Talk:Zhang-Suen thinning algorithm: Difference between revisions
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==Inverted axis?== |
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Could someone help explain why, in the example image the following |
Could someone help explain why, in the example image the following |
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transformation occurs (in the periods separating the R and C, and after the |
transformation occurs (in the periods separating the R and C, and after the |
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::I'm considering what happens to this individual cell (not necessarily its neighbours). As far as I am concerned, the situation I describe above is the calculation for Step-1 of the first iteration. Nothing has changed (i.e. been removed) before this: all changes are stored, and applied `after` the analysis. So the step-1 rule should apply to cell '?'. And it should be blank (by my interpretation of the rules). But it ain't. |
::I'm considering what happens to this individual cell (not necessarily its neighbours). As far as I am concerned, the situation I describe above is the calculation for Step-1 of the first iteration. Nothing has changed (i.e. been removed) before this: all changes are stored, and applied `after` the analysis. So the step-1 rule should apply to cell '?'. And it should be blank (by my interpretation of the rules). But it ain't. |
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--[[User:Tim-brown|Tim-brown]] ([[User talk:Tim-brown|talk]]) 19:27, 15 October 2013 (UTC) |
--[[User:Tim-brown|Tim-brown]] ([[User talk:Tim-brown|talk]]) 19:27, 15 October 2013 (UTC) |
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You are right the table given of P1 to P9 in the task page is actually transformed but ins such a way that the output is thinned in the same way (but with an offset possibly). I am actually calculating with: |
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<table border="1"> |
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<tr><td>P7</td><td>P6</td><td>P5</td></tr> |
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<tr><td>P8</td><td><b>P1</b></td><td>P4</td></tr> |
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<tr><td>P9</td><td>P2</td><td>P3</td></tr> |
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</table> |
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I've got an inverted vertical axis it seems. --[[User:Paddy3118|Paddy3118]] ([[User talk:Paddy3118|talk]]) 22:32, 15 October 2013 (UTC) |
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:Now I'm a bit more confident in posting my Racket solution! ---[[User:Tim-brown|Tim-brown]] ([[User talk:Tim-brown|talk]]) 09:50, 16 October 2013 (UTC) |
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Is the thinned version of the big "R.C." in the task description correct? |
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I have it as: |
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<pre>........................................................... |
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........................................................... |
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....#.##########.......................#######............. |
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.....##........#...................####.......#............ |
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.....#..........#.................##....................... |
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.....#..........#................#......................... |
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.....#..........#................#......................... |
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.....#..........#................#......................... |
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.....############...............#.......................... |
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.....#..........#...............#.......................... |
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.....#..........#................#......................... |
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.....#..........#................#......................... |
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.....#..........#................#......................... |
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.....#............................##....................... |
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.....#.............................############............ |
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.......................###..........................###.... |
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........................................................... |
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...........................................................</pre> |
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---[[User:Tim-brown|Tim-brown]] ([[User talk:Tim-brown|talk]]) 10:05, 16 October 2013 (UTC) |
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: I've corrected that too Tim. Thanks. --[[User:Paddy3118|Paddy3118]] ([[User talk:Paddy3118|talk]]) 22:30, 18 October 2013 (UTC) |
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==Three Data Files== |
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To avoid every entry to needlessly contain the same large tables, I suggest to add to the task description three links to the tree files of the test cases. -[[User:Bearophile|bearophile]] ([[User talk:Bearophile|talk]]) |
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:I think if the outputs are typically in ./# form, the inputs probably should be too, rather than csv. --[[User:TimToady|TimToady]] ([[User talk:TimToady|talk]]) 15:53, 21 October 2013 (UTC) |
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==Mathematica's in-built algorithm== |
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I converted Mathematica's ones and zeros to something I could see and it seems that their in-built thinning algorithm does not give the same result as that of Zhang Suen. |
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I think that it is right and proper to show the output of the in-built function as I would expect that to be used, but I also asked for an implementation of the ZS algorithm as I maked it incomplete. |
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If you think that is wrong then just give your argument please. |
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The following is just a code dump of how I converted to ascii art for comparison using Python: |
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<lang python>>>> print(''' |
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{ |
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{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, |
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{0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0}, |
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{0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0}, |
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{0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0}, |
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{0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, |
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{0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, |
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{0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0}, |
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{0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0}, |
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{0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0}, |
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{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}}'''.replace('{', '(').replace('}', ')')) |
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( |
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(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0), |
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(0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0), |
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(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)) |
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>>> inp = ( |
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(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), |
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(0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0), |
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(0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0), |
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(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)) |
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>>> print('\n'.join(''.join(('#' if x else ' ') for x in line) for line in inp)) |
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######### ######## |
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### #### #### #### |
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### ### ### ### |
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### #### ### |
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######### ### |
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### #### ### ### |
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### #### ### #### #### ### |
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### #### ### ######## ### |
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>>> print('\n'.join(''.join(('#' if x else ' ') for x in line) for line in out)) |
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#### ### |
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# # # ### |
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# # # |
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# # # |
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## # # |
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# ### # |
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# # # ## |
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## ## #### ## |
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>>> </lang> |
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--[[User:Paddy3118|Paddy3118]] ([[User talk:Paddy3118|talk]]) 20:20, 18 June 2014 (UTC) |
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Latest revision as of 20:20, 18 June 2014
Inverted axis?
Could someone help explain why, in the example image the following transformation occurs (in the periods separating the R and C, and after the C):
..... ..... .###. -> ..... .#?#. ..#.. ..... .....
Surely the the cell labelled '?' will be culled at step 1:
- It is black with 8 neighbours - B = 5 (2 <= 5 <= 6) - A = 1 - At least one of P2 P4 P6 is white (P6 is white) - At least one of P4 P6 P8 is white (P6 is white)
Why isn't it whitened at step 1?
--Tim-brown (talk) 17:33, 15 October 2013 (UTC)
- I expect that there are removals of some of those surrounding cells before it gets to your '?' cell which affects the final outcome. --Paddy3118 (talk) 18:54, 15 October 2013 (UTC)
- I'm considering what happens to this individual cell (not necessarily its neighbours). As far as I am concerned, the situation I describe above is the calculation for Step-1 of the first iteration. Nothing has changed (i.e. been removed) before this: all changes are stored, and applied `after` the analysis. So the step-1 rule should apply to cell '?'. And it should be blank (by my interpretation of the rules). But it ain't.
--Tim-brown (talk) 19:27, 15 October 2013 (UTC)
You are right the table given of P1 to P9 in the task page is actually transformed but ins such a way that the output is thinned in the same way (but with an offset possibly). I am actually calculating with:
| P7 | P6 | P5 |
| P8 | P1 | P4 |
| P9 | P2 | P3 |
I've got an inverted vertical axis it seems. --Paddy3118 (talk) 22:32, 15 October 2013 (UTC)
- Now I'm a bit more confident in posting my Racket solution! ---Tim-brown (talk) 09:50, 16 October 2013 (UTC)
Is the thinned version of the big "R.C." in the task description correct? I have it as:
........................................................... ........................................................... ....#.##########.......................#######............. .....##........#...................####.......#............ .....#..........#.................##....................... .....#..........#................#......................... .....#..........#................#......................... .....#..........#................#......................... .....############...............#.......................... .....#..........#...............#.......................... .....#..........#................#......................... .....#..........#................#......................... .....#..........#................#......................... .....#............................##....................... .....#.............................############............ .......................###..........................###.... ........................................................... ...........................................................
---Tim-brown (talk) 10:05, 16 October 2013 (UTC)
Three Data Files
To avoid every entry to needlessly contain the same large tables, I suggest to add to the task description three links to the tree files of the test cases. -bearophile (talk)
- I think if the outputs are typically in ./# form, the inputs probably should be too, rather than csv. --TimToady (talk) 15:53, 21 October 2013 (UTC)
Mathematica's in-built algorithm
I converted Mathematica's ones and zeros to something I could see and it seems that their in-built thinning algorithm does not give the same result as that of Zhang Suen.
I think that it is right and proper to show the output of the in-built function as I would expect that to be used, but I also asked for an implementation of the ZS algorithm as I maked it incomplete.
If you think that is wrong then just give your argument please.
The following is just a code dump of how I converted to ascii art for comparison using Python:
<lang python>>>> print( { {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0}, {0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}}.replace('{', '(').replace('}', ')'))
( (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0), (0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0), (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)) >>> inp = ( (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), (0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0), (0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0), (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)) >>> print('\n'.join(.join(('#' if x else ' ') for x in line) for line in inp))
######### ########
### #### #### ####
### ### ### ###
### #### ###
######### ###
### #### ### ###
### #### ### #### #### ###
### #### ### ######## ###
>>> print('\n'.join(.join(('#' if x else ' ') for x in line) for line in out))
#### ###
# # # ###
# # #
# # #
## # #
# ### #
# # # ##
## ## #### ##