6
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Calculate the determinant of an n x n matrix.

Rules

  • read matrix from standard input
  • write to standard output
  • do not use ready solutions from library or language
  • input: columns separated by any space char
  • input: lines separated by cr and/or lf

Winner

  • The shortest code :)
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closed as unclear what you're asking by Wheat Wizard, caird coinheringaahing, Xcali, FryAmTheEggman, Rɪᴋᴇʀ Nov 5 '17 at 21:33

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 13
    \$\begingroup\$ I'm 4'6". Do I win? \$\endgroup\$ – Gareth Sep 21 '12 at 11:52
  • 4
    \$\begingroup\$ Looks like Chandra Bahadur Dangi wins this competition at 1 ft 11 in. First competition ever won by somebody who's probably never heard of the site :) \$\endgroup\$ – mellamokb Sep 21 '12 at 13:03
  • 2
    \$\begingroup\$ Is the matrix over any particular ring? \$\endgroup\$ – Peter Taylor Sep 24 '12 at 15:14
  • \$\begingroup\$ Just Q I assume. \$\endgroup\$ – scleaver Sep 24 '12 at 15:49
  • \$\begingroup\$ NB this question is superseded by codegolf.stackexchange.com/q/147668/194 \$\endgroup\$ – Peter Taylor Nov 27 '18 at 12:52
5
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GolfScript, 58 chars

n%{~]}%{[.!{(.,,0\{:^2$=-1^?*3${.^<\^)>+}%d*+}/}or])\;}:d~

A straightforward recursive algorithm using Laplace expansion.

Example input (random 5 × 5 matrix):

-562   40   43 -586  347
-229  177  305 -367   50
-434  343  241 -365  -86
-3   -384 -351   61 -214
-400   96 -339   25 -116

Output: 282416596900 (Online demo; Verify with Wolfram Alpha)

The code consists of three parts:

  • n%{~]}% parses the input,
  • {[.!{(.,,0\{:^2$=-1^?*3${.^<\^)>+}%d*+}/}or])\;}:d defines a recursive subroutine to calculate the determinant, and
  • the final ~ applies the subroutine to the input.

I'm sure there's room for improvement in this code, but at least I beat Gareth by three chars. Also, {:^2$= is a pretty nice GS smiley. :)

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  • \$\begingroup\$ Message from online demo: "Code took longer than 5 seconds to run, so it was aborted." \$\endgroup\$ – DavidC Sep 24 '12 at 20:18
  • \$\begingroup\$ @David: Try reloading or running the script locally; the server seems to be overloaded or something. GolfScript may be slow, but not that slow. \$\endgroup\$ – Ilmari Karonen Sep 24 '12 at 20:20
  • 1
    \$\begingroup\$ You were right. It ran now in 1.14 secs. \$\endgroup\$ – DavidC Sep 24 '12 at 20:30
  • \$\begingroup\$ @IlmariKaronen No need to beat my solution since it doesn't actually work. :-) \$\endgroup\$ – Gareth Sep 24 '12 at 21:11
4
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Haskell, 139 131

(a:r)%0=r;(a:r)%n=a:r%(n-1)
d[]=1;d(f:r)=foldr(\(j,x)->(x*d(map(%j)r)-))0$zip[0..]f
main=interact$show.d.map(map read.words).lines
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2
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Python 233

def d(x):
 l=len(x)
 if l<2:return x[0][0]
 return sum([(-1)**i*x[i][0]*d(m(x,i))for i in range(l)])
def m(x,i):y=x[:];del(y[i]);y=zip(*y);del(y[0]);return zip(*y)
x=[input()]
for i in (len(x[0])-1)*[1]:x+=[input()]
print d(x)

Ungolfed:

def det(x):
    l = len(x)
    if l == 1:
        return x[0][0]
    return sum([(-1)**i*x[i][0]*det(minor(x,i+1,1)) for i in range(l)])

def minor(x,i,j):
    y = x[:]
    del(y[i-1])
    y=zip(*y)
    del(y[j-1])
    return zip(*y)

def main():
    x = [input()]
    for i in range(len(x[0])-1):
        x += [input()]
    print det(x)

if __name__ == '__main__':
    main()

Usage

As requested, input is on stdin and output is to stdout.

I interpreted columns separated by any space char to mean that I can use comma delimited numbers. If this is not the case, I will rework my solution.

This could be about 30 characters shorter if I could specify my input matrix in the form [[a,b,c],[d,e,f],[g,h,i]].

./det.py
1,-4,9
-6,7,3
1,2,3

Result

-240

The determinant is found using Laplace Expansion

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  • \$\begingroup\$ I'm working on a python solution myself, but when I input that test matrix I get -240 as the determinant. Looking elsewhere online confirmed that -240 is correct. Perhaps you typed the test case wrong when you submitted? \$\endgroup\$ – scleaver Sep 24 '12 at 15:34
  • \$\begingroup\$ @scleaver, yes you are correct, I just typed that wrong, I will correct it. \$\endgroup\$ – Matt Sep 24 '12 at 15:52
  • \$\begingroup\$ One character can be golfed by removing the space: in (len...=>in(len \$\endgroup\$ – Zacharý Apr 16 '17 at 20:02
2
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Python, 369

This is ridiculously long, but I thought I'd provide a solution that implements the Leibniz formula.

def f(l):
    if len(l)==1:return [l]
    a=[]
    for s in l:
        for p in f(list(set(l)-set([s]))):a+=[[s]+p]
    return a
def s(a):
    n=1
    for i in range(len(a)):n*=(-1)**(a[i]>i);a[a.index(i)]=a[i];a[i]=i
    return n
m=[map(int,w.split(','))for w in raw_input().split(' ')]
r=range(len(m))
print sum([reduce(lambda x,y:x*y,[m[p[i]][i] for i in r],1)*s(p) for p in f(r)])

Values are read from stdin in a format like 1,2,-3 7,0,4 -1,-3,0 with a single space between columns. I wouldn't have posted this, but it was fun to write and contains a couple interesting sub-problems. The function f generates all permutations of n elements, and the function s determines the parity of those permutations.

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  • \$\begingroup\$ import is your friend... itertools implements permutations for you. \$\endgroup\$ – boothby Sep 25 '12 at 2:28
  • \$\begingroup\$ In these code golf challenges I prefer to use my own implementations, especially if I don't have a shot at winning as is. You're right though, itertools' permutations is exactly what I needed. \$\endgroup\$ – scleaver Sep 25 '12 at 13:40
  • \$\begingroup\$ Y'know, if the sole argument to a function is list comprehension, you can often just pass in a generator: sum([reduce... for p in f(r)]) can be sum(reduce... for p in f(r)). \$\endgroup\$ – boothby Nov 2 '12 at 6:48
  • \$\begingroup\$ You can eliminate three characters by removing some of the spaces: return [l]=>return[l], m[p[i]][i] for=>m[p[i]][i]for, and s(p) for=>s(p)for. \$\endgroup\$ – Zacharý Apr 16 '17 at 19:57
2
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Python, 198

This also uses the Liebniz formula. I'm using a heavily-modified version of ugoren's permutation solution to generate permutations and their inversion counts simultaneously. (edit: now correct!)

t=input()
e=enumerate
p=lambda t:t and((b+[a],j+i)for i,a in e(t)for b,j in p(t[:i]+t[i+1:]))or[([],0)]
print sum(reduce(lambda t,(i,r):t*r[i],e(p),1-i%2*2)for p,i in p([t]+[input()for x in t[1:]]))
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1
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J, 61 characters

-/>([:+/#(([{.<:@[}.])[:*//._2,\2#])])&.>(|.;])".];._2[1!:1[3

A big ugly bit of code that takes its input from stdin (if you're running this on Windows you may need to change ".];._2[1!:1[3 to ".}:@];._2[1!:1[3 to ensure both character return and line feed are removed).

   -/>([:+/#(([{.<:@[}.])[:*//._2,\2#])])&.>(|.;])".];._2[1!:1[3
1 8 5
2 7 3
9 9 4
_72
   -/>([:+/#(([{.<:@[}.])[:*//._2,\2#])])&.>(|.;])".];._2[1!:1[3
_2 2 3
_1 1 3
2 0 _1
6

Of course, no-one using J would ever bother to do this. They'd just use:

(-/ .*)".];._2[1!:1[3

making use of the . determinant verb.

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  • \$\begingroup\$ I'm not really familiar with J, so I could just be doing something wrong, but I'm getting the wrong answer for [[7,8,5,4],[3,9,-5,3],[-9,5,7,3],[6,4,2,1]] (I get -426) \$\endgroup\$ – Matt Sep 21 '12 at 16:41
  • \$\begingroup\$ @Matt Hmm...a good example of this I think. :-) I saw the way to calculate the determinant for a 3x3 matrix and used that without checking if that was the correct way. Oops. Still somehow managed to gain an upvote for it. I'll come back and either correct it or just give the invalid solution. \$\endgroup\$ – Gareth Sep 21 '12 at 17:57
  • \$\begingroup\$ Haven't checked this myself, but you may be able to use the ideas in this essay to golf it down further. \$\endgroup\$ – Jonah Nov 5 '17 at 20:13
1
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Dyalog APL, 53 58 59 characters

{1=≢⍵:⊃⍵⋄-/⍵[1;]×∇¨{w[f~1;f~⍵]}¨f←⍳⊃⍴w←⍵}U⍪↑{⎕}¨1↓⍳⍴U←⎕

I'm pretty late to the party here. Please, alert me if there are any issues with this code! Takes negatives as input via ¯ (the APL negative sign). I think values by default go to STDOUT in APL, but I may be wrong.

Uses Laplace expansion to find the determinant of the matrix.

This program is separated into three main parts: an input section, and the case of a 1x1 matrix, and the actual recursive function.

Input program:

U⍪↑{⎕}¨1↓⍳⍴U←⎕
             ⎕ - Takes evaluated input, space separated numbers are vectors in APL.
           U←  - Stores the input into variable `U`.
         ⍳⍴    - The array of integers from 1 to the length of `U` (inclusive).
       1↓      - Drops the first element of the result.
   {⎕}¨        - Receives a new input for each element in the vector. (Result is a vector of vectors)
  ↑            - Transforms the vector vector into a 2D-array.
U⍪             - The original line was stored then used for its length, so we stick it on top of the new array.

1x1 case:

{1=≢⍵:⊃⍵⋄...}
       ⍵      - The input
      ⊃       - The first element of the input
     :        - Return that if and only if ...
   ≢⍵         - The largest dimension of the input matrix
 1=           - Has equality with 1.
        ⋄     - Statement separator
{           } - Function enclosure brackets.

Actual determinant program:

{...-/⍵[1;]×∇¨{w[f~1;f~⍵]}¨f←⍳⊃⍴w←⍵}
                                w←⍵  - The input stored in `w`
                               ⍴     - The dimensions of `w`
                              ⊃      - The first dimension
                             ⍳       - 1 to the result of last operation
                           f←        - Store the result in `f`
              {          }¨          - Do what is inside the brackets to each member of `f`
               w[f~1;f~⍵]            - Return the minor matrix.
            ∇¨                       - Find determinant of each element (recursive)
      ⍵[1;]×                         - Multiply elementwise by the first row of the input matrix
    -/                               - Reduction using subtraction, since APL evaluates right to left, it comes out to be + - + - ...
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0
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Python, 181 characters

Late to the party, but here's a simple 181 character implementation of Laplace:

l,r=len,range
d=lambda A:sum((-1)**i*A[0][i]*d([[A[x][y]for x in r(1,l(A))]for y in r(i)+r(1+i,l(A))])for i in r(l(A)))if len(A)>1 else A[0][0]
print d(map(eval,iter(raw_input,'')))

This meets the specs if "any space character" includes commas, which seems to be the norm here. Otherwise,

l,r=len,range
d=lambda A:sum((-1)**i*A[0][i]*d([[A[x][y]for x in r(1,l(A))]for y in r(i)+r(1+i,l(A))])for i in r(l(A)))if len(A)>1 else A[0][0]
print d(map(lambda x:map(int,x.split(' ')),iter(raw_input,'')))

is 207 characters.

Input is terminated by entering a blank line.

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  • \$\begingroup\$ Hey, don't complain about being late to the party now. And you could save a character by changing len(A)>1 else to len(A)>1else. \$\endgroup\$ – Zacharý Apr 16 '17 at 19:35
0
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Jq 1.5, 171 bytes

This is a golf-ed version of det from rosettacode Matrix arithmetic - jq.

def d:if[paths]==[[0],[0,0]]then.[0][0]else. as$m|reduce range(length)as$i(0;.+({"0":1}["\($i%2)"]//-1)*$m[0][$i]*($m|reduce del(.[0])[]as$r([];.+[$r|del(.[$i])])|d))end;d

Original

def except(i;j):
  reduce del(.[i])[] as $row ([]; . + [$row | del(.[j]) ] );

def det:
  def parity(i): if i % 2 == 0 then 1 else -1 end;
  if length == 1 and (.[0] | length) == 1 then .[0][0]
  else . as $m
    | reduce range(0; length) as $i
        (0; . + parity($i) * $m[0][$i] * ( $m | except(0;$i) | det) )
  end ;

Alterations:

  • replace base case condition with [paths]==[[0],[0,0]]
  • replace parity($i) with ({"0":1}["\($i%2)"]//-1)
  • replace except(0;$i) with inline definition
  • replace det,row with d,r and remove whitespace

Try it online!

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