Le triangle de Pascal est généré en commençant par un 1 sur la première ligne. Sur les lignes suivantes, le nombre est déterminé par la somme des deux nombres situés directement au-dessus, à gauche et à droite.

Pour démontrer, voici les 5 premières lignes du triangle de Pascal:

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

Le défi

Étant donné une entrée n (à condition toutefois que cela soit plus pratique dans la langue de votre choix), générez les n premières lignes du triangle de Pascal. Vous pouvez supposer que n est un entier compris entre 1 et 25. Il doit y avoir un saut de ligne entre chaque ligne et un espace entre chaque nombre, mais en dehors de cela, vous pouvez le formater à votre guise.

C'est du code-golf , donc la solution la plus courte gagne.

Exemple 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

J , 12 caractères

":@(!{:)\@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 | + - + - + |||||| |||
|| ||| ||| || || ||! | {: ||||||| |||
|| ||| ||| || || | + - + - + |||||| |||
|| ||| ||| || | + - + ------ + ||||| |||
|| ||| ||| | + - + ---------- + |||| |||
|| ||| || + - + --------------- + ||| |||
|| ||| | + ------------------- + || |||
|| || + - + --------------------- + | |||
|| | + ------------------------- + - + ||
| + - + ------------------------------ + |
+ ---------------------------------- +

Python, 56 octets

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

Exemple d'utilisation:

echo 9 | python filename.py

Produit:

[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]

Python, 94 91 88 70 63 caractères

x=[1]
for i in input()*x:
 print x
 x=map(sum,zip([0]+x,x+[0]))

Mathematica: 36 (41?)

Mathematica a la Binomialfonction, mais cela enlève du plaisir à cela. Je propose:

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

La ligne ci-dessus rendra un tableau fragmenté tel que:

{{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}}

Comme il s’agit d’un format de base dans Mathematica, j’ai pensé que ce serait acceptable, mais j’ai lu les règles à nouveau, je pense que ce n’est peut-être pas le cas. L’ajout Grid@produira une sortie parfaitement acceptable, pour un total de 41 caractères:

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

C, 522

Une réponse auto-démonstrative en C Ne pourrait pas être plus clair! Points bonus pour trouver le personnage supplémentaire.

#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][0]= 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; }

Golfscript (21 caractères)

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

Depuis qu'une explication a été demandée:

# 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
;

Haskell, 94 92

f=[1]:[zipWith(+)(0:x)x++[1]|x<-f]
main=readLn>>=mapM_(putStrLn.unwords.map show).(`take`f)

Sortie:

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

Une version à 71 caractères qui n'imprime pas d'espace entre chaque numéro:

f=[1]:[zipWith(+)(0:x)x++[1]|x<-f]
main=readLn>>=mapM_ print.(`take`f)

Sortie:

[1]
[1,1]
[1,2,1]
[1,3,3,1]

Scala, 81 78 72 70 caractères

81 caractères: première tentative, copiée sans vergogne depuis la version Python :)

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

Exécutez-le en tant que script ou directement dans le REPL.

Couper à 70 caractères avec quelque chose de étonnamment lisible et idiomatique:

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

Ou 72 70 caractères avec une méthode totalement différente:

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

Ruby: 51 49 46 caractères

(Code de 45 caractères + option de ligne de commande de 1 caractère)

p=[];$_.to_i.times{n=0;p p.map!{|i|n+n=i}<<1}

Grâce à:

• jsvnm for suggesting an alternative for the value switching (2 characters)
• Go pour repérer une variable non utilisée après une amélioration précédente (4 caractères)

Échantillon échantillon:

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

Essayez-le en ligne!

JavaScript (90 85 83 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)

Demo: http://jsfiddle.net/tcRCS/3/

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

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)

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

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..<>

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.

Keg, 40 bytes

1®n1¿1.
,(|(©n|:©n$@MCƒℤ. ,⑨)©n⑨®n_01.
,

Explained

Perl, 77 Chars

$o[0]=1;for(1..<>){$"=" ";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[25][25],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

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

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)

[1]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]
[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}}

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.

Maple, 46

seq(print(seq(binomial(i,k),k=0..i)),i=0..n-1)

Usage:

> f:=n->seq(print(seq(binomial(i,k),k=0..i)),i=0..n-1);
> f(3)
    1
   1 1
  1 2 1

VBA, 162 142 102 80 bytes

_{^{Saved 22 bytes thanks to Taylor Scott.}}

_{This is an old question now but I saw a shorter solution for VBA.}

[B2].Resize([A1],[A1])="=IF(COLUMN()>ROW(),"""",IF(ROW()=2,1,IFERROR(A1+B1,1)))"

This is meant to be run in the immediate window. Input is in cell A1 of the active worksheet. Output is in the active worksheet starting at B2 and using however many cells are required based on the input. The COLUMN()>ROW() check keeps the top right of the triangle blank. The ROW()=2 check makes the first value 1 to initiate the triangle. I could have shifted the output down and dropped this check, but it introduces a lot of extraneous output before the actual triangle and I didn't feel that it was in the spirit of the challenge.

I originally posted a much more complicated method that calculated every value based on its row and column. All this method does, though, is to use in-cell formulas. I start at B2 so I can reference the row above it without #REF! errors. Then, it copies and pastes the same formula over a block of cells n wide and n tall. The input and output for n=25 looks like this:

05AB1E, 14 bytes

FN©>F®Ne})ˆ}¯»

Try it online!

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1