Floyd's triangle
You are encouraged to solve this task according to the task description, using any language you may know.
Floyd's triangle lists the natural numbers in a right triangle aligned to the left where
- the first row is just 1
- successive rows start towards the left with the next number followed by successive naturals listing one more number than the line above.
The first few lines of a Floyd triangle looks like this:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
The task is to:
- Write a program to generate and display here the first n lines of a Floyd triangle.
(Use n=5 and n=14 rows). - Ensure that when displayed in a monospace font, the numbers line up in vertical columns as shown and that only one space separates numbers of the last row.
Ada
<lang Ada>with Ada.Text_IO, Ada.Integer_Text_IO, Ada.Command_Line;
procedure Floyd_Triangle is
Rows: constant Positive := Integer'Value(Ada.Command_Line.Argument(1)); Current: Positive := 1; Width: array(1 .. Rows) of Positive;
begin
-- compute the width for the different columns
for I in Width'Range loop
Width(I) := Integer'Image(I + (Rows * (Rows-1))/2)'Length;
end loop;
-- output the triangle
for Line in 1 .. Rows loop
for Column in 1 .. Line loop
Ada.Integer_Text_IO.Put(Current, Width => Width(Column));
Current := Current + 1;
end loop;
Ada.Text_IO.New_Line;
end loop;
end Floyd_Triangle;</lang>
- Output:
> ./floyd_triangle 5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 > ./floyd_triangle 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
AWK
<lang AWK>
- syntax: GAWK -f FLOYDS_TRIANGLE.AWK rows
BEGIN {
rows = ARGV[1]
if (rows !~ /^[0-9]+$/) {
print("rows invalid or missing from command line")
exit(1)
}
width = length(rows * (rows + 1) / 2) + 1 # width of last n
for (i=1; i<=rows; i++) {
cols++
for (j=1; j<=cols; j++) {
printf("%*d",width,++n)
}
printf("\n")
}
exit(0)
} </lang>
output from: GAWK -f FLOYDS_TRIANGLE.AWK 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
output from: GAWK -f FLOYDS_TRIANGLE.AWK 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
BBC BASIC
<lang bbcbasic> n = 14
num = 1
last = (n^2 - n + 2) DIV 2
FOR row = 1 TO n
col = last
FOR num = num TO num + row - 1
@% = LEN(STR$(col)) + 1 : REM set column width
PRINT num ;
col += 1
NEXT
PRINT
NEXT row</lang>
Output for n = 5:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Output for n = 14:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Bracmat
<lang bracmat> ( ( floyd
= lowerLeftCorner lastInColumn lastInRow row i W w
. put$(str$("Floyd " !arg ":\n"))
& !arg*(!arg+-1)*1/2+1
: ?lowerLeftCorner
: ?lastInColumn
& 1:?lastInRow:?row:?i
& whl
' ( !row:~>!arg
& @(!lastInColumn:? [?W)
& @(!i:? [?w)
& whl'(!w+1:~>!W:?w&put$" ")
& put$!i
& ( !i:<!lastInRow
& put$" "
& 1+!lastInColumn:?lastInColumn
| put$\n
& (1+!row:?row)+!lastInRow:?lastInRow
& !lowerLeftCorner:?lastInColumn
)
& 1+!i:?i
)
)
& floyd$5
& floyd$14
);</lang>
Output:
Floyd 5: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Floyd 14: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
C
<lang c>#include <stdio.h>
void t(int n) { int i, j, c, len;
i = n * (n - 1) / 2; for (len = c = 1; c < i; c *= 10, len++); c -= i; // c is the col where width changes
- define SPEED_MATTERS 0
- if SPEED_MATTERS // in case we really, really wanted to print huge triangles often
char tmp[32], s[4096], *p;
sprintf(tmp, "%*d", len, 0);
inline void inc_numstr(void) { int k = len;
redo: if (!k--) return;
if (tmp[k] == '9') { tmp[k] = '0'; goto redo; }
if (++tmp[k] == '!') tmp[k] = '1'; }
for (p = s, i = 1; i <= n; i++) { for (j = 1; j <= i; j++) { inc_numstr(); __builtin_memcpy(p, tmp + 1 - (j >= c), len - (j < c)); p += len - (j < c);
*(p++) = (i - j)? ' ' : '\n';
if (p - s + len >= 4096) { fwrite(s, 1, p - s, stdout); p = s; } } }
fwrite(s, 1, p - s, stdout);
- else // NO_IT_DOESN'T
int num; for (num = i = 1; i <= n; i++) for (j = 1; j <= i; j++) printf("%*d%c", len - (j < c), num++, i - j ? ' ':'\n');
- endif
}
int main(void) { t(5), t(14);
// maybe not // t(10000); return 0; }</lang> Output identical to D's.
D
<lang d>import std.stdio, std.conv;
void floydTriangle(in uint n) {
immutable lowerLeftCorner = n * (n - 1) / 2 + 1;
foreach (r; 0 .. n)
foreach (c; 0 .. r + 1)
writef("%*d%c",
text(lowerLeftCorner + c).length,
r * (r + 1) / 2 + c + 1,
c == r ? '\n' : ' ');
}
void main() {
floydTriangle(5); floydTriangle(14);
}</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Go
<lang go>package main
import "fmt"
func main() {
floyd(5) floyd(14)
}
func floyd(n int) {
fmt.Printf("Floyd %d:\n", n)
lowerLeftCorner := n*(n-1)/2 + 1
lastInColumn := lowerLeftCorner
lastInRow := 1
for i, row := 1, 1; row <= n; i++ {
w := len(fmt.Sprint(lastInColumn))
if i < lastInRow {
fmt.Printf("%*d ", w, i)
lastInColumn++
} else {
fmt.Printf("%*d\n", w, i)
row++
lastInRow += row
lastInColumn = lowerLeftCorner
}
}
}</lang>
- Output:
Floyd 5: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Floyd 14: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Haskell
Program <lang haskell>import Data.List import Control.Monad import Control.Arrow
alignR :: Int -> Integer -> String alignR n = (\s -> replicate (n - length s) ' ' ++ s). show
floydTriangle = liftM2 (zipWith (liftM2 (.) enumFromTo ((pred.). (+)))) (scanl (+) 1) id [1..]
formatFT n = mapM_ (putStrLn. unwords. zipWith alignR ws) t where
t = take n floydTriangle ws = map (length. show) $ last t</lang>
Output: <lang haskell>*Main> formatFT 5
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15
- Main> formatFT 14
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105</lang>
J
Note: require 'strings' does nothing in J7, but is harmless (strings is already incorporated in J7).
<lang J>require 'strings' floyd=: [: rplc&(' 0';' ')"1@":@(* ($ $ +/\@,)) >:/~@:i.</lang>
Note, the parenthesis around ($ $ +/\@,) is optional, and only included for emphasis.
Example use:
<lang J> floyd 5
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15
floyd 14 1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105</lang>
How it works:
First, we create a square lower triangular matrix with our argument as the length of one side. We have 1s along the diagonal and the lower triangle, and 0s for the upper triangle.
Second, we create a running sum of these values (treating rows as being adjacent horizontally for this purpose). Then, we multiply this result by our lower triangular matrix (forcing the upper triangle to be 0s).
Then, we format the matrix as text (which gives us the required vertical alignment), and in each row we replace each space followed by a zero with two spaces.
Efficiency note: In a measurement of time used: in floyd 100, 80% the time here goes into the string manipulations -- sequential additions and multiplications are cheap. In floyd 1000 this jumps to 98% of the time. Here's a faster version (about 3x on floyd 1000) courtesy of Aai of the J forums:
<lang J>floyd=: [: ({.~ i.&1@E.~&' 0')"1@":@(* ($ $ +/\@,)) >:/~@:i.</lang>
Java
<lang java> public class Floyd { public static void main(String[] args){ System.out.println("5 rows:"); printTriangle(5); System.out.println("14 rows:"); printTriangle(14); }
private static void printTriangle(int n){ for(int rowNum = 1, printMe = 1, numsPrinted = 0; rowNum <= n; printMe++){ int cols = (int)Math.ceil(Math.log10(n*(n-1)/2 + numsPrinted + 2)); System.out.printf("%"+cols+"d ", printMe); if(++numsPrinted == rowNum){ System.out.println(); rowNum++; numsPrinted = 0; } } } }</lang> Output:
5 rows: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 rows: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
JavaScript
Spidermonkey
(Used TCL example as a starting point.)
<lang javascript>#!/usr/bin/env js
function main() {
print('Floyd 5:');
floyd(5);
print('\nFloyd 14:');
floyd(14);
}
function padLeft(s, w) {
for (s = String(s); s.length < w; s = ' ' + s); return s;
}
function floyd(nRows) {
var lowerLeft = nRows * (nRows - 1) / 2 + 1;
var lowerRight = nRows * (nRows + 1) / 2;
var colWidths = [];
for (var col = lowerLeft; col <= lowerRight; col++) {
colWidths.push(String(col).length);
}
var num = 1;
for (var row = 0; row < nRows; row++) {
var line = [];
for (var col = 0; col <= row; col++, num++) {
line.push(padLeft(num, colWidths[col]));
}
print(line.join(' '));
}
}
main(); </lang>
Output:
Floyd 5: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Floyd 14: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Mathematica
<lang Mathematica> f=Function[n, Most/@(Range@@@Partition[FindSequenceFunction[{1,2,4,7,11}]/@Range[n+1],2,1])] TableForm[f@5,TableAlignments->Right,TableSpacing->{1,1}] TableForm[f@14,TableAlignments->Right,TableSpacing->{1,1}] </lang> Output:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Liberty BASIC
<lang lb> input "Number of rows needed:- "; rowsNeeded
dim colWidth(rowsNeeded) ' 5 rows implies 5 columns
for col=1 to rowsNeeded
colWidth(col) = len(str$(col + rowsNeeded*(rowsNeeded-1)/2))
next
currentNumber =1
for row=1 to rowsNeeded
for col=1 to row
print right$( " "+str$( currentNumber), colWidth(col)); " ";
currentNumber = currentNumber + 1
next
print
next
</lang> Output:
Number of rows needed:- 5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Number of rows needed:- 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
NetRexx
Both REXX versions lend themselves very well to conversion into NetRexx programs with few changes.
Version 1
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols binary /* REXX ***************************************************************
- 12.07.2012 Walter Pachl - translated from Python
- /
Parse Arg rowcount . if rowcount.length() == 0 then rowcount = 1 say 'Rows:' rowcount say col = 0 len = Rexx ll = -- last line of triangle Loop j = rowcount * (rowcount - 1) / 2 + 1 to rowcount * (rowcount + 1) / 2
col = col + 1 -- column number ll = ll j -- build last line len[col] = j.length() -- remember length of column End j
Loop i = 1 To rowcount - 1 -- now do and output the rest
ol = col = 0 Loop j = i * (i - 1) / 2 + 1 to i * (i + 1) / 2 -- elements of line i col = col + 1 ol=ol j.right(len[col]) -- element in proper length end Say ol -- output ith line end i
Say ll -- output last line </lang> Output:
Rows: 5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Rows: 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Version 2
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols binary /*REXX program constructs & displays Floyd's triangle for any number of rows.*/ parse arg numRows . if numRows == then numRows = 1 -- assume 1 row is not given maxVal = numRows * (numRows + 1) % 2 -- calculate the max value. say 'displaying a' numRows "row Floyd's triangle:" say digit = 1 loop row = 1 for numRows
col = 0 output = loop digit = digit for row col = col + 1 colMaxDigit = maxVal - numRows + col output = output Rexx(digit).right(colMaxDigit.length()) end digit say output end row
</lang>
Output:
displaying a 5 row Floyd's triangle: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 displaying a 14 row Floyd's triangle: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
OCaml
<lang ocaml>let ( |> ) f g x = g (f x) let rec last = function x::[] -> x | _::tl -> last tl | [] -> raise Not_found let rec list_map2 f l1 l2 =
match (l1, l2) with | ([], _) | (_, []) -> [] | (x::xs, y::ys) -> (f x y) :: list_map2 f xs ys
let floyd n =
let rec aux acc cur len i j =
if (List.length acc) = n then (List.rev acc) else
if j = len
then aux ((List.rev cur)::acc) [] (succ len) i 0
else aux acc (i::cur) len (succ i) (succ j)
in
aux [] [] 1 1 0
let print_floyd f =
let lens = List.map (string_of_int |> String.length) (last f) in
List.iter (fun row ->
print_endline (
String.concat " " (
list_map2 (Printf.sprintf "%*d") lens row))
) f
let () =
print_floyd (floyd (int_of_string Sys.argv.(1)))</lang>
OxygenBasic
<lang oxygenbasic>
function Floyd(sys n) as string
sys i,t
for i=1 to n
t+=i
next string s=str t sys le=1+len s string cr=chr(13,10) sys lc=len cr string buf=space(le*t+n*lc) sys j,o,p=1 t=0 for i=1 to n
for j=1 to i t++ s=str t o=le-len(s)-1 'right justify mid buf,p+o,str t p+=le next mid buf,p,cr p+=lc
next return left buf,p-1 end function
putfile "s.txt",Floyd(5)+floyd(14) </lang>
Pascal
<lang pascal>Program FloydDemo (input, output);
function digits(number: integer): integer;
begin digits := trunc(ln(number) / ln(10)) + 1; end;
procedure floyd1 (numberOfLines: integer); { variant with repeat .. until loop }
var
i, j, numbersInLine, startOfLastlLine: integer;
begin
startOfLastlLine := (numberOfLines - 1) * numberOfLines div 2 + 1;
i := 1;
j := 1;
numbersInLine := 1;
repeat
repeat
write(i: digits(startOfLastlLine - 1 + j), ' ');
inc(i);
inc(j);
until (j > numbersInLine);
writeln;
j := 1;
inc(numbersInLine);
until (numbersInLine > numberOfLines);
end;
procedure floyd2 (numberOfLines: integer); { Variant with for .. do loop }
var
i, j, numbersInLine, startOfLastlLine: integer;
begin
startOfLastlLine := (numberOfLines - 1) * numberOfLines div 2 + 1;
i := 1;
for numbersInLine := 1 to numberOfLines do
begin
for j := 1 to numbersInLine do
begin
write(i: digits(startOfLastlLine - 1 + j), ' ');
inc(i);
end;
writeln;
end;
end;
begin
writeln ('*** Floyd 5 ***');
floyd1(5);
writeln;
writeln ('*** Floyd 14 ***');
floyd2(14);
end.</lang> Output:
% ./Floyd *** Floyd 5 *** 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 *** Floyd 14 *** 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Perl
<lang perl>#!/usr/bin/env perl use strict; use warnings;
sub displayFloydTriangle {
my $numRows = shift;
print "\ndisplaying a $numRows row Floyd's triangle:\n\n";
my $maxVal = int($numRows * ($numRows + 1) / 2); # calculate the max value.
my $digit = 0;
foreach my $row (1 .. $numRows) {
my $col = 0;
my $output = ;
foreach (1 .. $row) {
++$digit;
++$col;
my $colMaxDigit = $maxVal - $numRows + $col;
$output .= sprintf " %*d", length($colMaxDigit), $digit;
}
print "$output\n";
}
return;
}
- ==== Main ================================================
my @counts; @counts = @ARGV; @counts = (5, 14) unless @ARGV;
foreach my $count (@counts) {
displayFloydTriangle($count);
}
0; __END__ </lang> Output:
displaying a 5 row Floyd's triangle: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 displaying a 14 row Floyd's triangle: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Perl 6
<lang perl6>sub chunk(@flat is copy, *@size) {
gather for @size -> $s { take [@flat.shift xx $s] }
}
constant @floyd = chunk 1..*, 1..*;
sub say-floyd($n) {
my @fmt = @floyd[$n-1].map: {"%{.chars}s"}
for @floyd[^$n] -> @i {
say join ' ', (@i Z @fmt).map: -> $i, $f { $i.fmt($f) }
}
}
say-floyd 5; say-floyd 14;</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
PicoLisp
Calculate widths relative to lower left corner
<lang PicoLisp>(de floyd (N)
(let LLC (/ (* N (dec N)) 2)
(for R N
(for C R
(prin
(align
(length (+ LLC C))
(+ C (/ (* R (dec R)) 2)) ) )
(if (= C R) (prinl) (space)) ) ) ) )</lang>
Pre-calculate all rows, and take format from last one
<lang PicoLisp>(de floyd (N)
(let
(Rows
(make
(for ((I . L) (range 1 (/ (* N (inc N)) 2)) L)
(link (cut I 'L)) ) )
Fmt (mapcar length (last Rows)) )
(map inc (cdr Fmt))
(for R Rows
(apply tab R Fmt) ) ) )</lang>
Output in both cases:
: (floyd 5) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 : (floyd 14) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
PL/I
<lang PL/I> (fofl, size): floyd: procedure options (main); /* Floyd's Triangle. Wiki 12 July 2012 */
declare (i, m, n) fixed (10), (j, k, w, nr) fixed binary;
put list ('How many rows do you want?');
get list (nr); /* the number of rows */
n = nr*(nr+1)/2; /* the total number of values */
j,k = 1; m = n - nr + 1;
do i = 1 to n;
put edit (i) ( x(1), f(length(trim(m))) );
if k > 1 then do; k = k - 1; m = m + 1; end;
else do; k,j = j + 1; m = n - nr + 1; put skip; end;
end;
end floyd; </lang>
How many rows do you want? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 How many rows do you want? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 Final row for n=45: 991 992 993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035
PureBasic
<lang PureBasic>Procedure.i sumTo(n)
Protected r,i For i=1 To n r+i Next ProcedureReturn r.i
EndProcedure
- [1]
- array rsA(n)... string-lengths of the numbers
- in the bottom row
- [2]
- sumTo(i-1)+1 to sumTo(i)
; 11 12 13 14 15 ; here k is the column-index for array rsA(k)
Procedure.s FloydsTriangle(n)
Protected r.s,s.s,t.s,i,j,k
; [1]
Dim rsA(n)
i=0
For j=sumTo(n-1)+1 To sumTo(n)
i+1
rsA(i)=Len(Str(j))
Next
; [2]
For i=1 To n
t.s="":k=0
For j=sumTo(i-1)+1 To sumTo(i)
k+1:t.s+RSet(Str(j),rsA(k)," ")+" "
Next
r.s+RTrim(t.s)+Chr(13)+Chr(10)
Next
r.s=Left(r.s,Len(r.s)-2)
ProcedureReturn r.s
EndProcedure
If OpenConsole()
n=5 r.s=FloydsTriangle(n) PrintN(r.s) n=14 r.s=FloydsTriangle(n) PrintN(r.s) Print(#crlf$ + #crlf$ + "Press ENTER to exit"): Input() CloseConsole()
EndIf</lang>
Sample Output
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Python
<lang python>>>> def floyd(rowcount=5): rows = 1 while len(rows) < rowcount: n = rows[-1][-1] + 1 rows.append(list(range(n, n + len(rows[-1]) + 1))) return rows
>>> floyd() [[1], [2, 3], [4, 5, 6], [7, 8, 9, 10], [11, 12, 13, 14, 15]] >>> def pfloyd(rows=[[1], [2, 3], [4, 5, 6], [7, 8, 9, 10]]): colspace = [len(str(n)) for n in rows[-1]] for row in rows: print( ' '.join('%*i' % space_n for space_n in zip(colspace, row)))
>>> pfloyd()
1
2 3
4 5 6
7 8 9 10
>>> pfloyd(floyd(5))
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 >>> pfloyd(floyd(14))
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 >>> </lang>
Alternately (using the mathematical formula for each row directly): <lang python>def floyd(rowcount=5): return [list(range(i*(i-1)/2+1, i*(i+1)/2+1))
for i in range(1, rowcount+1)]</lang>
REXX
version 1
<lang rexx> /* REXX ***************************************************************
- Parse Arg rowcount
- 12.07.2012 Walter Pachl - translated from Python
- /
Parse Arg rowcount col=0 ll= /* last line of triangle */ Do j=rowcount*(rowcount-1)/2+1 to rowcount*(rowcount+1)/2
col=col+1 /* column number */ ll=ll j /* build last line */ len.col=length(j) /* remember length of column */ End
Do i=1 To rowcount-1 /* now do and output the rest */
ol= col=0 Do j=i*(i-1)/2+1 to i*(i+1)/2 /* elements of line i */ col=col+1 ol=ol right(j,len.col) /* element in proper length */ end Say ol /* output ith line */ end
Say ll /* output last line */ </lang> Output:
n=5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 n=14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
version 2
<lang rexx>/*REXX program constructs & displays Floyd's triangle for any # of rows.*/ parse arg rows .; if rows== then rows=1 /*assume 1 row is not given*/ mV=rows * (rows+1) % 2 /*calculate the max value. */ say 'displaying a' rows "row Floyd's triangle:"; say
- =1
do r=1 for rows; i=0; _=
do #=# for r; i=i+1
_=_ right(#, length(mV-rows+i))
end /*#*/
say _
end /*r*/
/*stick a fork in it, we're done.*/</lang>
output when using the input of: 5
displaying a 5 row Floyd's triangle: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
output when using the input of: 14
displaying a 14 row Floyd's triangle: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
output (only showing the last row) when using the input of: 45
991 992 993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035
Ruby
<lang ruby>raise ArgumentError, "Usage: #{$0} ROWS" unless
ARGV.length == 1 && ARGV[0].to_i > 0
rows = ARGV[0].to_i max = (rows * (rows + 1)) / 2 widths = ((max - rows + 1)..max).map {|n| n.to_s.length + 1}
n = 0 rows.times do |r|
(r+1).times do |i|
n += 1
print "%#{widths[i]}d" % n
end
print "\n"
end </lang>
Run BASIC
<lang runbasic>input " Number of rows:"; rows for r = 1 to rows ' r = rows
for c = 1 to r ' c = columns j = j + 1 print j;" "; next c print
next r</lang>
Scala
<lang scala>def floydstriangle( n:Int ) {
val s = (1 to n)
val t = s map {i => (s take(i-1) sum) + 1}
(s zip t) foreach { n =>
var m = n._2;
for( i <- 0 until n._1 ) {
val w = (t.last + i).toString.length + 1 // Column width from last row
print(" " + m takeRight w )
m+=1
}
print("\n")
}
}
// Test floydstriangle(5) floydstriangle(14)</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Tcl
<lang tcl>proc floydTriangle n {
# Compute the column widths
for {set i [expr {$n*($n-1)/2+1}]} {$i <= $n*($n+1)/2} {incr i} {
lappend w [string length $i]
}
# Print the triangle
for {set i 0; set j 1} {$j <= $n} {incr j} {
for {set p -1; set k 0} {$k < $j} {incr k} { puts -nonewline [format "%*d " [lindex $w [incr p]] [incr i]] } puts ""
}
}
- Demonstration
puts "Floyd 5:" floydTriangle 5 puts "Floyd 14:" floydTriangle 14</lang>
- Output:
Floyd 5: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Floyd 14: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
XPL0
<lang XPL0>include c:\cxpl\codes; \include 'code' declarations
func IntLen(N); \Return number of digits in a positive integer int N; int I; for I:= 1 to 20 do
[N:= N/10; if N=0 then return I];
proc Floyd(N); \Display Floyd's triangle int N; int M, Row, Col; real F; [M:= (N-1+1)*(N-1)/2; \last Floyd number on second to last row F:= 1.0; \Floyd number counter for Row:= 1 to N do
[for Col:= 1 to Row do
[Format(IntLen(M+Col)+1, 0); RlOut(0, F); F:= F+1.0];
CrLf(0);
];
]; \Floyd
[Floyd(5); Floyd(14); ]</lang>
Output:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105