I writing a program to numerically find the roots of functions with irrational roots by various methods.
For methods such as linear interpolation, you need to find the approximate range in which a root lies, for this I wrote this code:
bool fxn1 = false;
bool fxn2 = false;
vector<float> root_list;
if(f_x(-100) < 0)
{
fxn2 = true;
}
for(float i = -99.99; i < 100.01; i += 0.01)
{
fxn1 = fxn2;
if(f_x(i) < 0)
{
fxn2 = true;
}
else
{
fxn2 = false;
}
if((fxn1 == false && fxn2 == true) || (fxn1 == true && fxn2 == false))
{
root_list.push_back(i-0.01);
root_list.push_back(i);
}
}
However, for non-continuous functions (i.e. functions with asymptotes), this code will also be triggered when the function swaps from positive to negative values either side of the asymptote.
Is there a way to get the program to tell the difference between a root and an asymptote?
Thanks in advance
If the function, f(x), is converging on a point inside [a,b] then the half-way point (a + b) / 2 should be closer to zero than a or b.
This observation leads to the following procedure:
Let mid = (a + b) / 2
If |f(mid)| < |f(a)| AND |f(mid)| < |f(b)| Then
Algorithm has converged to a root
Else
Algorithm has converged to an asymptote
End
In this pseudo code |.| denotes floating-point absolute value.
Finding numerically a root only make sense if the function has nice properties, and at least is continuous. What would you think about this one:
f: x -> f(x) defined by:
2 * i < x < 2 * i + 1 (i element of Z) : f(x) = x
2 - i + 1 < x < 2 * i (i element of Z) : f(x) = -x
x = i (i element of Z) : f(x) = 1
It is perfectly defined on R, is bounded on any bounded interval, has positive and negative values on any interval of size > 1, and is continuous on any non integer point, but it has no root.
It is simply because the rule that a root must exist on segment ]x, y[ if x < 0 < y or y < 0 < x only applies if the function is continuous on the interval.
And good luck if you want to numerically test for continuity of a function...
i am working on an algorithm and want to make my code more efficient.my code uses simple arithmetic and comparison statements.however,i want to replace if statements,as they could be time consuming.this code would be run over a million times,so even the slightest improvement is appreciated.please answer!here is the code-
int_1024 sqcalc(int_1024 s,int_1024 f){
f=f*20;
s=s-81;
s=s-(f*9);
if(s>=0){
return 9;
}
s=s+f;
s=s+17;
if(s>=0){
return 8;
}
s=s+f;
s=s+15;
if(s>=0){
return 7;
}
s=s+f;
s=s+13;
if(s>=0){
return 6;
}
s=s+f;
s=s+11;
if(s>=0){
return 5;
}
s=s+f;
s=s+9;
if(s>=0){
return 4;
}
s=s+f;
s=s+7;
if(s>=0){
return 3;
}
s=s+f;
s=s+5;
if(s>=0){
return 2;
}
s=s+f;
s=s+3;
if(s>=0){
return 1;
}
s=s+f;
s=s+1;
if(s>=0){
return 0;
}
}
i wish to replace if checks,since i 'think' they make the algorithm slow.any suggestions?int_1024 is a ttmath variable with 1000's of bits,so saving on it might be a good option?division or multiplication for such a large number might be slow,so i tried using addition,but to no avail.help please.
I don't know if it is any faster, but it is considerably shorter (and easier to analyze).
int k[] = { 17, 15, 13, 11, 9, 7, 5, 3, 1 };
int r = 0;
f *= 20;
s -= 81;
s -= f * 9;
while (s < 0) {
s += f;
s += k[r];
if (++r == 9) break;
}
if (s >= 0) return 9-r;
Edit:
In fact, the original poster came up with a clever way to optimize this loop by pre-computing the sum of constants in the k array, and compared s against the sums, rather than incrementally adding them to s.
Edit:
I followed moonshadow's analysis technique, but arrived at a different equation. Original TeX formatting replaced with ASCII art (I tried to get MathJax to render the TeX for me, but it wasn't working):
S[0] = s >= 0 => 9 - 0
S[1] = S[0] + f + 19 - 2*1 >= 0 => 9 - 1
S[2] = S[1] + f + 19 - 2*2 >= 0 => 9 - 2
...
S[i] = S[i-1] + f + 19 - 2*i >= 0 => 9 - i
So to calculate S[n]:
S[n] = S[n-1] + f + 19 - 2n
.-- n
=> S[n] = s + > (f + 19 - 2*i)
`-- i=1 .-- n
=> S[n] = s + n(f + 19) - 2 > i
`-- i=1
=> S[n] = s + n(f + 19) - n(n+1)
2
=> S[n] = s + n(f + 18) - n
So, the inequality S[n] >= 0 is a quadratic equation in n. Assuming s < 0, we want n to be the ceiling of the solution to the quadratic.
+-- --+
| _____________ |
| / 2 |
| f + 18 - . / (f + 18) + 4s |
| ` |
n = | --------------------------- |
| 2 |
So the routine would look something like:
f *= 180;
s -= 81;
s -= f;
if (s >= 0) return 9;
f /= 9;
f += 18;
s *= 4;
int1024_t ff = f;
ff *= f;
ff += s;
ff = ff.Sqrt();
f -= ff;
f += f.Mod2();
return 9 - f/2;
However, I am not sure the expense of performing these operations on your big integer objects is worth implementing to replace the simple loop shown above. (Unless you expect to extend the function and would require a much longer loop.)
To be faster than the loop, the big integer square root implementation would have to always converge within 4 iterations to beat the average expected 4.5 iterations of the existing while loop. However the ttmath implementation does not seem to be calculating an integer square root. It seems to calculate a floating point square root and then rounding the result, which I would guess would be much slower than the loop.
First of all, I note that if the condition of the final if() is false, the return value is undefined. You probably want to fix that.
Now, the function starts with
f=f*20;
s=s-81;
s=s-(f*9);
if(s>=0){
return 9;
}
and the rest looks incredibly repetitive. Let's see if we can use that repetition. Let's build a table of inequalities - values of s vs the eventual result:
s + (f+17) >= 0: 8
s + (f+17) + (f+15) >= 0: 7
s + (f+17) + (f+15) + (f+13) >= 0: 6
.
.
s + (f+17) + (f+15) + (f+13) + ... + (f+1) >= 0: 0
So, each line tests to see if s + some multiple of f + some constant is greater than 0. The value returned, the constant and the multiple of f all look related. Let's try expressing the relationship:
(s + ((9-n)*f) + (2*n)-1 >= 0)
Let's rearrange that so n is on one side.
(s + (9*f) - (n*f) + (2*n)-1 >= 0)
(s + (9*f) +1 >= (n*f) - (2*n))
(s + (9*f) +1 >= n*(f - 2))
n <= ((s + (9*f) +1) / (f - 2)
Now, the function has a range of return values for different inputs. In fact, we are interested in values of n in the range 0..8: the function supplied is undefined for inputs that would yield n < 0 (see above). The preamble ensures we never see inputs that would yield n > 8.
So we can just say
int_1024 sqcalc(int_1024 s,int_1024 f){
f=f*20;
s=s-81;
s=s-(f*9);
if(s>=0){
return 9;
}
return (s + (9*f) +1) / (f - 2);
}
and for all cases where the result is not undefined, the behaviour should be the same as the old version, without needing tons of conditionals or a loop.
Demonstration of accuracy is at http://ideone.com/UzMZs.
According to the OP's comment, the function is trying to find all values that satisfy the inequality:
N * ((20 * F) + N) <= S
For all N, given an F and S.
Using algebra, this comes out to:
1) N^2 + 20Fn - S <= 0 (where N^2 is N*N or sqr(N))
The OP should use some constants for F and N and solve algebraically (sp?) or search the web for "C++ find root quadratic equation".
One a function is selected, then profile the function and optimize if necessary.
i tried solving the quadratics,and it makes the function slower for larger digits.following the answer by #user315052,i made this code.
int_1024 sqcalc(int_1024 s,int_1024 f){
int k[] = { 0, 17, 32, 45, 56, 65, 72, 77, 80, 81 };
f=f*20;
s=((f*9)+81)-s;
int i=0;
while(s>k[i]){
s-=f;
i++;
}
return 9-i;
}
in this code,instead of subtracting a number and then comparing with zero,i directly compare it with the number.by far,this produces the fastest results.i could do binary search though....
I recently came across the following interview question, I was wondering if a dynamic programming approach would work, or/and if there was some kind of mathematical insight that would make the solution easier... Its very similar to how ieee754 doubles are constructed.
Question:
There is vector V of N double values. Where the value at the ith index of the vector is equal to 1/2^(i+1). eg: 1/2, 1/4, 1/8, 1/16 etc...
You're to write a function that takes one double 'r' as input, where 0 < r < 1, and output the indexes of V to stdout that when summed will give a value closest to the value 'r' than any other combination of indexes from the vector V.
Furthermore the number of indexes should be a minimum, and in the event there are two solutions, the solution closest to zero should be preferred.
void getIndexes(std::vector<double>& V, double r)
{
....
}
int main()
{
std::vector<double> V;
// populate V...
double r = 0.3;
getIndexes(V,r);
return 0;
}
Note: It seems like there are a few SO'ers that aren't in the mood of reading the question completely. So lets all note the following:
The solution, aka the sum may be larger than r - hence any strategy incrementally subtracting fractions from r, until it hits zero or near zero is wrong
There are examples of r, where there will be 2 solutions, that is |r-s0| == |r-s1| and s0 < s1 - in this case s0 should be selected, this makes the problem slightly more difficult, as the knapsack style solutions tend to greedy overestimates first.
If you believe this problem is trivial, you most likely haven't understood it. Hence it would be a good idea to read the question again.
EDIT (Matthieu M.): 2 examples for V = {1/2, 1/4, 1/8, 1/16, 1/32}
r = 0.3, S = {1, 3}
r = 0.256652, S = {1}
Algorithm
Consider a target number r and a set F of fractions {1/2, 1/4, ... 1/(2^N)}. Let the smallest fraction, 1/(2^N), be denoted P.
Then the optimal sum will be equal to:
S = P * round(r/P)
That is, the optimal sum S will be some integer multiple of the smallest fraction available, P. The maximum error, err = r - S, is ± 1/2 * 1/(2^N). No better solution is possible because this would require the use of a number smaller than 1/(2^N), which is the smallest number in the set F.
Since the fractions F are all power-of-two multiples of P = 1/(2^N), any integer multiple of P can be expressed as a sum of the fractions in F. To obtain the list of fractions that should be used, encode the integer round(r/P) in binary and read off 1 in the kth binary place as "include the kth fraction in the solution".
Example:
Take r = 0.3 and F as {1/2, 1/4, 1/8, 1/16, 1/32}.
Multiply the entire problem by 32.
Take r = 9.6, and F as {16, 8, 4, 2, 1}.
Round r to the nearest integer.
Take r = 10.
Encode 10 as a binary integer (five places)
10 = 0b 0 1 0 1 0 ( 8 + 2 )
^ ^ ^ ^ ^
| | | | |
| | | | 1
| | | 2
| | 4
| 8
16
Associate each binary bit with a fraction.
= 0b 0 1 0 1 0 ( 1/4 + 1/16 = 0.3125 )
^ ^ ^ ^ ^
| | | | |
| | | | 1/32
| | | 1/16
| | 1/8
| 1/4
1/2
Proof
Consider transforming the problem by multiplying all the numbers involved by 2**N so that all the fractions become integers.
The original problem:
Consider a target number r in the range 0 < r < 1, and a list of fractions {1/2, 1/4, .... 1/(2**N). Find the subset of the list of fractions that sums to S such that error = r - S is minimised.
Becomes the following equivalent problem (after multiplying by 2**N):
Consider a target number r in the range 0 < r < 2**N and a list of integers {2**(N-1), 2**(N-2), ... , 4, 2, 1}. Find the subset of the list of integers that sums to S such that error = r - S is minimised.
Choosing powers of two that sum to a given number (with as little error as possible) is simply binary encoding of an integer. This problem therefore reduces to binary encoding of a integer.
Existence of solution: Any positive floating point number r, 0 < r < 2**N, can be cast to an integer and represented in binary form.
Optimality: The maximum error in the integer version of the solution is the round-off error of ±0.5. (In the original problem, the maximum error is ±0.5 * 1/2**N.)
Uniqueness: for any positive (floating point) number there is a unique integer representation and therefore a unique binary representation. (Possible exception of 0.5 = see below.)
Implementation (Python)
This function converts the problem to the integer equivalent, rounds off r to an integer, then reads off the binary representation of r as an integer to get the required fractions.
def conv_frac (r,N):
# Convert to equivalent integer problem.
R = r * 2**N
S = int(round(R))
# Convert integer S to N-bit binary representation (i.e. a character string
# of 1's and 0's.) Note use of [2:] to trim leading '0b' and zfill() to
# zero-pad to required length.
bin_S = bin(S)[2:].zfill(N)
nums = list()
for index, bit in enumerate(bin_S):
k = index + 1
if bit == '1':
print "%i : 1/%i or %f" % (index, 2**k, 1.0/(2**k))
nums.append(1.0/(2**k))
S = sum(nums)
e = r - S
print """
Original number `r` : %f
Number of fractions `N` : %i (smallest fraction 1/%i)
Sum of fractions `S` : %f
Error `e` : %f
""" % (r,N,2**N,S,e)
Sample output:
>>> conv_frac(0.3141,10)
1 : 1/4 or 0.250000
3 : 1/16 or 0.062500
8 : 1/512 or 0.001953
Original number `r` : 0.314100
Number of fractions `N` : 10 (smallest fraction 1/1024)
Sum of fractions `S` : 0.314453
Error `e` : -0.000353
>>> conv_frac(0.30,5)
1 : 1/4 or 0.250000
3 : 1/16 or 0.062500
Original number `r` : 0.300000
Number of fractions `N` : 5 (smallest fraction 1/32)
Sum of fractions `S` : 0.312500
Error `e` : -0.012500
Addendum: the 0.5 problem
If r * 2**N ends in 0.5, then it could be rounded up or down. That is, there are two possible representations as a sum-of-fractions.
If, as in the original problem statement, you want the representation that uses fewest fractions (i.e. the least number of 1 bits in the binary representation), just try both rounding options and pick whichever one is more economical.
Perhaps I am dumb...
The only trick I can see here is that the sum of (1/2)^(i+1) for i in [0..n) where n tends towards infinity gives 1. This simple fact proves that (1/2)^i is always superior to sum (1/2)^j for j in [i+1, n), whatever n is.
So, when looking for our indices, it does not seem we have much choice. Let's start with i = 0
either r is superior to 2^-(i+1) and thus we need it
or it is inferior and we need to choose whether 2^-(i+1) OR sum 2^-j for j in [i+2, N] is closest (deferring to the latter in case of equality)
The only step that could be costly is obtaining the sum, but it can be precomputed once and for all (and even precomputed lazily).
// The resulting vector contains at index i the sum of 2^-j for j in [i+1, N]
// and is padded with one 0 to get the same length as `v`
static std::vector<double> partialSums(std::vector<double> const& v) {
std::vector<double> result;
// When summing doubles, we need to start with the smaller ones
// because of the precision of representations...
double sum = 0;
BOOST_REVERSE_FOREACH(double d, v) {
sum += d;
result.push_back(sum);
}
result.pop_back(); // there is a +1 offset in the indexes of the result
std::reverse(result.begin(), result.end());
result.push_back(0); // pad the vector to have the same length as `v`
return result;
}
// The resulting vector contains the indexes elected
static std::vector<size_t> getIndexesImpl(std::vector<double> const& v,
std::vector<double> const& ps,
double r)
{
std::vector<size_t> indexes;
for (size_t i = 0, max = v.size(); i != max; ++i) {
if (r >= v[i]) {
r -= v[i];
indexes.push_back(i);
continue;
}
// We favor the closest to 0 in case of equality
// which is the sum of the tail as per the theorem above.
if (std::fabs(r - v[i]) < std::fabs(r - ps[i])) {
indexes.push_back(i);
return indexes;
}
}
return indexes;
}
std::vector<size_t> getIndexes(std::vector<double>& v, double r) {
std::vector<double> const ps = partialSums(v);
return getIndexesImpl(v, ps, r);
}
The code runs (with some debug output) at ideone. Note that for 0.3 it gives:
0.3:
1: 0.25
3: 0.0625
=> 0.3125
which is slightly different from the other answers.
At the risk of downvotes, this problem seems to be rather straightforward. Just start with the largest and smallest numbers you can produce out of V, adjust each index in turn until you have the two possible closest answers. Then evaluate which one is the better answer.
Here is untested code (in a language that I don't write):
void getIndexes(std::vector<double>& V, double r)
{
double v_lower = 0;
double v_upper = 1.0 - 0.5**V.size();
std::vector<int> index_lower;
std::vector<int> index_upper;
if (v_upper <= r)
{
// The answer is trivial.
for (int i = 0; i < V.size(); i++)
cout << i;
return;
}
for (int i = 0; i < N; i++)
{
if (v_lower + V[i] <= r)
{
v_lower += V[i];
index_lower.push_back(i);
}
if (r <= v_upper - V[i])
v_upper -= V[i];
else
index_upper.push_back(i);
}
if (r - v_lower < v_upper - r)
printIndexes(index_lower);
else if (v_upper - r < r - v_lower)
printIndexes(index_upper);
else if (v_upper.size() < v_lower.size())
printIndexes(index_upper);
else
printIndexes(index_lower);
}
void printIndexes(std::vector<int>& ind)
{
for (int i = 0; i < ind.size(); i++)
{
cout << ind[i];
}
}
Did I get the job! :D
(Please note, this is horrible code that relies on our knowing exactly what V has in it...)
I will start by saying that I do believe that this problem is trivial...
(waits until all stones have been thrown)
Yes, I did read the OP's edit that says that I have to re-read the question if I think so. Therefore I might be missing something that I fail to see - in this case please excuse my ignorance and feel free to point out my mistakes.
I don't see this as a dynamic programming problem. At the risk of sounding naive, why not try keeping two estimations of r while searching for indices - namely an under-estimation and an over-estimation. After all, if r does not equal any sum that can be computed from elements of V, it will lie between some two sums of the kind. Our goal is to find these sums and to report which is closer to r.
I threw together some quick-and-dirty Python code that does the job. The answer it reports is correct for the two test cases that the OP provided. Note that if the return is structured such that at least one index always has to be returned - even if the best estimation is no indices at all.
def estimate(V, r):
lb = 0 # under-estimation (lower-bound)
lbList = []
ub = 1 - 0.5**len(V) # over-estimation = sum of all elements of V
ubList = range(len(V))
# calculate closest under-estimation and over-estimation
for i in range(len(V)):
if r == lb + V[i]:
return (lbList + [i], lb + V[i])
elif r == ub:
return (ubList, ub)
elif r > lb + V[i]:
lb += V[i]
lbList += [i]
elif lb + V[i] < ub:
ub = lb + V[i]
ubList = lbList + [i]
return (ubList, ub) if ub - r < r - lb else (lbList, lb) if lb != 0 else ([len(V) - 1], V[len(V) - 1])
# populate V
N = 5 # number of elements
V = []
for i in range(1, N + 1):
V += [0.5**i]
# test
r = 0.484375 # this value is equidistant from both under- and over-estimation
print "r:", r
estimate = estimate(V, r)
print "Indices:", estimate[0]
print "Estimate:", estimate[1]
Note: after finishing writing my answer I noticed that this answer follows the same logic. Alas!
I don't know if you have test cases, try the code below. It is a dynamic-programming approach.
1] exp: given 1/2^i, find the largest i as exp. Eg. 1/32 returns 5.
2] max: 10^exp where exp=i.
3] create an array of size max+1 to hold all possible sums of the elements of V.
Actually the array holds the indexes, since that's what you want.
4] dynamically compute the sums (all invalids remain null)
5] the last while loop finds the nearest correct answer.
Here is the code:
public class Subset {
public static List<Integer> subsetSum(double[] V, double r) {
int exp = exponent(V);
int max = (int) Math.pow(10, exp);
//list to hold all possible sums of the elements in V
List<Integer> indexes[] = new ArrayList[max + 1];
indexes[0] = new ArrayList();//base case
//dynamically compute the sums
for (int x=0; x<V.length; x++) {
int u = (int) (max*V[x]);
for(int i=max; i>=u; i--) if(null != indexes[i-u]) {
List<Integer> tmp = new ArrayList<Integer>(indexes[i - u]);
tmp.add(x);
indexes[i] = tmp;
}
}
//find the best answer
int i = (int)(max*r);
int j=i;
while(null == indexes[i] && null == indexes[j]) {
i--;j++;
}
return indexes[i]==null || indexes[i].isEmpty()?indexes[j]:indexes[i];
}// subsetSum
private static int exponent(double[] V) {
double d = V[V.length-1];
int i = (int) (1/d);
String s = Integer.toString(i,2);
return s.length()-1;
}// summation
public static void main(String[] args) {
double[] V = {1/2.,1/4.,1/8.,1/16.,1/32.};
double r = 0.6, s=0.3,t=0.256652;
System.out.println(subsetSum(V,r));//[0, 3, 4]
System.out.println(subsetSum(V,s));//[1, 3]
System.out.println(subsetSum(V,t));//[1]
}
}// class
Here are results of running the code:
For 0.600000 get 0.593750 => [0, 3, 4]
For 0.300000 get 0.312500 => [1, 3]
For 0.256652 get 0.250000 => [1]
For 0.700000 get 0.687500 => [0, 2, 3]
For 0.710000 get 0.718750 => [0, 2, 3, 4]
The solution implements Polynomial time approximate algorithm. Output of the program is the same as outputs of another solutions.
#include <math.h>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <functional>
void populate(std::vector<double> &vec, int count)
{
double val = .5;
vec.clear();
for (int i = 0; i < count; i++) {
vec.push_back(val);
val *= .5;
}
}
void remove_values_with_large_error(const std::vector<double> &vec, std::vector<double> &res, double r, double max_error)
{
std::vector<double>::const_iterator iter;
double min_err, err;
min_err = 1.0;
for (iter = vec.begin(); iter != vec.end(); ++iter) {
err = fabs(*iter - r);
if (err < max_error) {
res.push_back(*iter);
}
min_err = std::min(err, min_err);
}
}
void find_partial_sums(const std::vector<double> &vec, std::vector<double> &res, double r)
{
std::vector<double> svec, tvec, uvec;
std::vector<double>::const_iterator iter;
int step = 0;
svec.push_back(0.);
for (iter = vec.begin(); iter != vec.end(); ++iter) {
step++;
printf("step %d, svec.size() %d\n", step, svec.size());
tvec.clear();
std::transform(svec.begin(), svec.end(), back_inserter(tvec),
std::bind2nd(std::plus<double>(), *iter));
uvec.clear();
uvec.insert(uvec.end(), svec.begin(), svec.end());
uvec.insert(uvec.end(), tvec.begin(), tvec.end());
sort(uvec.begin(), uvec.end());
uvec.erase(unique(uvec.begin(), uvec.end()), uvec.end());
svec.clear();
remove_values_with_large_error(uvec, svec, r, *iter * 4);
}
sort(svec.begin(), svec.end());
svec.erase(unique(svec.begin(), svec.end()), svec.end());
res.clear();
res.insert(res.end(), svec.begin(), svec.end());
}
double find_closest_value(const std::vector<double> &sums, double r)
{
std::vector<double>::const_iterator iter;
double min_err, res, err;
min_err = fabs(sums.front() - r);
res = sums.front();
for (iter = sums.begin(); iter != sums.end(); ++iter) {
err = fabs(*iter - r);
if (err < min_err) {
min_err = err;
res = *iter;
}
}
printf("found value %lf with err %lf\n", res, min_err);
return res;
}
void print_indexes(const std::vector<double> &vec, double value)
{
std::vector<double>::const_iterator iter;
int index = 0;
printf("indexes: [");
for (iter = vec.begin(); iter != vec.end(); ++iter, ++index) {
if (value >= *iter) {
printf("%d, ", index);
value -= *iter;
}
}
printf("]\n");
}
int main(int argc, char **argv)
{
std::vector<double> vec, sums;
double r = .7;
int n = 5;
double value;
populate(vec, n);
find_partial_sums(vec, sums, r);
value = find_closest_value(sums, r);
print_indexes(vec, value);
return 0;
}
Sort the vector and search for the closest fraction available to r. store that index, subtract the value from r, and repeat with the remainder of r. iterate until r is reached, or no such index can be found.
Example :
0.3 - the biggest value available would be 0.25. (index 2). the remainder now is 0.05
0.05 - the biggest value available would be 0.03125 - the remainder will be 0.01875
etc.
etc. every step would be an O(logN) search in a sorted array. the number of steps will also be O(logN) total complexity will be than O(logN^2).
This is not dynamic programming question
The output should rather be vector of ints (indexes), not vector of doubles
This might by off 0-2 in exact values, this is just concept:
A) output zero index until the r0 (r - index values already outputded) is bigger than 1/2
B) Inspect the internal representation of r0 double and:
x (1st bit shift) = -Exponent; // The bigger exponent, the smallest numbers (bigger x in 1/2^(x) you begin with)
Inspect bit representation of the fraction part of float in cycle with body:
(direction depends on little/big endian)
{
if (bit is 1)
output index x;
x++;
}
Complexity of each step is constant, so overall it is O(n) where n is size of output.
To paraphrase the question, what are the one bits in the binary representation of r (after the binary point)? N is the 'precision', if you like.
In Cish pseudo-code
for (int i=0; i<N; i++) {
if (r>V[i]) {
print(i);
r -= V[i];
}
}
You could add an extra test for r == 0 to terminate the loop early.
Note that this gives the least binary number closest to 'r', i.e. the one closer to zero if there are two equally 'right' answers.
If the Nth digit was a one, you'll need to add '1' to the 'binary' number obtained and check both against the original 'r'. (Hint: construct vectors a[N], b[N] of 'bits', set '1' bits instead of 'print'ing above. Set b = a and do a manual add, digit by digit from the end of 'b' until you stop carrying. Convert to double and choose whichever is closer.
Note that a[] <= r <= a[] + 1/2^N and that b[] = a[] + 1/2^N.
The 'least number of indexes [sic]' is a red-herring.