Learn X in Y minutes

> anything undefined technically evaluates to nothing and so is also a comment
# but commenting it out explicitly anyway is probably a good idea because if
# you use a reserved keyword or any punctuation you'll run into trouble.
[]
> ######################################################################
#       datatypes
########################################################################
> 1 # Here we add 1 to the stack.  Any object entry adds things to the stack
[1]
> 'abc' # here we are adding a string. The only difference between single and
# double quotes is that double lets you escape more things other than \' and \n
# it won't matter for the sake of this tutorial.
[1 "abc"]
> {+} # the third type of object you can put on the stack is a block
[1 "abc" {+}]
> ] # this takes everything prior and puts it into an array, the fourth type
# of object. (besides bug exploits like [2-1?] those are the only four types)
[[1 "abc" {+}]]
> ; # let's clear the stack by executing the discard function on this array.
# if you type the characters  ]; it always clears the stack.
[]
> 1"abc"{+}]; # newlines are whitespaces. Everything we did up to this point
# can be put into one line and it all works the exact same.
########################################################################
#       operators and math
########################################################################
[]
> 1 1 # we add two 1s to the stack.  We could also duplicate the first with .
[1 1]
> + # math is done by executing an operation on the top of the stack. This
# can be a standalone character. The way to read this is that we put a 1 on
# the stack, another one one the stack, and then executed a + operation which
# takes the top two elements off of the stack, sums them up, and returns them
# to the stack.  This is typically referred to as postfix notation.  It can be
# a bit jarring, but this is the way to think about things.  You're adding to
# the stack with objects and modifying the top of the stack with operators.
[2]
> 8 1- # minus works the same way. N.B. that we still have that 2 on the stack
# from earlier
[2 7]
> 10 2* # multiplication works the same way. The product is added to the stack
[2 7 20]
> 35 4/ # all division is integer division
[2 7 20 8]
> 35 4%  # modulo operation
[2 7 20 8 3]
> 2 3? # exponentiation
[2 7 20 8 3 8]
> 8~ # bitwise "not" function on signed integers
[2 7 20 8 3 8 -9]
> -1~ # this yields 0, which is useful to know for the ? operator
[2 7 20 8 3 8 -9 0]
> 5 3| # or: yields 7, since [1 0 1] | [0 1 1] => [1 1 1]
[2 7 20 8 3 8 -9 0 7]
> 5 3^ # xor: yields 6, since the parity differs at [1 1 0]
[2 7 20 8 3 8 -9 0 7 6]
> 5 3& # and: yields 1, since it's the only bit active in both: [0 0 1]
[2 7 20 8 3 8 -9 0 7 6 1]
> ]; ###################################################################
#       booleans
########################################################################
[]
> 5 3
[5 3]
> < #add two numbers to the stack, and then perform a lessthan operation
# booleans are False if 0, [], {}, '', and true if anything else.
[0]
> 5 3> # greater than operation.
[0 1]
> 5 3= #single equal is the operator. Again, before the equals is executed,
# the stack reads [0 1 5 3], and then the equals operator checks the top 2
# values and yields:
[0 1 0]
> ! #not, returns 1 if 0 else 0.
[0 1 1]
> ) #increments the last number
[0 1 2]
> ( #decrements the last number
[0 1 1]
> ]; ###################################################################
#       stack control
########################################################################
[]
> 1 # put a number on the stack
[1]
> . # duplicate the number
[1 1]
> ) # increment
[1 2]
> \ # flip the top two items
[2 1]
> 1$ # $ copies the nth-to-last item on the stack at the index preceding.
# Here we get the 1-indexed item.
[2 1 2]
> 0$ # to copy the 0-indexed item we use the appropriate index.
# This is identical to . operation
[2 1 2 2]
> ) # increment
[2 1 2 3]
> @ # pulls the third item up to the top
[2 2 3 1]
> [@] # use this trick to flip the top 3 items and put them into an array
# if you wrap any operation in brackets it flips the results into an array.
# even math operations like, [+] and [-]
[2 [3 1 2]]
> ]; # also, using at most two strokes you can orient the top three items
# in any permutation. Below are shown the results on 3,~
  #      => 0 1 2    (i.e. doing nothing)
  #   \  => 0 2 1
  #   @\ => 1 0 2
  #   @  => 1 2 0
  #   @@ => 2 0 1
  #   \@ => 2 1 0
[]
> ######################################################################
#       using arrays
########################################################################
[]
> 2, # comma is the range() function
[[0 1]]
> , # and also the length() function
[2]
> ;4, # let's get an array of four items together
[[0 1 2 3]]
> ) # we can pop off the last value
[[0 1 2] 3]
> + # and put it back
[[0 1 2 3]]
> ( # we can pop off the first value
[[1 2 3] 0]
> \+ # and put it back
[[0 1 2 3]]
> 2- # we can subtract a particular value
[[0 1 3]]
> [1 3] # or a list of values
[[0 1 3] [1 3]]
> -
[[0]]
> ! # boolean operations also work on lists, strings, and blocks. If it's
# empty it's a 1, otherwise 0. Here, the list has a zero, but it's not zero-
# length, so the array as a whole is still True... and hence "not" is False
[0]
> ;4,(+ # let's make a range, pop the first value, and tack it on the end
[[1 2 3 0]]
> $ # we can also restore order by sorting the array
[[0 1 2 3]]
> 1 >  # we can also use < > and = to get the indeces that match. Note this
# is not a filter! This is an index match. Filtering items greater than one
# is done with {1>},
[[1 2 3]]
> 2 < # remember it's zero-indexed, so everything in this array is at an index
# less than 2, the indeces are 0 and 1.
[[1 2]]
> 1= # < and > return an array, even if it's one item.  Equals always drops
# it out of the array
[2]
> ;6,2% # the modulo operator works on lists as the step.
[[0 2 4]]
> ;4,2,-:a 3,2+:b # booleans also work on lists. lets define two lists
[[2 3] [0 1 2 2]]
> | # "or" - returns set of items that appear in either list i.e. "union set"
[[2 3 0 1]]
> ;a b& # returns set of items that appear in 1 AND 2, e.g. "intersection set"
[[2]]
> ;a b^ # returns the symmetric difference set between two lists,
[[3 0 1]]
> ~ # tilde unpacks the items from a list
[3 0 1]
> ]; a
[2 3]
> 2? # finds the index of an item
[0]
> ;3a?
[1]
> 4a? # returns -1 if the item doesn't exist. Note: Order of element and array
# doesn't matter for searching. it can be [item list?] or [list item?].
[1 -1]
> ]; # clear
[]
> 3,[4]* # join or intersperse: puts items in between the items
[[0 4 1 4 2]]
> ; 3,4* # multiplication of lists
[[0 1 2 0 1 2 0 1 2 0 1 2]]
> ;[1 2 3 2 3 5][2 3]/ # "split at"
[[[1] [] [5]]]
> ;[1 2 3 2 3 5][2 3]% # modulo is "split at... and drop empty"
[[[1] [5]]]
> ];####################################################################
#       strings
########################################################################
# strings work just like arrays
[]
> "use arch, am vegan, drive a stick" ', '/ # split
[["use arch" "am vegan" "drive a stick"]]
> {'I '\+', BTW.'+}% # map
[["I use arch, BTW." "I am vegan, BTW." "I drive a stick, BTW."]]
> n* # join.  Note the variable n is defined as a newline char by default
["I use arch, BTW.\nI am vegan, BTW.\nI drive a stick, BTW."]
> n/ # to replace, use split, and join with the replacement string.
[n "Also, not sure if I mentioned this, but" n]{+}* # fold sum 3-item array
* # and use join to get the result
n+ print # and then pop/print the results prettily
I use arch, BTW.
Also, not sure if I mentioned this, but
I am vegan, BTW.
Also, not sure if I mentioned this, but
I drive a stick, BTW.
[]
> '22222'{+}* # note that if you fold-sum a string not in an array, you'll
# get the sum of the ascii values. '2' is 50, so five times that is:
[250]
> ]; # this actually is a clever trick to get ascii values into an array.
[]
> "aabc" [{""+~}*] # if you fold over addition and drop it into a string: 
[[97 97 98 99]]
> {[.]""^}%""+ # which can be returned to a string as such using a ""^ map.
# and an empty string join.
["aabc"]
> {32-}% # note that most mapping operations work on the ascii values as
# you would expect, for instance with the difference between A and a being
# 32, you can just subtract that from the ascii value to get:
["AABC"] 
> ]; ###################################################################
#       blocks
########################################################################
[]
> 3,~ # start with an unpacked array
[0 1 2]
> {+-} # brackets define a block which can comprise multiple functions
[0 1 2 {+-}]
> ~ # blocks are functions waiting for execution. tilde does a single
# execution of the block in this case, we added the top two values, 1 and 2,
# and subtracted from 0
[-3]
> ;10,~{+}5* # multiplication works on executing blocks multiple times
# in this case we added the last 6 values together by running "add" 5 times
[0 1 2 3 39]
> ];10,4> # we can achieve the same result by just grabbing the last 6 items
[[4 5 6 7 8 9]]
> {+}* # and using the "fold" function for addition.
[39]
> # "fold" sequentially applies the operation pairwise from the left
# and then dumps the results.  Watch what happens when we use the duplicate
# operator to fold. it's clear what happens when we duplicate and then negate
# the duplicated item:
> ;4,{.-1*}*
[0 1 -1 2 -2 3 -3]
> ]{3%}, # we can filter a list based on applying the block to each element
# in this case we get the numbers that do NOT give 0 mod 3
[[1 -1 2 -2]]
> ;10,{3%0}, # note that only the last element matters for retaining in the
# array.  Here we take 0..9, calculate x mod 3, and then return a 0. The
# intermediate generated values are dumped out sequentially.
[0 1 2 0 1 2 0 1 2 0 []]
> ]; # clear
[]
> 5,{5*}% # map performs each operation on the array and returns the result
# to an array
[[0 5 10 15 20]]
> {.}% # watch what happens when you map duplicate on each item
[[0 0 5 5 10 10 15 15 20 20]]
> ]; ###################################################################
#       Control Flow!
########################################################################
# This is the most important part of scripting. Most languages have
# two main types of loops, for loops and while loops. Even though golfscript
# has many possible loops, only a few are generally useful and terse. For loops
# are implemented using mapping, filtering, folding, and sorting over lists.
# For instance, we can take the factorial of 6 by:
6, # get 0..5
{)}% # increment the list, i.e. "i++ for i in list" to get 1..6
{*}* # fold by multiplication , 9 characters for the operator itself.
[720]
> 6),(;{*}* # but can we get shorter? We can save some space by incrementing
# the 6, dropping the zero, and folding. 8 characters.
> # we can also use fold to do the same thing with unfold
1 6        # accumlator and multiplicand, we'll call A and M
{}{        # while M
  .        # copy M, so now the stack is A M M
    @      # bring A to the top, so now M M A
     *     # apply M to the accumulator, so M A
       \(  # flip the order, so it's A M, and M--
}/;        # "end", drop the list of multiplicands
# this is effectively a while-loop factorial
[720 720]
> 1.{6>!}{.@*\)}/; # we can also do the same thing with M++ while M not > 6
> 1 6{.@*\(.}do; # works the same way as the decrementing fold.
[720 720 720]
> ]; #obviously a for loop is ideal for factorials, since it naturally lends
# itself to running over a finite set of items.
########################################################################
#       Writing code
########################################################################
# Let's go through the process for writing a script. There are some tricks and
# ways to think about things. Let's take a simple example: a prime sieve.
# There are a few strategies for sieving. First, there's a strategy that
# uses two lists, candidates and primes. We pop a value from candidates,
# remove all the candidates divisible by it, and then add it to the primes.
# Second, there's just a filtering operation on numbers. I think it's
# probably shorter to write a program that just checks if a number has no
# numbers mod zero besides 0, 1, and itself. Slower, but shorter is king.
# Let's try this second strategy first.
[]
> 10 # we're probably going to filter a list using this strategy. It's easiest
# to start working with one element of the list. So let's take some example
# where we know the answer that we want to get.
[10]
> .,2> # let's duplicate it and take a list of values, and drop the first two
[10 [2 3 4 5 6 7 8 9]]
> {1$\%!}, # duplicate the ten, and scoot it behind the element, and then run
# 10 element %, and then ! the answer, so we are left with even multiples
[10 [2 5]]
> \; # we want to get rid of the intermediate so it doesn't show up in our
# solution.
[[2 5]]
> 10.,2,-{1$\%!},\; # Okay, let's put our little function together on one line
[[2 5] [2 5]]
> ;; # now we just filter the list using this strategy. We need to negate the
# result with ! so when we get a number with a factor, ! evaluates to 0, and
# the number is filtered out.
[]
> 10,{.,2,-{1$\%!},\;!}, # let's try filtering on the first 10 numbers
[[0 1 2 3 5 7]]
> 2> # now we can just drop 0 and 1.
[[2 3 5 7]]
> 4.?,{.,2,-{1$\%!},\;!},2> # trick: an easy way to generate large numbers in
# a few bytes is duplicate and exponentiate. 4.? is 256, and 9.? is 387420489
[[2 3 5 7] [2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89
97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193
197 199 211 223 227 229 233 239 241 251]]
> ];'4.?,{.,2,-{1$\%!},\;!},2>', # how long is our code for p<256 ?
[25]
> ; # this is 25 characters. Can we do better?!
[]
> []99,2> # let's go with the first strategy.  We'll start with an empty list
# of primes and a list of candidates
[[] [2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98]]
> (\ # pop left and leave left, we're going to copy this value with the filter
[[] 2 [3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98]]
> {1$%}, # filter out anything that is 0 mod by the popped item one back on the
# stack
[[] 2 [3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51
53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97]]
> @@+ # great, all the 2-divisible values are off the list! now we need to add
# it to the running list of primes
[[3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55
57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97] [2]]
> \ # swap back. Now it seems pretty clear when our candidates list is empty
# we're done. So let's try it with a do loop. Remember we need to duplicate
# the final value for the pop check. So we add a dot
[[2] [3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53
55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97]]
> {(\{1$%},@@+\.}do;
[[2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97]]
> ; # ok that worked. So let's start with our initialization as well.
[]4.?,2>{(\{1$%},@@+\.}do; # and let's check our work
[[2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199
211 223 227 229 233 239 241 251]]
> ,'[]99,2>{(\{1$%},@@+\.}do;', # how long is this?
[26]
> ]; # wow this solution is only 26 long, and much more effective. I don't see
# a way to get any smaller here. I wonder if with unfold we can do better?  The
# strategy here is to use unfold and then at the end grab the first value from
# each table.
[]
> 99,2> # start with the candidates list
[[2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81
82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98]]
> (\{1$%}, # pop left and filter
[2 [3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53
55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97]]
> (\{1$%}, # again
[2 3 [5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59 61 65 67 71 73 77
79 83 85 89 91 95 97]]
89 91 95 97]]
> {}{(\{1$%},}/ # ok I think it'll work. let's try to put it into an unfold.
[2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 [[5 7
11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59 61 65 67 71 73 77 79 83 85
89 91 95 97] [7 11 13 17 19 23 29 31 37 41 43 47 49 53 59 61 67 71 73 77 79 83
89 91 97] [11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97] [13
17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97] [17 19 23 29 31 37 41
43 47 53 59 61 67 71 73 79 83 89 97] [19 23 29 31 37 41 43 47 53 59 61 67 71 73
79 83 89 97] [23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97] [29 31 37 41
43 47 53 59 61 67 71 73 79 83 89 97] [31 37 41 43 47 53 59 61 67 71 73 79 83 89
97] [37 41 43 47 53 59 61 67 71 73 79 83 89 97] [41 43 47 53 59 61 67 71 73 79
83 89 97] [43 47 53 59 61 67 71 73 79 83 89 97] [47 53 59 61 67 71 73 79 83 89
97] [53 59 61 67 71 73 79 83 89 97] [59 61 67 71 73 79 83 89 97] [61 67 71 73
79 83 89 97] [67 71 73 79 83 89 97] [71 73 79 83 89 97] [73 79 83 89 97] [79 83
89 97] [83 89 97] [89 97] [97]]]
> ;] # drop that list of candidates generated at each step and put the items
# left behind by the unfold at each step (which is the primes) into a list
[[2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97]]
> ]; # clear and let's try with larger numbers
[]
> 4.?,2>{}{(\{1$%},}/;]
[[2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199
211 223 227 229 233 239 241 251]]
>;'4.?,2>{}{(\{1$%},}/;]', # find the length of our solution.
[21]
> ]; # only 21 characters for the primes! Let's see if we actually can use this
# strategy of leaving items behind, now using the do loop to get even shorter!
> 3.?,2> # candidates
[[2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26]]
> (\{1$%}, # pop and filter
[2 [3 5 7 9 11 13 15 17 19 21 23 25]]
> (\{1$%}, # again!
[2 3 [5 7 11 13 17 19 23 25]]
> {(\{1$%},.}do;] # try in a do loop and drop the empty list of candidates at
# the end of the do loop.  Don't forget the dot before the closing brace!
[[2 3 5 7 11 13 17 19 23]]
> ;4.?,2>{(\{1$%},.}do;] # check our work
[[2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199
211 223 227 229 233 239 241 251]]
> ;'4.?,2>{(\{1$%},.}do;]',
[21]
>]; # Still 21 characters. there's one other thing to try, which is the prime
# test known as Wilson's theorem. We can try filtering the items down using
# this test.
[]
> '4.?,2>{.,(;{*}*.*\%},'.~\, # let's run it and take the length
[[2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199
211 223 227 229 233 239 241 251] 21]
> ; # Still 21 characters! I think this number is quite good and it's not
# obvious how to beat it. The problem with GolfScript is there's always someone
# out there who thinks of some trick you didn't. For instance, you might think
# you're doing well with a Collatz seq generator of {(}{.2%{3*)}{2/}if}/ until
# you find that someone figured out {(}{3*).2%6\?/}/ which is so much shorter
# and cleaner - the unfold operation is nearly half the length!
########################################################################
#       How to read GolfScript
########################################################################
# let's take the gcd from the GolfScript banner. It starts with:
[]
> '2706 410'~ # so that's pretty straightforward, that it just evals the list
# and dumps the results on the stack. It's common to read from stdin which
# necessitates unpacking with ~
[2706 410]
> . # we want to know what that do loop does. the best way to do that is to
# drop the braces and run the loop one command at a time. We duplicate
[2706 410 410]
> @\ # We rearrange
[410 2706 410]
> % # we take the modulo
[410 246]
> .@\% # repeat. Note we don't need to run the final dot before the closing
# brace since this is just a value that is popped to check the loop condition
# you can also replicate the loop end with a semicolon to pop it yourself.
[246 164]
> .@\% # again!
[164 82]
> .@\% # and finally we hit zero. The loop would exit and ; would pop the zero,
# leaving you with the gcd of 82.
[82 0]
> ;; 2706 410{1$1$%.}do # Clearly this involves knowing about Euclid's method.
# you can also try a more obvious method like this one here which shows the
# numbers in sequence.
[2706 410 246 164 82 0]
>]; # so sometimes it pays dividends to know the math and you can write short
# algorithms that rely on easy tricks that aren't immediately obvious.
[]
> # let's try looking at the sudoku solver that is on the examples page. I'll
# skip the unpack step.
[2 8 4 3 7 5 1 6 9 0 0 9 2 0 0 0 0 7 0 0 1 0 0 4 0 0 2 0 5 0 0 0 0 8 0 0 0 0 8
0 0 0 9 0 0 0 0 6 0 0 0 0 4 0 9 0 0 1 0 0 5 0 0 8 0 0 0 0 7 6 0 4 4 2 5 6 8 9 7
3 1]:a 0?:b # again the grid is put into an array. Now, the next step
# is to define the "@" symbol as the working grid. This is because "@9" is
# interpreted as two symbols, whereas if you used something like "a" as the
# variable "a9" is interpreted as a single symbol, and this is not defined,
# so it will not get run at execution time. You would need a space which is an
# additional char. On the other hand, redefining built-ins is confusing so I
# will use "a" and "b" for the "@" and "^" definitions respectively. So the
# grid is "a" and the zero-index location of the first zero is "b", at index 9.
[9]
> ~! # this makes sure that the value is not -1 for find, i.e. -1~ evaluates to
# 0 so a ! makes it nonzero.  ?~! is a great trick for "isn't in the list"
[0]
> {@p}* # this prints out the grid the number of times as the previous value,
# which is how this thing "finishes". So if 0 isn't in the grid, it prints.
> 10, # let's get the digits 0-9. Zero will be eliminated because our original
# value is zero so when we look in any row or column, zero is guaranteed to be
# there.
[[0 1 2 3 4 5 6 7 8 9]]
> a 9/  # split the original grid row-wise
b 9/    # get the row of our checked value, in this case the second row
=       # and we get that row and
-       # take those numbers off the candidates
[[1 3 4 5 6 8]]
> a     # put the grid on the stack
b 9%    # get the column of the zero
>       # drop the first x values of the grid
9%      # take every ninth digit. We now have the column the zero is in
> -     # pull those items off the candidates list
[[1 3 5 6]]
> a 3/  # split the grid into three-long arrays
b 9%    # get the column of the zero
3/      # is the column in the left (0), middle (1), or right (2) triad?
 >      # pull that many three-groups off
3%      # get every third. Now we have 9 groups - the left side of the grid
3/      # divide those 9 groups it into thirds
b 27/   # was the zero on top (0), middle (1), or bottom (2) third of the grid?
=       # since it's the top, grab the top group of triads. You now have the
        # 1/9th of The sudoku grid where the zero sits
[[1 3 5 6] [[2 8 4] [0 0 9] [0 0 1]]]
> {+}*- # flatten those lists and remove those items from the candidates
# We now have the possible values for the position in question that work given
# the current state of the grid! if this list is empty then we've hit a
# contradiction given our previous values.
[[3 5 6]]
> 0= # {a b<\+a 1 b+>+}/ # now we've hit this unfold operation. If you run it
# you'll find we get the grids back. How does that work?! Let's take the first
# value in the "each" []{}/ operation. This is the best way to figure out what
# is happening in a mapping situation.
[3]
> a     # get the grid
b<      # get the grid up to the zero
\+      # and tack on our value of 3.
[[2 8 4 3 7 5 1 6 9 3]]
> a 1b+>+ # and we add on the rest of the grid. Note: we could do 1 char better
# because 1b+ is equivalent to but, longer than, just b)
[[2 8 4 3 7 5 1 6 9 3 0 9 2 0 0 0 0 7 0 0 1 0 0 4 0 0 2 0 5 0 0 0 0 8 0 0 0 0 8
0 0 0 9 0 0 0 0 6 0 0 0 0 4 0 9 0 0 1 0 0 5 0 0 8 0 0 0 0 7 6 0 4 4 2 5 6 8 9 7
3 1]]
> 1;; # and the do block runs again no matter what. So it's now clear why this
# thing exists with an error: if you solve the last digit, then this loop just
# keeps on rolling! You could add some bytes for some control flow but if it
# works it works and short is king.
[]

# Closing Tips for getting to the next level:
# 0. using lookback might be more effective than swapping around the values.
#    for instance, 1$1$ and \.@.@.\ do the same thing: duplicate last two items
#    but the former is more obvious and shorter.
# 1. golfscript can be fun to use for messing around with integer sequences or
#    do other cool math. So, don't be afraid to define your own functions to
#    make your life easier, like
> {$0=}:min; {$-1=}:max; {.,(;{*}*.*\%}:isprime; {.|}:set; # etc.
# 2. write pseudocode in another language or port a script over to figure out
#    what's going on. Especially useful when you combine this strategy with
#    algebra engines. For instance, you can port the examples-page 1000 digits
#    of pi to python and get:
#        import sympy as sp
#        a, k = sp.var('a'), list(range(20))[1::2]
#        for _ in range(len(k)-1):
#            m = k.pop()
#            l = k.pop()
#            k.append(((l+1)//2*m)//(l+2)+2*a)
#        print(str(k[0]))
#    which gives "2*a + floor(2*a/3 + floor(4*a/5 + 2*floor(6*a/7 + 3*floor(
#    8*a/9 + 4*floor(10*a/11 + 5*floor(12*a/13 + 6*floor(14*a/15 + 7*floor(16*
#    a/17 + 72/17)/15)/13)/11)/9)/7)/5)/3)"... which makes it much more obvious
#    what's going on than 10.3??:a;20,-2%{2+.2/@*\/a 2*+}* especially when
#    you're new to the language
# 3. a little math goes a long way. The above prime test uses Wilson's theorem
#    a comparable program testing for factors {:i,(;{i\%!},(;!}:isprime is
#    longer and slower. Also, as discussed above, Collatz is much shorter if
#    you recognize that you can do (3x+1) and then divide by 6 to the power
#    ((3x+1) mod 2).  (If x was even, (3x+1) is now odd, so 3x+1 div 6 is x/2.)
#    avoiding conditionals and redundancy can sometimes require such insight.
#    And of course, unless you know this continued fraction of pi it's hard to
#    calculate it in a terse block of code.
# 4. don't be afraid to define variables and use arrays! particularly if you
#    have 4 or more items to shuffle.
# 5. don't be afraid to use [some_long_script] to pack a bunch of items in an
#    array after the fact, rather than gathering or adding them later or
#    forcing yourself to use a datastructure that keeps the items in an array
# 6. sometimes you might get in a jam with - followed by an int that can be
#    solved with ^ to do a symmetric set difference without adding a space
# 7. "#{require 'net/http';Net::HTTP.get_response(URI.parse(address)).body}"
#    can get any page source from the internet, substituting 'address' for your
#    URL. Try it with an OEIS b-file or wordlists, etc. You can also use the
#    shorter "#{File.open('filename.txt').read}" to read in a file. GolfScript
#    can run "#{any_ruby_code_here}" and add the results to the stack.
# 8. you can set anything to mean anything, which can be useful for golf:
#       3:^;^2?  => 9 because this set ^ to 3, and 3 2 ? => 9
#       3:a;a2?  => Warning: pop on empty stack - because a2 doesn't exist
#       3:a;a 2? => 9 - it works again, but takes an extra character over ^2
#    usually you will only want to do this once you're trying to squeeze the
#    last few chars out of your code because it ruins your environment.