Recognize handwritten digits

Question

Your task is to read an image containing a handwritten digit, recognize and print out the digit.

Input: A 28*28 grayscale image, given as a sequence of 784 plain-text numbers from 0 to 255, separated by space. 0 means white and 255 means black.

Output: The recognized digit.

Scoring: I will test your program with 1000 of the images from the MNIST database training set (converted to ASCII form). I have already selected the images (randomly), but will not publish the list. The test must finish within 1 hour, and will determine n - the number of correct answers.
n must be at least 200 for your program to qualify. If the size of your source code is s, then your score will be calculated as s * (1200 - n) / 1000. Lowest score wins.

Rules:

• Your program must read the image from the standard input and write the digit to the standard output
• No built-in OCR function
• No third-party libraries
• No external resources (files, programs, web sites)
• Your program must be runnable in Linux using freely available software (Wine is acceptable if necessary)
• The source code must use only ASCII characters
• Please post your estimated score and a unique version number every time you modify your answer

Example input:

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 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 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 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 18 18 18 126 136 175 26 166 255 247 127 0 0 0 0 0 0 0 0 0 0 0 0 30 36 94 154 170 253 253 253 253 253 225 172 253 242 195 64 0 0 0 0 0 0 0 0 0 0 0 49 238 253 253 253 253 253 253 253 253 251 93 82 82 56 39 0 0 0 0 0 0 0 0 0 0 0 0 18 219 253 253 253 253 253 198 182 247 241 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 80 156 107 253 253 205 11 0 43 154 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 14 1 154 253 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 139 253 190 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 11 190 253 70 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 35 241 225 160 108 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 81 240 253 253 119 25 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 45 186 253 253 150 27 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 16 93 252 253 187 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 249 253 249 64 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 46 130 183 253 253 207 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 39 148 229 253 253 253 250 182 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 24 114 221 253 253 253 253 201 78 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 23 66 213 253 253 253 253 198 81 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 18 171 219 253 253 253 253 195 80 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 55 172 226 253 253 253 253 244 133 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 136 253 253 253 212 135 132 16 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 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 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

By the way, if you prepend this line to the input:

P2 28 28 255

you will obtain a valid image file in pgm format, with inverted/negated colors.

This is what it looks like with correct colors:

Example output:

5

Standings:

No.| Name         | Language   | Alg | Ver | n   | s   |  Score
----------------------------------------------------------------
 1 | Peter Taylor | GolfScript | 6D  | v2  | 567 | 101 |  63.933
 2 | Peter Taylor | GolfScript | 3x3 | v1  | 414 | 207 | 162.702

Answer 1

GolfScript 6D (v2: estimated score 101 * 0.63 ~= 64)

This is a very different approach to my earlier GolfScript answer, so it makes more sense to post it as a separate answer at v1 than to edit the other answer and make this v2.

~]:B;569'!EM,R.==|%NL2+^=1'{{32-}%95{base}:^~\^}:&~2/{~B=<}%2^10'#]8Y,;KiZfnnRsDzPsvQ!%4C&..z,g,$m'&=

Ungolfed

~]:B;
[30 183 21 378 31 381 7 461 113 543 15 568]
2/{~B=<}%2base
7060456576664262556515119565486100005262700292623582181233639882 10base
=

Explanation

The raw problem is classification of points in a 784-dimensional space. One standard approach is dimension reduction: identifying a small subset of dimensions which provide sufficient distinguishing power to do the classification. I evaluated each dimension and each possible threshold to identify 18 pairs of (dimension, range of threshold) which looked promising. I then picked the centre of each range of threshold, and evaluated 6-element subsets of the 18 pairs. Finally I optimised the threshold for each dimension of the best 6-D projection, improving its accuracy from 56.3% to 56.6%.

Because the projection is into 6 dimensions and for each dimension I apply a simple threshold, the final lookup table only needs 64 elements. It doesn't seem to be particularly compressible, so the main golfing is to base-convert both lookup tables (the list of dimension and thresholds; and the halfspace vector to digit map) and share the base-conversion code.

Answer 2

GolfScript 3x3 (v1: estimated score 207 * 0.8 ~= 166)

~]28/10:?/{zip?/{[]*0-!!}/}%2{base}:^~'"yN(YZ5B 7k{&w,M`f>wMb>}F2A#.{E6T9kNP_s 3Q?V`;Z\'C-z*kA5M@?l=^3ASH/@*@HeI@A<^)YN_bDI^hgD>jI"OUWiGct%7/U($*;h*<"r@xdTz6x~,/M:gT|\\:#cII8[lBr<%0r&y4'{32-}%95^?^2/{))*~}%=

Or in overview,

~]28/10:?/{zip?/{[]*0-!!}/}%2{base}:^~'MAGIC STRING'{32-}%95^?^2/{))*~}%=

Explanation

My approach at a high level is:

1. Threshold the pixels: if the pixel is above t1 then set it to 1; otherwise to 0.
2. Group the pixels. Initially I broke the 28x28 grid into a 4x4 grid (each subgrid being 7x7 pixels); but breaking it into a 3x3 grid (subgrids being 10x10, 10x8, or 8x8 pixels) gives a massive reduction in lookup table size while dropping accuracy rate from about 56% to about 40%.
3. Sum the pixels in each group and threshold again: if the number of set pixels is above t2 then score the group as 1; otherwise as 0.
4. Do a table lookup by the vector of group scores. (The table is compressed using run-length encoding and the standard base-conversion trick. Most choices of t1 and t2 leave between 50% and 63% of the table as "don't care" values, which can be combined with adjacent values to increase run lengths; the average run length in my v1 table is 3.6).

It turns out that setting t1=t2=0, although not optimal, isn't far off the best values of t1 and t2 in terms of accuracy; is pretty good in terms of table compressibility; and allows me to combine the two thresholding operations into []*0-!! (flatten 2D array to 1D; remove 0s; check whether it's empty).

The lookup table gives the most likely candidate for the given vector of group scores. It may well be possible to improve the score by identifying table entries which can be changed such that the improved compressibility of the table outweighs the decreased accuracy.