1 // Copyright 2009 The Go Authors. All rights reserved.
2 // Use of this source code is governed by a BSD-style
3 // license that can be found in the LICENSE file.
4 5 /*
6 7 Multi-precision division. Here be dragons.
8 9 Given u and v, where u is n+m digits, and v is n digits (with no leading zeros),
10 the goal is to return quo, rem such that u = quo*v + rem, where 0 ≤ rem < v.
11 That is, quo = ⌊u/v⌋ where ⌊x⌋ denotes the floor (truncation to integer) of x,
12 and rem = u - quo·v.
13 14 15 Long Division
16 17 Division in a computer proceeds the same as long division in elementary school,
18 but computers are not as good as schoolchildren at following vague directions,
19 so we have to be much more precise about the actual steps and what can happen.
20 21 We work from most to least significant digit of the quotient, doing:
22 23 • Guess a digit q, the number of v to subtract from the current
24 section of u to zero out the topmost digit.
25 • Check the guess by multiplying q·v and comparing it against
26 the current section of u, adjusting the guess as needed.
27 • Subtract q·v from the current section of u.
28 • Add q to the corresponding section of the result quo.
29 30 When all digits have been processed, the final remainder is left in u
31 and returned as rem.
32 33 For example, here is a sketch of dividing 5 digits by 3 digits (n=3, m=2).
34 35 q₂ q₁ q₀
36 _________________
37 v₂ v₁ v₀ ) u₄ u₃ u₂ u₁ u₀
38 ↓ ↓ ↓ | |
39 [u₄ u₃ u₂]| |
40 - [ q₂·v ]| |
41 ----------- ↓ |
42 [ rem | u₁]|
43 - [ q₁·v ]|
44 ----------- ↓
45 [ rem | u₀]
46 - [ q₀·v ]
47 ------------
48 [ rem ]
49 50 Instead of creating new storage for the remainders and copying digits from u
51 as indicated by the arrows, we use u's storage directly as both the source
52 and destination of the subtractions, so that the remainders overwrite
53 successive overlapping sections of u as the division proceeds, using a slice
54 of u to identify the current section. This avoids all the copying as well as
55 shifting of remainders.
56 57 Division of u with n+m digits by v with n digits (in base B) can in general
58 produce at most m+1 digits, because:
59 60 • u < B^(n+m) [B^(n+m) has n+m+1 digits]
61 • v ≥ B^(n-1) [B^(n-1) is the smallest n-digit number]
62 • u/v < B^(n+m) / B^(n-1) [divide bounds for u, v]
63 • u/v < B^(m+1) [simplify]
64 65 The first step is special: it takes the top n digits of u and divides them by
66 the n digits of v, producing the first quotient digit and an n-digit remainder.
67 In the example, q₂ = ⌊u₄u₃u₂ / v⌋.
68 69 The first step divides n digits by n digits to ensure that it produces only a
70 single digit.
71 72 Each subsequent step appends the next digit from u to the remainder and divides
73 those n+1 digits by the n digits of v, producing another quotient digit and a
74 new n-digit remainder.
75 76 Subsequent steps divide n+1 digits by n digits, an operation that in general
77 might produce two digits. However, as used in the algorithm, that division is
78 guaranteed to produce only a single digit. The dividend is of the form
79 rem·B + d, where rem is a remainder from the previous step and d is a single
80 digit, so:
81 82 • rem ≤ v - 1 [rem is a remainder from dividing by v]
83 • rem·B ≤ v·B - B [multiply by B]
84 • d ≤ B - 1 [d is a single digit]
85 • rem·B + d ≤ v·B - 1 [add]
86 • rem·B + d < v·B [change ≤ to <]
87 • (rem·B + d)/v < B [divide by v]
88 89 90 Guess and Check
91 92 At each step we need to divide n+1 digits by n digits, but this is for the
93 implementation of division by n digits, so we can't just invoke a division
94 routine: we _are_ the division routine. Instead, we guess at the answer and
95 then check it using multiplication. If the guess is wrong, we correct it.
96 97 How can this guessing possibly be efficient? It turns out that the following
98 statement (let's call it the Good Guess Guarantee) is true.
99 100 If
101 102 • q = ⌊u/v⌋ where u is n+1 digits and v is n digits,
103 • q < B, and
104 • the topmost digit of v = vₙ₋₁ ≥ B/2,
105 106 then q̂ = ⌊uₙuₙ₋₁ / vₙ₋₁⌋ satisfies q ≤ q̂ ≤ q+2. (Proof below.)
107 108 That is, if we know the answer has only a single digit and we guess an answer
109 by ignoring the bottom n-1 digits of u and v, using a 2-by-1-digit division,
110 then that guess is at least as large as the correct answer. It is also not
111 too much larger: it is off by at most two from the correct answer.
112 113 Note that in the first step of the overall division, which is an n-by-n-digit
114 division, the 2-by-1 guess uses an implicit uₙ = 0.
115 116 Note that using a 2-by-1-digit division here does not mean calling ourselves
117 recursively. Instead, we use an efficient direct hardware implementation of
118 that operation.
119 120 Note that because q is u/v rounded down, q·v must not exceed u: u ≥ q·v.
121 If a guess q̂ is too big, it will not satisfy this test. Viewed a different way,
122 the remainder r̂ for a given q̂ is u - q̂·v, which must be positive. If it is
123 negative, then the guess q̂ is too big.
124 125 This gives us a way to compute q. First compute q̂ with 2-by-1-digit division.
126 Then, while u < q̂·v, decrement q̂; this loop executes at most twice, because
127 q̂ ≤ q+2.
128 129 130 Scaling Inputs
131 132 The Good Guess Guarantee requires that the top digit of v (vₙ₋₁) be at least B/2.
133 For example in base 10, ⌊172/19⌋ = 9, but ⌊18/1⌋ = 18: the guess is wildly off
134 because the first digit 1 is smaller than B/2 = 5.
135 136 We can ensure that v has a large top digit by multiplying both u and v by the
137 right amount. Continuing the example, if we multiply both 172 and 19 by 3, we
138 now have ⌊516/57⌋, the leading digit of v is now ≥ 5, and sure enough
139 ⌊51/5⌋ = 10 is much closer to the correct answer 9. It would be easier here
140 to multiply by 4, because that can be done with a shift. Specifically, we can
141 always count the number of leading zeros i in the first digit of v and then
142 shift both u and v left by i bits.
143 144 Having scaled u and v, the value ⌊u/v⌋ is unchanged, but the remainder will
145 be scaled: 172 mod 19 is 1, but 516 mod 57 is 3. We have to divide the remainder
146 by the scaling factor (shifting right i bits) when we finish.
147 148 Note that these shifts happen before and after the entire division algorithm,
149 not at each step in the per-digit iteration.
150 151 Note the effect of scaling inputs on the size of the possible quotient.
152 In the scaled u/v, u can gain a digit from scaling; v never does, because we
153 pick the scaling factor to make v's top digit larger but without overflowing.
154 If u and v have n+m and n digits after scaling, then:
155 156 • u < B^(n+m) [B^(n+m) has n+m+1 digits]
157 • v ≥ B^n / 2 [vₙ₋₁ ≥ B/2, so vₙ₋₁·B^(n-1) ≥ B^n/2]
158 • u/v < B^(n+m) / (B^n / 2) [divide bounds for u, v]
159 • u/v < 2 B^m [simplify]
160 161 The quotient can still have m+1 significant digits, but if so the top digit
162 must be a 1. This provides a different way to handle the first digit of the
163 result: compare the top n digits of u against v and fill in either a 0 or a 1.
164 165 166 Refining Guesses
167 168 Before we check whether u < q̂·v, we can adjust our guess to change it from
169 q̂ = ⌊uₙuₙ₋₁ / vₙ₋₁⌋ into the refined guess ⌊uₙuₙ₋₁uₙ₋₂ / vₙ₋₁vₙ₋₂⌋.
170 Although not mentioned above, the Good Guess Guarantee also promises that this
171 3-by-2-digit division guess is more precise and at most one away from the real
172 answer q. The improvement from the 2-by-1 to the 3-by-2 guess can also be done
173 without n-digit math.
174 175 If we have a guess q̂ = ⌊uₙuₙ₋₁ / vₙ₋₁⌋ and we want to see if it also equal to
176 ⌊uₙuₙ₋₁uₙ₋₂ / vₙ₋₁vₙ₋₂⌋, we can use the same check we would for the full division:
177 if uₙuₙ₋₁uₙ₋₂ < q̂·vₙ₋₁vₙ₋₂, then the guess is too large and should be reduced.
178 179 Checking uₙuₙ₋₁uₙ₋₂ < q̂·vₙ₋₁vₙ₋₂ is the same as uₙuₙ₋₁uₙ₋₂ - q̂·vₙ₋₁vₙ₋₂ < 0,
180 and
181 182 uₙuₙ₋₁uₙ₋₂ - q̂·vₙ₋₁vₙ₋₂ = (uₙuₙ₋₁·B + uₙ₋₂) - q̂·(vₙ₋₁·B + vₙ₋₂)
183 [splitting off the bottom digit]
184 = (uₙuₙ₋₁ - q̂·vₙ₋₁)·B + uₙ₋₂ - q̂·vₙ₋₂
185 [regrouping]
186 187 The expression (uₙuₙ₋₁ - q̂·vₙ₋₁) is the remainder of uₙuₙ₋₁ / vₙ₋₁.
188 If the initial guess returns both q̂ and its remainder r̂, then checking
189 whether uₙuₙ₋₁uₙ₋₂ < q̂·vₙ₋₁vₙ₋₂ is the same as checking r̂·B + uₙ₋₂ < q̂·vₙ₋₂.
190 191 If we find that r̂·B + uₙ₋₂ < q̂·vₙ₋₂, then we can adjust the guess by
192 decrementing q̂ and adding vₙ₋₁ to r̂. We repeat until r̂·B + uₙ₋₂ ≥ q̂·vₙ₋₂.
193 (As before, this fixup is only needed at most twice.)
194 195 Now that q̂ = ⌊uₙuₙ₋₁uₙ₋₂ / vₙ₋₁vₙ₋₂⌋, as mentioned above it is at most one
196 away from the correct q, and we've avoided doing any n-digit math.
197 (If we need the new remainder, it can be computed as r̂·B + uₙ₋₂ - q̂·vₙ₋₂.)
198 199 The final check u < q̂·v and the possible fixup must be done at full precision.
200 For random inputs, a fixup at this step is exceedingly rare: the 3-by-2 guess
201 is not often wrong at all. But still we must do the check. Note that since the
202 3-by-2 guess is off by at most 1, it can be convenient to perform the final
203 u < q̂·v as part of the computation of the remainder r = u - q̂·v. If the
204 subtraction underflows, decremeting q̂ and adding one v back to r is enough to
205 arrive at the final q, r.
206 207 That's the entirety of long division: scale the inputs, and then loop over
208 each output position, guessing, checking, and correcting the next output digit.
209 210 For a 2n-digit number divided by an n-digit number (the worst size-n case for
211 division complexity), this algorithm uses n+1 iterations, each of which must do
212 at least the 1-by-n-digit multiplication q̂·v. That's O(n) iterations of
213 O(n) time each, so O(n²) time overall.
214 215 216 Recursive Division
217 218 For very large inputs, it is possible to improve on the O(n²) algorithm.
219 Let's call a group of n/2 real digits a (very) “wide digit”. We can run the
220 standard long division algorithm explained above over the wide digits instead of
221 the actual digits. This will result in many fewer steps, but the math involved in
222 each step is more work.
223 224 Where basic long division uses a 2-by-1-digit division to guess the initial q̂,
225 the new algorithm must use a 2-by-1-wide-digit division, which is of course
226 really an n-by-n/2-digit division. That's OK: if we implement n-digit division
227 in terms of n/2-digit division, the recursion will terminate when the divisor
228 becomes small enough to handle with standard long division or even with the
229 2-by-1 hardware instruction.
230 231 For example, here is a sketch of dividing 10 digits by 4, proceeding with
232 wide digits corresponding to two regular digits. The first step, still special,
233 must leave off a (regular) digit, dividing 5 by 4 and producing a 4-digit
234 remainder less than v. The middle steps divide 6 digits by 4, guaranteed to
235 produce two output digits each (one wide digit) with 4-digit remainders.
236 The final step must use what it has: the 4-digit remainder plus one more,
237 5 digits to divide by 4.
238 239 q₆ q₅ q₄ q₃ q₂ q₁ q₀
240 _______________________________
241 v₃ v₂ v₁ v₀ ) u₉ u₈ u₇ u₆ u₅ u₄ u₃ u₂ u₁ u₀
242 ↓ ↓ ↓ ↓ ↓ | | | | |
243 [u₉ u₈ u₇ u₆ u₅]| | | | |
244 - [ q₆q₅·v ]| | | | |
245 ----------------- ↓ ↓ | | |
246 [ rem |u₄ u₃]| | |
247 - [ q₄q₃·v ]| | |
248 -------------------- ↓ ↓ |
249 [ rem |u₂ u₁]|
250 - [ q₂q₁·v ]|
251 -------------------- ↓
252 [ rem |u₀]
253 - [ q₀·v ]
254 ------------------
255 [ rem ]
256 257 An alternative would be to look ahead to how well n/2 divides into n+m and
258 adjust the first step to use fewer digits as needed, making the first step
259 more special to make the last step not special at all. For example, using the
260 same input, we could choose to use only 4 digits in the first step, leaving
261 a full wide digit for the last step:
262 263 q₆ q₅ q₄ q₃ q₂ q₁ q₀
264 _______________________________
265 v₃ v₂ v₁ v₀ ) u₉ u₈ u₇ u₆ u₅ u₄ u₃ u₂ u₁ u₀
266 ↓ ↓ ↓ ↓ | | | | | |
267 [u₉ u₈ u₇ u₆]| | | | | |
268 - [ q₆·v ]| | | | | |
269 -------------- ↓ ↓ | | | |
270 [ rem |u₅ u₄]| | | |
271 - [ q₅q₄·v ]| | | |
272 -------------------- ↓ ↓ | |
273 [ rem |u₃ u₂]| |
274 - [ q₃q₂·v ]| |
275 -------------------- ↓ ↓
276 [ rem |u₁ u₀]
277 - [ q₁q₀·v ]
278 ---------------------
279 [ rem ]
280 281 Today, the code in divRecursiveStep works like the first example. Perhaps in
282 the future we will make it work like the alternative, to avoid a special case
283 in the final iteration.
284 285 Either way, each step is a 3-by-2-wide-digit division approximated first by
286 a 2-by-1-wide-digit division, just as we did for regular digits in long division.
287 Because the actual answer we want is a 3-by-2-wide-digit division, instead of
288 multiplying q̂·v directly during the fixup, we can use the quick refinement
289 from long division (an n/2-by-n/2 multiply) to correct q to its actual value
290 and also compute the remainder (as mentioned above), and then stop after that,
291 never doing a full n-by-n multiply.
292 293 Instead of using an n-by-n/2-digit division to produce n/2 digits, we can add
294 (not discard) one more real digit, doing an (n+1)-by-(n/2+1)-digit division that
295 produces n/2+1 digits. That single extra digit tightens the Good Guess Guarantee
296 to q ≤ q̂ ≤ q+1 and lets us drop long division's special treatment of the first
297 digit. These benefits are discussed more after the Good Guess Guarantee proof
298 below.
299 300 301 How Fast is Recursive Division?
302 303 For a 2n-by-n-digit division, this algorithm runs a 4-by-2 long division over
304 wide digits, producing two wide digits plus a possible leading regular digit 1,
305 which can be handled without a recursive call. That is, the algorithm uses two
306 full iterations, each using an n-by-n/2-digit division and an n/2-by-n/2-digit
307 multiplication, along with a few n-digit additions and subtractions. The standard
308 n-by-n-digit multiplication algorithm requires O(n²) time, making the overall
309 algorithm require time T(n) where
310 311 T(n) = 2T(n/2) + O(n) + O(n²)
312 313 which, by the Bentley-Haken-Saxe theorem, ends up reducing to T(n) = O(n²).
314 This is not an improvement over regular long division.
315 316 When the number of digits n becomes large enough, Karatsuba's algorithm for
317 multiplication can be used instead, which takes O(n^log₂3) = O(n^1.6) time.
318 (Karatsuba multiplication is implemented in func karatsuba in nat.go.)
319 That makes the overall recursive division algorithm take O(n^1.6) time as well,
320 which is an improvement, but again only for large enough numbers.
321 322 It is not critical to make sure that every recursion does only two recursive
323 calls. While in general the number of recursive calls can change the time
324 analysis, in this case doing three calls does not change the analysis:
325 326 T(n) = 3T(n/2) + O(n) + O(n^log₂3)
327 328 ends up being T(n) = O(n^log₂3). Because the Karatsuba multiplication taking
329 time O(n^log₂3) is itself doing 3 half-sized recursions, doing three for the
330 division does not hurt the asymptotic performance. Of course, it is likely
331 still faster in practice to do two.
332 333 334 Proof of the Good Guess Guarantee
335 336 Given numbers x, y, let us break them into the quotients and remainders when
337 divided by some scaling factor S, with the added constraints that the quotient
338 x/y and the high part of y are both less than some limit T, and that the high
339 part of y is at least half as big as T.
340 341 x₁ = ⌊x/S⌋ y₁ = ⌊y/S⌋
342 x₀ = x mod S y₀ = y mod S
343 344 x = x₁·S + x₀ 0 ≤ x₀ < S x/y < T
345 y = y₁·S + y₀ 0 ≤ y₀ < S T/2 ≤ y₁ < T
346 347 And consider the two truncated quotients:
348 349 q = ⌊x/y⌋
350 q̂ = ⌊x₁/y₁⌋
351 352 We will prove that q ≤ q̂ ≤ q+2.
353 354 The guarantee makes no real demands on the scaling factor S: it is simply the
355 magnitude of the digits cut from both x and y to produce x₁ and y₁.
356 The guarantee makes only limited demands on T: it must be large enough to hold
357 the quotient x/y, and y₁ must have roughly the same size.
358 359 To apply to the earlier discussion of 2-by-1 guesses in long division,
360 we would choose:
361 362 S = Bⁿ⁻¹
363 T = B
364 x = u
365 x₁ = uₙuₙ₋₁
366 x₀ = uₙ₋₂...u₀
367 y = v
368 y₁ = vₙ₋₁
369 y₀ = vₙ₋₂...u₀
370 371 These simpler variables avoid repeating those longer expressions in the proof.
372 373 Note also that, by definition, truncating division ⌊x/y⌋ satisfies
374 375 x/y - 1 < ⌊x/y⌋ ≤ x/y.
376 377 This fact will be used a few times in the proofs.
378 379 Proof that q ≤ q̂:
380 381 q̂·y₁ = ⌊x₁/y₁⌋·y₁ [by definition, q̂ = ⌊x₁/y₁⌋]
382 > (x₁/y₁ - 1)·y₁ [x₁/y₁ - 1 < ⌊x₁/y₁⌋]
383 = x₁ - y₁ [distribute y₁]
384 385 So q̂·y₁ > x₁ - y₁.
386 Since q̂·y₁ is an integer, q̂·y₁ ≥ x₁ - y₁ + 1.
387 388 q̂ - q = q̂ - ⌊x/y⌋ [by definition, q = ⌊x/y⌋]
389 ≥ q̂ - x/y [⌊x/y⌋ < x/y]
390 = (1/y)·(q̂·y - x) [factor out 1/y]
391 ≥ (1/y)·(q̂·y₁·S - x) [y = y₁·S + y₀ ≥ y₁·S]
392 ≥ (1/y)·((x₁ - y₁ + 1)·S - x) [above: q̂·y₁ ≥ x₁ - y₁ + 1]
393 = (1/y)·(x₁·S - y₁·S + S - x) [distribute S]
394 = (1/y)·(S - x₀ - y₁·S) [-x = -x₁·S - x₀]
395 > -y₁·S / y [x₀ < S, so S - x₀ > 0; drop it]
396 ≥ -1 [y₁·S ≤ y]
397 398 So q̂ - q > -1.
399 Since q̂ - q is an integer, q̂ - q ≥ 0, or equivalently q ≤ q̂.
400 401 Proof that q̂ ≤ q+2:
402 403 x₁/y₁ - x/y = x₁·S/y₁·S - x/y [multiply left term by S/S]
404 ≤ x/y₁·S - x/y [x₁S ≤ x]
405 = (x/y)·(y/y₁·S - 1) [factor out x/y]
406 = (x/y)·((y - y₁·S)/y₁·S) [move -1 into y/y₁·S fraction]
407 = (x/y)·(y₀/y₁·S) [y - y₁·S = y₀]
408 = (x/y)·(1/y₁)·(y₀/S) [factor out 1/y₁]
409 < (x/y)·(1/y₁) [y₀ < S, so y₀/S < 1]
410 ≤ (x/y)·(2/T) [y₁ ≥ T/2, so 1/y₁ ≤ 2/T]
411 < T·(2/T) [x/y < T]
412 = 2 [T·(2/T) = 2]
413 414 So x₁/y₁ - x/y < 2.
415 416 q̂ - q = ⌊x₁/y₁⌋ - q [by definition, q̂ = ⌊x₁/y₁⌋]
417 = ⌊x₁/y₁⌋ - ⌊x/y⌋ [by definition, q = ⌊x/y⌋]
418 ≤ x₁/y₁ - ⌊x/y⌋ [⌊x₁/y₁⌋ ≤ x₁/y₁]
419 < x₁/y₁ - (x/y - 1) [⌊x/y⌋ > x/y - 1]
420 = (x₁/y₁ - x/y) + 1 [regrouping]
421 < 2 + 1 [above: x₁/y₁ - x/y < 2]
422 = 3
423 424 So q̂ - q < 3.
425 Since q̂ - q is an integer, q̂ - q ≤ 2.
426 427 Note that when x/y < T/2, the bounds tighten to x₁/y₁ - x/y < 1 and therefore
428 q̂ - q ≤ 1.
429 430 Note also that in the general case 2n-by-n division where we don't know that
431 x/y < T, we do know that x/y < 2T, yielding the bound q̂ - q ≤ 4. So we could
432 remove the special case first step of long division as long as we allow the
433 first fixup loop to run up to four times. (Using a simple comparison to decide
434 whether the first digit is 0 or 1 is still more efficient, though.)
435 436 Finally, note that when dividing three leading base-B digits by two (scaled),
437 we have T = B² and x/y < B = T/B, a much tighter bound than x/y < T.
438 This in turn yields the much tighter bound x₁/y₁ - x/y < 2/B. This means that
439 ⌊x₁/y₁⌋ and ⌊x/y⌋ can only differ when x/y is less than 2/B greater than an
440 integer. For random x and y, the chance of this is 2/B, or, for large B,
441 approximately zero. This means that after we produce the 3-by-2 guess in the
442 long division algorithm, the fixup loop essentially never runs.
443 444 In the recursive algorithm, the extra digit in (2·⌊n/2⌋+1)-by-(⌊n/2⌋+1)-digit
445 division has exactly the same effect: the probability of needing a fixup is the
446 same 2/B. Even better, we can allow the general case x/y < 2T and the fixup
447 probability only grows to 4/B, still essentially zero.
448 449 450 References
451 452 There are no great references for implementing long division; thus this comment.
453 Here are some notes about what to expect from the obvious references.
454 455 Knuth Volume 2 (Seminumerical Algorithms) section 4.3.1 is the usual canonical
456 reference for long division, but that entire series is highly compressed, never
457 repeating a necessary fact and leaving important insights to the exercises.
458 For example, no rationale whatsoever is given for the calculation that extends
459 q̂ from a 2-by-1 to a 3-by-2 guess, nor why it reduces the error bound.
460 The proof that the calculation even has the desired effect is left to exercises.
461 The solutions to those exercises provided at the back of the book are entirely
462 calculations, still with no explanation as to what is going on or how you would
463 arrive at the idea of doing those exact calculations. Nowhere is it mentioned
464 that this test extends the 2-by-1 guess into a 3-by-2 guess. The proof of the
465 Good Guess Guarantee is only for the 2-by-1 guess and argues by contradiction,
466 making it difficult to understand how modifications like adding another digit
467 or adjusting the quotient range affects the overall bound.
468 469 All that said, Knuth remains the canonical reference. It is dense but packed
470 full of information and references, and the proofs are simpler than many other
471 presentations. The proofs above are reworkings of Knuth's to remove the
472 arguments by contradiction and add explanations or steps that Knuth omitted.
473 But beware of errors in older printings. Take the published errata with you.
474 475 Brinch Hansen's “Multiple-length Division Revisited: a Tour of the Minefield”
476 starts with a blunt critique of Knuth's presentation (among others) and then
477 presents a more detailed and easier to follow treatment of long division,
478 including an implementation in Pascal. But the algorithm and implementation
479 work entirely in terms of 3-by-2 division, which is much less useful on modern
480 hardware than an algorithm using 2-by-1 division. The proofs are a bit too
481 focused on digit counting and seem needlessly complex, especially compared to
482 the ones given above.
483 484 Burnikel and Ziegler's “Fast Recursive Division” introduced the key insight of
485 implementing division by an n-digit divisor using recursive calls to division
486 by an n/2-digit divisor, relying on Karatsuba multiplication to yield a
487 sub-quadratic run time. However, the presentation decisions are made almost
488 entirely for the purpose of simplifying the run-time analysis, rather than
489 simplifying the presentation. Instead of a single algorithm that loops over
490 quotient digits, the paper presents two mutually-recursive algorithms, for
491 2n-by-n and 3n-by-2n. The paper also does not present any general (n+m)-by-n
492 algorithm.
493 494 The proofs in the paper are remarkably complex, especially considering that
495 the algorithm is at its core just long division on wide digits, so that the
496 usual long division proofs apply essentially unaltered.
497 */
498 499 package big
500 501 import "math/bits"
502 503 // rem returns r such that r = u%v.
504 // It uses z as the storage for r.
505 func (z nat) rem(stk *stack, u, v nat) (r nat) {
506 if alias(z, u) {
507 z = nil
508 }
509 defer stk.restore(stk.save())
510 q := stk.nat(max(1, len(u)-(len(v)-1)))
511 _, r = q.div(stk, z, u, v)
512 return r
513 }
514 515 // div returns q, r such that q = ⌊u/v⌋ and r = u%v = u - q·v.
516 // It uses z and z2 as the storage for q and r.
517 // The caller may pass stk == nil to request that div obtain and release one itself.
518 func (z nat) div(stk *stack, z2, u, v nat) (q, r nat) {
519 if len(v) == 0 {
520 panic("division by zero")
521 }
522 523 if len(v) == 1 {
524 // Short division: long optimized for a single-word divisor.
525 // In that case, the 2-by-1 guess is all we need at each step.
526 var r2 Word
527 q, r2 = z.divW(u, v[0])
528 r = z2.setWord(r2)
529 return
530 }
531 532 if u.cmp(v) < 0 {
533 q = z[:0]
534 r = z2.set(u)
535 return
536 }
537 538 if stk == nil {
539 stk = getStack()
540 defer stk.free()
541 }
542 543 q, r = z.divLarge(stk, z2, u, v)
544 return
545 }
546 547 // divW returns q, r such that q = ⌊x/y⌋ and r = x%y = x - q·y.
548 // It uses z as the storage for q.
549 // Note that y is a single digit (Word), not a big number.
550 func (z nat) divW(x nat, y Word) (q nat, r Word) {
551 m := len(x)
552 switch {
553 case y == 0:
554 panic("division by zero")
555 case y == 1:
556 q = z.set(x) // result is x
557 return
558 case m == 0:
559 q = z[:0] // result is 0
560 return
561 }
562 // m > 0
563 z = z.make(m)
564 r = divWVW(z, 0, x, y)
565 q = z.norm()
566 return
567 }
568 569 // modW returns x % d.
570 func (x nat) modW(d Word) (r Word) {
571 // TODO(agl): we don't actually need to store the q value.
572 var q nat
573 q = q.make(len(x))
574 return divWVW(q, 0, x, d)
575 }
576 577 // divWVW overwrites z with ⌊x/y⌋, returning the remainder r.
578 // The caller must ensure that len(z) = len(x).
579 func divWVW(z []Word, xn Word, x []Word, y Word) (r Word) {
580 r = xn
581 if len(x) == 1 {
582 qq, rr := bits.Div(uint(r), uint(x[0]), uint(y))
583 z[0] = Word(qq)
584 return Word(rr)
585 }
586 rec := reciprocalWord(y)
587 for i := len(z) - 1; i >= 0; i-- {
588 z[i], r = divWW(r, x[i], y, rec)
589 }
590 return r
591 }
592 593 // div returns q, r such that q = ⌊uIn/vIn⌋ and r = uIn%vIn = uIn - q·vIn.
594 // It uses z and u as the storage for q and r.
595 // The caller must ensure that len(vIn) ≥ 2 (use divW otherwise)
596 // and that len(uIn) ≥ len(vIn) (the answer is 0, uIn otherwise).
597 func (z nat) divLarge(stk *stack, u, uIn, vIn nat) (q, r nat) {
598 n := len(vIn)
599 m := len(uIn) - n
600 601 // Scale the inputs so vIn's top bit is 1 (see “Scaling Inputs” above).
602 // vIn is treated as a read-only input (it may be in use by another
603 // goroutine), so we must make a copy.
604 // uIn is copied to u.
605 defer stk.restore(stk.save())
606 shift := nlz(vIn[n-1])
607 v := stk.nat(n)
608 u = u.make(len(uIn) + 1)
609 if shift == 0 {
610 copy(v, vIn)
611 copy(u[:len(uIn)], uIn)
612 u[len(uIn)] = 0
613 } else {
614 lshVU(v, vIn, shift)
615 u[len(uIn)] = lshVU(u[:len(uIn)], uIn, shift)
616 }
617 618 // The caller should not pass aliased z and u, since those are
619 // the two different outputs, but correct just in case.
620 if alias(z, u) {
621 z = nil
622 }
623 q = z.make(m + 1)
624 625 // Use basic or recursive long division depending on size.
626 if n < divRecursiveThreshold {
627 q.divBasic(stk, u, v)
628 } else {
629 q.divRecursive(stk, u, v)
630 }
631 632 q = q.norm()
633 634 // Undo scaling of remainder.
635 if shift != 0 {
636 rshVU(u, u, shift)
637 }
638 r = u.norm()
639 640 return q, r
641 }
642 643 // divBasic implements long division as described above.
644 // It overwrites q with ⌊u/v⌋ and overwrites u with the remainder r.
645 // q must be large enough to hold ⌊u/v⌋.
646 func (q nat) divBasic(stk *stack, u, v nat) {
647 n := len(v)
648 m := len(u) - n
649 650 defer stk.restore(stk.save())
651 qhatv := stk.nat(n + 1)
652 653 // Set up for divWW below, precomputing reciprocal argument.
654 vn1 := v[n-1]
655 rec := reciprocalWord(vn1)
656 657 // Invent a leading 0 for u, for the first iteration.
658 // Invariant: ujn == u[j+n] in each iteration.
659 ujn := Word(0)
660 661 // Compute each digit of quotient.
662 for j := m; j >= 0; j-- {
663 // Compute the 2-by-1 guess q̂.
664 qhat := Word(_M)
665 666 // ujn ≤ vn1, or else q̂ would be more than one digit.
667 // For ujn == vn1, we set q̂ to the max digit M above.
668 // Otherwise, we compute the 2-by-1 guess.
669 if ujn != vn1 {
670 var rhat Word
671 qhat, rhat = divWW(ujn, u[j+n-1], vn1, rec)
672 673 // Refine q̂ to a 3-by-2 guess. See “Refining Guesses” above.
674 vn2 := v[n-2]
675 x1, x2 := mulWW(qhat, vn2)
676 ujn2 := u[j+n-2]
677 for greaterThan(x1, x2, rhat, ujn2) { // x1x2 > r̂ u[j+n-2]
678 qhat--
679 prevRhat := rhat
680 rhat += vn1
681 // If r̂ overflows, then
682 // r̂ u[j+n-2]v[n-1] is now definitely > x1 x2.
683 if rhat < prevRhat {
684 break
685 }
686 // TODO(rsc): No need for a full mulWW.
687 // x2 += vn2; if x2 overflows, x1++
688 x1, x2 = mulWW(qhat, vn2)
689 }
690 }
691 692 // Compute q̂·v.
693 qhatv[n] = mulAddVWW(qhatv[0:n], v, qhat, 0)
694 qhl := len(qhatv)
695 if j+qhl > len(u) && qhatv[n] == 0 {
696 qhl--
697 }
698 699 // Subtract q̂·v from the current section of u.
700 // If it underflows, q̂·v > u, which we fix up
701 // by decrementing q̂ and adding v back.
702 c := subVV(u[j:j+qhl], u[j:j+qhl], qhatv[:qhl])
703 if c != 0 {
704 c := addVV(u[j:j+n], u[j:j+n], v)
705 // If n == qhl, the carry from subVV and the carry from addVV
706 // cancel out and don't affect u[j+n].
707 if n < qhl {
708 u[j+n] += c
709 }
710 qhat--
711 }
712 713 ujn = u[j+n-1]
714 715 // Save quotient digit.
716 // Caller may know the top digit is zero and not leave room for it.
717 if j == m && m == len(q) && qhat == 0 {
718 continue
719 }
720 q[j] = qhat
721 }
722 }
723 724 // greaterThan reports whether the two digit numbers x1 x2 > y1 y2.
725 // TODO(rsc): In contradiction to most of this file, x1 is the high
726 // digit and x2 is the low digit. This should be fixed.
727 func greaterThan(x1, x2, y1, y2 Word) bool {
728 return x1 > y1 || x1 == y1 && x2 > y2
729 }
730 731 // divRecursiveThreshold is the number of divisor digits
732 // at which point divRecursive is faster than divBasic.
733 var divRecursiveThreshold = 40 // see calibrate_test.go
734 735 // divRecursive implements recursive division as described above.
736 // It overwrites z with ⌊u/v⌋ and overwrites u with the remainder r.
737 // z must be large enough to hold ⌊u/v⌋.
738 // This function is just for allocating and freeing temporaries
739 // around divRecursiveStep, the real implementation.
740 func (z nat) divRecursive(stk *stack, u, v nat) {
741 clear(z)
742 z.divRecursiveStep(stk, u, v, 0)
743 }
744 745 // divRecursiveStep is the actual implementation of recursive division.
746 // It adds ⌊u/v⌋ to z and overwrites u with the remainder r.
747 // z must be large enough to hold ⌊u/v⌋.
748 // It uses temps[depth] (allocating if needed) as a temporary live across
749 // the recursive call. It also uses tmp, but not live across the recursion.
750 func (z nat) divRecursiveStep(stk *stack, u, v nat, depth int) {
751 // u is a subsection of the original and may have leading zeros.
752 // TODO(rsc): The v = v.norm() is useless and should be removed.
753 // We know (and require) that v's top digit is ≥ B/2.
754 u = u.norm()
755 v = v.norm()
756 if len(u) == 0 {
757 clear(z)
758 return
759 }
760 761 // Fall back to basic division if the problem is now small enough.
762 n := len(v)
763 if n < divRecursiveThreshold {
764 z.divBasic(stk, u, v)
765 return
766 }
767 768 // Nothing to do if u is shorter than v (implies u < v).
769 m := len(u) - n
770 if m < 0 {
771 return
772 }
773 774 // We consider B digits in a row as a single wide digit.
775 // (See “Recursive Division” above.)
776 //
777 // TODO(rsc): rename B to Wide, to avoid confusion with _B,
778 // which is something entirely different.
779 // TODO(rsc): Look into whether using ⌈n/2⌉ is better than ⌊n/2⌋.
780 B := n / 2
781 782 // Allocate a nat for qhat below.
783 defer stk.restore(stk.save())
784 qhat0 := stk.nat(B + 1)
785 786 // Compute each wide digit of the quotient.
787 //
788 // TODO(rsc): Change the loop to be
789 // for j := (m+B-1)/B*B; j > 0; j -= B {
790 // which will make the final step a regular step, letting us
791 // delete what amounts to an extra copy of the loop body below.
792 j := m
793 for j > B {
794 // Divide u[j-B:j+n] (3 wide digits) by v (2 wide digits).
795 // First make the 2-by-1-wide-digit guess using a recursive call.
796 // Then extend the guess to the full 3-by-2 (see “Refining Guesses”).
797 //
798 // For the 2-by-1-wide-digit guess, instead of doing 2B-by-B-digit,
799 // we use a (2B+1)-by-(B+1) digit, which handles the possibility that
800 // the result has an extra leading 1 digit as well as guaranteeing
801 // that the computed q̂ will be off by at most 1 instead of 2.
802 803 // s is the number of digits to drop from the 3B- and 2B-digit chunks.
804 // We drop B-1 to be left with 2B+1 and B+1.
805 s := (B - 1)
806 807 // uu is the up-to-3B-digit section of u we are working on.
808 uu := u[j-B:]
809 810 // Compute the 2-by-1 guess q̂, leaving r̂ in uu[s:B+n].
811 qhat := qhat0
812 clear(qhat)
813 qhat.divRecursiveStep(stk, uu[s:B+n], v[s:], depth+1)
814 qhat = qhat.norm()
815 816 // Extend to a 3-by-2 quotient and remainder.
817 // Because divRecursiveStep overwrote the top part of uu with
818 // the remainder r̂, the full uu already contains the equivalent
819 // of r̂·B + uₙ₋₂ from the “Refining Guesses” discussion.
820 // Subtracting q̂·vₙ₋₂ from it will compute the full-length remainder.
821 // If that subtraction underflows, q̂·v > u, which we fix up
822 // by decrementing q̂ and adding v back, same as in long division.
823 824 // TODO(rsc): Instead of subtract and fix-up, this code is computing
825 // q̂·vₙ₋₂ and decrementing q̂ until that product is ≤ u.
826 // But we can do the subtraction directly, as in the comment above
827 // and in long division, because we know that q̂ is wrong by at most one.
828 mark := stk.save()
829 qhatv := stk.nat(3 * n)
830 clear(qhatv)
831 qhatv = qhatv.mul(stk, qhat, v[:s])
832 for i := 0; i < 2; i++ {
833 e := qhatv.cmp(uu.norm())
834 if e <= 0 {
835 break
836 }
837 subVW(qhat, qhat, 1)
838 c := subVV(qhatv[:s], qhatv[:s], v[:s])
839 if len(qhatv) > s {
840 subVW(qhatv[s:], qhatv[s:], c)
841 }
842 addTo(uu[s:], v[s:])
843 }
844 if qhatv.cmp(uu.norm()) > 0 {
845 panic("impossible")
846 }
847 c := subVV(uu[:len(qhatv)], uu[:len(qhatv)], qhatv)
848 if c > 0 {
849 subVW(uu[len(qhatv):], uu[len(qhatv):], c)
850 }
851 addTo(z[j-B:], qhat)
852 j -= B
853 stk.restore(mark)
854 }
855 856 // TODO(rsc): Rewrite loop as described above and delete all this code.
857 858 // Now u < (v<<B), compute lower bits in the same way.
859 // Choose shift = B-1 again.
860 s := B - 1
861 qhat := qhat0
862 clear(qhat)
863 qhat.divRecursiveStep(stk, u[s:].norm(), v[s:], depth+1)
864 qhat = qhat.norm()
865 qhatv := stk.nat(3 * n)
866 clear(qhatv)
867 qhatv = qhatv.mul(stk, qhat, v[:s])
868 // Set the correct remainder as before.
869 for i := 0; i < 2; i++ {
870 if e := qhatv.cmp(u.norm()); e > 0 {
871 subVW(qhat, qhat, 1)
872 c := subVV(qhatv[:s], qhatv[:s], v[:s])
873 if len(qhatv) > s {
874 subVW(qhatv[s:], qhatv[s:], c)
875 }
876 addTo(u[s:], v[s:])
877 }
878 }
879 if qhatv.cmp(u.norm()) > 0 {
880 panic("impossible")
881 }
882 c := subVV(u[:len(qhatv)], u[:len(qhatv)], qhatv)
883 if c > 0 {
884 c = subVW(u[len(qhatv):], u[len(qhatv):], c)
885 }
886 if c > 0 {
887 panic("impossible")
888 }
889 890 // Done!
891 addTo(z, qhat.norm())
892 }
893