input+1]
Σ1=    Create function that takes a list checks if sum(list) == 1.
░   Push values of the Cartesian product where f returns a truthy value.


# Ruby, 77 74 bytes

Another answer using the algorithm in TheBikingViking's Python answer. Golfing suggestions welcome.

->n{a=[*0.step(1,i=1.0/(n+1))];a.product(a,a).reject{|j|j.reduce(&:+)!=1}}


Another 74-byte algorithm based on Not that Charles's Ruby answer.

->x{y=x+1;z=1.0/y;[*0..y].map{|a|[*0..y-a].map{|b|p [a*z,b*z,(y-a-b)*z]}}}


## JavaScript (Firefox 30-57), 88 81 bytes

n=>[for(x of a=[...Array(++n+1).keys()])for(y of a)if(x+y<=n)[x/n,y/n,(n-x-y)/n]]


Returns an array of arrays of floating-point numbers. Edit: Saved 7 bytes by computing the third coordinate directly. I tried eliminating the if by calculating the range of y directly but it cost an extra byte:

n=>[for(x of a=[...Array(++n+1).keys()])for(y of a.slice(x))[x/n,(y-x)/n,(n-y)/n]]

• At the end, you wrote [x/n,y/n/z/n], did you forget a comma? Commented Aug 24, 2016 at 3:44
• @kamoroso94 You're right, I typoed the last comma, thanks for letting me know.
– Neil
Commented Aug 24, 2016 at 7:28