# Talk:Zhang-Suen thinning algorithm

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## 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)

I've corrected that too Tim. Thanks. --Paddy3118 (talk) 22:30, 18 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))

```  ####            ###
#    #          #   ###
#     #        #
#     #        #
##  #         #
#  ###         #
#     #         #    ##
## ##     ####    ##

```

>>> </lang> --Paddy3118 (talk) 20:20, 18 June 2014 (UTC)