# Pascal's Triangle

Hello,

some time ago I started to learn Turbo Pascal.
Now I've got a complex question, for me that is.

I want to write a program that prints out Pascal's Triangle.
But I'm stuck since I started to figure out how to write this program.

The program has to ask the user how many rows it has to display, after that the program should give Pascal's Triangle up to the number of rows the user has determined.

I hope someone can help me with this problem.
I'm too stuck

Thanks for helping me!

Pepijn de Brouwer

• : Hello,
:
: some time ago I started to learn Turbo Pascal.
: Now I've got a complex question, for me that is.
:
: I want to write a program that prints out Pascal's Triangle.
: But I'm stuck since I started to figure out how to write this program.
:
: The program has to ask the user how many rows it has to display, after that the program should give Pascal's Triangle up to the number of rows the user has determined.
:
: I hope someone can help me with this problem.
: I'm too stuck
:
: Thanks for helping me!
:
:
: Pepijn de Brouwer
:

Hi Pepijn!

In the first line will 1, in the second will 1 and 1. In all lines the first and last element will be 1.

In pursuance (in counting ...) there are 3 versions for getting non-winger elements:

a) count them as sum of two elements "above" them:
1
1 1
1 2 1
1 3 3 1
...
But for this You need store at one time at least half of elements of the last line. For this You need static or dinamic array, and for large line numbers large memory.

b) for these elements one-by-one use the binomial coefficients' formula by expression "k under n", that is (you have see two-lines-height parentheses ...),
/n
k/
where "k under n" = n!/(k!*(n-k)!); here x! (i.e. factorial of x) = 1*2*3*4*...*(x-1)*x, while 0!=1.
In our case k = 0...n is the index of elements in one (in the nth) line.
This method isn't require many memory, but counting of factorials on the one hand is time-waster, on the second it results very large numbers while the elements still aren't large.

c) the simplest way is based on equality
(k under n) = (k-1 under n) * (n - (k-1)) / k.
This formula follows from previous formula in section b.

The program, using this method:

[code]
Program Pascal_Triangle;

var
NMax,
N,
K,
E : longint;

begin
Write ('How many lines: ');

for N := 0 to NMax-1 do begin
E := 1; (* for K=0; this is "0 under N" *)
Write (E);
for K := 1 to N do begin
(* (K u N) = (K-1 u N) * (N - (K-1)) / K : *)
E := E * (N - K + 1) div K;
Write (' ', E);
end;
WriteLn;
end;
end.
[/code]

Best regards:

gyapesz

• Thanks for the reply and helping me out.

But this brings me to another question.
Can someone help me figure out what the variables stand for?
What do Nmax, N, K and E mean?
I guess 'Nmax' gives the number of rows, but I'm not sure.

Thanks for helping me out in advance!

Pepijn de Brouwer
• : Thanks for the reply and helping me out.
:
: But this brings me to another question.
: Can someone help me figure out what the variables stand for?
: What do Nmax, N, K and E mean?
: I guess 'Nmax' gives the number of rows, but I'm not sure.
:
: Thanks for helping me out in advance!
:
:
: Pepijn de Brouwer
:

Hi Pepijn!

NMax
is the requested number of rows.

N = 0..NMax-1
is the index (running counter) for the rows.

K = 0..N (N is other for each row)
is the index (running counter) for elements in the Nth row; but because of algorithm's method, for K=0 we calculate the element separately and the rows' _cycles_ will run for K=1..N only.

E
is the running element; it's value is recalculated (and printed) for each element of each row, one by one.

in example below NMax = 5:

[code]
1 N=0 K=0 (=N) E= 1
1 1 N=1 K=0..1 (=N) E= 1, then 1
1 2 1 N=2 K=0..2 (=N) E= 1, then 2, then 1
1 3 3 1 N=3 K=0..3 (=N) E= 1, 3, 3, 1
1 4 6 4 1 N=4 (=NMax-1) K=0..4 (=N) E= 1, 4, 6, 4, 1
[/code]

Best regards

gyapesz