1: #!perl w
2:
3: # BinomialExpansion.pl
4:
5: # usage perl BinomialExpansion.pl n, where n is an integer > 0
6:
7: # ** the memory goes crazy after the 170th power
8: # the expansion of (x+y)^170 peaks at 9.14484184513157e+049 * x^85 * y^85
9: # so you can imagine what happens if you put int n>170
10:
11: my $n = ($_=shift) > 0 ? int $_ :
12: die "'usage perl BinomialExpansion.pl 9, where n > 0'";
13:
14: print _titled_hr(' Binomial Expansion For (x+y)^',$n,' ');
15:
16: for my $j (0 .. $n)
17: {
18: my $coefficient = nCr($n,$j);
19: my $nj=$n$j;
20: print $coefficient;
21: print $_ = ($nj!=0)?( ($nj>1)?(' * x^'.$nj):(' * x') ):'';
22: print $_ = ($j!=0)?( ($j==1)?(' * y'):(' * y^'.$j) ):'';
23: print $_ = ($j!=$n)?(" +\n"):("\n");
24: }
25:
26: print ' 'x 25,' = (x + y)^',$n, "\n"x 3;
27:
28: # returns n!/r!(nr)!
29: sub nCr
30: {
31: my $n=shift;
32: my $r=shift;
33:
34: return int nFactorial($n) / int nFactorial($r) * int nFactorial($n$r);
35: }
36:
37: # like the name says, n!
38: sub nFactorial
39: {
40: my $n=shift;
41: my $product = 1;
42:
43: while($n>0)
44: {
45: $product *= $n;
46: }
47:
48: return $product;
49: }
50:
51: # neat little titled hr, that does a < 80 chars since int rounds down
52: # i really, really, like it
53: sub _titled_hr
54: {
55: my $string = join('', @_);
56: my $oy = int (80 (length $string) )/ 2;
57: return "\n","" x $oy, $string, "" x $oy,"\n";
58: }
59:
60: __END__
61: # some random things i say
62: #1
63: ". .. . .  .. . .. . ... ..."
64:
65: #2
66: "... ..    .. ... ..    .."
67:
68: #3
69:  ... . ... ...  .... .  .... .. . . ... .. ...  ... 
70: ..
71: ..  .. ... ...  .  .. . ..  .... .   ...  ...
72: .... ....  .. .. .. .  ... ...  .. .. . .
73:
74: # a lil sample from my machine
75:
76: F:\>perl BinomialExpansion.pl 9
77: 1 * x^9 +
78: 9 * x^8 * y +
79: 36 * x^7 * y^2 +
80: 84 * x^6 * y^3 +
81: 126 * x^5 * y^4 +
82: 126 * x^4 * y^5 +
83: 84 * x^3 * y^6 +
84: 36 * x^2 * y^7 +
85: 9 * x * y^8 +
86: 1 * * y^9
87: = (x + y)^9
88: F:\>
89:
90: # Notice a pattern? Fleet attack!
91: # >
92: # >
93: # >
94: # >
95: # >
96: # >
97: # >
98: # >
99: # >
100: # >
Sierpinskij (Re: Binomial Expansion)
by larsen (Parson) on Mar 29, 2001 at 13:48 UTC

#!/usr/bin/perl
use strict;
my $n = 10;
for my $i (0 .. $n) {
for my $j (0 .. $i) {
my $coefficient = nCr($i, $j);
if ($coefficient % 2) {
print '#';
} else {
print ' ';
}
}
print "\n";
}
# returns n!/r!(nr)!
sub nCr
{
my $n=shift;
my $r=shift;
return int nFactorial($n) / int nFactorial($r) * int nFactorial($n
+$r);
}
# like the name says, n!
sub nFactorial
{
my $n=shift;
my $product = 1;
while($n>0)
{
$product *= $n;
}
return $product;
}
 [reply] [Watch: Dir/Any] [d/l] 

my @line= (1);
while( 1 ) {
print "@line\n";
push @line, 0;
for( reverse 1..$#line ) {
$line[$_] += $line[$_1];
}
}
Be sure to pipe the output to more.

tye
(but my friends call me "Tye")  [reply] [Watch: Dir/Any] [d/l] [select] 

 [reply] [Watch: Dir/Any] 

i found that code some times ago, unfortunately i don't know the author.
i post here, maybe someone don't know.
#!/usr/bin/perl l
s@{[(gE^Ge)=~/[^g^e]/g]}[g^e]x((!!+~~g^e^g^e)<<pop).!gEge,
s[^ge^ge]s,,,,s,@{[(g^';').(e^'?')]},(G^'/').(E^'')^Ge,ge,
print,s,(?<=/[^g^e])[^g^e][^g^e],$&^(G^'/').(E^'')^gE,gege
 [reply] [Watch: Dir/Any] [d/l] 
Re: Binomial Expansion
by ariels (Curate) on Mar 29, 2001 at 14:41 UTC

The nCr function above is horribly inaccurate. While n!/r!(nr)! is indeed a valid formula for binomial coefficients, it's hardly a good way to calculate them. Here's a better way:
sub nCr {
my ($n, $r) = @_;
my $res = 1;
for my $i (1..$r) {
$res *= $n;
$res /= $i;
}
$res;
}
Why $res remains an integer (for "reasonable" parameter values) remains an exercise for the reader...  [reply] [Watch: Dir/Any] [d/l] [select] 

Ok, for fun, and because I like mathematical proofs (I made it to the USA Mathematic Olympiad twice), here's the proof I use.
In pseudocode, the nchooser function looks like:
nCr (n,r) {
res = 1;
for i in 1 .. r {
res *= n;
n;
res /= i;
}
return res;
}
The question is, why does res remain an integer? It is not difficult to show that res /= 1 is an integer, but how can we prove that by the time we get to res /= 7, res is a multiple of 7?
The proof is that by the time we have gotten to res /= X, we have multiplied res's original value, 1, by X continuous integer, and in every series of X continuous integers, there will be one number divisible by X:
X = 2, series = (y, y+1); one is div. by 2
X = 3, series = (y, y+1, y+2); one is div. by 3
...
in nCr(17,7),
n is of the series (17,16,15,14,13,12,11)
i is of the series ( 1, 2, 3, 4, 5, 6, 7)
(17/1)
(17/1) (16/2)
(17/1) (16/2) (15/3)
(17/1) (16/4) (15/3) (14/2)
(17/1) (16/4) (15/(3*5)) (14/2) (13/1)
(17/1) (16/4) (15/(3*5)) (14/2) (13/1) (12/6)
(17/1) (16/4) (15/(3*5)) (14/(2*7)) (13/1) (12/6) (11/1)
We're allowed to move the denominators around like I did, because we've already showed the product will already be an integer. The product looks like:
r

  n  (i1)
  
  i
i=1
japhy 
Perl and Regex Hacker  [reply] [Watch: Dir/Any] [d/l] [select] 

I take great offense in calling my nCr function horribly inaccurate.
You do say right away that is a valid formula for binomial coefficients, but then you say it's hardly a good one?
It is a formula for calculating rcombinations, not binomial coefficients.
The formula for binomial expansion is:
n
_
(x+y)^n = \ C(n,j) * x^(nj) *y^j
/
¯
j=0
And btw, it is the definitive formula for calculating rcombinations. It's like saying 4/2 is not the same as 2. While 2 is a better way to write 4/2, not every one recognizes 2 right away, and I feel, for the benefit of the reader of me 'craft', n!/(r!(nr)! is a better way to go.
As for your function, it is shorter, but a litle mysterious when it comes to the math.
9! 9*8*7* 6! 9*8*7 504
nCr(9,3) =  =  =  =  = 84
3!*(93)! 3*2*1*(6!) 3*2*1 6
which follows from the fact that nCr(n,r) = nCr(n,nr)
n! n! n! _ n!
 =  =  = 
r!(nr) (nr)!(n(nr))! (nr)!(nn+r)! (nr)!(r)!
This however only works for n>r, as long as n and r are both nonnegative integers, but shouldn't be a problem in this case.
As for things being integer, since the input is an integer, it follows that any integer multiple of that integer, will also be an integer.
update:
I give, I give, theory v. practice, there is a difference.
___crazyinsomniac_______________________________________
Disclaimer: Don't blame. It came from inside the void
perl e "$q=$_;map({chr unpack qq;H*;,$_}split(q;;,q*H*));print;$q/$q;"  [reply] [Watch: Dir/Any] 

 [reply] [Watch: Dir/Any] [d/l] 

If you want to understand the math, I fully agree
with your use of the formula. Furthermore for more complex
combinatorial problems, understand how to get those
formulas is extremely important.
That said, tye is right as well. While in theory that
calculation gets the right answer, in practice it starts
to be inaccurate for fairly small numbers. You can get
around this by using a big integer package. (Or a language
that understands integers of arbitrary size.) But that
will be somewhat slow.
If you want to understand this, go with the formula. If
you want to calculate it, go with the tricks. If you want
to calculate answers to arbitrary problems, be prepared to
manipulate big integers and hope that performance doesn't
matter.
And if you want an even better example some day, bug me
about why you should always use rowreduction to find the
inverses of matrices rather than Cramer's formula. (Hint:
Think numerically unstable and exponential running time,
vs n**3 and much better behaved.)
 [reply] [Watch: Dir/Any] 





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