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+// Copyright 2009 The Go Authors. All rights reserved.
+// Use of this source code is governed by a BSD-style
+// license that can be found in the LICENSE file.
+
+/*
+
+Multi-precision division. Here be dragons.
+
+Given u and v, where u is n+m digits, and v is n digits (with no leading zeros),
+the goal is to return quo, rem such that u = quo*v + rem, where 0 ≤ rem < v.
+That is, quo = ⌊u/v⌋ where ⌊x⌋ denotes the floor (truncation to integer) of x,
+and rem = u - quo·v.
+
+
+Long Division
+
+Division in a computer proceeds the same as long division in elementary school,
+but computers are not as good as schoolchildren at following vague directions,
+so we have to be much more precise about the actual steps and what can happen.
+
+We work from most to least significant digit of the quotient, doing:
+
+ • Guess a digit q, the number of v to subtract from the current
+ section of u to zero out the topmost digit.
+ • Check the guess by multiplying q·v and comparing it against
+ the current section of u, adjusting the guess as needed.
+ • Subtract q·v from the current section of u.
+ • Add q to the corresponding section of the result quo.
+
+When all digits have been processed, the final remainder is left in u
+and returned as rem.
+
+For example, here is a sketch of dividing 5 digits by 3 digits (n=3, m=2).
+
+ q₂ q₁ q₀
+ _________________
+ v₂ v₁ v₀ ) u₄ u₃ u₂ u₁ u₀
+ ↓ ↓ ↓ | |
+ [u₄ u₃ u₂]| |
+ - [ q₂·v ]| |
+ ----------- ↓ |
+ [ rem | u₁]|
+ - [ q₁·v ]|
+ ----------- ↓
+ [ rem | u₀]
+ - [ q₀·v ]
+ ------------
+ [ rem ]
+
+Instead of creating new storage for the remainders and copying digits from u
+as indicated by the arrows, we use u's storage directly as both the source
+and destination of the subtractions, so that the remainders overwrite
+successive overlapping sections of u as the division proceeds, using a slice
+of u to identify the current section. This avoids all the copying as well as
+shifting of remainders.
+
+Division of u with n+m digits by v with n digits (in base B) can in general
+produce at most m+1 digits, because:
+
+ • u < B^(n+m) [B^(n+m) has n+m+1 digits]
+ • v ≥ B^(n-1) [B^(n-1) is the smallest n-digit number]
+ • u/v < B^(n+m) / B^(n-1) [divide bounds for u, v]
+ • u/v < B^(m+1) [simplify]
+
+The first step is special: it takes the top n digits of u and divides them by
+the n digits of v, producing the first quotient digit and an n-digit remainder.
+In the example, q₂ = ⌊u₄u₃u₂ / v⌋.
+
+The first step divides n digits by n digits to ensure that it produces only a
+single digit.
+
+Each subsequent step appends the next digit from u to the remainder and divides
+those n+1 digits by the n digits of v, producing another quotient digit and a
+new n-digit remainder.
+
+Subsequent steps divide n+1 digits by n digits, an operation that in general
+might produce two digits. However, as used in the algorithm, that division is
+guaranteed to produce only a single digit. The dividend is of the form
+rem·B + d, where rem is a remainder from the previous step and d is a single
+digit, so:
+
+ • rem ≤ v - 1 [rem is a remainder from dividing by v]
+ • rem·B ≤ v·B - B [multiply by B]
+ • d ≤ B - 1 [d is a single digit]
+ • rem·B + d ≤ v·B - 1 [add]
+ • rem·B + d < v·B [change ≤ to <]
+ • (rem·B + d)/v < B [divide by v]
+
+
+Guess and Check
+
+At each step we need to divide n+1 digits by n digits, but this is for the
+implementation of division by n digits, so we can't just invoke a division
+routine: we _are_ the division routine. Instead, we guess at the answer and
+then check it using multiplication. If the guess is wrong, we correct it.
+
+How can this guessing possibly be efficient? It turns out that the following
+statement (let's call it the Good Guess Guarantee) is true.
+
+If
+
+ • q = ⌊u/v⌋ where u is n+1 digits and v is n digits,
+ • q < B, and
+ • the topmost digit of v = vₙ₋₁ ≥ B/2,
+
+then q̂ = ⌊uₙuₙ₋₁ / vₙ₋₁⌋ satisfies q ≤ q̂ ≤ q+2. (Proof below.)
+
+That is, if we know the answer has only a single digit and we guess an answer
+by ignoring the bottom n-1 digits of u and v, using a 2-by-1-digit division,
+then that guess is at least as large as the correct answer. It is also not
+too much larger: it is off by at most two from the correct answer.
+
+Note that in the first step of the overall division, which is an n-by-n-digit
+division, the 2-by-1 guess uses an implicit uₙ = 0.
+
+Note that using a 2-by-1-digit division here does not mean calling ourselves
+recursively. Instead, we use an efficient direct hardware implementation of
+that operation.
+
+Note that because q is u/v rounded down, q·v must not exceed u: u ≥ q·v.
+If a guess q̂ is too big, it will not satisfy this test. Viewed a different way,
+the remainder r̂ for a given q̂ is u - q̂·v, which must be positive. If it is
+negative, then the guess q̂ is too big.
+
+This gives us a way to compute q. First compute q̂ with 2-by-1-digit division.
+Then, while u < q̂·v, decrement q̂; this loop executes at most twice, because
+q̂ ≤ q+2.
+
+
+Scaling Inputs
+
+The Good Guess Guarantee requires that the top digit of v (vₙ₋₁) be at least B/2.
+For example in base 10, ⌊172/19⌋ = 9, but ⌊18/1⌋ = 18: the guess is wildly off
+because the first digit 1 is smaller than B/2 = 5.
+
+We can ensure that v has a large top digit by multiplying both u and v by the
+right amount. Continuing the example, if we multiply both 172 and 19 by 3, we
+now have ⌊516/57⌋, the leading digit of v is now ≥ 5, and sure enough
+⌊51/5⌋ = 10 is much closer to the correct answer 9. It would be easier here
+to multiply by 4, because that can be done with a shift. Specifically, we can
+always count the number of leading zeros i in the first digit of v and then
+shift both u and v left by i bits.
+
+Having scaled u and v, the value ⌊u/v⌋ is unchanged, but the remainder will
+be scaled: 172 mod 19 is 1, but 516 mod 57 is 3. We have to divide the remainder
+by the scaling factor (shifting right i bits) when we finish.
+
+Note that these shifts happen before and after the entire division algorithm,
+not at each step in the per-digit iteration.
+
+Note the effect of scaling inputs on the size of the possible quotient.
+In the scaled u/v, u can gain a digit from scaling; v never does, because we
+pick the scaling factor to make v's top digit larger but without overflowing.
+If u and v have n+m and n digits after scaling, then:
+
+ • u < B^(n+m) [B^(n+m) has n+m+1 digits]
+ • v ≥ B^n / 2 [vₙ₋₁ ≥ B/2, so vₙ₋₁·B^(n-1) ≥ B^n/2]
+ • u/v < B^(n+m) / (B^n / 2) [divide bounds for u, v]
+ • u/v < 2 B^m [simplify]
+
+The quotient can still have m+1 significant digits, but if so the top digit
+must be a 1. This provides a different way to handle the first digit of the
+result: compare the top n digits of u against v and fill in either a 0 or a 1.
+
+
+Refining Guesses
+
+Before we check whether u < q̂·v, we can adjust our guess to change it from
+q̂ = ⌊uₙuₙ₋₁ / vₙ₋₁⌋ into the refined guess ⌊uₙuₙ₋₁uₙ₋₂ / vₙ₋₁vₙ₋₂⌋.
+Although not mentioned above, the Good Guess Guarantee also promises that this
+3-by-2-digit division guess is more precise and at most one away from the real
+answer q. The improvement from the 2-by-1 to the 3-by-2 guess can also be done
+without n-digit math.
+
+If we have a guess q̂ = ⌊uₙuₙ₋₁ / vₙ₋₁⌋ and we want to see if it also equal to
+⌊uₙuₙ₋₁uₙ₋₂ / vₙ₋₁vₙ₋₂⌋, we can use the same check we would for the full division:
+if uₙuₙ₋₁uₙ₋₂ < q̂·vₙ₋₁vₙ₋₂, then the guess is too large and should be reduced.
+
+Checking uₙuₙ₋₁uₙ₋₂ < q̂·vₙ₋₁vₙ₋₂ is the same as uₙuₙ₋₁uₙ₋₂ - q̂·vₙ₋₁vₙ₋₂ < 0,
+and
+
+ uₙuₙ₋₁uₙ₋₂ - q̂·vₙ₋₁vₙ₋₂ = (uₙuₙ₋₁·B + uₙ₋₂) - q̂·(vₙ₋₁·B + vₙ₋₂)
+ [splitting off the bottom digit]
+ = (uₙuₙ₋₁ - q̂·vₙ₋₁)·B + uₙ₋₂ - q̂·vₙ₋₂
+ [regrouping]
+
+The expression (uₙuₙ₋₁ - q̂·vₙ₋₁) is the remainder of uₙuₙ₋₁ / vₙ₋₁.
+If the initial guess returns both q̂ and its remainder r̂, then checking
+whether uₙuₙ₋₁uₙ₋₂ < q̂·vₙ₋₁vₙ₋₂ is the same as checking r̂·B + uₙ₋₂ < q̂·vₙ₋₂.
+
+If we find that r̂·B + uₙ₋₂ < q̂·vₙ₋₂, then we can adjust the guess by
+decrementing q̂ and adding vₙ₋₁ to r̂. We repeat until r̂·B + uₙ₋₂ ≥ q̂·vₙ₋₂.
+(As before, this fixup is only needed at most twice.)
+
+Now that q̂ = ⌊uₙuₙ₋₁uₙ₋₂ / vₙ₋₁vₙ₋₂⌋, as mentioned above it is at most one
+away from the correct q, and we've avoided doing any n-digit math.
+(If we need the new remainder, it can be computed as r̂·B + uₙ₋₂ - q̂·vₙ₋₂.)
+
+The final check u < q̂·v and the possible fixup must be done at full precision.
+For random inputs, a fixup at this step is exceedingly rare: the 3-by-2 guess
+is not often wrong at all. But still we must do the check. Note that since the
+3-by-2 guess is off by at most 1, it can be convenient to perform the final
+u < q̂·v as part of the computation of the remainder r = u - q̂·v. If the
+subtraction underflows, decremeting q̂ and adding one v back to r is enough to
+arrive at the final q, r.
+
+That's the entirety of long division: scale the inputs, and then loop over
+each output position, guessing, checking, and correcting the next output digit.
+
+For a 2n-digit number divided by an n-digit number (the worst size-n case for
+division complexity), this algorithm uses n+1 iterations, each of which must do
+at least the 1-by-n-digit multiplication q̂·v. That's O(n) iterations of
+O(n) time each, so O(n²) time overall.
+
+
+Recursive Division
+
+For very large inputs, it is possible to improve on the O(n²) algorithm.
+Let's call a group of n/2 real digits a (very) “wide digit”. We can run the
+standard long division algorithm explained above over the wide digits instead of
+the actual digits. This will result in many fewer steps, but the math involved in
+each step is more work.
+
+Where basic long division uses a 2-by-1-digit division to guess the initial q̂,
+the new algorithm must use a 2-by-1-wide-digit division, which is of course
+really an n-by-n/2-digit division. That's OK: if we implement n-digit division
+in terms of n/2-digit division, the recursion will terminate when the divisor
+becomes small enough to handle with standard long division or even with the
+2-by-1 hardware instruction.
+
+For example, here is a sketch of dividing 10 digits by 4, proceeding with
+wide digits corresponding to two regular digits. The first step, still special,
+must leave off a (regular) digit, dividing 5 by 4 and producing a 4-digit
+remainder less than v. The middle steps divide 6 digits by 4, guaranteed to
+produce two output digits each (one wide digit) with 4-digit remainders.
+The final step must use what it has: the 4-digit remainder plus one more,
+5 digits to divide by 4.
+
+ q₆ q₅ q₄ q₃ q₂ q₁ q₀
+ _______________________________
+ v₃ v₂ v₁ v₀ ) u₉ u₈ u₇ u₆ u₅ u₄ u₃ u₂ u₁ u₀
+ ↓ ↓ ↓ ↓ ↓ | | | | |
+ [u₉ u₈ u₇ u₆ u₅]| | | | |
+ - [ q₆q₅·v ]| | | | |
+ ----------------- ↓ ↓ | | |
+ [ rem |u₄ u₃]| | |
+ - [ q₄q₃·v ]| | |
+ -------------------- ↓ ↓ |
+ [ rem |u₂ u₁]|
+ - [ q₂q₁·v ]|
+ -------------------- ↓
+ [ rem |u₀]
+ - [ q₀·v ]
+ ------------------
+ [ rem ]
+
+An alternative would be to look ahead to how well n/2 divides into n+m and
+adjust the first step to use fewer digits as needed, making the first step
+more special to make the last step not special at all. For example, using the
+same input, we could choose to use only 4 digits in the first step, leaving
+a full wide digit for the last step:
+
+ q₆ q₅ q₄ q₃ q₂ q₁ q₀
+ _______________________________
+ v₃ v₂ v₁ v₀ ) u₉ u₈ u₇ u₆ u₅ u₄ u₃ u₂ u₁ u₀
+ ↓ ↓ ↓ ↓ | | | | | |
+ [u₉ u₈ u₇ u₆]| | | | | |
+ - [ q₆·v ]| | | | | |
+ -------------- ↓ ↓ | | | |
+ [ rem |u₅ u₄]| | | |
+ - [ q₅q₄·v ]| | | |
+ -------------------- ↓ ↓ | |
+ [ rem |u₃ u₂]| |
+ - [ q₃q₂·v ]| |
+ -------------------- ↓ ↓
+ [ rem |u₁ u₀]
+ - [ q₁q₀·v ]
+ ---------------------
+ [ rem ]
+
+Today, the code in divRecursiveStep works like the first example. Perhaps in
+the future we will make it work like the alternative, to avoid a special case
+in the final iteration.
+
+Either way, each step is a 3-by-2-wide-digit division approximated first by
+a 2-by-1-wide-digit division, just as we did for regular digits in long division.
+Because the actual answer we want is a 3-by-2-wide-digit division, instead of
+multiplying q̂·v directly during the fixup, we can use the quick refinement
+from long division (an n/2-by-n/2 multiply) to correct q to its actual value
+and also compute the remainder (as mentioned above), and then stop after that,
+never doing a full n-by-n multiply.
+
+Instead of using an n-by-n/2-digit division to produce n/2 digits, we can add
+(not discard) one more real digit, doing an (n+1)-by-(n/2+1)-digit division that
+produces n/2+1 digits. That single extra digit tightens the Good Guess Guarantee
+to q ≤ q̂ ≤ q+1 and lets us drop long division's special treatment of the first
+digit. These benefits are discussed more after the Good Guess Guarantee proof
+below.
+
+
+How Fast is Recursive Division?
+
+For a 2n-by-n-digit division, this algorithm runs a 4-by-2 long division over
+wide digits, producing two wide digits plus a possible leading regular digit 1,
+which can be handled without a recursive call. That is, the algorithm uses two
+full iterations, each using an n-by-n/2-digit division and an n/2-by-n/2-digit
+multiplication, along with a few n-digit additions and subtractions. The standard
+n-by-n-digit multiplication algorithm requires O(n²) time, making the overall
+algorithm require time T(n) where
+
+ T(n) = 2T(n/2) + O(n) + O(n²)
+
+which, by the Bentley-Haken-Saxe theorem, ends up reducing to T(n) = O(n²).
+This is not an improvement over regular long division.
+
+When the number of digits n becomes large enough, Karatsuba's algorithm for
+multiplication can be used instead, which takes O(n^log₂3) = O(n^1.6) time.
+(Karatsuba multiplication is implemented in func karatsuba in nat.go.)
+That makes the overall recursive division algorithm take O(n^1.6) time as well,
+which is an improvement, but again only for large enough numbers.
+
+It is not critical to make sure that every recursion does only two recursive
+calls. While in general the number of recursive calls can change the time
+analysis, in this case doing three calls does not change the analysis:
+
+ T(n) = 3T(n/2) + O(n) + O(n^log₂3)
+
+ends up being T(n) = O(n^log₂3). Because the Karatsuba multiplication taking
+time O(n^log₂3) is itself doing 3 half-sized recursions, doing three for the
+division does not hurt the asymptotic performance. Of course, it is likely
+still faster in practice to do two.
+
+
+Proof of the Good Guess Guarantee
+
+Given numbers x, y, let us break them into the quotients and remainders when
+divided by some scaling factor S, with the added constraints that the quotient
+x/y and the high part of y are both less than some limit T, and that the high
+part of y is at least half as big as T.
+
+ x₁ = ⌊x/S⌋ y₁ = ⌊y/S⌋
+ x₀ = x mod S y₀ = y mod S
+
+ x = x₁·S + x₀ 0 ≤ x₀ < S x/y < T
+ y = y₁·S + y₀ 0 ≤ y₀ < S T/2 ≤ y₁ < T
+
+And consider the two truncated quotients:
+
+ q = ⌊x/y⌋
+ q̂ = ⌊x₁/y₁⌋
+
+We will prove that q ≤ q̂ ≤ q+2.
+
+The guarantee makes no real demands on the scaling factor S: it is simply the
+magnitude of the digits cut from both x and y to produce x₁ and y₁.
+The guarantee makes only limited demands on T: it must be large enough to hold
+the quotient x/y, and y₁ must have roughly the same size.
+
+To apply to the earlier discussion of 2-by-1 guesses in long division,
+we would choose:
+
+ S = Bⁿ⁻¹
+ T = B
+ x = u
+ x₁ = uₙuₙ₋₁
+ x₀ = uₙ₋₂...u₀
+ y = v
+ y₁ = vₙ₋₁
+ y₀ = vₙ₋₂...u₀
+
+These simpler variables avoid repeating those longer expressions in the proof.
+
+Note also that, by definition, truncating division ⌊x/y⌋ satisfies
+
+ x/y - 1 < ⌊x/y⌋ ≤ x/y.
+
+This fact will be used a few times in the proofs.
+
+Proof that q ≤ q̂:
+
+ q̂·y₁ = ⌊x₁/y₁⌋·y₁ [by definition, q̂ = ⌊x₁/y₁⌋]
+ > (x₁/y₁ - 1)·y₁ [x₁/y₁ - 1 < ⌊x₁/y₁⌋]
+ = x₁ - y₁ [distribute y₁]
+
+ So q̂·y₁ > x₁ - y₁.
+ Since q̂·y₁ is an integer, q̂·y₁ ≥ x₁ - y₁ + 1.
+
+ q̂ - q = q̂ - ⌊x/y⌋ [by definition, q = ⌊x/y⌋]
+ ≥ q̂ - x/y [⌊x/y⌋ < x/y]
+ = (1/y)·(q̂·y - x) [factor out 1/y]
+ ≥ (1/y)·(q̂·y₁·S - x) [y = y₁·S + y₀ ≥ y₁·S]
+ ≥ (1/y)·((x₁ - y₁ + 1)·S - x) [above: q̂·y₁ ≥ x₁ - y₁ + 1]
+ = (1/y)·(x₁·S - y₁·S + S - x) [distribute S]
+ = (1/y)·(S - x₀ - y₁·S) [-x = -x₁·S - x₀]
+ > -y₁·S / y [x₀ < S, so S - x₀ < 0; drop it]
+ ≥ -1 [y₁·S ≤ y]
+
+ So q̂ - q > -1.
+ Since q̂ - q is an integer, q̂ - q ≥ 0, or equivalently q ≤ q̂.
+
+Proof that q̂ ≤ q+2:
+
+ x₁/y₁ - x/y = x₁·S/y₁·S - x/y [multiply left term by S/S]
+ ≤ x/y₁·S - x/y [x₁S ≤ x]
+ = (x/y)·(y/y₁·S - 1) [factor out x/y]
+ = (x/y)·((y - y₁·S)/y₁·S) [move -1 into y/y₁·S fraction]
+ = (x/y)·(y₀/y₁·S) [y - y₁·S = y₀]
+ = (x/y)·(1/y₁)·(y₀/S) [factor out 1/y₁]
+ < (x/y)·(1/y₁) [y₀ < S, so y₀/S < 1]
+ ≤ (x/y)·(2/T) [y₁ ≥ T/2, so 1/y₁ ≤ 2/T]
+ < T·(2/T) [x/y < T]
+ = 2 [T·(2/T) = 2]
+
+ So x₁/y₁ - x/y < 2.
+
+ q̂ - q = ⌊x₁/y₁⌋ - q [by definition, q̂ = ⌊x₁/y₁⌋]
+ = ⌊x₁/y₁⌋ - ⌊x/y⌋ [by definition, q = ⌊x/y⌋]
+ ≤ x₁/y₁ - ⌊x/y⌋ [⌊x₁/y₁⌋ ≤ x₁/y₁]
+ < x₁/y₁ - (x/y - 1) [⌊x/y⌋ > x/y - 1]
+ = (x₁/y₁ - x/y) + 1 [regrouping]
+ < 2 + 1 [above: x₁/y₁ - x/y < 2]
+ = 3
+
+ So q̂ - q < 3.
+ Since q̂ - q is an integer, q̂ - q ≤ 2.
+
+Note that when x/y < T/2, the bounds tighten to x₁/y₁ - x/y < 1 and therefore
+q̂ - q ≤ 1.
+
+Note also that in the general case 2n-by-n division where we don't know that
+x/y < T, we do know that x/y < 2T, yielding the bound q̂ - q ≤ 4. So we could
+remove the special case first step of long division as long as we allow the
+first fixup loop to run up to four times. (Using a simple comparison to decide
+whether the first digit is 0 or 1 is still more efficient, though.)
+
+Finally, note that when dividing three leading base-B digits by two (scaled),
+we have T = B² and x/y < B = T/B, a much tighter bound than x/y < T.
+This in turn yields the much tighter bound x₁/y₁ - x/y < 2/B. This means that
+⌊x₁/y₁⌋ and ⌊x/y⌋ can only differ when x/y is less than 2/B greater than an
+integer. For random x and y, the chance of this is 2/B, or, for large B,
+approximately zero. This means that after we produce the 3-by-2 guess in the
+long division algorithm, the fixup loop essentially never runs.
+
+In the recursive algorithm, the extra digit in (2·⌊n/2⌋+1)-by-(⌊n/2⌋+1)-digit
+division has exactly the same effect: the probability of needing a fixup is the
+same 2/B. Even better, we can allow the general case x/y < 2T and the fixup
+probability only grows to 4/B, still essentially zero.
+
+
+References
+
+There are no great references for implementing long division; thus this comment.
+Here are some notes about what to expect from the obvious references.
+
+Knuth Volume 2 (Seminumerical Algorithms) section 4.3.1 is the usual canonical
+reference for long division, but that entire series is highly compressed, never
+repeating a necessary fact and leaving important insights to the exercises.
+For example, no rationale whatsoever is given for the calculation that extends
+q̂ from a 2-by-1 to a 3-by-2 guess, nor why it reduces the error bound.
+The proof that the calculation even has the desired effect is left to exercises.
+The solutions to those exercises provided at the back of the book are entirely
+calculations, still with no explanation as to what is going on or how you would
+arrive at the idea of doing those exact calculations. Nowhere is it mentioned
+that this test extends the 2-by-1 guess into a 3-by-2 guess. The proof of the
+Good Guess Guarantee is only for the 2-by-1 guess and argues by contradiction,
+making it difficult to understand how modifications like adding another digit
+or adjusting the quotient range affects the overall bound.
+
+All that said, Knuth remains the canonical reference. It is dense but packed
+full of information and references, and the proofs are simpler than many other
+presentations. The proofs above are reworkings of Knuth's to remove the
+arguments by contradiction and add explanations or steps that Knuth omitted.
+But beware of errors in older printings. Take the published errata with you.
+
+Brinch Hansen's “Multiple-length Division Revisited: a Tour of the Minefield”
+starts with a blunt critique of Knuth's presentation (among others) and then
+presents a more detailed and easier to follow treatment of long division,
+including an implementation in Pascal. But the algorithm and implementation
+work entirely in terms of 3-by-2 division, which is much less useful on modern
+hardware than an algorithm using 2-by-1 division. The proofs are a bit too
+focused on digit counting and seem needlessly complex, especially compared to
+the ones given above.
+
+Burnikel and Ziegler's “Fast Recursive Division” introduced the key insight of
+implementing division by an n-digit divisor using recursive calls to division
+by an n/2-digit divisor, relying on Karatsuba multiplication to yield a
+sub-quadratic run time. However, the presentation decisions are made almost
+entirely for the purpose of simplifying the run-time analysis, rather than
+simplifying the presentation. Instead of a single algorithm that loops over
+quotient digits, the paper presents two mutually-recursive algorithms, for
+2n-by-n and 3n-by-2n. The paper also does not present any general (n+m)-by-n
+algorithm.
+
+The proofs in the paper are remarkably complex, especially considering that
+the algorithm is at its core just long division on wide digits, so that the
+usual long division proofs apply essentially unaltered.
+*/
+
+package big
+
+import "math/bits"
+
+// div returns q, r such that q = ⌊u/v⌋ and r = u%v = u - q·v.
+// It uses z and z2 as the storage for q and r.
+func (z nat) div(z2, u, v nat) (q, r nat) {
+ if len(v) == 0 {
+ panic("division by zero")
+ }
+
+ if u.cmp(v) < 0 {
+ q = z[:0]
+ r = z2.set(u)
+ return
+ }
+
+ if len(v) == 1 {
+ // Short division: long optimized for a single-word divisor.
+ // In that case, the 2-by-1 guess is all we need at each step.
+ var r2 Word
+ q, r2 = z.divW(u, v[0])
+ r = z2.setWord(r2)
+ return
+ }
+
+ q, r = z.divLarge(z2, u, v)
+ return
+}
+
+// divW returns q, r such that q = ⌊x/y⌋ and r = x%y = x - q·y.
+// It uses z as the storage for q.
+// Note that y is a single digit (Word), not a big number.
+func (z nat) divW(x nat, y Word) (q nat, r Word) {
+ m := len(x)
+ switch {
+ case y == 0:
+ panic("division by zero")
+ case y == 1:
+ q = z.set(x) // result is x
+ return
+ case m == 0:
+ q = z[:0] // result is 0
+ return
+ }
+ // m > 0
+ z = z.make(m)
+ r = divWVW(z, 0, x, y)
+ q = z.norm()
+ return
+}
+
+// modW returns x % d.
+func (x nat) modW(d Word) (r Word) {
+ // TODO(agl): we don't actually need to store the q value.
+ var q nat
+ q = q.make(len(x))
+ return divWVW(q, 0, x, d)
+}
+
+// divWVW overwrites z with ⌊x/y⌋, returning the remainder r.
+// The caller must ensure that len(z) = len(x).
+func divWVW(z []Word, xn Word, x []Word, y Word) (r Word) {
+ r = xn
+ if len(x) == 1 {
+ qq, rr := bits.Div(uint(r), uint(x[0]), uint(y))
+ z[0] = Word(qq)
+ return Word(rr)
+ }
+ rec := reciprocalWord(y)
+ for i := len(z) - 1; i >= 0; i-- {
+ z[i], r = divWW(r, x[i], y, rec)
+ }
+ return r
+}
+
+// div returns q, r such that q = ⌊uIn/vIn⌋ and r = uIn%vIn = uIn - q·vIn.
+// It uses z and u as the storage for q and r.
+// The caller must ensure that len(vIn) ≥ 2 (use divW otherwise)
+// and that len(uIn) ≥ len(vIn) (the answer is 0, uIn otherwise).
+func (z nat) divLarge(u, uIn, vIn nat) (q, r nat) {
+ n := len(vIn)
+ m := len(uIn) - n
+
+ // Scale the inputs so vIn's top bit is 1 (see “Scaling Inputs” above).
+ // vIn is treated as a read-only input (it may be in use by another
+ // goroutine), so we must make a copy.
+ // uIn is copied to u.
+ shift := nlz(vIn[n-1])
+ vp := getNat(n)
+ v := *vp
+ shlVU(v, vIn, shift)
+ u = u.make(len(uIn) + 1)
+ u[len(uIn)] = shlVU(u[0:len(uIn)], uIn, shift)
+
+ // The caller should not pass aliased z and u, since those are
+ // the two different outputs, but correct just in case.
+ if alias(z, u) {
+ z = nil
+ }
+ q = z.make(m + 1)
+
+ // Use basic or recursive long division depending on size.
+ if n < divRecursiveThreshold {
+ q.divBasic(u, v)
+ } else {
+ q.divRecursive(u, v)
+ }
+ putNat(vp)
+
+ q = q.norm()
+
+ // Undo scaling of remainder.
+ shrVU(u, u, shift)
+ r = u.norm()
+
+ return q, r
+}
+
+// divBasic implements long division as described above.
+// It overwrites q with ⌊u/v⌋ and overwrites u with the remainder r.
+// q must be large enough to hold ⌊u/v⌋.
+func (q nat) divBasic(u, v nat) {
+ n := len(v)
+ m := len(u) - n
+
+ qhatvp := getNat(n + 1)
+ qhatv := *qhatvp
+
+ // Set up for divWW below, precomputing reciprocal argument.
+ vn1 := v[n-1]
+ rec := reciprocalWord(vn1)
+
+ // Compute each digit of quotient.
+ for j := m; j >= 0; j-- {
+ // Compute the 2-by-1 guess q̂.
+ // The first iteration must invent a leading 0 for u.
+ qhat := Word(_M)
+ var ujn Word
+ if j+n < len(u) {
+ ujn = u[j+n]
+ }
+
+ // ujn ≤ vn1, or else q̂ would be more than one digit.
+ // For ujn == vn1, we set q̂ to the max digit M above.
+ // Otherwise, we compute the 2-by-1 guess.
+ if ujn != vn1 {
+ var rhat Word
+ qhat, rhat = divWW(ujn, u[j+n-1], vn1, rec)
+
+ // Refine q̂ to a 3-by-2 guess. See “Refining Guesses” above.
+ vn2 := v[n-2]
+ x1, x2 := mulWW(qhat, vn2)
+ ujn2 := u[j+n-2]
+ for greaterThan(x1, x2, rhat, ujn2) { // x1x2 > r̂ u[j+n-2]
+ qhat--
+ prevRhat := rhat
+ rhat += vn1
+ // If r̂ overflows, then
+ // r̂ u[j+n-2]v[n-1] is now definitely > x1 x2.
+ if rhat < prevRhat {
+ break
+ }
+ // TODO(rsc): No need for a full mulWW.
+ // x2 += vn2; if x2 overflows, x1++
+ x1, x2 = mulWW(qhat, vn2)
+ }
+ }
+
+ // Compute q̂·v.
+ qhatv[n] = mulAddVWW(qhatv[0:n], v, qhat, 0)
+ qhl := len(qhatv)
+ if j+qhl > len(u) && qhatv[n] == 0 {
+ qhl--
+ }
+
+ // Subtract q̂·v from the current section of u.
+ // If it underflows, q̂·v > u, which we fix up
+ // by decrementing q̂ and adding v back.
+ c := subVV(u[j:j+qhl], u[j:], qhatv)
+ if c != 0 {
+ c := addVV(u[j:j+n], u[j:], v)
+ // If n == qhl, the carry from subVV and the carry from addVV
+ // cancel out and don't affect u[j+n].
+ if n < qhl {
+ u[j+n] += c
+ }
+ qhat--
+ }
+
+ // Save quotient digit.
+ // Caller may know the top digit is zero and not leave room for it.
+ if j == m && m == len(q) && qhat == 0 {
+ continue
+ }
+ q[j] = qhat
+ }
+
+ putNat(qhatvp)
+}
+
+// greaterThan reports whether the two digit numbers x1 x2 > y1 y2.
+// TODO(rsc): In contradiction to most of this file, x1 is the high
+// digit and x2 is the low digit. This should be fixed.
+func greaterThan(x1, x2, y1, y2 Word) bool {
+ return x1 > y1 || x1 == y1 && x2 > y2
+}
+
+// divRecursiveThreshold is the number of divisor digits
+// at which point divRecursive is faster than divBasic.
+const divRecursiveThreshold = 100
+
+// divRecursive implements recursive division as described above.
+// It overwrites z with ⌊u/v⌋ and overwrites u with the remainder r.
+// z must be large enough to hold ⌊u/v⌋.
+// This function is just for allocating and freeing temporaries
+// around divRecursiveStep, the real implementation.
+func (z nat) divRecursive(u, v nat) {
+ // Recursion depth is (much) less than 2 log₂(len(v)).
+ // Allocate a slice of temporaries to be reused across recursion,
+ // plus one extra temporary not live across the recursion.
+ recDepth := 2 * bits.Len(uint(len(v)))
+ tmp := getNat(3 * len(v))
+ temps := make([]*nat, recDepth)
+
+ z.clear()
+ z.divRecursiveStep(u, v, 0, tmp, temps)
+
+ // Free temporaries.
+ for _, n := range temps {
+ if n != nil {
+ putNat(n)
+ }
+ }
+ putNat(tmp)
+}
+
+// divRecursiveStep is the actual implementation of recursive division.
+// It adds ⌊u/v⌋ to z and overwrites u with the remainder r.
+// z must be large enough to hold ⌊u/v⌋.
+// It uses temps[depth] (allocating if needed) as a temporary live across
+// the recursive call. It also uses tmp, but not live across the recursion.
+func (z nat) divRecursiveStep(u, v nat, depth int, tmp *nat, temps []*nat) {
+ // u is a subsection of the original and may have leading zeros.
+ // TODO(rsc): The v = v.norm() is useless and should be removed.
+ // We know (and require) that v's top digit is ≥ B/2.
+ u = u.norm()
+ v = v.norm()
+ if len(u) == 0 {
+ z.clear()
+ return
+ }
+
+ // Fall back to basic division if the problem is now small enough.
+ n := len(v)
+ if n < divRecursiveThreshold {
+ z.divBasic(u, v)
+ return
+ }
+
+ // Nothing to do if u is shorter than v (implies u < v).
+ m := len(u) - n
+ if m < 0 {
+ return
+ }
+
+ // We consider B digits in a row as a single wide digit.
+ // (See “Recursive Division” above.)
+ //
+ // TODO(rsc): rename B to Wide, to avoid confusion with _B,
+ // which is something entirely different.
+ // TODO(rsc): Look into whether using ⌈n/2⌉ is better than ⌊n/2⌋.
+ B := n / 2
+
+ // Allocate a nat for qhat below.
+ if temps[depth] == nil {
+ temps[depth] = getNat(n) // TODO(rsc): Can be just B+1.
+ } else {
+ *temps[depth] = temps[depth].make(B + 1)
+ }
+
+ // Compute each wide digit of the quotient.
+ //
+ // TODO(rsc): Change the loop to be
+ // for j := (m+B-1)/B*B; j > 0; j -= B {
+ // which will make the final step a regular step, letting us
+ // delete what amounts to an extra copy of the loop body below.
+ j := m
+ for j > B {
+ // Divide u[j-B:j+n] (3 wide digits) by v (2 wide digits).
+ // First make the 2-by-1-wide-digit guess using a recursive call.
+ // Then extend the guess to the full 3-by-2 (see “Refining Guesses”).
+ //
+ // For the 2-by-1-wide-digit guess, instead of doing 2B-by-B-digit,
+ // we use a (2B+1)-by-(B+1) digit, which handles the possibility that
+ // the result has an extra leading 1 digit as well as guaranteeing
+ // that the computed q̂ will be off by at most 1 instead of 2.
+
+ // s is the number of digits to drop from the 3B- and 2B-digit chunks.
+ // We drop B-1 to be left with 2B+1 and B+1.
+ s := (B - 1)
+
+ // uu is the up-to-3B-digit section of u we are working on.
+ uu := u[j-B:]
+
+ // Compute the 2-by-1 guess q̂, leaving r̂ in uu[s:B+n].
+ qhat := *temps[depth]
+ qhat.clear()
+ qhat.divRecursiveStep(uu[s:B+n], v[s:], depth+1, tmp, temps)
+ qhat = qhat.norm()
+
+ // Extend to a 3-by-2 quotient and remainder.
+ // Because divRecursiveStep overwrote the top part of uu with
+ // the remainder r̂, the full uu already contains the equivalent
+ // of r̂·B + uₙ₋₂ from the “Refining Guesses” discussion.
+ // Subtracting q̂·vₙ₋₂ from it will compute the full-length remainder.
+ // If that subtraction underflows, q̂·v > u, which we fix up
+ // by decrementing q̂ and adding v back, same as in long division.
+
+ // TODO(rsc): Instead of subtract and fix-up, this code is computing
+ // q̂·vₙ₋₂ and decrementing q̂ until that product is ≤ u.
+ // But we can do the subtraction directly, as in the comment above
+ // and in long division, because we know that q̂ is wrong by at most one.
+ qhatv := tmp.make(3 * n)
+ qhatv.clear()
+ qhatv = qhatv.mul(qhat, v[:s])
+ for i := 0; i < 2; i++ {
+ e := qhatv.cmp(uu.norm())
+ if e <= 0 {
+ break
+ }
+ subVW(qhat, qhat, 1)
+ c := subVV(qhatv[:s], qhatv[:s], v[:s])
+ if len(qhatv) > s {
+ subVW(qhatv[s:], qhatv[s:], c)
+ }
+ addAt(uu[s:], v[s:], 0)
+ }
+ if qhatv.cmp(uu.norm()) > 0 {
+ panic("impossible")
+ }
+ c := subVV(uu[:len(qhatv)], uu[:len(qhatv)], qhatv)
+ if c > 0 {
+ subVW(uu[len(qhatv):], uu[len(qhatv):], c)
+ }
+ addAt(z, qhat, j-B)
+ j -= B
+ }
+
+ // TODO(rsc): Rewrite loop as described above and delete all this code.
+
+ // Now u < (v<<B), compute lower bits in the same way.
+ // Choose shift = B-1 again.
+ s := B - 1
+ qhat := *temps[depth]
+ qhat.clear()
+ qhat.divRecursiveStep(u[s:].norm(), v[s:], depth+1, tmp, temps)
+ qhat = qhat.norm()
+ qhatv := tmp.make(3 * n)
+ qhatv.clear()
+ qhatv = qhatv.mul(qhat, v[:s])
+ // Set the correct remainder as before.
+ for i := 0; i < 2; i++ {
+ if e := qhatv.cmp(u.norm()); e > 0 {
+ subVW(qhat, qhat, 1)
+ c := subVV(qhatv[:s], qhatv[:s], v[:s])
+ if len(qhatv) > s {
+ subVW(qhatv[s:], qhatv[s:], c)
+ }
+ addAt(u[s:], v[s:], 0)
+ }
+ }
+ if qhatv.cmp(u.norm()) > 0 {
+ panic("impossible")
+ }
+ c := subVV(u[0:len(qhatv)], u[0:len(qhatv)], qhatv)
+ if c > 0 {
+ c = subVW(u[len(qhatv):], u[len(qhatv):], c)
+ }
+ if c > 0 {
+ panic("impossible")
+ }
+
+ // Done!
+ addAt(z, qhat.norm(), 0)
+}
generated by cgit v1.1 at 2024-08-21 17:41:47 +0000