Catalan numbers/Pascal's triangle: Difference between revisions
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</pre> |
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=={{header|Perl 6}}== |
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<lang perl6>constant @pascal = [1], -> @p { [0, @p Z+ @p, 0] } ... *; |
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constant @catalan = gather for 2, 4 ... * -> $ix { |
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my @row := @pascal[$ix]; |
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my $mid = +@row div 2; |
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take [-] @row[$mid, $mid+1] |
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} |
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.say for @catalan[^20];</lang> |
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{{out}} |
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<pre>1 |
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2 |
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5 |
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14 |
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42 |
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132 |
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429 |
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1430 |
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4862 |
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16796 |
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58786 |
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208012 |
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742900 |
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2674440 |
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9694845 |
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35357670 |
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129644790 |
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477638700 |
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1767263190 |
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6564120420</pre> |
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=={{header|Run BASIC}}== |
=={{header|Run BASIC}}== |
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<lang runbasic>n = 15 |
<lang runbasic>n = 15 |
Revision as of 06:51, 8 May 2013
Catalan numbers/Pascal's triangle
You are encouraged to solve this task according to the task description, using any language you may know.
You are encouraged to solve this task according to the task description, using any language you may know.
The task is to print out the first 15 Catalan numbers by extracting them from Pascal's triangle, see Catalan Numbers and the Pascal Triangle.
C++
<lang cpp>// Generate Catalan Numbers // // Nigel Galloway: June 9th., 2012 //
- include <iostream>
int main() {
const int N = 15; int t[N+2] = {0,1}; for(int i = 1; i<=N; i++){ for(int j = i; j>1; j--) t[j] = t[j] + t[j-1]; t[i+1] = t[i]; for(int j = i+1; j>1; j--) t[j] = t[j] + t[j-1]; std::cout << t[i+1] - t[i] << " "; } return 0;
}</lang>
- Produces:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Perl 6
<lang perl6>constant @pascal = [1], -> @p { [0, @p Z+ @p, 0] } ... *;
constant @catalan = gather for 2, 4 ... * -> $ix {
my @row := @pascal[$ix]; my $mid = +@row div 2; take [-] @row[$mid, $mid+1]
}
.say for @catalan[^20];</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 35357670 129644790 477638700 1767263190 6564120420
Run BASIC
<lang runbasic>n = 15 dim t(n+2) t(1) = 1 for i = 1 to n
for j = i to 1 step -1 : t(j) = t(j) + t(j-1): next j t(i+1) = t(i) for j = i+1 to 1 step -1: t(j) = t(j) + t(j-1 : next j
print t(i+1) - t(i);" "; next i</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Tcl
<lang tcl>proc catalan n {
set result {} array set t {0 0 1 1} for {set i 1} {[set k $i] <= $n} {incr i} {
for {set j $i} {$j > 1} {} {incr t($j) $t([incr j -1])} set t([incr k]) $t($i) for {set j $k} {$j > 1} {} {incr t($j) $t([incr j -1])} lappend result [expr {$t($k) - $t($i)}]
} return $result
}
puts [catalan 15]</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845