# Source file src/math/big/natdiv.go

```     1  // Copyright 2009 The Go Authors. All rights reserved.
2  // Use of this source code is governed by a BSD-style
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
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(u, v nat) (r nat) {
506  	if alias(z, u) {
507  		z = nil
508  	}
509  	qp := getNat(0)
510  	q, r := qp.div(z, u, v)
511  	*qp = q
512  	putNat(qp)
513  	return r
514  }
515
516  // div returns q, r such that q = ⌊u/v⌋ and r = u%v = u - q·v.
517  // It uses z and z2 as the storage for q and r.
518  func (z nat) div(z2, u, v nat) (q, r nat) {
519  	if len(v) == 0 {
520  		panic("division by zero")
521  	}
522
523  	if u.cmp(v) < 0 {
524  		q = z[:0]
525  		r = z2.set(u)
526  		return
527  	}
528
529  	if len(v) == 1 {
530  		// Short division: long optimized for a single-word divisor.
531  		// In that case, the 2-by-1 guess is all we need at each step.
532  		var r2 Word
533  		q, r2 = z.divW(u, v[0])
534  		r = z2.setWord(r2)
535  		return
536  	}
537
538  	q, r = z.divLarge(z2, u, v)
539  	return
540  }
541
542  // divW returns q, r such that q = ⌊x/y⌋ and r = x%y = x - q·y.
543  // It uses z as the storage for q.
544  // Note that y is a single digit (Word), not a big number.
545  func (z nat) divW(x nat, y Word) (q nat, r Word) {
546  	m := len(x)
547  	switch {
548  	case y == 0:
549  		panic("division by zero")
550  	case y == 1:
551  		q = z.set(x) // result is x
552  		return
553  	case m == 0:
554  		q = z[:0] // result is 0
555  		return
556  	}
557  	// m > 0
558  	z = z.make(m)
559  	r = divWVW(z, 0, x, y)
560  	q = z.norm()
561  	return
562  }
563
564  // modW returns x % d.
565  func (x nat) modW(d Word) (r Word) {
566  	// TODO(agl): we don't actually need to store the q value.
567  	var q nat
568  	q = q.make(len(x))
569  	return divWVW(q, 0, x, d)
570  }
571
572  // divWVW overwrites z with ⌊x/y⌋, returning the remainder r.
573  // The caller must ensure that len(z) = len(x).
574  func divWVW(z []Word, xn Word, x []Word, y Word) (r Word) {
575  	r = xn
576  	if len(x) == 1 {
577  		qq, rr := bits.Div(uint(r), uint(x[0]), uint(y))
578  		z[0] = Word(qq)
579  		return Word(rr)
580  	}
581  	rec := reciprocalWord(y)
582  	for i := len(z) - 1; i >= 0; i-- {
583  		z[i], r = divWW(r, x[i], y, rec)
584  	}
585  	return r
586  }
587
588  // div returns q, r such that q = ⌊uIn/vIn⌋ and r = uIn%vIn = uIn - q·vIn.
589  // It uses z and u as the storage for q and r.
590  // The caller must ensure that len(vIn) ≥ 2 (use divW otherwise)
591  // and that len(uIn) ≥ len(vIn) (the answer is 0, uIn otherwise).
592  func (z nat) divLarge(u, uIn, vIn nat) (q, r nat) {
593  	n := len(vIn)
594  	m := len(uIn) - n
595
596  	// Scale the inputs so vIn's top bit is 1 (see “Scaling Inputs” above).
597  	// vIn is treated as a read-only input (it may be in use by another
598  	// goroutine), so we must make a copy.
599  	// uIn is copied to u.
600  	shift := nlz(vIn[n-1])
601  	vp := getNat(n)
602  	v := *vp
603  	shlVU(v, vIn, shift)
604  	u = u.make(len(uIn) + 1)
605  	u[len(uIn)] = shlVU(u[0:len(uIn)], uIn, shift)
606
607  	// The caller should not pass aliased z and u, since those are
608  	// the two different outputs, but correct just in case.
609  	if alias(z, u) {
610  		z = nil
611  	}
612  	q = z.make(m + 1)
613
614  	// Use basic or recursive long division depending on size.
615  	if n < divRecursiveThreshold {
616  		q.divBasic(u, v)
617  	} else {
618  		q.divRecursive(u, v)
619  	}
620  	putNat(vp)
621
622  	q = q.norm()
623
624  	// Undo scaling of remainder.
625  	shrVU(u, u, shift)
626  	r = u.norm()
627
628  	return q, r
629  }
630
631  // divBasic implements long division as described above.
632  // It overwrites q with ⌊u/v⌋ and overwrites u with the remainder r.
633  // q must be large enough to hold ⌊u/v⌋.
634  func (q nat) divBasic(u, v nat) {
635  	n := len(v)
636  	m := len(u) - n
637
638  	qhatvp := getNat(n + 1)
639  	qhatv := *qhatvp
640
641  	// Set up for divWW below, precomputing reciprocal argument.
642  	vn1 := v[n-1]
643  	rec := reciprocalWord(vn1)
644
645  	// Compute each digit of quotient.
646  	for j := m; j >= 0; j-- {
647  		// Compute the 2-by-1 guess q̂.
648  		// The first iteration must invent a leading 0 for u.
649  		qhat := Word(_M)
650  		var ujn Word
651  		if j+n < len(u) {
652  			ujn = u[j+n]
653  		}
654
655  		// ujn ≤ vn1, or else q̂ would be more than one digit.
656  		// For ujn == vn1, we set q̂ to the max digit M above.
657  		// Otherwise, we compute the 2-by-1 guess.
658  		if ujn != vn1 {
659  			var rhat Word
660  			qhat, rhat = divWW(ujn, u[j+n-1], vn1, rec)
661
662  			// Refine q̂ to a 3-by-2 guess. See “Refining Guesses” above.
663  			vn2 := v[n-2]
664  			x1, x2 := mulWW(qhat, vn2)
665  			ujn2 := u[j+n-2]
666  			for greaterThan(x1, x2, rhat, ujn2) { // x1x2 > r̂ u[j+n-2]
667  				qhat--
668  				prevRhat := rhat
669  				rhat += vn1
670  				// If r̂  overflows, then
671  				// r̂ u[j+n-2]v[n-1] is now definitely > x1 x2.
672  				if rhat < prevRhat {
673  					break
674  				}
675  				// TODO(rsc): No need for a full mulWW.
676  				// x2 += vn2; if x2 overflows, x1++
677  				x1, x2 = mulWW(qhat, vn2)
678  			}
679  		}
680
681  		// Compute q̂·v.
682  		qhatv[n] = mulAddVWW(qhatv[0:n], v, qhat, 0)
683  		qhl := len(qhatv)
684  		if j+qhl > len(u) && qhatv[n] == 0 {
685  			qhl--
686  		}
687
688  		// Subtract q̂·v from the current section of u.
689  		// If it underflows, q̂·v > u, which we fix up
690  		// by decrementing q̂ and adding v back.
691  		c := subVV(u[j:j+qhl], u[j:], qhatv)
692  		if c != 0 {
693  			c := addVV(u[j:j+n], u[j:], v)
694  			// If n == qhl, the carry from subVV and the carry from addVV
695  			// cancel out and don't affect u[j+n].
696  			if n < qhl {
697  				u[j+n] += c
698  			}
699  			qhat--
700  		}
701
702  		// Save quotient digit.
703  		// Caller may know the top digit is zero and not leave room for it.
704  		if j == m && m == len(q) && qhat == 0 {
705  			continue
706  		}
707  		q[j] = qhat
708  	}
709
710  	putNat(qhatvp)
711  }
712
713  // greaterThan reports whether the two digit numbers x1 x2 > y1 y2.
714  // TODO(rsc): In contradiction to most of this file, x1 is the high
715  // digit and x2 is the low digit. This should be fixed.
716  func greaterThan(x1, x2, y1, y2 Word) bool {
717  	return x1 > y1 || x1 == y1 && x2 > y2
718  }
719
720  // divRecursiveThreshold is the number of divisor digits
721  // at which point divRecursive is faster than divBasic.
722  const divRecursiveThreshold = 100
723
724  // divRecursive implements recursive division as described above.
725  // It overwrites z with ⌊u/v⌋ and overwrites u with the remainder r.
726  // z must be large enough to hold ⌊u/v⌋.
727  // This function is just for allocating and freeing temporaries
728  // around divRecursiveStep, the real implementation.
729  func (z nat) divRecursive(u, v nat) {
730  	// Recursion depth is (much) less than 2 log₂(len(v)).
731  	// Allocate a slice of temporaries to be reused across recursion,
732  	// plus one extra temporary not live across the recursion.
733  	recDepth := 2 * bits.Len(uint(len(v)))
734  	tmp := getNat(3 * len(v))
735  	temps := make([]*nat, recDepth)
736
737  	z.clear()
738  	z.divRecursiveStep(u, v, 0, tmp, temps)
739
740  	// Free temporaries.
741  	for _, n := range temps {
742  		if n != nil {
743  			putNat(n)
744  		}
745  	}
746  	putNat(tmp)
747  }
748
749  // divRecursiveStep is the actual implementation of recursive division.
750  // It adds ⌊u/v⌋ to z and overwrites u with the remainder r.
751  // z must be large enough to hold ⌊u/v⌋.
752  // It uses temps[depth] (allocating if needed) as a temporary live across
753  // the recursive call. It also uses tmp, but not live across the recursion.
754  func (z nat) divRecursiveStep(u, v nat, depth int, tmp *nat, temps []*nat) {
755  	// u is a subsection of the original and may have leading zeros.
756  	// TODO(rsc): The v = v.norm() is useless and should be removed.
757  	// We know (and require) that v's top digit is ≥ B/2.
758  	u = u.norm()
759  	v = v.norm()
760  	if len(u) == 0 {
761  		z.clear()
762  		return
763  	}
764
765  	// Fall back to basic division if the problem is now small enough.
766  	n := len(v)
767  	if n < divRecursiveThreshold {
768  		z.divBasic(u, v)
769  		return
770  	}
771
772  	// Nothing to do if u is shorter than v (implies u < v).
773  	m := len(u) - n
774  	if m < 0 {
775  		return
776  	}
777
778  	// We consider B digits in a row as a single wide digit.
779  	// (See “Recursive Division” above.)
780  	//
781  	// TODO(rsc): rename B to Wide, to avoid confusion with _B,
782  	// which is something entirely different.
783  	// TODO(rsc): Look into whether using ⌈n/2⌉ is better than ⌊n/2⌋.
784  	B := n / 2
785
786  	// Allocate a nat for qhat below.
787  	if temps[depth] == nil {
788  		temps[depth] = getNat(n) // TODO(rsc): Can be just B+1.
789  	} else {
790  		*temps[depth] = temps[depth].make(B + 1)
791  	}
792
793  	// Compute each wide digit of the quotient.
794  	//
795  	// TODO(rsc): Change the loop to be
796  	//	for j := (m+B-1)/B*B; j > 0; j -= B {
797  	// which will make the final step a regular step, letting us
798  	// delete what amounts to an extra copy of the loop body below.
799  	j := m
800  	for j > B {
801  		// Divide u[j-B:j+n] (3 wide digits) by v (2 wide digits).
802  		// First make the 2-by-1-wide-digit guess using a recursive call.
803  		// Then extend the guess to the full 3-by-2 (see “Refining Guesses”).
804  		//
805  		// For the 2-by-1-wide-digit guess, instead of doing 2B-by-B-digit,
806  		// we use a (2B+1)-by-(B+1) digit, which handles the possibility that
807  		// the result has an extra leading 1 digit as well as guaranteeing
808  		// that the computed q̂ will be off by at most 1 instead of 2.
809
810  		// s is the number of digits to drop from the 3B- and 2B-digit chunks.
811  		// We drop B-1 to be left with 2B+1 and B+1.
812  		s := (B - 1)
813
814  		// uu is the up-to-3B-digit section of u we are working on.
815  		uu := u[j-B:]
816
817  		// Compute the 2-by-1 guess q̂, leaving r̂ in uu[s:B+n].
818  		qhat := *temps[depth]
819  		qhat.clear()
820  		qhat.divRecursiveStep(uu[s:B+n], v[s:], depth+1, tmp, temps)
821  		qhat = qhat.norm()
822
823  		// Extend to a 3-by-2 quotient and remainder.
824  		// Because divRecursiveStep overwrote the top part of uu with
825  		// the remainder r̂, the full uu already contains the equivalent
826  		// of r̂·B + uₙ₋₂ from the “Refining Guesses” discussion.
827  		// Subtracting q̂·vₙ₋₂ from it will compute the full-length remainder.
828  		// If that subtraction underflows, q̂·v > u, which we fix up
829  		// by decrementing q̂ and adding v back, same as in long division.
830
831  		// TODO(rsc): Instead of subtract and fix-up, this code is computing
832  		// q̂·vₙ₋₂ and decrementing q̂ until that product is ≤ u.
833  		// But we can do the subtraction directly, as in the comment above
834  		// and in long division, because we know that q̂ is wrong by at most one.
835  		qhatv := tmp.make(3 * n)
836  		qhatv.clear()
837  		qhatv = qhatv.mul(qhat, v[:s])
838  		for i := 0; i < 2; i++ {
839  			e := qhatv.cmp(uu.norm())
840  			if e <= 0 {
841  				break
842  			}
843  			subVW(qhat, qhat, 1)
844  			c := subVV(qhatv[:s], qhatv[:s], v[:s])
845  			if len(qhatv) > s {
846  				subVW(qhatv[s:], qhatv[s:], c)
847  			}
849  		}
850  		if qhatv.cmp(uu.norm()) > 0 {
851  			panic("impossible")
852  		}
853  		c := subVV(uu[:len(qhatv)], uu[:len(qhatv)], qhatv)
854  		if c > 0 {
855  			subVW(uu[len(qhatv):], uu[len(qhatv):], c)
856  		}
858  		j -= B
859  	}
860
861  	// TODO(rsc): Rewrite loop as described above and delete all this code.
862
863  	// Now u < (v<<B), compute lower bits in the same way.
864  	// Choose shift = B-1 again.
865  	s := B - 1
866  	qhat := *temps[depth]
867  	qhat.clear()
868  	qhat.divRecursiveStep(u[s:].norm(), v[s:], depth+1, tmp, temps)
869  	qhat = qhat.norm()
870  	qhatv := tmp.make(3 * n)
871  	qhatv.clear()
872  	qhatv = qhatv.mul(qhat, v[:s])
873  	// Set the correct remainder as before.
874  	for i := 0; i < 2; i++ {
875  		if e := qhatv.cmp(u.norm()); e > 0 {
876  			subVW(qhat, qhat, 1)
877  			c := subVV(qhatv[:s], qhatv[:s], v[:s])
878  			if len(qhatv) > s {
879  				subVW(qhatv[s:], qhatv[s:], c)
880  			}
882  		}
883  	}
884  	if qhatv.cmp(u.norm()) > 0 {
885  		panic("impossible")
886  	}
887  	c := subVV(u[0:len(qhatv)], u[0:len(qhatv)], qhatv)
888  	if c > 0 {
889  		c = subVW(u[len(qhatv):], u[len(qhatv):], c)
890  	}
891  	if c > 0 {
892  		panic("impossible")
893  	}
894
895  	// Done!