Code Golf: Zigzag pattern scanning
https://stackoverflow.com/questions/3025595
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26-09-2019 - |
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The Challenge
The shortest code by character count that takes a single input integer N (N >= 3) and returns an array of indices that when iterated would traverse an NxN matrix according to the JPEG "zigzag" scan pattern. The following is an example traversal over an 8x8 matrixsrc:
Examples
(The middle matrix is not part of the input or output, just a representation of the NxN matrix the input represents.)
1 2 3
(Input) 3 --> 4 5 6 --> 1 2 4 7 5 3 6 8 9 (Output)
7 8 9
1 2 3 4
(Input) 4 --> 5 6 7 8 --> 1 2 5 9 6 3 4 7 10 13 14 11 8 12 15 16 (Output)
9 10 11 12
13 14 15 16
Notes
- The resulting array's base should be appropriate for your language (e.g., Matlab arrays are 1-based, C++ arrays are 0-based).
- This is related to this question.
Bonus
Extend your answer to take two inputs N and M (N, M >=3) and perform the same scan over an NxM matrix. (In this case N would be the number of columns and M the number of rows.)
Bonus Examples
1 2 3 4
(Input) 4 3 --> 5 6 7 8 --> 1 2 5 9 6 3 4 7 10 11 8 12 (Output)
9 10 11 12
1 2 3
(Input) 3 4 --> 4 5 6 --> 1 2 4 7 5 3 6 8 10 11 9 12 (Output)
7 8 9
10 11 12
Solution
J, 13 15 characters
;<@|.`</.i.2$
Usage:
;<@|.`</.i.2$ 3
0 1 3 6 4 2 5 7 8
;<@|.`</.i.2$ 4
0 1 4 8 5 2 3 6 9 12 13 10 7 11 14 15
Explanation
(NB. is J's comment indicator)
; NB. Link together...
<@|.`< NB. ... 'take the reverse of' and 'take normally'
/. NB. ... applied to alternating diagonals of...
i. NB. ... successive integers starting at 0 and counting up to fill an array with dimensions of...
2$ NB. ... the input extended cyclically to a list of length two.
J, bonus, 13 characters
;<@|.`</.i.|.
Usage:
;<@|.`</.i.|. 3 4
0 1 3 6 4 2 5 7 9 10 8 11
;<@|.`</.i.|. 9 6
0 1 9 18 10 2 3 11 19 27 36 28 20 12 4 5 13 21 29 37 45 46 38 30 22 14 6 7 15 23 31 39 47 48 40 32 24 16 8 17 25 33 41 49 50 42 34 26 35 43 51 52 44 53
OTHER TIPS
Python, 92, 95, 110, 111, 114, 120, 122, 162, 164 chars
N=input()
for a in sorted((p%N+p/N,(p%N,p/N)[(p%N-p/N)%2],p)for p in range(N*N)):print a[2],
Testing:
$ echo 3 | python ./code-golf.py
0 1 3 6 4 2 5 7 8
$ echo 4 | python ./code-golf.py
0 1 4 8 5 2 3 6 9 12 13 10 7 11 14 15
This solution easily generalizes for NxM boards: tweak the input processing and replace N*N with N*M:
N,M=map(int,raw_input().split())
for a in sorted((p%N+p/N,(p%N,p/N)[(p%N-p/N)%2],p)for p in range(N*M)):print a[2],
I suspect there's some easier/shorter way to read two numbers.
Testing:
$ echo 4 3 | python ./code-golf.py
0 1 4 8 5 2 3 6 9 10 7 11
Ruby, 69 89 chars
n=gets.to_i
puts (0...n*n).sort_by{|p|[t=p%n+p/n,[p%n,p/n][t%2]]}*' '
89 chars
n=gets.to_i
puts (0...n*n).map{|p|[t=p%n+p/n,[p%n,p/n][t%2],p]}.sort.map{|i|i[2]}.join' '
Run
> zigzag.rb
3
0 1 3 6 4 2 5 7 8
> zigzag.rb
4
0 1 4 8 5 2 3 6 9 12 13 10 7 11 14 15
Credits to doublep for the sort method.
F#, 126 chars
let n=stdin.ReadLine()|>int
for i=0 to 2*n do for j in[id;List.rev].[i%2][0..i]do if i-j<n&&j<n then(i-j)*n+j|>printf"%d "
Examples:
$ echo 3 | fsi --exec Program.fsx
0 1 3 6 4 2 5 7 8
$ echo 4 | fsi --exec Program.fsx
0 1 4 8 5 2 3 6 9 12 13 10 7 11 14 15
Golfscript, 26/30 32/36 45 59 characters
Shortest non-J solution so far:
Updated sort (don't tell the others!) - 30 chars:
~:1.*,{..1/\1%+.2%.+(@*]}$' '* #solution 1
#~:\.*,{.\/1$\%+.1&@.~if]}$' '* #solution 2
#~\:1*,{..1/\1%+.2%.+(@*]}$' '* #(bonus)
#~\:\*,{.\/1$\%+.1&@.~if]}$' '* #(bonus)
Straight implementation - 36 chars:
~:@.*,{[.@%:|\@/:^+|^- 2%^|if]}$' '*
#~\:@*,{[.@%:|\@/:^+|^- 2%^|if]}$' '* #(bonus)
If you can provide output as "013642578" instead of "0 1 3 6 4 2 5 7 8", then you can remove the last 4 characters.
Credit to doublep for the sorting technique.
Explanation:
~\:@* #read input, store first number into @, multiply the two
, #put range(@^2) on the stack
{...}$ #sort array using the key in ...
" "* #join array w/ spaces
and for the key:
[ #put into an array whatever is left on the stack until ]
.@%:| #store @%n on the stack, also save it as |
\@/:^ #store @/n on the stack, also save it as ^
+ #add them together. this remains on the stack.
|^- 2%^|if #if (| - ^) % 2 == 1, then put ^ on stack, else put | on stack.
] #collect them into an array
MATLAB, 101/116 chars
Its basically a condensed version of the same answer given here, to be run directly on the command prompt:
N=input('');i=fliplr(spdiags(fliplr(reshape(1:N*N,N,N)')));i(:,1:2:end)=flipud(i(:,1:2:end));i(i~=0)'
and an extended one that read two values from the user:
S=str2num(input('','s'));i=fliplr(spdiags(fliplr(reshape(1:prod(S),S)')));i(:,1:2:end)=flipud(i(:,1:2:end));i(i~=0)'
Testing:
3
ans =
1 2 4 7 5 3 6 8 9
and
4 3
ans =
1 2 5 9 6 3 4 7 10 11 8 12
Ruby 137 130 138 characters
n=gets.to_i
def g(a,b,r,t,s);x=[s*r]*t;t==r ?[a,x,a]:[a,x,g(b,a,r,t+1,-s),x,a];end
q=0;puts ([1]+g(1,n,n-1,1,1)).flatten.map{|s|q+=s}*' '
$ zz.rb
3
1 2 4 7 5 3 6 8 9
$ zz.rb
4
1 2 5 9 6 3 4 7 10 13 14 11 8 12 15 16
C89 (280 bytes)
I guess this can still be optimized - I use four arrays to store the possible movement vectors when hitting a wall. I guess it can be done, saving some chars at the definition, but I think it will cost more to implement the logic further down. Anyway, here you go:
t,l,b,r,i,v,n;main(int c,char**a){n=atoi(*++a);b=n%2;int T[]={n-1,1},L[]={1-n,n}
,B[]={1-n,1},R[]={n-1,n};for(c=n*n;c--;){printf("%d%c",i,c?32:10);if(i>=n*(n-1))
v=B[b=!b];else if(i%n>n-2){if(!(n%2)&&i<n)goto g;v=R[r=!r];}else if(i<n)g:v=T[t=
!t];else if(!(i%n))v=L[l=!l];i+=v;}}
compiles with a few warnings, but as far as I know it is portable C89. I'm actually not sure whether my algorithm is clever at all, maybe you can get way shorter with a better one (haven't taken the time to understand the other solutions yet).
Haskell 117 characters
i s=concatMap(\q->d q++(reverse.d$q+1))[0,2..s+s]
where d r=[x+s*(r-x)|x<-[0..r],x<s&&(r-x)<s]
main=readLn>>=print.i
Runs:
$ echo 3 | ./Diagonals
[0,1,3,6,4,2,5,7,8]
$ echo 4 | ./Diagonals
[0,1,4,8,5,2,3,6,9,12,13,10,7,11,14,15]
The rectangular variant is a little longer, at 120 characters:
j(w,h)=concatMap(\q->d q++(reverse.d$q+1))[0,2..w+h]
where d r=[x+w*(r-x)|x<-[0..r],x<w&&(r-x)<h]
main=readLn>>=print.j
Input here requires a tuple:
$ echo '(4,3)' | ./Diagonals
[0,1,4,8,5,2,3,6,9,10,7,11]
$ echo '(3,4)' | ./Diagonals
[0,1,3,6,4,2,5,7,9,10,8,11]
Answers are all 0-based, and returned as lists (natural forms for Haskell).
Perl 102 characters
<>=~/ /;print map{$_%1e6," "}sort map{($x=($%=$_%$`)+($/=int$_/$`))*1e9+($x%2?$/:$%)*1e6+$_}0..$'*$`-1
usage :
echo 3 4 | perl zigzag.pl
0 1 3 6 4 2 5 7 9 10 8 11
