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Reduced row echelon form: Difference between revisions
→{{header|OCaml}}
(→TI-89 BASIC: new example) |
|||
Line 451:
{{1, 0, 0, -8}, {0, 1, 0, 1}, {0, 0, 1, -2}}
</lang>
=={{header|OCaml}}==
<lang ocaml>let swap_rows m i j =
let tmp = m.(i) in
m.(i) <- m.(j);
m.(j) <- tmp;
;;
let rref m =
try
let lead = ref 0
and rows = Array.length m
and cols = Array.length m.(0) in
for r = 0 to pred rows do
if cols <= !lead then
raise Exit;
let i = ref r in
while m.(!i).(!lead) = 0 do
incr i;
if rows = !i then begin
i := r;
incr lead;
if cols = !lead then
raise Exit;
end
done;
swap_rows m !i r;
let lv = m.(r).(!lead) in
m.(r) <- Array.map (fun v -> v / lv) m.(r);
for i = 0 to pred rows do
if i <> r then
let lv = m.(i).(!lead) in
m.(i) <- Array.mapi (fun i iv -> iv - lv * m.(r).(i)) m.(i);
done;
incr lead;
done
with Exit -> ()
;;
let () =
let m =
[| [| 1; 2; -1; -4 |];
[| 2; 3; -1; -11 |];
[| -2; 0; -3; 22 |]; |]
in
rref m;
Array.iter (fun row ->
Array.iter (fun v ->
Printf.printf " %d" v
) row;
print_newline()
) m</lang>
=={{header|Octave}}==
Line 457 ⟶ 510:
refA = rref(A);
disp(refA);</lang>
=={{header|Perl}}==
| |||