Display a linear combination
- Task
Display a finite linear combination in an infinite vector basis .
Write a function that, when given a finite list of scalars ,
creates a string representing the linear combination in an explicit format often used in mathematics, that is:
where
The output must comply to the following rules:
- don't show null terms, unless the whole combination is null.
- e(1) is fine, e(1) + 0*e(3) or e(1) + 0 is wrong.
- don't show scalars when they are equal to one or minus one.
- e(3) is fine, 1*e(3) is wrong.
- don't prefix by a minus sign if it follows a preceding term. Instead you use subtraction.
- e(4) - e(5) is fine, e(4) + -e(5) is wrong.
Show here output for the following lists of scalars:
1) 1, 2, 3 2) 0, 1, 2, 3 3) 1, 0, 3, 4 4) 1, 2, 0 5) 0, 0, 0 6) 0 7) 1, 1, 1 8) -1, -1, -1 9) -1, -2, 0, -3 10) -1
C
Accepts vector coefficients from the command line, prints usage syntax if invoked with no arguments. This implementation can handle floating point values but displays integer values as integers. All test case results shown with invocation. <lang C> /*Abhishek Ghosh, 24th September 2017*/
- include<stdlib.h>
- include<stdio.h>
- include<math.h> /*Optional, but better if included as fabs, labs and abs functions are being used. */
int main(int argC, char* argV[]) {
int i,zeroCount= 0,firstNonZero = -1; double* vector;
if(argC == 1){ printf("Usage : %s <Vector component coefficients seperated by single space>",argV[0]); }
else{
printf("Vector for ["); for(i=1;i<argC;i++){ printf("%s,",argV[i]); } printf("\b] -> ");
vector = (double*)malloc((argC-1)*sizeof(double));
for(i=1;i<=argC;i++){ vector[i-1] = atof(argV[i]); if(vector[i-1]==0.0) zeroCount++; if(vector[i-1]!=0.0 && firstNonZero==-1) firstNonZero = i-1; }
if(zeroCount == argC){ printf("0"); }
else{ for(i=0;i<argC;i++){ if(i==firstNonZero && vector[i]==1) printf("e%d ",i+1); else if(i==firstNonZero && vector[i]==-1) printf("- e%d ",i+1); else if(i==firstNonZero && vector[i]<0 && fabs(vector[i])-abs(vector[i])>0.0) printf("- %lf e%d ",fabs(vector[i]),i+1); else if(i==firstNonZero && vector[i]<0 && fabs(vector[i])-abs(vector[i])==0.0) printf("- %ld e%d ",labs(vector[i]),i+1); else if(i==firstNonZero && vector[i]>0 && fabs(vector[i])-abs(vector[i])>0.0) printf("%lf e%d ",vector[i],i+1); else if(i==firstNonZero && vector[i]>0 && fabs(vector[i])-abs(vector[i])==0.0) printf("%ld e%d ",vector[i],i+1); else if(fabs(vector[i])==1.0 && i!=0) printf("%c e%d ",(vector[i]==-1)?'-':'+',i+1); else if(i!=0 && vector[i]!=0 && fabs(vector[i])-abs(vector[i])>0.0) printf("%c %lf e%d ",(vector[i]<0)?'-':'+',fabs(vector[i]),i+1); else if(i!=0 && vector[i]!=0 && fabs(vector[i])-abs(vector[i])==0.0) printf("%c %ld e%d ",(vector[i]<0)?'-':'+',labs(vector[i]),i+1); } } }
free(vector);
return 0; } </lang> Output:
C:\rossetaCode>vectorDisplay.exe 1 2 3 Vector for [1,2,3] -> e1 + 2 e2 + 3 e3 C:\rossetaCode>vectorDisplay.exe 0 0 0 Vector for [0,0,0] -> 0 C:\rossetaCode>vectorDisplay.exe 0 1 2 3 Vector for [0,1,2,3] -> e2 + 2 e3 + 3 e4 C:\rossetaCode>vectorDisplay.exe 1 0 3 4 Vector for [1,0,3,4] -> e1 + 3 e3 + 4 e4 C:\rossetaCode>vectorDisplay.exe 1 2 0 Vector for [1,2,0] -> e1 + 2 e2 C:\rossetaCode>vectorDisplay.exe 0 0 0 Vector for [0,0,0] -> 0 C:\rossetaCode>vectorDisplay.exe 0 Vector for [0] -> 0 C:\rossetaCode>vectorDisplay.exe 1 1 1 Vector for [1,1,1] -> e1 + e2 + e3 C:\rossetaCode>vectorDisplay.exe -1 -1 -1 Vector for [-1,-1,-1] -> - e1 - e2 - e3 C:\rossetaCode>vectorDisplay.exe -1 -2 0 -3 Vector for [-1,-2,0,-3] -> - e1 - 2 e2 - 3 e4 C:\rossetaCode>vectorDisplay.exe -1 Vector for [-1] -> - e1
EchoLisp
<lang scheme>
- build an html string from list of coeffs
(define (linear->html coeffs)
(define plus #f)
(or*
(for/fold (html "") ((a coeffs) (i (in-naturals 1)))
(unless (zero? a)
(set! plus (if plus "+" "")))
(string-append html
(cond ((= a 1) (format "%a e%d " plus i)) ((= a -1) (format "- e%d " i)) ((> a 0) (format "%a %d*e%d " plus a i)) ((< a 0) (format "- %d*e%d " (abs a) i)) (else ""))))
"0"))
(define linears '((1 2 3)
(0 1 2 3) (1 0 3 4) (1 2 0) (0 0 0) (0) (1 1 1) (-1 -1 -1) (-1 -2 0 -3) (-1)))
(define (task linears)
(html-print ;; send string to stdout
(for/string ((linear linears))
(format "%a -> %a
" linear (linear->html linear)))))
</lang>
- Output:
(1 2 3) -> e1 + 2*e2 + 3*e3
(0 1 2 3) -> e2 + 2*e3 + 3*e4
(1 0 3 4) -> e1 + 3*e3 + 4*e4
(1 2 0) -> e1 + 2*e2
(0 0 0) -> 0
(0) -> 0
(1 1 1) -> e1 + e2 + e3
(-1 -1 -1) -> - e1 - e2 - e3
(-1 -2 0 -3) -> - e1 - 2*e2 - 3*e4
(-1) -> - e1
Elixir
<lang elixir>defmodule Linear_combination do
def display(coeff) do
Enum.with_index(coeff)
|> Enum.map_join(fn {n,i} ->
{m,s} = if n<0, do: {-n,"-"}, else: {n,"+"}
case {m,i} do
{0,_} -> ""
{1,i} -> "#{s}e(#{i+1})"
{n,i} -> "#{s}#{n}*e(#{i+1})"
end
end)
|> String.trim_leading("+")
|> case do
"" -> IO.puts "0"
str -> IO.puts str
end
end
end
coeffs =
[ [1, 2, 3], [0, 1, 2, 3], [1, 0, 3, 4], [1, 2, 0], [0, 0, 0], [0], [1, 1, 1], [-1, -1, -1], [-1, -2, 0, -3], [-1] ]
Enum.each(coeffs, &Linear_combination.display(&1))</lang>
- Output:
e(1)+2*e(2)+3*e(3) e(2)+2*e(3)+3*e(4) e(1)+3*e(3)+4*e(4) e(1)+2*e(2) 0 0 e(1)+e(2)+e(3) -e(1)-e(2)-e(3) -e(1)-2*e(2)-3*e(4) -e(1)
J
Implementation:
<lang J>fourbanger=:3 :0
e=. ('e(',')',~])@":&.> 1+i.#y
firstpos=. 0< {.y-.0
if. */0=y do. '0' else. firstpos}.;y gluedto e end.
)
gluedto=:4 :0 each
pfx=. '+-' {~ x<0
select. |x
case. 0 do.
case. 1 do. pfx,y
case. do. pfx,(":|x),'*',y
end.
)</lang>
Example use:
<lang J> fourbanger 1 2 3 e(1)+2*e(2)+3*e(3)
fourbanger 0 1 2 3
e(2)+2*e(3)+3*e(4)
fourbanger 1 0 3 4
e(1)+3*e(3)+4*e(4)
fourbanger 0 0 0
0
fourbanger 0
0
fourbanger 1 1 1
e(1)+e(2)+e(3)
fourbanger _1 _1 _1
-e(1)-e(2)-e(3)
fourbanger _1 _2 0 _3
-e(1)-2*e(2)-3*e(4)
fourbanger _1
-e(1)</lang>
Julia
<lang julia># v0.6
linearcombination(coef::Array) = join(collect("$c * e($i)" for (i, c) in enumerate(coef) if c != 0), " + ")
for c in [[1, 2, 3], [0, 1, 2, 3], [1, 0, 3, 4], [1, 2, 0], [0, 0, 0], [0], [1, 1, 1],
[-1, -1, -1], [-1, -2, 0, -3], [-1]]
@printf("%20s -> %s\n", c, linearcombination(c))
end</lang>
- Output:
[1, 2, 3] -> 1 * e(1) + 2 * e(2) + 3 * e(3)
[0, 1, 2, 3] -> 1 * e(2) + 2 * e(3) + 3 * e(4)
[1, 0, 3, 4] -> 1 * e(1) + 3 * e(3) + 4 * e(4)
[1, 2, 0] -> 1 * e(1) + 2 * e(2)
[0, 0, 0] ->
[0] ->
[1, 1, 1] -> 1 * e(1) + 1 * e(2) + 1 * e(3)
[-1, -1, -1] -> -1 * e(1) + -1 * e(2) + -1 * e(3)
[-1, -2, 0, -3] -> -1 * e(1) + -2 * e(2) + -3 * e(4)
[-1] -> -1 * e(1)
Kotlin
<lang scala>// version 1.1.2
fun linearCombo(c: IntArray): String {
val sb = StringBuilder()
for ((i, n) in c.withIndex()) {
if (n == 0) continue
val op = when {
n < 0 && sb.isEmpty() -> "-"
n < 0 -> " - "
n > 0 && sb.isEmpty() -> ""
else -> " + "
}
val av = Math.abs(n)
val coeff = if (av == 1) "" else "$av*"
sb.append("$op${coeff}e(${i + 1})")
}
return if(sb.isEmpty()) "0" else sb.toString()
}
fun main(args: Array<String>) {
val combos = arrayOf(
intArrayOf(1, 2, 3),
intArrayOf(0, 1, 2, 3),
intArrayOf(1, 0, 3, 4),
intArrayOf(1, 2, 0),
intArrayOf(0, 0, 0),
intArrayOf(0),
intArrayOf(1, 1, 1),
intArrayOf(-1, -1, -1),
intArrayOf(-1, -2, 0, -3),
intArrayOf(-1)
)
for (c in combos) {
println("${c.contentToString().padEnd(15)} -> ${linearCombo(c)}")
}
}</lang>
- Output:
[1, 2, 3] -> e(1) + 2*e(2) + 3*e(3) [0, 1, 2, 3] -> e(2) + 2*e(3) + 3*e(4) [1, 0, 3, 4] -> e(1) + 3*e(3) + 4*e(4) [1, 2, 0] -> e(1) + 2*e(2) [0, 0, 0] -> 0 [0] -> 0 [1, 1, 1] -> e(1) + e(2) + e(3) [-1, -1, -1] -> -e(1) - e(2) - e(3) [-1, -2, 0, -3] -> -e(1) - 2*e(2) - 3*e(4) [-1] -> -e(1)
Perl 6
<lang perl6>sub linear-combination(@coeff) {
(@coeff Z=> map { "e($_)" }, 1 .. *)
.grep(+*.key)
.map({ .key ~ '*' ~ .value })
.join(' + ')
.subst('+ -', '- ', :g)
.subst(/<|w>1\*/, , :g)
|| '0'
}
say linear-combination($_) for [1, 2, 3], [0, 1, 2, 3], [1, 0, 3, 4], [1, 2, 0], [0, 0, 0], [0], [1, 1, 1], [-1, -1, -1], [-1, -2, 0, -3], [-1 ]
- </lang>
- Output:
e(1) + 2*e(2) + 3*e(3) e(2) + 2*e(3) + 3*e(4) e(1) + 3*e(3) + 4*e(4) e(1) + 2*e(2) 0 0 e(1) + e(2) + e(3) -e(1) - e(2) - e(3) -e(1) - 2*e(2) - 3*e(4) -e(1)
Python
<lang python> def linear(x):
return ' + '.join(['{}e({})'.format('-' if v == -1 else if v == 1 else str(v) + '*', i + 1)
for i, v in enumerate(x) if v] or ['0']).replace(' + -', ' - ')
list(map(lambda x: print(linear(x)), [[1, 2, 3], [0, 1, 2, 3], [1, 0, 3, 4], [1, 2, 0],
[0, 0, 0], [0], [1, 1, 1], [-1, -1, -1], [-1, -2, 0, 3], [-1]]))
</lang>
- Output:
e(1) + 2*e(2) + 3*e(3) e(2) + 2*e(3) + 3*e(4) e(1) + 3*e(3) + 4*e(4) e(1) + 2*e(2) 0 0 e(1) + e(2) + e(3) -e(1) - e(2) - e(3) -e(1) - 2*e(2) + 3*e(4) -e(1)
Racket
<lang racket>#lang racket/base (require racket/match racket/string)
(define (linear-combination->string es)
(let inr ((es es) (i 1) (rv ""))
(match* (es rv)
[((list) "") "0"]
[((list) rv) rv]
[((list (? zero?) t ...) rv)
(inr t (add1 i) rv)]
[((list n t ...) rv)
(define ±n
(match* (n rv)
;; zero is handled above
[(1 "") ""]
[(1 _) "+"]
[(-1 _) "-"]
[((? positive? n) (not "")) (format "+~a*" n)]
[(n _) (format "~a*" n)]))
(inr t (add1 i) (string-append rv ±n "e("(number->string i)")"))])))
(for-each
(compose displayln linear-combination->string) '((1 2 3) (0 1 2 3) (1 0 3 4) (1 2 0) (0 0 0) (0) (1 1 1) (-1 -1 -1) (-1 -2 0 -3) (-1)))
</lang>
- Output:
e(1)+2*e(2)+3*e(3) e(2)+2*e(3)+3*e(4) e(1)+3*e(3)+4*e(4) e(1)+2*e(2) 0 0 e(1)+e(2)+e(3) -e(1)-e(2)-e(3) -e(1)-2*e(2)-3*e(4) -e(1)
REXX
<lang rexx>/*REXX program displays a finite liner combination in an infinite vector basis. */ @.=.; @.1 = '1, 2, 3'
@.2 = '0, 1, 2, 3'
@.3 = '1, 0, 3, 4'
@.4 = '1, 2, 0'
@.5 = '0, 0, 0'
@.6 = 0
@.7 = '1, 1, 1'
@.8 = '-1, -1, -1'
@.9 = '-1, -2, 0, -3'
@.10 = -1
do j=1 while @.j\==.; n=0 /*process each vector; zero element cnt*/
y=space( translate(@.j, ,',') ) /*elide commas and superfluous blanks. */
$= /*nullify output (liner combination).*/
do k=1 for words(y); #=word(y, k) /* ◄───── process each of the elements.*/
if #=0 then iterate; a=abs(# / 1) /*if the value is zero, then ignore it.*/
s='+ '; if #<0 then s='- ' /*define the sign: plus(+) or minus(-)*/
n=n+1; if n==1 then s=strip(s) /*if the 1st element used, remove plus.*/
if a\==1 then s=s || a'*' /*if multiplier is unity, then ignore #*/
$=$ s'e('k")" /*construct a liner combination element*/
end /*k*/
$=strip( strip($), 'L', "+") /*strip leading plus sign (1st element)*/
if $== then $=0 /*handle special case of no elements. */
say right( space(@.j), 20) ' ──► ' strip($) /*align the output for presentation. */
end /*j*/
/*stick a fork in it, we're all done. */</lang>
output when using the default inputs:
1, 2, 3 ──► e(1) + 2*e(2) + 3*e(3)
0, 1, 2, 3 ──► e(2) + 2*e(3) + 3*e(4)
1, 0, 3, 4 ──► e(1) + 3*e(3) + 4*e(4)
1, 2, 0 ──► e(1) + 2*e(2)
0, 0, 0 ──► 0
0 ──► 0
1, 1, 1 ──► e(1) + e(2) + e(3)
-1, -1, -1 ──► -e(1) - e(2) - e(3)
-1, -2, 0, -3 ──► -e(1) - 2*e(2) - 3*e(4)
-1 ──► -e(1)
Ring
<lang ring>
- Project : Display a linear combination
- Date : 2017/09/22
- Author : Gal Zsolt (~ CalmoSoft ~)
- Email : <calmosoft@gmail.com>
scalars = [[1, 2, 3], [0, 1, 2, 3], [1, 0, 3, 4], [1, 2, 0], [0, 0, 0], [0], [1, 1, 1], [-1, -1, -1], [-1, -2, 0, -3], [-1]] for n=1 to len(scalars)
str = ""
for m=1 to len(scalars[n])
scalar = scalars[n] [m]
if scalar != "0"
if scalar = 1
str = str + "+e" + m
elseif scalar = -1
str = str + "" + "-e" + m
else
if scalar > 0
str = str + char(43) + scalar + "*e" + m
else
str = str + "" + scalar + "*e" + m
ok
ok
ok
next
if str = ""
str = "0"
ok
if left(str, 1) = "+"
str = right(str, len(str)-1)
ok
see str + nl
next </lang> Output:
e1+2*e2+3*e3 e2+2*e3+3*e4 e1+3*e3+4*e4 e1+2*e2 0 0 e1+e2+e3 -e1-e2-e3 -e1-2*e2-3*e4 -e1
Sidef
<lang ruby>func linear_combination(coeffs) {
var res = ""
for e,f in (coeffs.kv) {
given(f) {
when (1) {
res += "+e(#{e+1})"
}
when (-1) {
res += "-e(#{e+1})"
}
case (.> 0) {
res += "+#{f}*e(#{e+1})"
}
case (.< 0) {
res += "#{f}*e(#{e+1})"
}
}
}
res -= /^\+/
res || 0
}
var tests = [
%n{1 2 3},
%n{0 1 2 3},
%n{1 0 3 4},
%n{1 2 0},
%n{0 0 0},
%n{0},
%n{1 1 1},
%n{-1 -1 -1},
%n{-1 -2 0 -3},
%n{-1},
]
tests.each { |t|
printf("%10s -> %-10s\n", t.join(' '), linear_combination(t))
}</lang>
- Output:
1 2 3 -> e(1)+2*e(2)+3*e(3)
0 1 2 3 -> e(2)+2*e(3)+3*e(4)
1 0 3 4 -> e(1)+3*e(3)+4*e(4)
1 2 0 -> e(1)+2*e(2)
0 0 0 -> 0
0 -> 0
1 1 1 -> e(1)+e(2)+e(3)
-1 -1 -1 -> -e(1)-e(2)-e(3)
-1 -2 0 -3 -> -e(1)-2*e(2)-3*e(4)
-1 -> -e(1)
Tcl
This solution strives for legibility rather than golf.
<lang Tcl>proc lincom {factors} {
set exp 0
set res ""
foreach f $factors {
incr exp
if {$f == 0} {
continue
} elseif {$f == 1} {
append res "+e($exp)"
} elseif {$f == -1} {
append res "-e($exp)"
} elseif {$f > 0} {
append res "+$f*e($exp)"
} else {
append res "$f*e($exp)"
}
}
if {$res eq ""} {set res 0}
regsub {^\+} $res {} res
return $res
}
foreach test {
{1 2 3}
{0 1 2 3}
{1 0 3 4}
{1 2 0}
{0 0 0}
{0}
{1 1 1}
{-1 -1 -1}
{-1 -2 0 -3}
{-1}
} {
puts [format "%10s -> %-10s" $test [lincom $test]]
}</lang>
- Output:
1 2 3 -> e(1)+2*e(2)+3*e(3)
0 1 2 3 -> e(2)+2*e(3)+3*e(4)
1 0 3 4 -> e(1)+3*e(3)+4*e(4)
1 2 0 -> e(1)+2*e(2)
0 0 0 -> 0
0 -> 0
1 1 1 -> e(1)+e(2)+e(3)
-1 -1 -1 -> -e(1)-e(2)-e(3)
-1 -2 0 -3 -> -e(1)-2*e(2)-3*e(4)
-1 -> -e(1)
zkl
<lang zkl>fcn linearCombination(coeffs){
[1..].zipWith(fcn(n,c){ if(c==0) "" else "%s*e(%s)".fmt(c,n) },coeffs)
.filter().concat("+").replace("+-","-").replace("1*","")
or 0
}</lang> <lang zkl>T(T(1,2,3),T(0,1,2,3),T(1,0,3,4),T(1,2,0),T(0,0,0),T(0),T(1,1,1),T(-1,-1,-1),
T(-1,-2,0,-3),T(-1),T)
.pump(Console.println,linearCombination);</lang>
- Output:
e(1)+2*e(2)+3*e(3) e(2)+2*e(3)+3*e(4) e(1)+3*e(3)+4*e(4) e(1)+2*e(2) 0 0 e(1)+e(2)+e(3) -e(1)-e(2)-e(3) -e(1)-2*e(2)-3*e(4) -e(1) 0