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5 | <title>The Burro Programming Language, v2.0</title> | |

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7 | </head> | |

8 | <body> | |

9 | <h1>The Burro Programming Language</h1> | |

10 | ||

11 | <p>Version 2.0 | |

12 | June 2010, Chris Pressey, Cat's Eye Technologies</p> | |

13 | ||

14 | <h2>Introduction</h2> | |

15 | ||

16 | <p>Burro is a programming language whose programs form an algebraic group under | |

17 | the operation of concatenation and over the equivalence relation of "computes | |

18 | the same function." This means that, for every Burro program, we can | |

19 | construct a corresponding antiprogram that, when appended onto the first | |

20 | program, results in a "no-op" program (a program with no effect — the | |

21 | identity function.)</p> | |

22 | ||

23 | <p>(In fact, for every set of Burro programs that compute the same function, | |

24 | there is a corresponding set of antiprograms, any of which can "cancel out" | |

25 | any program in the first set. From our proof that Burro programs form a | |

26 | group, we obtain a constructive algorithm which, for any given program, will | |

27 | derive only one corresponding antiprogram, a kind of syntactic inverse.)</p> | |

28 | ||

29 | <p>This is a kind of reversible computing, but Burro differs from most reversible | |

30 | languages in that it is not the execution trace that is being "undone", but | |

31 | the program itself that is being annihilated.</p> | |

32 | ||

33 | <p>This document describes version 2.0 of the Burro language, a reformulation | |

34 | which addresses several issues with Burro 1.0. An update to the language was | |

35 | desired by the author after it was pointed out by Alex Smith that the set of | |

36 | Burro version 1.0 programs do not, in fact, form a proper group (the inverse | |

37 | of (/) is {}, but no inverse of {} is defined; also, the implementations | |

38 | (at least) did not support moving the tape head left past the "start" of the | |

39 | tape, so <> was not a well-defined program.)</p> | |

40 | ||

41 | <p>Additionally in this document, we construct a Burro 2.0 program equivalent to | |

42 | a certain Turing machine. While this Turing machine is not universal, the | |

43 | translation method we use demonstrates how it would be possible to map an | |

44 | arbitrary Turing machine to a Burro program, hopefully making uncontroversial | |

45 | the idea that Burro qualifies as universal.</p> | |

46 | ||

47 | <p>For further background information on the Burro project, you may also wish | |

48 | to read the <a href="burro-1.0.html">Burro 1.0 article</a>, with the understanding that | |

49 | the language description given there is obsolete.</p> | |

50 | ||

51 | <h2>Changes from Burro 1.0</h2> | |

52 | ||

53 | <p>The {} construct does not appear in Burro 2.0. Instead, the (/) construct | |

54 | serves as its own inverse. The tree-like structure of decision continuations | |

55 | is not present in Burro 2.0 either. Instead, decision information is kept on | |

56 | a second tape, called the "stack tape".</p> | |

57 | ||

58 | <p>Henceforth in this document, the term Burro refers to Burro 2.0.</p> | |

59 | ||

60 | <h2>About this Document</h2> | |

61 | ||

62 | <p>This document is a reference implementation of Burro in literate Haskell, | |

63 | using Markdown syntax for the textual prose portions (although the version | |

64 | you are reading may have been converted to another format, such as HTML, | |

65 | for presentation.) As such, this document serves as an "executable | |

66 | semantics", both defining the language and providing a ready tool.</p> | |

67 | ||

68 | <pre><code> module Main where | |

69 | import System | |

70 | </code></pre> | |

71 | ||

72 | <h2>Inductive Definition of a Burro Program</h2> | |

73 | ||

74 | <p>The symbol e is a Burro program. <br /> | |

75 | The symbol ! is a Burro program. <br /> | |

76 | The symbol + is a Burro program. <br /> | |

77 | The symbol - is a Burro program. <br /> | |

78 | The symbol < is a Burro program. <br /> | |

79 | The symbol > is a Burro program. <br /> | |

80 | If a and b are Burro programs, then (a/b) is a Burro program. <br /> | |

81 | If a and b are Burro programs, then ab is a Burro program. <br /> | |

82 | Nothing else is a Burro program. </p> | |

83 | ||

84 | <pre><code> data Burro = Null | |

85 | | ToggleHalt | |

86 | | Inc | |

87 | | Dec | |

88 | | GoLeft | |

89 | | GoRight | |

90 | | Test Burro Burro | |

91 | | Seq Burro Burro | |

92 | deriving (Read, Eq) | |

93 | </code></pre> | |

94 | ||

95 | <h2>Representation of Burro Programs</h2> | |

96 | ||

97 | <p>For a concrete representation, the symbols in the inductive definition | |

98 | given above can be taken to be a subset of a character set; for the | |

99 | purposes of this semantics, we will use the ASCII character set. Parsing | |

100 | a given string of symbols into a Burro program is straightforward; all | |

101 | symbols which are not Burro symbols are simply ignored.</p> | |

102 | ||

103 | <pre><code> instance Show Burro where | |

104 | show Null = "e" | |

105 | show ToggleHalt = "!" | |

106 | show Inc = "+" | |

107 | show Dec = "-" | |

108 | show GoLeft = "<" | |

109 | show GoRight = ">" | |

110 | show (Test a b) = "(" ++ (show a) ++ "/" ++ (show b) ++ ")" | |

111 | show (Seq a b) = (show a) ++ (show b) | |

112 | ||

113 | parse string = | |

114 | let | |

115 | (rest, acc) = parseProgram string Null | |

116 | in | |

117 | trim acc | |

118 | ||

119 | parseProgram [] acc = | |

120 | ([], acc) | |

121 | parseProgram ('e':rest) acc = | |

122 | parseProgram rest (Seq acc Null) | |

123 | parseProgram ('+':rest) acc = | |

124 | parseProgram rest (Seq acc Inc) | |

125 | parseProgram ('-':rest) acc = | |

126 | parseProgram rest (Seq acc Dec) | |

127 | parseProgram ('<':rest) acc = | |

128 | parseProgram rest (Seq acc GoLeft) | |

129 | parseProgram ('>':rest) acc = | |

130 | parseProgram rest (Seq acc GoRight) | |

131 | parseProgram ('!':rest) acc = | |

132 | parseProgram rest (Seq acc ToggleHalt) | |

133 | parseProgram ('(':rest) acc = | |

134 | let | |

135 | (rest', thenprog) = parseProgram rest Null | |

136 | (rest'', elseprog) = parseProgram rest' Null | |

137 | test = Test thenprog elseprog | |

138 | in | |

139 | parseProgram rest'' (Seq acc test) | |

140 | parseProgram ('/':rest) acc = | |

141 | (rest, acc) | |

142 | parseProgram (')':rest) acc = | |

143 | (rest, acc) | |

144 | parseProgram (_:rest) acc = | |

145 | parseProgram rest acc | |

146 | ||

147 | trim (Seq Null a) = trim a | |

148 | trim (Seq a Null) = trim a | |

149 | trim (Seq a b) = Seq (trim a) (trim b) | |

150 | trim (Test a b) = Test (trim a) (trim b) | |

151 | trim x = x | |

152 | </code></pre> | |

153 | ||

154 | <h2>Group Properties of Burro Programs</h2> | |

155 | ||

156 | <p>We assert these first, and when we describe the program semantics we will | |

157 | show that the semantics do not violate them.</p> | |

158 | ||

159 | <p>The inverse of e is e: ee = e <br /> | |

160 | The inverse of ! is !: !! = e <br /> | |

161 | The inverse of + is -: +- = e <br /> | |

162 | The inverse of - is +: -+ = e <br /> | |

163 | The inverse of < is >: <> = e <br /> | |

164 | The inverse of > is <: >< = e <br /> | |

165 | If aa' = e and bb' = e, (a/b)(b'/a') = e. <br /> | |

166 | If aa' = e and bb' = e, abb'a' = e. </p> | |

167 | ||

168 | <pre><code> inverse Null = Null | |

169 | inverse ToggleHalt = ToggleHalt | |

170 | inverse Inc = Dec | |

171 | inverse Dec = Inc | |

172 | inverse GoLeft = GoRight | |

173 | inverse GoRight = GoLeft | |

174 | inverse (Test a b) = Test (inverse b) (inverse a) | |

175 | inverse (Seq a b) = Seq (inverse b) (inverse a) | |

176 | </code></pre> | |

177 | ||

178 | <p>For every Burro program x, annihilationOf x is always equivalent | |

179 | computationally to e.</p> | |

180 | ||

181 | <pre><code> annihilationOf x = Seq x (inverse x) | |

182 | </code></pre> | |

183 | ||

184 | <h2>State Model for Burro Programs</h2> | |

185 | ||

186 | <p>Central to the state of a Burro program is an object called a tape. | |

187 | A tape consists of a sequence of cells in a one-dimensional array, | |

188 | unbounded in both directions. Each cell contains an integer of unbounded | |

189 | extent, both positive and negative. The initial value of each cell is | |

190 | zero. One of the cells of the tape is distinguished as the "current cell"; | |

191 | this is the cell that we think of as having the "tape head" hovering over it | |

192 | at the moment.</p> | |

193 | ||

194 | <p>In this semantics, we represent a tape as two lists, which we treat as | |

195 | stacks. The first list contains the cell under the tape head, and | |

196 | everything to the left of the tape head (in the reverse order from how it | |

197 | appears on the tape.) The second list contains everything to the right of | |

198 | the tape head, in the same order as it appears on the tape.</p> | |

199 | ||

200 | <pre><code> data Tape = Tape [Integer] [Integer] | |

201 | deriving (Read) | |

202 | ||

203 | instance Show Tape where | |

204 | show t@(Tape l r) = | |

205 | let | |

206 | (Tape l' r') = strip t | |

207 | in | |

208 | show (reverse l') ++ "<" ++ (show r') | |

209 | </code></pre> | |

210 | ||

211 | <p>When comparing two tapes for equality, we must disregard any zero cells | |

212 | farther to the left/right than the outermost non-zero cells. Specifically, | |

213 | we strip leading/trailing zeroes from tapes before comparison. We don't | |

214 | strip out a zero that a tape head is currently over, however.</p> | |

215 | ||

216 | <p>Also, the current cell must be the same for both tapes (that is, tape heads | |

217 | must be in the same location) for two tapes to be considered equal.</p> | |

218 | ||

219 | <pre><code> stripzeroes list = (reverse (sz (reverse list))) | |

220 | where sz [] = [] | |

221 | sz (0:rest) = sz rest | |

222 | sz x = x | |

223 | ||

224 | ensurecell [] = [0] | |

225 | ensurecell x = x | |

226 | ||

227 | strip (Tape l r) = Tape (ensurecell (stripzeroes l)) (stripzeroes r) | |

228 | ||

229 | tapeeq :: Tape -> Tape -> Bool | |

230 | tapeeq t1 t2 = | |

231 | let | |

232 | (Tape t1l t1r) = strip t1 | |

233 | (Tape t2l t2r) = strip t2 | |

234 | in | |

235 | (t1l == t2l) && (t1r == t2r) | |

236 | ||

237 | instance Eq Tape where | |

238 | t1 == t2 = tapeeq t1 t2 | |

239 | </code></pre> | |

240 | ||

241 | <p>A convenience function for creating an inital tape is also provided.</p> | |

242 | ||

243 | <pre><code> tape :: [Integer] -> Tape | |

244 | tape x = Tape [head x] (tail x) | |

245 | </code></pre> | |

246 | ||

247 | <p>We now define some operations on tapes that we will use in the semantics. | |

248 | First, operations on tapes that alter or access the cell under the tape head.</p> | |

249 | ||

250 | <pre><code> inc (Tape (cell:left) right) = Tape (cell + 1 : left) right | |

251 | dec (Tape (cell:left) right) = Tape (cell - 1 : left) right | |

252 | get (Tape (cell:left) right) = cell | |

253 | set (Tape (_:left) right) value = Tape (value : left) right | |

254 | </code></pre> | |

255 | ||

256 | <p>Next, operations on tapes that move the tape head.</p> | |

257 | ||

258 | <pre><code> left (Tape (cell:[]) right) = Tape [0] (cell:right) | |

259 | left (Tape (cell:left) right) = Tape left (cell:right) | |

260 | right (Tape left []) = Tape (0:left) [] | |

261 | right (Tape left (cell:right)) = Tape (cell:left) right | |

262 | </code></pre> | |

263 | ||

264 | <p>Finally, an operation on two tapes that swaps the current cell between | |

265 | them.</p> | |

266 | ||

267 | <pre><code> swap t1 t2 = (set t1 (get t2), set t2 (get t1)) | |

268 | </code></pre> | |

269 | ||

270 | <p>A program state consists of:</p> | |

271 | ||

272 | <ul> | |

273 | <li>A "data tape";</li> | |

274 | <li>A "stack tape"; and</li> | |

275 | <li>A flag called the "halt flag", which may be 0 or 1.</li> | |

276 | </ul> | |

277 | ||

278 | <p>The 0 and 1 are represented by False and True boolean values in this | |

279 | semantics.</p> | |

280 | ||

281 | <pre><code> data State = State Tape Tape Bool | |

282 | deriving (Show, Read, Eq) | |

283 | ||

284 | newstate = State (tape [0]) (tape [0]) True | |

285 | </code></pre> | |

286 | ||

287 | <h2>Semantics of Burro Programs</h2> | |

288 | ||

289 | <p>Each instruction is defined as a function from program state to program | |

290 | state. Concatenation of instructions is defined as composition of | |

291 | functions, like so:</p> | |

292 | ||

293 | <p>If ab is a Burro program, and a maps state S to state S', and b maps | |

294 | state S' to S'', then ab maps state S to state S''.</p> | |

295 | ||

296 | <pre><code> exec (Seq a b) t = exec b (exec a t) | |

297 | </code></pre> | |

298 | ||

299 | <p>The e instruction is the identity function on states.</p> | |

300 | ||

301 | <pre><code> exec Null s = s | |

302 | </code></pre> | |

303 | ||

304 | <p>The ! instruction toggles the halt flag. If it is 0 in the input state, it | |

305 | is 1 in the output state, and vice versa.</p> | |

306 | ||

307 | <pre><code> exec ToggleHalt (State dat stack halt) = (State dat stack (not halt)) | |

308 | </code></pre> | |

309 | ||

310 | <p>The + instruction increments the current data cell, while - decrements the | |

311 | current data cell.</p> | |

312 | ||

313 | <pre><code> exec Inc (State dat stack halt) = (State (inc dat) stack halt) | |

314 | exec Dec (State dat stack halt) = (State (dec dat) stack halt) | |

315 | </code></pre> | |

316 | ||

317 | <p>The instruction < makes the cell to the left of the current data cell, the | |

318 | new current data cell. The instruction > makes the cell to the right of the | |

319 | current data cell, the new current data cell.</p> | |

320 | ||

321 | <pre><code> exec GoLeft (State dat stack halt) = (State (left dat) stack halt) | |

322 | exec GoRight (State dat stack halt) = (State (right dat) stack halt) | |

323 | </code></pre> | |

324 | ||

325 | <p>(a/b) is the conditional construct, which is quite special.</p> | |

326 | ||

327 | <p>First, the current data cell is remembered for the duration of the execution | |

328 | of this construct — let's call it x.</p> | |

329 | ||

330 | <p>Second, the current data cell and the current stack cell are swapped.</p> | |

331 | ||

332 | <p>Third, the current stack cell is negated.</p> | |

333 | ||

334 | <p>Fourth, the stack cell to the right of the current stack cell is made | |

335 | the new current stack cell.</p> | |

336 | ||

337 | <p>Fifth, if x is positive, a is evaluated; if x is negative, b is evaluated; | |

338 | otherwise x = 0 and neither is evaluated. Evaluation occurs in the state | |

339 | established by the preceding four steps.</p> | |

340 | ||

341 | <p>Sixth, the stack cell to the left of the current stack cell is made | |

342 | the new current stack cell.</p> | |

343 | ||

344 | <p>Seventh, the current data cell and the current stack cell are swapped again.</p> | |

345 | ||

346 | <pre><code> exec (Test thn els) (State dat stack halt) = | |

347 | let | |

348 | x = get dat | |

349 | (dat', stack') = swap dat stack | |

350 | stack'' = right (set stack' (0 - (get stack'))) | |

351 | f = if x > 0 then thn else if x < 0 then els else Null | |

352 | (State dat''' stack''' halt') = exec f (State dat' stack'' halt) | |

353 | (dat'''', stack'''') = swap dat''' (left stack''') | |

354 | in | |

355 | (State dat'''' stack'''' halt') | |

356 | </code></pre> | |

357 | ||

358 | <p>We observe an invariant here: because only the (a/b) construct affects the | |

359 | stack tape, and because it does so in a monotonic way — that is, both a | |

360 | and b inside (a/b) have access only to the portion of the stack tape to the | |

361 | right of what (a/b) has access to — the current stack cell in step seven | |

362 | always holds the same value as the current stack cell in step two, except | |

363 | negated.</p> | |

364 | ||

365 | <h2>Repetition</h2> | |

366 | ||

367 | <p>The repetition model of Burro 2.0 is identical to that of Burro 1.0. | |

368 | The program text is executed, resulting in a final state, S. If, in | |

369 | S, the halt flag is 1, execution terminates with state S. On the other | |

370 | hand, if the halt flag is 0, the program text is executed once more, | |

371 | this time on state S, and the whole procedure repeats. Initially the | |

372 | halt flag is 1, so if ! is never executed, the program never repeats.</p> | |

373 | ||

374 | <p>Additionally, each time the program repeats, the stack tape is cleared.</p> | |

375 | ||

376 | <pre><code> run program state = | |

377 | let | |

378 | state'@(State dat' stack' halt') = exec program state | |

379 | in | |

380 | if | |

381 | not halt' | |

382 | then | |

383 | run program (State dat' (tape [0]) True) | |

384 | else | |

385 | state' | |

386 | </code></pre> | |

387 | ||

388 | <h2>Central theorem of Burro</h2> | |

389 | ||

390 | <p>We now have established enough definitions to give a proof of the central | |

391 | theorem of Burro, which is:</p> | |

392 | ||

393 | <p><em>Theorem: The set of all Burro programs forms a group over computational | |

394 | equivalence under the operation of concatenation.</em></p> | |

395 | ||

396 | <p>As covered in the Burro 1.0 article, a "group over an equivalence relation" | |

397 | captures the notion of replacing the concept of equality in the group | |

398 | axioms with the concept of equivalency. Our particular equivalence relation | |

399 | here is that two programs are equivalent if they compute the same function.</p> | |

400 | ||

401 | <p>In order to show that a set G is a group, it is sufficient to show the | |

402 | following four properties hold:</p> | |

403 | ||

404 | <ol> | |

405 | <li><p>Closure: For all a, b in G, ab is also in G.</p> | |

406 | ||

407 | <p>This follows from the inductive definition of Burro programs.</p></li> | |

408 | <li><p>Associativity: For all a, b and c in G, (ab)c ≡ a(bc).</p> | |

409 | ||

410 | <p>This follows from the definition of concatenation (sequential composition); | |

411 | it doesn't matter if we concatenate a with b first, then concatenate that | |

412 | with c, or if we concatenate b with c first, then concatenate a with that. | |

413 | Either way the result is the same string (or in this case, the same Burro | |

414 | program.)</p></li> | |

415 | <li><p>Identity element: There exists an element e in G, such that for every | |

416 | element a in G, ea ≡ ae ≡ a.</p> | |

417 | ||

418 | <p>The instruction e in Burro has no effect on the program state. Therefore | |

419 | concatenating it to any existing program, or concatenating any existing | |

420 | program to it, results in a computationally equivalent program.</p></li> | |

421 | <li><p>Inverse element: For each a in G, there exists an element b in G such | |

422 | that ab ≡ ba ≡ e.</p> | |

423 | ||

424 | <p>This is the key property. We first show that it holds for each element of | |

425 | the inductive definition of Burro programs. We can then conclude, through | |

426 | structural induction, that all Burro programs have this property.</p> | |

427 | ||

428 | <ol> | |

429 | <li><p>Since e is the identity function on states, e is trivially its own | |

430 | inverse.</p></li> | |

431 | <li><p>Since toggling the flag twice is the same as not changing it at all, | |

432 | the inverse of ! is !.</p></li> | |

433 | <li><p>By the definitions of incrementation and decrementation, and because | |

434 | data cells cannot overflow, the inverse of + is -, and the inverse | |

435 | of - is +.</p></li> | |

436 | <li><p>By the definitions of left and right, and because the data tape is | |

437 | unbounded (never reaches an end,) the inverse of > is <, and the inverse | |

438 | of < is >.</p></li> | |

439 | <li><p>The inverse of ab is b'a' where b' is the inverse of b and a' is the | |

440 | inverse of a. This is because abb'a' ≡ aea' ≡ aa' ≡ e.</p></li> | |

441 | <li><p>The inverse of (a/b) is (b'/a'). (This is the key case of the key | |

442 | property.) Going back to the definition of (/), we see there are three | |

443 | sub-cases to consider. Before execution of (a/b)(b'/a'), the data tape | |

444 | may be in one of three possible states:</p> | |

445 | ||

446 | <ol> | |

447 | <li><p>The current data cell is zero. So in (a/b), x is 0, which goes on | |

448 | the stack and the current data cell becomes whatever was on the | |

449 | stack (call it k.) The 0 on the stack is negated, thus stays 0 | |

450 | (because 0 - 0 = 0). The stack head is moved right. Neither a nor | |

451 | b is evaluated. The stack head is moved back left. The stack and | |

452 | data cells are swapped again, so 0 is back in the current data cell | |

453 | and k is back in the current stack cell. This is the same as the | |

454 | initial configuration, so (a/b) is equivalent to e. By the same | |

455 | reasoning, (b'/a') is equivalent to e, and (a/b)(b'/a') ≡ ee ≡ e.</p></li> | |

456 | <li><p>The current data cell is positive (x > 0). We first evaluate (a/b). | |

457 | The data and stack cells are swapped: the current data cell becomes | |

458 | k, and the current stack cell becomes x > 0. The current stack cell | |

459 | is negated, so becomes -x < 0. The stack head is moved to the right.</p> | |

460 | ||

461 | <p>Because x > 0, the first of the sub-programs, a, is now evaluated. | |

462 | The current data cell could be anything — call it k'.</p> | |

463 | ||

464 | <p>The stack head is moved back to the left, so that the current stack | |

465 | cell is once again -x < 0, and it is swapped with the current data | |

466 | cell, making it -x and making the current stack cell k'.</p> | |

467 | ||

468 | <p>We are now to evaluate (b'/a'). This time, we know the current data | |

469 | cell is negative (-x < 0). The data and stack cells are swapped: | |

470 | the current data cell becomes k', and the current stack cell becomes | |

471 | -x < 0. The current stack cell is negated, so becomes x > 0. The | |

472 | stack head is moved to the right.</p> | |

473 | ||

474 | <p>Because -x < 0, the second of the sub-programs, a', is now evaluated. | |

475 | Because a' is the inverse of a, and it is being applied to a state | |

476 | that is the result of executing a, it will reverse this state to | |

477 | what it was before a was executed (inside (a/b).) This means the | |

478 | current data cell will become k once more.</p> | |

479 | ||

480 | <p>The stack head is moved back to the left, so that the current stack | |

481 | cell is once again x > 0, and it is swapped with the current data | |

482 | cell, making it x and making the current stack cell k. This is | |

483 | the state we started from, so (a/b)(b'/a') ≡ e.</p></li> | |

484 | <li><p>Case 3 is an exact mirror image of case 2 — the only difference | |

485 | is that the first time through, x < 0 and b is evaluated, thus the | |

486 | second time through, -x > 0 and b' is evaluated. Therefore | |

487 | (a/b)(b'/a') ≡ e in this instance as well.</p></li> | |

488 | </ol></li> | |

489 | </ol></li> | |

490 | </ol> | |

491 | ||

492 | <p>QED.</p> | |

493 | ||

494 | <h2>Driver and Unit Tests</h2> | |

495 | ||

496 | <p>We define a few more convenience functions to cater for the execution | |

497 | of Burro programs on an initial state.</p> | |

498 | ||

499 | <pre><code> interpret text = run (parse text) newstate | |

500 | ||

501 | main = do | |

502 | [fileName] <- getArgs | |

503 | burroText <- readFile fileName | |

504 | putStrLn $ show $ interpret burroText | |

505 | </code></pre> | |

506 | ||

507 | <p>Although we have proved that Burro programs form a group, it is not a | |

508 | mechanized proof, and only goes so far in helping us tell if the | |

509 | implementation (which, for an executable semantics, is one and the same | |

510 | as the formal semantic formulation) is correct. Unit tests can't tell us | |

511 | definitively that there are no errors in the formulation, but they can | |

512 | help us catch a class of errors, if there is one present.</p> | |

513 | ||

514 | <p>For the first set of test cases, we give a set of pairs of Burro programs. | |

515 | In each of these pairs, both programs should be equivalent in the sense of | |

516 | evaluating to the same tape given an initial blank tape.</p> | |

517 | ||

518 | <p>For the second set, we simply give a list of Burro programs. We test | |

519 | each one by applying the annihilationOf function to it and checking that | |

520 | the result of executing it on a blank tape is equivalent to e.</p> | |

521 | ||

522 | <pre><code> testCases = [ | |

523 | ("+++", "-++-++-++"), | |

524 | ("+(>+++</---)", "->+++<"), | |

525 | ("-(+++/>---<)", "+>---<"), | |

526 | ("(!/!)", "e"), | |

527 | ("+(--------!/e)", "+(/)+"), | |

528 | ("+++(/)", "---"), | |

529 | ("---(/)", "+++"), | |

530 | ("+> +++ --(--(--(/>>>>>+)+/>>>+)+/>+)+", | |

531 | "+> >>> +(---(/+)/)+") | |

532 | ] | |

533 | ||

534 | annihilationTests = [ | |

535 | "e", "+", "-", "<", ">", "!", | |

536 | "++", "--", "<+<-", "-->>--", | |

537 | "(+/-)", "+(+/-)", "-(+/-)", | |

538 | "+(--------!/e)" | |

539 | ] | |

540 | ||

541 | allTestCases = testCases ++ map nihil annihilationTests | |

542 | where | |

543 | nihil x = ((show (annihilationOf (parse x))), "e") | |

544 | </code></pre> | |

545 | ||

546 | <p>Our unit test harness evaluates to a list of tests which did | |

547 | not pass. If all went well, it will evaluate to the empty list.</p> | |

548 | ||

549 | <pre><code> test [] = | |

550 | [] | |

551 | test ((a, b):cases) = | |

552 | let | |

553 | resultA = interpret a | |

554 | resultB = interpret b | |

555 | in | |

556 | if | |

557 | resultA == resultB | |

558 | then | |

559 | test cases | |

560 | else | |

561 | ((a, b):(test cases)) | |

562 | </code></pre> | |

563 | ||

564 | <p>Finally, some miscellaneous functions for helping analyze why the | |

565 | Burro tests you've written aren't working :)</p> | |

566 | ||

567 | <pre><code> debug (a, b) = ((a, interpret a), (b, interpret b)) | |

568 | ||

569 | debugTests = map debug (test allTestCases) | |

570 | </code></pre> | |

571 | ||

572 | <h2>Implementing a Turing Machine in Burro</h2> | |

573 | ||

574 | <p>First we note a Burro idiom. Assume the current data cell (which | |

575 | we'll call C) contains an odd positive integer (which we'll call x.) | |

576 | We can test if x = 1, write a zero into C, write -x into the cell | |

577 | to the right of C, with the following construct:</p> | |

578 | ||

579 | <pre><code>--(F>/T>)< | |

580 | </code></pre> | |

581 | ||

582 | <p>T if executed if x = 1 and F is executed otherwise. (Remember, | |

583 | we're assuming x is odd and positive.) To make the idiom hold, we | |

584 | also insist that T and F both leave the data tape head in the | |

585 | position they found it. If you are wondering where the zero came | |

586 | from — it came from the stack.</p> | |

587 | ||

588 | <p>We now note that this idiom can be nested to detect larger odd | |

589 | numbers. For example, to determine if x is 1 or 3 or 5:</p> | |

590 | ||

591 | <pre><code>--(--(--(F>/T5>)<>/T3>)<>/T1>)< | |

592 | </code></pre> | |

593 | ||

594 | <p>We can of course optimize that a bit:</p> | |

595 | ||

596 | <pre><code>--(--(--(F>/T5>)/T3>)/T1>)< | |

597 | </code></pre> | |

598 | ||

599 | <p>Our basic strategy is to encode the state of the Turing machine's finite | |

600 | control as a positive odd integer, which we will call a "finite control | |

601 | index." We make sure to always keep the current finite control index | |

602 | available in the current data cell at the start of each loop, and we | |

603 | dispatch on its contents using the above idiom to simulate the finite | |

604 | control. We may use odd integers to encode the symbols on the Turing | |

605 | Machine's tape as well.</p> | |

606 | ||

607 | <p>We map the Turing machine state onto the Burro data tape in an interleaved | |

608 | fashion, where three adjacent cells are used to represent one TM tape | |

609 | cell. The first (leftmost) cell in the triple contains the finite control | |

610 | index described above. The second cell is a "junk cell" where we can write | |

611 | stuff and never care about it again. The third cell contains our | |

612 | representation of the contents of the TM tape cell. Moving the TM tape | |

613 | head one cell is simulated by moving the Burro data tape head three cells, | |

614 | skipping over the interim finite control cell and junk cell.</p> | |

615 | ||

616 | <p>As we always want to be on an up-to-date finite control cell at the start | |

617 | of each iteration, we must make sure to copy it to the new tape cell triple | |

618 | each time we move the TM tape head to a new position. If we are | |

619 | transitioning to a different TM state as well as moving the TM tape head, | |

620 | we can even just write in the new state at the new finite control cell. | |

621 | None of this copying requires intermediate loops, as these value are all | |

622 | fixed constants. The only subtlety is that we must "erase" any finite | |

623 | control cell we move off of (set it back to zero) so that we can get it to | |

624 | the desired value by incrementation and decrementation only. The idiom | |

625 | given above supplies that functionality for us.</p> | |

626 | ||

627 | <p>Note that the junk cell is used to store the end result of (/), which | |

628 | we don't care to predict, and don't care to use again. Note also that | |

629 | the junk cell in which the value is stored belongs to the triple of the | |

630 | destination TM tape cell, the one to which the TM tape head is moving | |

631 | on this transition.</p> | |

632 | ||

633 | <p>A concrete example follows. We consider the TM tape to be over an | |

634 | alphabet of two symbols, 1 and 3 (or in fact any odd integer greater | |

635 | than 1), and the finite control to have three states, denoted 1, 3, | |

636 | and 5. In addition, state 7 (or in fact any odd integer greater than 5) | |

637 | is a halt state.</p> | |

638 | ||

639 | <p>In state 1, <br /> | |

640 | - If the symbol is 1, enter state 3; <br /> | |

641 | - If the symbol is 3, move head right one square, and remain in state 1. </p> | |

642 | ||

643 | <pre><code>>>--(+++>+>/+<<+++>)< | |

644 | </code></pre> | |

645 | ||

646 | <p>In state 3, <br /> | |

647 | - If the symbol is 1, write 3, move head left one square, and remain in | |

648 | state 3; <br /> | |

649 | - If the symbol is 3, move head right one square, and enter state 5. </p> | |

650 | ||

651 | <pre><code>>>--(+++>+++++>/+++<<<<<+++>)< | |

652 | </code></pre> | |

653 | ||

654 | <p>In state 5, <br /> | |

655 | - If the symbol is 1, write 3, move head right one square, and remain in | |

656 | state 5; <br /> | |

657 | - If the symbol is 3, write 1 and enter state 7. </p> | |

658 | ||

659 | <pre><code>>>--(+<<+++++++>/+++>+++++>)< | |

660 | </code></pre> | |

661 | ||

662 | <p>Putting it all together, including toggling the halt flag so that, unless | |

663 | we reach state 7 or higher, we loop through this sequence indefinitely:</p> | |

664 | ||

665 | <pre><code>!--(--(--(!>/ | |

666 | >>--(+<<+++++++>/+++>+++++>)< | |

667 | >)/ | |

668 | >>--(+++>+++++>/+++<<<<<+++>)< | |

669 | >)/ | |

670 | >>--(+++>+>/+<<+++>)< | |

671 | >)< | |

672 | </code></pre> | |

673 | ||

674 | <p>It is not a very interesting Turing machine, but by following this | |

675 | construction, it should be apparent how any arbitrary Turing machine | |

676 | could be mapped to a Burro program in the same way.</p> | |

677 | ||

678 | <p>Happy annihilating (for reals this time)!</p> | |

679 | ||

680 | <p>-Chris Pressey <br /> | |

681 | Cat's Eye Technologies <br /> | |

682 | June 7, 2010 <br /> | |

683 | Evanston, Illinois, USA</p> | |

684 | </body> | |

685 | </html> |

0 | #!/usr/bin/perl -w | |

1 | ||

2 | $title = "The Burro Programming Language, v2.0"; | |

3 | $lhs_input = "../src/Burro.lhs"; | |

4 | $lhs_temp = "lhs.tmp"; | |

5 | $html_temp = "html.tmp"; | |

6 | $html_output = "burro.html"; | |

7 | ||

8 | open INFILE, "<$lhs_input"; | |

9 | open OUTFILE, ">$lhs_temp"; | |

10 | ||

11 | while (defined ($line = <INFILE>)) { | |

12 | chomp $line; | |

13 | if ($line =~ /^\>(.*?)coding\:(.*?)$/) { | |

14 | # pass | |

15 | } elsif ($line =~ /^\>(.*?)$/) { | |

16 | print OUTFILE " $1\n"; | |

17 | } else { | |

18 | print OUTFILE "$line\n"; | |

19 | } | |

20 | } | |

21 | ||

22 | close INFILE; | |

23 | close OUTFILE; | |

24 | ||

25 | system "Markdown.pl <$lhs_temp >$html_temp"; | |

26 | ||

27 | open INFILE, "<$html_temp"; | |

28 | open OUTFILE, ">$html_output"; | |

29 | ||

30 | print OUTFILE <<"EOH"; | |

31 | <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> | |

32 | <!-- encoding: UTF-8 --> | |

33 | <html xmlns="http://www.w3.org/1999/xhtml" lang="en"> | |

34 | <head> | |

35 | <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> | |

36 | <title>$title</title> | |

37 | <link rel="stylesheet" type="text/css" href="markdown-lhs.css" /> | |

38 | </head> | |

39 | <body> | |

40 | EOH | |

41 | ||

42 | while (defined ($line = <INFILE>)) { | |

43 | print OUTFILE $line; | |

44 | } | |

45 | ||

46 | print OUTFILE <<"EOF"; | |

47 | </body> | |

48 | </html> | |

49 | EOF | |

50 | ||

51 | close INFILE; | |

52 | close OUTFILE; | |

53 | ||

54 | system "rm *.tmp"; |

0 | blockquote { | |

1 | font-family: monospace; | |

2 | white-space: pre; | |

3 | border-style: solid; | |

4 | border-width: 1px; | |

5 | border-color: black; | |

6 | padding-left: 3em; | |

7 | } | |

8 | ||

9 | pre { | |

10 | border-style: solid; | |

11 | border-width: 1px; | |

12 | border-color: black; | |

13 | padding-left: 3em; | |

14 | padding-top: 1em; | |

15 | padding-bottom: 1em; | |

16 | } |