问题
There are a couple of ways to find integer square roots using only integer arithmetic. For example this one. It makes for interesting reading and also a very interesting theory, particularly for my generation where such techniques aren't so useful any more.
The main thing is that it can't use floating point arithmetic, so that rules out newtons method and it's derivations. The only other way I know of to find roots is binomial expansion, but that also requires floating point arithmetic.
What techniques/algorithms are there for computing integral nth roots using only integer arithmetic?
Edit: Thanks for all the answers so far. They all seem to be slightly more intelligent trial and improvement. Is there no better way?
Edit2: Ok, so it would seem there is no smart way to do this without trial/improvement and either newtons method or a binary search. Can anyone provide a comparison of the two in theory? I have run a number of benchmarks between the two and found them quite similar.
回答1:
You can use Newton's method using only integer arithmetic, the step is the same as for floating point arithmetic, except you have to replace floating point operators with the corresponding integer operators in languages which have different operators for these.
Let's say you want to find the integer-k-th root of a > 0, which should be the largest integer r such that r^k <= a. You start with any positive integer (of course a good starting point helps).
int_type step(int_type k, int_type a, int_type x) {
return ((k-1)*x + a/x^(k-1))/k;
}
int_type root(int_type k, int_type a) {
int_type x = 1, y = step(k,a,x);
do {
x = y;
y = step(k,a,x);
}while(y < x);
return x;
}
Except for the very first step, you have x == r <==> step(k,a,x) >= x.
回答2:
One obvious way would be to use binary search together with exponentiation by squaring. This will allow you to find nthRoot(x, n) in O(log (x + n)): binary search in [0, x] for the largest integer k such that k^n <= x. For some k, if k^n <= x, reduce the search to [k + 1, x], otherwise reduce it to [0, k].
Do you require something smarter or faster?
回答3:
It seems to me that the Shifting nth root algorithm provides exactly what you want:
The shifting nth root algorithm is an algorithm for extracting the nth root of a positive real number which proceeds iteratively by shifting in n digits of the radicand, starting with the most significant, and produces one digit of the root on each iteration, in a manner similar to long division.
There are worked examples on the linked wikipedia page.
回答4:
One easy solution is to use the binary search.
Assume we are finding nth root of x.
Function GetRange(x,n):
y=1
While y^n < x:
y*2
return (y/2,y)
Function BinSearch(a,b,x,):
if a == b+1:
if x-a^n < b^n - x:
return a
else:
return b
c = (a+b)/2
if n< c^n:
return BinSearch(a,c,x,n)
else:
return BinSearch(c,b,x,n)
a,b = GetRange(x,n)
print BinSearch(a,b,x,n)
===Faster Version===
Function BinSearch(a,b,x,):
w1 = x-a^n
w2 = b^n - x
if a <= b+1:
if w1 < w2:
return a
else:
return b
c = (w2*a+w1*b)/(w1+w2)
if n< c^n:
return BinSearch(a,c,x,n)
else:
return BinSearch(c,b,x,n)
回答5:
Algorithm more simple in VBA.
Public Function RootNth(radicand As Double, degree As Long) As Double
Dim countDigits As Long, digit As Long, potency As Double
Dim minDigit As Long, maxDigit As Long, partialRadicand As String
Dim totalRadicand As String, remainder As Double
radicand = Int(radicand)
degree = Abs(degree)
RootNth = 0
partialRadicand = ""
totalRadicand = CStr(radicand)
countDigits = Len(totalRadicand) Mod degree
countDigits = IIf(countDigits = 0, degree, countDigits)
Do While totalRadicand <> ""
partialRadicand = partialRadicand + Left(totalRadicand, countDigits)
totalRadicand = Mid(totalRadicand, countDigits + 1)
countDigits = degree
minDigit = 0
maxDigit = 9
Do While minDigit <= maxDigit
digit = Int((minDigit + maxDigit) / 2)
potency = (RootNth * 10 + digit) ^ degree
If potency = Val(partialRadicand) Then
maxDigit = digit
Exit Do
End If
If potency < Val(partialRadicand) Then
minDigit = digit + 1
Else
maxDigit = digit - 1
End If
Loop
RootNth = RootNth * 10 + maxDigit
Loop
End Function
回答6:
I made the algorithm in VBA in Excel. For now it only calculates roots of integers. It is easy to implement the decimals as well.
Just copy and paste the code into an EXCEL module and type the name of the function into some cell, passing the parameters.
Public Function RootShift(ByVal radicand As Double, degree As Long, Optional ByRef remainder As Double = 0) As Double
Dim fullRadicand As String, partialRadicand As String, missingZeroes As Long, digit As Long
Dim minimalPotency As Double, minimalRemainder As Double, potency As Double
radicand = Int(radicand)
degree = Abs(degree)
fullRadicand = CStr(radicand)
missingZeroes = degree - Len(fullRadicand) Mod degree
If missingZeroes < degree Then
fullRadicand = String(missingZeroes, "0") + fullRadicand
End If
remainder = 0
RootShift = 0
Do While fullRadicand <> ""
partialRadicand = Left(fullRadicand, degree)
fullRadicand = Mid(fullRadicand, degree + 1)
minimalPotency = (RootShift * 10) ^ degree
minimalRemainder = remainder * 10 ^ degree + Val(partialRadicand)
For digit = 9 To 0 Step -1
potency = (RootShift * 10 + digit) ^ degree - minimalPotency
If potency <= minimalRemainder Then
Exit For
End If
Next
RootShift = RootShift * 10 + digit
remainder = minimalRemainder - potency
Loop
End Function
来源：https://stackoverflow.com/questions/8826822/calculate-nth-root-with-integer-arithmetic