Clark's Triangle

Question

Ok I've been on a bit of a triangle kick recently so here's another one.

Clark's Triangle is a triangle where the leftmost entry of each row is 1 and the rightmost entries are made up of multiples of 6 which increase as the row number increases. Here's a visualization

     1 6
    1 . 12
   1 . . 18
  1 . . . 24
 1 . . . . 30
1 . . . . . 36

Just like Pascal's Triangle all other entries are the sum of the numbers to their upper right and upper left.

Here are the first few rows filled in

          1   6
        1   7  12
      1   8  19  18
    1   9  27  37  24
  1  10  36  64  61  30
1  11  46  100 125 91  36

Task

Given a row number (starting from the top) and an column number (starting from the first non-zero item on that row) output the value at that particular cell. Both inputs may be either 1 or 0 indexed (you may mix and match if you desire). Out of the bounds of the triangle is undefined and you may do whatever you wish when queried for these values.

This is code-golf, the goal is to minimize the number of bytes in your solution.

OEIS A046902

Can we build a solution with zero in the first row? like in the OEIS sequence — Jörg Hülsermann, Commented Jul 6, 2017 at 14:42
@JörgHülsermann Since that is out of bounds for the triangle defined here you may do whatever you want. — Wheat Wizard, Commented Jul 6, 2017 at 14:43

Community · Accepted Answer · 2020-06-17 09:04:33Z

7

MATL, 15 bytes

[lBB]i:"TTY+]i)

First input is 0-based row; second is 1-based column.

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Explanation

[lBB]   % Push [1 6 6]
i       % Input: row number (0-based)
:"      % Repeat that many times
  TT    %   Push [1 1]
  Y+    %   Convolution, increasing size. This computes the sum of overlapping
        %   pairs, including the endpoints. So for example [1 6 6] becomes
        %   [1 7 12 6], which will later become [1 8 19 18 6], ...
]       % End
i       % Input: column number (1-based)
)       % Use as index. Implicit display

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 6, 2017 at 14:25

Luis Mendo

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Uriel · Accepted Answer · 2017-07-10 16:28:25Z

7

Pascal, 132 bytes

function f(n,k:integer):integer;begin if k=1 then f:=1 else if k>n then f:=6*n else if k<0 then f:=0 else f:=f(n-1,k-1)+f(n-1,k)end;

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1-indexed.

edited Jul 10, 2017 at 16:28

answered Jul 6, 2017 at 14:19

Uriel

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\$\begingroup\$ Pascal's triangle! \$\endgroup\$
– Henry
Commented Jul 6, 2017 at 20:57

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Business Cat · Accepted Answer · 2017-07-06 15:16:02Z

5

CJam, 22 18 bytes

-4 bytes thanks to Martin Ender

X6_]ri{0X$+.+}*ri=

Input is (0-based row) (0-based column)

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Explanation

X6_]  e# Push the list [1 6 6]. This is the first row, but each row will have an extra 6 at
      e# the end, which is out of bounds.
ri    e# Push the first input as an integer.
{     e# The following block calculates the next row given a row on top of the stack:
 0X$+ e#  Copy the top list on the stack and prepend 0.
 .+   e#  Element-wise addition with the list before prepending 0. This adds each element of
      e#  with the one to its left, except the initial 1 gets added to 0 and the final number
      e#  gets added to the out-of-bounds 6. The out-of-bounds 6 is unchanged since one list
      e#  is longer.
}*    e# Run this block (row index) times.
ri=   e# Get the (column index)th item of the final list.

edited Jul 6, 2017 at 15:16

answered Jul 6, 2017 at 15:00

Business Cat

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\$\begingroup\$ A different technique for getting pairwise sums is to shift one copy left and use .+. Normally that has the problem that it retains the trailing element without summing it (which costs bytes to remove), but in this case that actually saves bytes because then you don't need to add a 6 on every iteration. You can save even more bytes because shifting left is free if you only prepend the 0 to one copy: X6_]ri{0X$+.+}*ri= \$\endgroup\$
– Martin Ender
Commented Jul 6, 2017 at 15:08
\$\begingroup\$ _0\+ instead of 0X$+ is the same byte count if you prefer. \$\endgroup\$
– Martin Ender
Commented Jul 6, 2017 at 15:08
\$\begingroup\$ @MartinEnder Oh I see, you get an extra 6 at the end of each row that's out of bounds so it doesn't matter. Clever, thanks. \$\endgroup\$
– Business Cat
Commented Jul 6, 2017 at 15:11

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LiefdeWen · Accepted Answer · 2017-07-06 14:45:49Z

4

C#, 157 bytes

using System.Linq;(b,c)=>{var f=new[]{1,6};for(;c>0;c--){int s=f.Length;f=new int[s+1].Select((e,i)=>i<1?1:i==s?f[s-1]+6:f[i-1]+f[i]).ToArray();}return f[b];

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edited Jul 6, 2017 at 14:45

answered Jul 6, 2017 at 14:37

LiefdeWen

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Rod · Accepted Answer · 2017-07-06 14:28:17Z

3

Python 2, 67 bytes

a,b=input()
x=[1,6]
exec"x=map(sum,zip([0]+x,x+[6]));"*a
print x[b]

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Brute-force approach, calculate the ath row, and then print the bth number, both inputs are 0-based

edited Jul 6, 2017 at 14:28

answered Jul 6, 2017 at 14:15

Rod

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Chase · Accepted Answer · 2017-07-06 15:36:12Z

3

Python 3, 64 60 52 bytes

f=lambda r,c:c<2or c>r and r*6or f(r-1,c-1)+f(r-1,c)

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Recursive solution using 1-indexing. Outputs "True" instead of 1 for the sake of golfing.

Thanks to:

@totallyhuman for saving 4 bytes!
@Rod for saving 8 bytes!

edited Jul 6, 2017 at 15:36

answered Jul 6, 2017 at 14:38

Chase

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2

\$\begingroup\$ 60 bytes \$\endgroup\$
– totallyhuman
Commented Jul 6, 2017 at 14:44
2

\$\begingroup\$ 52 bytes replacing the if/else with boolean operators and with a more flexible output \$\endgroup\$
– Rod
Commented Jul 6, 2017 at 14:54
\$\begingroup\$ @Rod, this is a brilliant solution. I'm still trying to wrap my head around why it works. I am still fairly new here (this is only my second answer on the site), so I'm unsure on the protocol: should I include your revision in my answer even though you switched from Python 3 to 2? \$\endgroup\$
– Chase
Commented Jul 6, 2017 at 15:03
3

\$\begingroup\$ @icosahedron the python version is irrelevent in that case, so you don't have to mind it. generally, switching between python versions to exploit features is considered OK. \$\endgroup\$
– Uriel
Commented Jul 6, 2017 at 15:20
\$\begingroup\$ @Uriel thank you for the clarification. \$\endgroup\$
– Chase
Commented Jul 6, 2017 at 15:32

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Julian Wolf · Accepted Answer · 2017-07-06 15:52:03Z

3

Haskell, 41 bytes

n#1=1
n#m|m>n=6*n
n#m=(n-1)#(m-1)+(n-1)#m

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Call using n # m where n is the row number and m is the column number, both 1-indexed.

answered Jul 6, 2017 at 15:52

Julian Wolf

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Razetime · Accepted Answer · 2020-10-22 12:43:45Z

3

Husk, ¹⁵ 12 bytes

!!¡S`JẊ+d166

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-3 bytes from Dominic Van Essen.

Same method as the MATL answer, except it uses an infinite list of rows of Clark's triangle.

edited Oct 22, 2020 at 12:43

answered Oct 21, 2020 at 12:50

Razetime

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\$\begingroup\$ 12 bytes using J and recycling the d166... \$\endgroup\$
– Dominic van Essen
Commented Oct 22, 2020 at 12:41

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Mathematica, 32 bytes

b=Binomial;b[#,#2-1]6+b[#-1,#2]&

input

[row,column]
[1-indexed,0-indexed]

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 6, 2017 at 14:15

ZaMoC

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Tomer Shetah · Accepted Answer · 2020-10-22 14:10:10Z

2

Scala, 75 bytes - Brute-force solution

n=>k=>2.to(n)./:(Seq(1,6))((s,i)=>1+:s.sliding(2).map(_.sum).toSeq:+6*i)(k)

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Scala, 88 bytes - Recursive solution

n=>k=>{def g(m:Int,l:Int):Int=if(m==l)m*6 else if(l<1)1 else g(m-1,l-1)+g(m-1,l)
g(n,k)}

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edited Oct 22, 2020 at 14:10

answered Oct 22, 2020 at 13:02

Tomer Shetah

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Wheat Wizard · Accepted Answer · 2020-10-22 14:57:17Z

2

Haskell, 40 36 bytes

Thanks to Lynn for pointing out the obvious and saving me 4 bytes.

n#1=1
n#m|m>n=6*n|q<-n-1=q#(m-1)+q#m

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Works like this answer but ~~1 byte~~ 5 bytes shorter.

edited Oct 22, 2020 at 14:57

answered Oct 21, 2020 at 21:39

Wheat Wizard♦

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\$\begingroup\$ Hint: You can combine the last two lines into one with guards, and save some bytes. \$\endgroup\$
– lynn
Commented Oct 21, 2020 at 22:06
\$\begingroup\$ @Lynn Gah! That feels so obvious now. \$\endgroup\$
– Wheat Wizard ♦
Commented Oct 22, 2020 at 14:58

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coltim · Accepted Answer · 2020-10-22 15:11:37Z

2

K (oK), 28 18 bytes

{(x{+':x,6}/1 6)y}

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Rows and columns are both 0-indexed. Out of bounds return null.

edited Oct 22, 2020 at 15:11

answered Oct 22, 2020 at 0:14

coltim

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user · Accepted Answer · 2020-10-25 00:28:09Z

2

Desmos, 118 71 55 bytes

Saved a whopping 47 bytes thanks to @ovs!

\$f(r,c)=\sum_{k=c-2}^c(6-5*0^{c-k})/k!\prod_{n=r-k+1}^rn\$

f(r,c)=\sum_{k=c-2}^c(6-5*0^{c-k})/k!\prod_{n=r-k+1}^rn

Both the row and column are 0-indexed.

edited Oct 25, 2020 at 0:28

answered Oct 24, 2020 at 21:28

user

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1

\$\begingroup\$ This can be expressed quite a bit shorter as a sum. I think this works: f(r,c)=\sum_{k=0}^2\frac{(6-5*0^k)\prod_{n=0}^{c-1-k}(r-n)}{(c-k)!} \$\endgroup\$
– ovs
Commented Oct 24, 2020 at 23:23
\$\begingroup\$ @ovs Nice! I was struggling with getting the 1, 6, 6, and didn't think about 0^k \$\endgroup\$
– user
Commented Oct 24, 2020 at 23:32

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Giuseppe · Accepted Answer · 2017-07-06 15:15:02Z

1

R, 77 bytes

Reduce(function(x,y)zoo::rollsum(c(0,x,6),2),double(scan()-1),c(1,6))[scan()]

Requires the zoo library; reads from stdin (the inputs separated by two newlines) and returns the value, with NA for out of bounds selections.

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answered Jul 6, 2017 at 15:15

Giuseppe

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\$\begingroup\$ I know this is >3 years old, but you could save 7 bytes by taking the first input value as 0-indexed instead of 1-indexed, and replacing double(scan()-1) with !1:scan(). \$\endgroup\$
– Robin Ryder
Commented Oct 22, 2020 at 13:34

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Neil · Accepted Answer · 2017-07-06 16:19:17Z

1

JavaScript (ES6), 38 bytes

f=(r,c)=>c?r>c?f(--r,c)+f(r,--c):r*6:1

Crashes for negative columns, and returns multiples of six for negative rows or overlarge columns.

edited Jul 6, 2017 at 16:19

answered Jul 6, 2017 at 15:50

Neil

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Kamil Drakari · Accepted Answer · 2017-07-06 16:55:38Z

1

C# (.NET Core), 44 bytes

f=(c,r)=>c<=1?1:c>r?6*r:f(c-1,r-1)+f(c,r-1);

Takes column then row, both 1-indexed. Can take row then column by swapping the inputs: (r,c). Will return row * 6 for coordinates outside the bounds on the right (i.e. column > row + 1), and 1 for coordinates outside the bounds on the left (i.e. column < 1).

answered Jul 6, 2017 at 16:55

Kamil Drakari

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Jörg Hülsermann · Accepted Answer · 2017-07-06 17:15:51Z

1

PHP, 64 bytes

recursive function

rows 1-indexing columns 0-indexing

Output for row=0 and column=0 is 0 like in the OEIS sequence

function f($r,$c){return$c-$r?$c?f($r-=1,$c-1)+f($r,$c):1:$r*6;}

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PHP, 126 bytes

rows 1-indexing columns 0-indexing

Output for row=0 and column=0 is 0 like in the OEIS sequence

for(;$r<=$argv[1];$r++)for($z++,$c=~0;++$c<$z;)$t[+$r][$c]=$c<$r?$c?$t[$r-1][$c-1]+$t[$r-1][$c]:1:$r*6;echo$t[$r-1][$argv[2]];

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edited Jul 6, 2017 at 17:15

answered Jul 6, 2017 at 15:00

Jörg Hülsermann

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Jonathan Allan · Accepted Answer · 2017-07-06 21:01:30Z

Jelly, 13 bytes

,"’U0¦c/x6,1S

A monadic link taking a list of [row, entry] (0-indexing for entries, 1-indexing for rows), returning the value.

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How?

,"’U0¦c/x6,1S - Link: list of numbers, [row, entry]
  ’           - decrement     -> [row-1, entry-1]
 "            - zip with:
,             -   pair        -> [[row, row-1], [entry, entry-1]]
     ¦        - sparse application of:
   U          -   upend
    0         - for indexes: 0 -> [[row, row-1], [entry-1, entry]]
       /      - reduce by:
      c       -   choose       -> [(row choose entry-1), (row-1 choose entry)]
         6,1  - 6 paired with 1 = [6,1]
        x     - times        i.e. [a, a, a, a, a, a, a, b]
            S - sum            -> 6*(row choose entry-1) + (row-1 choose entry)

Stack Exchange Network

Clark's Triangle

Task

18 Answers 18

MATL, 15 bytes

Explanation

Pascal, 132 bytes

CJam, 22 18 bytes

Explanation

C#, 157 bytes

Python 2, 67 bytes

Python 3, 64 60 52 bytes

Haskell, 41 bytes

Husk, ¹⁵ 12 bytes

Mathematica, 32 bytes

Scala, 75 bytes - Brute-force solution

Scala, 88 bytes - Recursive solution

Haskell, 40 36 bytes

K (oK), 28 18 bytes

Desmos, 118 71 55 bytes

R, 77 bytes

JavaScript (ES6), 38 bytes

C# (.NET Core), 44 bytes

PHP, 64 bytes

PHP, 126 bytes

Jelly, 13 bytes

How?

Your Answer

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code-golf
math
number
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Hot Network Questions

Clark's Triangle

Task

18 Answers 18

MATL, 15 bytes

Explanation

Pascal, 132 bytes

CJam, 22 18 bytes

Explanation

C#, 157 bytes

Python 2, 67 bytes

Python 3, 64 60 52 bytes

Haskell, 41 bytes

Husk, 15 12 bytes

Mathematica, 32 bytes

Scala, 75 bytes - Brute-force solution

Scala, 88 bytes - Recursive solution

Haskell, 40 36 bytes

K (oK), 28 18 bytes

Desmos, 118 71 55 bytes

R, 77 bytes

JavaScript (ES6), 38 bytes

C# (.NET Core), 44 bytes

PHP, 64 bytes

PHP, 126 bytes

Jelly, 13 bytes

How?

Your Answer

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Husk, ¹⁵ 12 bytes

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