# Generate Pascal's triangle

Pascal's triangle is generated by starting with a 1 on the first row. On subsequent rows, the number is determined by the sum of the two numbers directly above it to the left and right.

To demonstrate, here are the first 5 rows of Pascal's triangle:

    1
1 1
1 2 1
1 3 3 1
1 4 6 4 1


## The Challenge

Given an input n (provided however is most convenient in your chosen language), generate the first n rows of Pascal's triangle. You may assume that n is an integer inclusively between 1 and 25. There must be a line break between each row and a space between each number, but aside from that, you may format it however you like.

This is code-golf, so the shortest solution wins.

## Example I/O

> 1
1
> 9
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1

• NB In a sense this is a simplified version of Distributing the balls – Peter Taylor Oct 21 '11 at 11:03
• @Peter Olson: What's your opinion of ratchet freak's interpretation of "you may format it however you like"? If I followed his interpretation I could shave 18 characters. – Steven Rumbalski Oct 21 '11 at 20:57
• @StevenRumbalski He's fine. There's a newline between each row, and there is a space between each number, so it meets the criteria. – Peter Olson Oct 21 '11 at 21:32
• @Peter Olson: Thanks for the clarification. What about Tomas T's assumption that n is defined already? – Steven Rumbalski Oct 21 '11 at 21:50
• @Gaffi Probably not, accepting an answer makes me feel like I'm ending the contest and discouraging new and possibly better answers. – Peter Olson Jun 8 '12 at 4:05

## J, 12 characters

":@(!{:)\@i.


   i.5
0 1 2 3 4
{:i.5
4
(i.5)!{:i.5
1 4 6 4 1
(!{:)i.5
1 4 6 4 1
(!{:)\i.5
1 0 0 0 0
1 1 0 0 0
1 2 1 0 0
1 3 3 1 0
1 4 6 4 1
":@(!{:)\i.5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
(":@(!{:)\@i.)''
+----------------------------------+
|+-+------------------------------+|
||@|+-------------------------+--+||
|| ||+-+---------------------+|i.|||
|| |||\|+-------------------+||  |||
|| ||| ||+-+---------------+|||  |||
|| ||| |||@|+--+----------+||||  |||
|| ||| ||| ||":|+-+------+|||||  |||
|| ||| ||| ||  ||2|+-+--+||||||  |||
|| ||| ||| ||  || ||!|{:|||||||  |||
|| ||| ||| ||  || |+-+--+||||||  |||
|| ||| ||| ||  |+-+------+|||||  |||
|| ||| ||| |+--+----------+||||  |||
|| ||| ||+-+---------------+|||  |||
|| ||| |+-------------------+||  |||
|| ||+-+---------------------+|  |||
|| |+-------------------------+--+||
|+-+------------------------------+|
+----------------------------------+

• J beats GolfScript? Interesting. I would like to see an explanation for this code, if you have time. – Mr.Wizard Oct 26 '11 at 9:31
• It's already split down, but here's a line by line if you'd like additional english. Line 1 i.5 returns the first five naturals. Line 2 adds {: "Tail" (return last). Line 3 combines them with ! "Out Of" (number of combinations). Line 4 (!{:)i.5 is the same. factoring the hook out. So (!:) is an operation that transforms the first n naturals to the nth line of Pascal's triangle. Line 5 applies it to all Prefixes (backslash) of 0..4, but J fills in the unused spots with 0, so the operation is combined (@) with the string formatting operation ":. Very cool J, upvoted. – J B Nov 2 '11 at 14:51
• @JB Isn't ! means factorial here? Also we can get rid of @ at the right. – defhlt Aug 15 '12 at 14:35
• @ArtemIce Monadic ! means factorial; dyadic ! counts combinations. The final @ in ":@(!{:)\@i. is just there to make this a stand-alone verb. – ephemient Aug 15 '12 at 14:38

# Python, 56 Bytes

a=;exec"print a;a=map(sum,zip(+a,a+));"*input()


Sample usage:

echo 9 | python filename.py


Produces:


[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
[1, 8, 28, 56, 70, 56, 28, 8, 1]

• +1 Clever way to use exec avoid a for loop. – Steven Rumbalski Nov 23 '11 at 18:00

# C, 522

A self demonstrating C answer. Couldn't be clearer! Bonus points for finding the extra character.

#define returns return 0
#define fr for
#define twentyonechexpressis0 0
i
, x
[ 52 ]
[ 52] ,j, y
; main (c){fr (;i< c
; i++){ x[i][i]=x[ i]= 1
; }for(i =2;i<c;i++){for (j=1;j<i;j++){x [i][j] =
1 +x[i][j ]+x[i-1][j-1]+x[i-1] [j]+1-1+1-1+1-1+1-1+1-1+111-11- twentyonechexpressis0 -100-1; }
} ;for(i=0 ;i<c;i++){for(j=0;j<=i;j++){ printf("%3d%c",x[i][j],(1+1+1+1)*(1+1+1+1+1+1+1+1)) ;}putchar(1+1+(1<<1+1)+1+1+1+1+1+111111-111111-1);} /*thiscomment_takes28chars*/ returns; }

• I can't help but feel that this misses the point of code golf. (I also can't help pointing out that the extra character is in the \binom{5}{4} position). – Peter Taylor Jun 6 '12 at 16:11
• It was fun to write. That's generally what I come to codegolf for. – walpen Jun 6 '12 at 21:03
• Clever :) Have an upvote. Maybe not a winner candidate but a creative one! – Accatyyc Aug 10 '12 at 13:02

## Python, 94918870 63 characters

x=
for i in input()*x:
print x
x=map(sum,zip(+x,x+))


## Mathematica: 36 (41?)

Mathematica has the Binomial function, but that takes the fun out of this. I propose:

NestList[{0,##}+{##,0}&@@#&,{1},n-1]


The line above will render a ragged array such as:

{{1}, {1, 1}, {1, 2, 1}, {1, 3, 3, 1}, {1, 4, 6, 4, 1},
{1, 5, 10, 10, 5, 1}, {1, 6, 15, 20, 15, 6, 1}}


Since this is a basic format in Mathematica I thought it would be acceptable, but as I read the rules again, I think it may not be. Adding Grid@ will produce unequivocally acceptable output, for a total of 41 characters:

Grid@NestList[{0,##}+{##,0}&@@#&,{1},n-1]


n = 6:

1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10  10  5   1
1   6   15  20  15  6   1


### Golfscript (21 chars)

~]({0\{.@+\}/;1].p}*;


Since an explanation was requested:

# Stack contains 'n'
~](
# Stack: [] n
{
# prev_row is [\binom{i,0} ... \binom{i,i}]
# We loop to generate almost all of the next row as
#     [(\binom{i,-1} + \binom{i,0}) ... (\binom{i,i-1} + \binom{i,i})]
# \binom{i,-1} is, of course, 0
# Stack: prev_row
0\
# Stack: 0 prev_row
{
# Stack: ... \binom{i,j-1} \binom{i,j}
.@+\
# Stack: ... (\binom{i,j-1} + \binom{i,j}) \binom{i,j}
}/
# Stack: \binom{i+1,0} ... \binom{i+1,i} \binom{i,i}
# unless it's the first time round, when we still have 0
# so we need to pop and then push a 1 for \binom{i+1,i+1}
;1]
# next_row
.p
}*
# final_row
;

• You might want to try golf.shinh.org/p.rb?pascal+triangle – Nabb Oct 22 '11 at 3:31
• Could you please provide some pseudo-code or explanation? I kind of understand what's going on, but I'm not entirely understanding the swapping part. – Rob Aug 10 '12 at 2:50
• Thank you for the detailed explanation and excellent answer (+1), but I'm even more confused now. The logic (process) isn't sitting right. – Rob Aug 10 '12 at 19:03
• @MikeDtrick, there was a slight error in the explanation. There's also a subtle point which needed explaining, but which I'd missed because it's so long since I wrote the code. – Peter Taylor Aug 10 '12 at 19:23
• Okay, it's starting to make sense. My final question be does the printing and executing process work from the top down or the bottom up (1, 1 1, 1 2 1: top down, 1 2 1, 1 1, 1: bottom up)? – Rob Aug 11 '12 at 2:40

## Scala, 817872 70 characters

81 chars: first attempt, shamelessly copied from the Python version :)

var x=Seq(1)
for(i<-1 to args(0).toInt){println(x)
x=(0+:x,x:+0).zipped.map(_+_)}


Run it as a script, or directly in the REPL.

Cut to 70 chars with something surprisingly readable and idiomatic:

Seq.iterate(Seq(1),readInt)(a=>(0+:a,a:+0).zipped.map(_+_))map println


Or 72 70 characters with a totally different method:

0 to(readInt-1)map(i=>println(0 to i map(1 to i combinations(_)size)))

• + 1 for shameless copying! – Steven Rumbalski Oct 25 '11 at 21:23
• The last version should be used carefully for huge values of readInt, like 50. ;) – user unknown May 31 '12 at 22:16
• @userunknown presumably that's why the question specifies an upper limit of 25... – Luigi Plinge May 31 '12 at 22:25
• It wasn't meant as critique, just as a warning for the curious. – user unknown May 31 '12 at 22:41

# Keg, 40 33 bytes

1:&¿1.
,(|(⑻|:⑻$@MCƒℤ. ,⑨)⑹_01. ,  Try it online! ## Old Program Explained  # Haskell, 94 92 f=:[zipWith(+)(0:x)x++|x<-f] main=readLn>>=mapM_(putStrLn.unwords.map show).(takef)  Output: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1  A 71 character version which does not print a space between each number: f=:[zipWith(+)(0:x)x++|x<-f] main=readLn>>=mapM_ print.(takef)  Output:  [1,1] [1,2,1] [1,3,3,1]  • You can save a character by using mapM instead of mapM_. – dfeuer Feb 25 '19 at 4:23 # Ruby: 5149 46 characters (45 characters code + 1 character command line option) p=[];$_.to_i.times{n=0;p p.map!{|i|n+n=i}<<1}


Thanks to:

• jsvnm for suggesting an alternative for the value switching (2 characters)
• G B for spotting out a variable unused after previous improvement (4 characters)

Sample run:

bash-4.4$ruby -ne 'p=[];$_.to_i.times{n=0;p p.map!{|i|n+n=i}<<1}' <<< 1


bash-4.4$ruby -ne 'p=[];$_.to_i.times{n=0;p p.map!{|i|n+n=i}<<1}' <<< 9

[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
[1, 8, 28, 56, 70, 56, 28, 8, 1]


Try it online!

• you can save 2 chars with p.map!{|i|(v=n)+n=i} – jsvnm Aug 10 '12 at 11:21
• Great one, @jsvnm! Man, how long I combined to shorten that part. Thanks. – manatwork Aug 10 '12 at 11:44
• Maybe a little late, but: why use the variable v? – G B Dec 15 '17 at 8:08
• Good catch, @GB! That left behind from 1st revision, where… where… doh. Where was also kind of useless. I guess it comes from an earlier attempt when used .map. Thank you. – manatwork Dec 15 '17 at 9:07

# convey, 39 bytes

convey is a new 2d esolang I made. It is based on conveyor belts that move values around in a factory. v<^> are explicit belt directions, other connections are inferred, as each function has a fixed number of inputs and outputs. For example, + takes 2 values in and returns 1 value.

v<<<.~/.]
v>,v}"!{
"/&v}11
\v>+/^
$,^ _  Try it online! A row enters the main part from the top left. Two tiles in, the row gets " copied down and right. The right part prepends a _, which is a so called 'default value'. The actual value of _ depends on the function it enters. It works like the neutral element, thus _ + 3 = 3, 4 * _ = 4. To understand the next part we have to take a look at port order. For example, , will join two paths, favoring the main input port. The order is as follows (lower takes precedence): For input: For output: 0 3 v ^ 1>f<3 2<f>0 ^ v 2 1  So the upper , will favor inputs from the left, and the lower , will favor inputs from the top. In the animation the arrow heads are drawn a bit thicker to indicate main ports. On / (rising flank) whenever a new list enters the tile, _ gets pushed to the side port. _ and the row then get joined, whereas the _ comes from the main port and thus has right of way. The bottom part appends a _ by similar means: \ (falling flank) pushes the length of the list to the side port, that will be set $ to _.

Finally, _ 1 2 1 and 1 2 1 _ get added in + and we get the next row 1 3 3 1! We print a newline /} (_ in } prints newlines) and then copy the row into the output "}. To prevent a row colliding with the next row when they get longer, we batch ~. the lists: we first let 1 element through, then 2, then 3 … by getting the indices /. of an list of n 1s ({!1}), that get discarded in a sink ].

• unrelated question: how did you get the λ.land domain? – Razetime Dec 7 '20 at 4:44
• @Razetime most generic TLDs support non-latin letters/IDNs which are rarely taken because realistically you have to use the puny code variant (xn--wxa.land) most of the time. My more fun domain ꙮ.world f.e. lies in the extended Cyrillic UTF-8 block. – xash Dec 7 '20 at 8:49

## JavaScript (908583 81)

for(n=prompt(o=i='');i++<n;o+='\n')for(s=j=1;j<=i;s=s*(i-j)/j++)o+=s+' ';alert(o)


NOTE: Doesn't work well in practice for about n > 30 because numbers overflow built-in integer data type and become floating-point numbers.

Edit 1: removed 5 characters by converting while to for and combining statements

Edit 2: move s= statement inside for and save 2 chars

Edit 3: combine s=1,j=1 initializer into s=j=1 and save 2 chars

• Nice! You can save one more character by changing "s=s*..." to "s*=..." – Derek Kurth Oct 25 '11 at 19:41
• @DerekKurth: I had thought that when I was first doing optimizations, but that would mess up the logic because it needs to be s*(i-j)/j, not s*((i-j)/j). – mellamokb Oct 25 '11 at 19:49
• Hmm, I tried it as s*=... in the jsfiddle and it seemed to work. Maybe I did something wrong, though. – Derek Kurth Oct 25 '11 at 20:16
• @DerekKurth: Technically it is the same, but the idea is that if you multiply by (i-j) before dividing by j, then there is no need for floating point arithmetic because the results should always be an integer. If you do ((i-j)/j) first, this will result in decimal values which can be a source of error, and at the very least will require extra code for rounding/truncating. You don't begin to see this until you get to about n>11, and you'll see decimal values in the output, i.e., 1 11 55 165 330 461.99999999999994 461.99999999999994... – mellamokb Oct 25 '11 at 21:47
• Ah, that makes sense! – Derek Kurth Oct 25 '11 at 22:10

# R, 39 chars

R seems to be the very right tool for this task :-)

x=1;for(i in 1:n)x=c(print(x),0)+c(0,x)

• You're missing one of the requirements: "Given an input n (provided however is most convenient in your chosen language)" – Steven Rumbalski Oct 21 '11 at 20:59
• @Steven, "Given an input n"... so may I assume the n is given? I corrected the code. Is this now OK? – Tomas Oct 21 '11 at 21:38
• I'm asked Peter Olson to clarify. – Steven Rumbalski Oct 21 '11 at 21:52
• @StevenRumbalski I don't think that's valid unless it takes input. I don't know R, so maybe the compiler makes it so that undefined variables prompt an input, so it might be ok, but if it's like most other languages in that regard, I don't think it is. – Peter Olson Oct 22 '11 at 0:04
• Basically, the n must be supplied from an external source at run time and the apparatus for capturing it is included in your program. Typically, that means by command line argument, or stdin, or file. By file is almost never used because it's invariably longer than the other two options. – Steven Rumbalski Oct 24 '11 at 13:01

## in Q (25 characters/20 with shorter version)

t:{(x-1) (p:{0+':x,0})\1}


Shorter

t:{(x-1){0+':x,0}\1}


Sample usage:

q)t 4
1
1 1
1 2 1
1 3 3 1

• Or alternatively, 20 characters t:{(x-1){0+':x,0}\1} – skeevey Mar 8 '12 at 20:35
• Nice, shorter than the GolfScript solution now. – sinedcm Mar 9 '12 at 12:10

# awk - 73 chars

fairly straightforward implementation:

{for(i=0;i<$1;++i)for(j=i;j>=0;)printf"%d%c",Y[j]+=i?Y[j-1]:1,j--?32:10}  sample run: % awk -f pascal.awk <<<10 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1  ## Perl, 52, 49 characters Edit: using say instead of print map{@_=(1,map$_[$_-1]+$_[$_],1..@_);say"@_"}1..<>  ## APL, 19 15 characters A bit late to the party, perhaps? {⍪{⍵!⍨⍳⍵+1}¨⍳⍵}  It doesn't beat the J entry, though. This assumes that the index origin (⎕IO) is set to 0. Unfortunately, with an index origin of 1, we need 25 18 characters: {⍪{⍵!⍨0,⍳⍵}¨1-⍨⍳⍵}  There are two ⍨s in the code to express my frustration. Demo:  {⍪{⍵!⍨⍳⍵+1}¨⍳⍵}5 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1  Explanations Short version: • ⍳⍵ (with an index origin of 0) produces an array of the numbers from 0 to ⍵-1 inclusive, where ⍵ is the right argument to the function. • ⍳⍵+1 generates all numbers from 0 to ⍵ • {⍵!⍨⍳⍵+1} generates ⍵ choose k for every element k in ⍳⍵+1. The ⍨ (commute) operator swaps the arguments to a function around, such that the right hand argument becomes the left, and vice versa. • {⍵!⍨⍳⍵+1}¨⍳⍵ passes each element in ⍳⍵ using the ¨ (each) operator. The result is a one dimensional array containing the first ⍵ rows of the Pascal's Triangle. • The one argument form of ⍪ takes a one dimensional vector, and makes it a column rather than a row. Each row of the triangle is put on its own line. Long answer: • Virtually the same as the other version, except that 1-⍨ is placed before an ⍳ to replicate an index origin of 0. • 0,⍳⍵ with an index origin of 1 replicates ⍳⍵+1 with an index origin of 0. # Perl, 47 54 characters $p=1;map{print"@{[split//,$p]}\n";$p*=11}1..<>


It takes a number from the command line, but doesn't perform any error checks.

Just realized it only works up to n=4. It was some old code I had on my hd.

This works though:

map{@a=(1,map$a[$_-1]+=$a[$_],1..@a);print"@a\n"}a..n


n has to be input into the script though, or it would be one character more.

# Husk, 13 10 bytes

-3 bytes thanks to Zgarb

mw↑¡Sż+Θ;1


Try it online!

### Explanation

Based on the same idea as my Pip answer: generate each row from the last one by tacking a 0 to the front of the row and adding it to itself element-wise. The Husk version creates the entire triangle as an infinite lazy list and then takes however many rows we need.

        ;1  Start with 
¡        Iterate this function:
S        S-combinator: Sfgx is fx(gx) (aka f(x,g(x)) in C-style syntax)
ż+       f: add a list itemwise to another list, keeping the extra value at
the end of the longer list unchanged
Θ      g: prepend a falsey value (given a list of integers, prepend 0)
The result of Sż+Θ is a 1-argument function that prepends a 0 to its
argument and adds that itemwise to another copy of the argument
↑         From the infinite list of results, take N (where N is the input)
mw          Join each sublist on spaces (which means the main list gets
autojoined on newlines)

• Nice first Husk answer! A couple of hints. Θ prepends a default falsy value, which is 0 for the number type. You can avoid creating a helper function with S. Sfg is equivalent to calling fg¹ from a separate line. – Zgarb Oct 5 '20 at 6:21

$o=1;for(1..<>){$"=" ";for(1..$_){$n[$_]=$o[$_]+$o[$_-1]}@o=@n;print"@o "}  Example input 5  Example output  1 1 1 1 2 1 1 3 3 1 1 4 6 4 1  ## C, 132 127 characters c,n,i,j;main(){for(scanf("%d",&n);i<n;i++)for(j=0;j<=i;j++)printf("%d%c",c[i][j]=j?c[i-1][j-1]+c[i-1][j]:1,i-j?32:10);}  # Pascal: 216 192 characters (Not a real competitor, just an honorific presence.) var p:array[0..1,0..25]of LongInt;i,j,n,u:Word;begin Read(n);u:=0;for i:=1to n do begin p[1,1]:=1;for j:=1to i do begin p[u,j]:=p[1-u,j-1]+p[1-u,j];Write(p[u,j],' ')end;u:=1-u;Writeln end end.  Sample run: bash-4.2$ fpc pascal.pas
/usr/bin/ld: warning: link.res contains output sections; did you forget -T?

bash-4.2$./pascal <<< 1 1 bash-4.2$ ./pascal <<< 9
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1


# MATL, 10 bytes

Language created after this challenge

1iq:"tTTY+


Try it online!

1       % Push a 1. This will be the first row
iq:     % Take input n. Generate range [1,2,...,n-1]
"       % For each (that is, repeat n-1 times)
t     %   Duplicate latest row
TT    %   Push [1 1]
Y+    %   Convolve latest row with [1 1] to produce next row
% Implicitly end for each
% Implicitly display stack contents

• non-competing but a holy disaster, none from previous submissions (even J) succeeded to reduce it up to how much Matl did !!! – Abr001am May 21 '16 at 9:09
• I'm pretty sure Jelly or 05AB1E would be shorter though :-) – Luis Mendo May 21 '16 at 10:34

# JavaScript, 70 69 bytes

Generating Pascal's Triangle in a golfy manner has always given me brain ache but every time it comes up, I give it another try. Last night, armed with a few beers, I finally cracked it and came up with a working solution I was happy with. Fitting, then, that this should be my 500th (undeleted) solution here.

0-indexed and includes a trailing newline and a trailing space on each line.

n=>(g=x=>x++>n?:(h=z=>y>x?:z+ +h(z*(x-y)/y++))(y=1)+
+g(x))


## Try it

o.innerText=(f=
n=>(g=x=>x++>n?:(h=z=>y>x?:z+ +h(z*(x-y)/y++))(y=1)+
+g(x)))(i.value=8);oninput=_=>o.innerText=f(+i.value)
<input id=i type=number><pre id=o></pre>

## D 134 128 chars

import std.stdio;void main(){int n,m;int[]l,k=[0,1];readf("%d",&n);foreach(i;0..n){writeln(l=k~0);k=[];foreach(e;l)k~=m+(m=e);}}


output for 9 is

>9
[0, 1, 0]
[0, 1, 1, 0]
[0, 1, 2, 1, 0]
[0, 1, 3, 3, 1, 0]
[0, 1, 4, 6, 4, 1, 0]
[0, 1, 5, 10, 10, 5, 1, 0]
[0, 1, 6, 15, 20, 15, 6, 1, 0]
[0, 1, 7, 21, 35, 35, 21, 7, 1, 0]
[0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0]


taking full advantage of "you may format it however you like"; there is a space between each number and a linebreak

edit repositioned the assignment to l to shave of some chars

## Scala, 131 characters

object P extends App{var x=List(1)
while(x.size<=args(0).toInt){println(x.mkString(" "))
x=(0+:x:+0).sliding(2).map(_.sum).toList}}


Takes the input from the command line.

Output for n=10:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1

• What's with all those 0s :-)? – mellamokb Oct 20 '11 at 22:23
• @mellamokb Bit of re-arranging made them go away and shortened the code. :-) – Gareth Oct 20 '11 at 22:28

## F♯ - 203 characters

My first attempt at a round of code golf, and first attempt at functional programming. There is probably some obvious way to shorten it I haven't quite figured out yet. It complies in VS2010s F♯ compiler (which has the effect of running #light by default unlike earlier versions), and also works in the F♯ interpreter. Accepts input via stdin. Wish there was a better way for the input/output though! Lots of characters!

open System
let rec C r m =if r=0||m<=0||m>=r then 1 else C(r-1)m+C(r-1)(m-1)
for j = 0 to Convert.ToInt32(Console.ReadLine ()) do (
[0..j]|>List.map(C j)|>List.iter(fun k->printf "%i " k)
printf "\n")


Why is there no accepted answer to this question?

## VBA - 249 chars

Sub t(n)
ReDim a(1 To n,1 To n*2)
a(1,n)=1:y=vbCr:z=" ":d=z & 1 & z & y:For b=2 To n:For c=1 To n*2:x=a(b-1,c)
If c>1 Then a(b,c)=a(b-1,c-1)+x
If c<n*2 Then a(b,c)=a(b-1,c+1)+x
d=IIf(a(b,c)<>0,d & z & a(b,c) & z,d):Next:d=d & y:Next:MsgBox d
End Sub


postscript - 59 chars (63 if you count -dn= to get the number of rows in)

n{dup ==[0 3 2 roll{dup 3 2 roll add exch}forall]}repeat


run with

gs -q -dn=10 -dBATCH pascal.ps


to get


[1 1]
[1 2 1]
[1 3 3 1]
[1 4 6 4 1]
[1 5 10 10 5 1]
[1 6 15 20 15 6 1]
[1 7 21 35 35 21 7 1]
[1 8 28 56 70 56 28 8 1]
[1 9 36 84 126 126 84 36 9 1]


# Mathematica 35 chars

Here is the dull and lazy way of slicing Pascal's triangle:

Table[n~Binomial~k,{n,0,5},{k,0,n}]

(* out *)
{{1}, {1, 1}, {1, 2, 1}, {1, 3, 3, 1}, {1, 4, 6, 4, 1}, {1, 5, 10, 10,5, 1}}
`