Project Euler #16 - C# 2.0

Question

I've been wrestling with Project Euler Problem #16 in C# 2.0. The crux of the question is that you have to calculate and then iterate through each digit in a number that is 604 digits long (or there-abouts). You then add up these digits to produce the answer.

This presents a problem: C# 2.0 doesn't have a built-in datatype that can handle this sort of calculation precision. I could use a 3rd party library, but that would defeat the purpose of attempting to solve it programmatically without external libraries. I can solve it in Perl; but I'm trying to solve it in C# 2.0 (I'll attempt to use C# 3.0 in my next run-through of the Project Euler questions).

Question

What suggestions (not answers!) do you have for solving project Euler #16 in C# 2.0? What methods would work?

NB: If you decide to post an answer, please prefix your attempt with a blockquote that has ###Spoiler written before it.

Answer 1

A number of a series of digits. A 32 bit unsigned int is 32 binary digits. The string "12345" is a series of 5 digits. Digits can be stored in many ways: as bits, characters, array elements and so on. The largest "native" datatype in C# with complete precision is probably the decimal type (128 bits, 28-29 digits). Just choose your own method of storing digits that allows you to store much bigger numbers.

As for the rest, this will give you a clue:

2¹ = 2
2² = 2¹ + 2¹
2³ = 2² + 2²

Example:

The sum of digits of 2^100000 is 135178
Ran in 4875 ms

The sum of digits of 2^10000 is 13561
Ran in 51 ms

The sum of digits of 2^1000 is 1366
Ran in 2 ms

SPOILER ALERT: Algorithm and solution in C# follows.

Basically, as alluded to a number is nothing more than an array of digits. This can be represented easily in two ways:

• As a string;
• As an array of characters or digits.

As others have mentioned, storing the digits in reverse order is actually advisable. It makes the calculations much easier. I tried both of the above methods. I found strings and the character arithmetic irritating (it's easier in C/C++; the syntax is just plain annoying in C#).

The first thing to note is that you can do this with one array. You don't need to allocate more storage at each iteration. As mentioned you can find a power of 2 by doubling the previous power of 2. So you can find 2¹⁰⁰⁰ by doubling 1 one thousand times. The doubling can be done in place with the general algorithm:

carry = 0
foreach digit in array
  sum = digit + digit + carry
  if sum > 10 then
    carry = 1
    sum -= 10
  else
    carry = 0
  end if
  digit = sum
end foreach

This algorithm is basically the same for using a string or an array. At the end you just add up the digits. A naive implementation might add the results into a new array or string with each iteration. Bad idea. Really slows it down. As mentioned, it can be done in place.

But how large should the array be? Well that's easy too. Mathematically you can convert 2^a to 10^f(a) where f(a) is a simple logarithmic conversion and the number of digits you need is the next higher integer from that power of 10. For simplicity, you can just use:

digits required = ceil(power of 2 / 3)

which is a close approximation and sufficient.

Where you can really optimise this is by using larger digits. A 32 bit signed int can store a number between +/- 2 billion (approximately. Well 9 digits equals a billion so you can use a 32 bit int (signed or unsigned) as basically a base one billion "digit". You can work out how many ints you need, create that array and that's all the storage you need to run the entire algorithm (being 130ish bytes) with everything being done in place.

Solution follows (in fairly rough C#):

    static void problem16a()
    {
        const int limit = 1000;
        int ints = limit / 29;
        int[] number = new int[ints + 1];
        number[0] = 2;
        for (int i = 2; i <= limit; i++)
        {
            doubleNumber(number);
        }
        String text = NumberToString(number);
        Console.WriteLine(text);
        Console.WriteLine("The sum of digits of 2^" + limit + " is " + sumDigits(text));
    }

    static void doubleNumber(int[] n)
    {
        int carry = 0;
        for (int i = 0; i < n.Length; i++)
        {
            n[i] <<= 1;
            n[i] += carry;
            if (n[i] >= 1000000000)
            {
                carry = 1;
                n[i] -= 1000000000;
            }
            else
            {
                carry = 0;
            }
        }
    }

    static String NumberToString(int[] n)
    {
        int i = n.Length;
        while (i > 0 && n[--i] == 0)
            ;
        String ret = "" + n[i--];
        while (i >= 0)
        {
            ret += String.Format("{0:000000000}", n[i--]);
        }
        return ret;
    }

Answer 2

I solved this one using C# also, much to my dismay when I discovered that Python can do this in one simple operation.

Your goal is to create an adding machine using arrays of int values.

Spoiler follows

I ended up using an array of int values to simulate an adding machine, but I represented the number backwards - which you can do because the problem only asks for the sum of the digits, this means order is irrelevant.

What you're essentially doing is doubling the value 1000 times, so you can double the value 1 stored in the 1st element of the array, and then continue looping until your value is over 10. This is where you will have to keep track of a carry value. The first power of 2 that is over 10 is 16, so the elements in the array after the 5th iteration are 6 and 1.

Now when you loop through the array starting at the 1st value (6), it becomes 12 (so you keep the last digit, and set a carry bit on the next index of the array) - which when that value is doubled you get 2 ... plus the 1 for the carry bit which equals 3. Now you have 2 and 3 in your array which represents 32.

Continues this process 1000 times and you'll have an array with roughly 600 elements that you can easily add up.

Answer 3

I have solved this one before, and now I re-solved it using C# 3.0. :)

I just wrote a Multiply extension method that takes an IEnumerable<int> and a multiplier and returns an IEnumerable<int>. (Each int represents a digit, and the first one it the least significant digit.) Then I just created a list with the item { 1 } and multiplied it by 2 a 1000 times. Adding the items in the list is simple with the Sum extension method.

19 lines of code, which runs in 13 ms. on my laptop. :)

Answer 4

If you wish to do the primary calculation in C#, you will need some sort of big integer implementation (Much like gmp for C/C++). Programming is about using the right tool for the right job. If you cannot find a good big integer library for C#, it's not against the rules to calculate the number in a language like Python which already has the ability to calculate large numbers. You could then put this number into your C# program via your method of choice, and iterate over each character in the number (you will have to store it as a string). For each character, convert it to an integer and add it to your total until you reach the end of the number. If you would like the big integer, I calculated it with python below. The answer is further down.

Partial Spoiler

10715086071862673209484250490600018105614048117055336074437503883703510511249361 22493198378815695858127594672917553146825187145285692314043598457757469857480393 45677748242309854210746050623711418779541821530464749835819412673987675591655439 46077062914571196477686542167660429831652624386837205668069376

Spoiler Below!

>>> val = str(2**1000)
>>> total = 0
>>> for i in range(0,len(val)): total += int(val[i])
>>> print total











1366

Answer 5

Pretend you are very young, with square paper. To me, that is like a list of numbers. Then to double it you double each number, then handle any "carries", by subtracting the 10s and adding 1 to the next index. So if the answer is 1366... something like (completely unoptimized, rot13):

hfvat Flfgrz;
hfvat Flfgrz.Pbyyrpgvbaf.Trarevp;    
pynff Cebtenz {
    fgngvp ibvq Pneel(Yvfg<vag> yvfg, vag vaqrk) {
        juvyr (yvfg[vaqrk] > 9) {
            yvfg[vaqrk] -= 10;
            vs (vaqrk == yvfg.Pbhag - 1) yvfg.Nqq(1);
            ryfr yvfg[vaqrk + 1]++;
        }
    }
    fgngvp ibvq Znva() {
        ine qvtvgf = arj Yvfg<vag> { 1 }; // 2^0
        sbe (vag cbjre = 1; cbjre <= 1000; cbjre++) {
            sbe (vag qvtvg = 0; qvtvg < qvtvgf.Pbhag; qvtvg++) {
                qvtvgf[qvtvg] *= 2;
            }
            sbe (vag qvtvg = 0; qvtvg < qvtvgf.Pbhag; qvtvg++) {
                Pneel(qvtvgf, qvtvg);
            }
        }

        qvtvgf.Erirefr();
        sbernpu (vag v va qvtvgf) {
            Pbafbyr.Jevgr(v);
        }
        Pbafbyr.JevgrYvar();

        vag fhz = 0;
        sbernpu (vag v va qvtvgf) fhz += v;
        Pbafbyr.Jevgr("fhz: ");
        Pbafbyr.JevgrYvar(fhz);
    }
}

Answer 6

If you've got ruby, you can easily calculate "2**1000" and get it as a string. Should be an easy cut/paste into a string in C#.

Spoiler

In Ruby: (2**1000).to_s.split(//).inject(0){|x,y| x+y.to_i}

Answer 7

spoiler

If you want to see a solution check out my other answer. This is in Java but it's very easy to port to C#

Here's a clue:

Represent each number with a list. That way you can do basic sums like:

[1,2,3,4,5,6]
+       [4,5]
_____________
[1,2,3,5,0,1]

Answer 8

Working solution that I have posted it here as well: http://www.mycoding.net/2012/01/solution-to-project-euler-problem-16/

The code:

import java.math.BigInteger;
public class Euler16 {

public static void main(String[] args) {
    int power = 1;
    BigInteger expo = new BigInteger("2");
    BigInteger num = new BigInteger("2");
    while(power < 1000){
        expo = expo.multiply(num);
        power++;
    }
    System.out.println(expo);   //Printing the value of 2^1000
    int sum = 0;
    char[] expoarr = expo.toString().toCharArray();
    int max_count = expoarr.length;
    int count = 0;
    while(count<max_count){ //While loop to calculate the sum of digits
        sum = sum + (expoarr[count]-48);
        count++;
    }
    System.out.println(sum);
}
}

Answer 9

Euler problem #16 has been discussed many times here, but I could not find an answer that gives a good overview of possible solution approaches, the lay of the land as it were. Here's my attempt at rectifying that.

This overview is intended for people who have already found a solution and want to get a more complete picture. It is basically language-agnostic even though the sample code is C#. There are some usages of features that are not available in C# 2.0 but they are not essential - their purpose is only to get boring stuff out of the way with a minimum of fuss.

Apart from using a ready-made BigInteger library (which doesn't count), straightforward solutions for Euler #16 fall into two fundamental categories: performing calculations natively - i.e. in a base that is a power of two - and converting to decimal in order to get at the digits, or performing the computations directly in a decimal base so that the digits are available without any conversion.

For the latter there are two reasonably simple options:

• repeated doubling
• powering by repeated squaring

Native Computation + Radix Conversion

This approach is the simplest and its performance exceeds that of naive solutions using .Net's builtin BigInteger type.

The actual computation is trivially achieved: just perform the moral equivalent of 1 << 1000, by storing 1000 binary zeroes and appending a single lone binary 1.

The conversion is also quite simple and can be done by coding the pencil-and-paper division method, with a suitably large choice of 'digit' for efficiency. Variables for intermediate results need to be able to hold two 'digits'; dividing the number of decimal digits that fit in a long by 2 gives 9 decimal digits for the maximum meta-digit (or 'limb', as it is usually called in bignum lore).

class E16_RadixConversion
{
    const int BITS_PER_WORD = sizeof(uint) * 8;
    const uint RADIX = 1000000000;  // == 10^9

    public static int digit_sum_for_power_of_2 (int exponent)
    {
        var dec = new List<int>();
        var bin = new uint[(exponent + BITS_PER_WORD) / BITS_PER_WORD];
        int top = bin.Length - 1;

        bin[top] = 1u << (exponent % BITS_PER_WORD);

        while (top >= 0)
        {
            ulong rest = 0;

            for (int i = top; i >= 0; --i)
            {
                ulong temp = (rest << BITS_PER_WORD) | bin[i];
                ulong quot = temp / RADIX;  // x64 uses MUL (sometimes), x86 calls a helper function

                rest = temp - quot * RADIX;         
                bin[i] = (uint)quot;
            }

            dec.Add((int)rest);

            if (bin[top] == 0)
                --top;
        }

        return E16_Common.digit_sum(dec);
    }
}

I wrote (rest << BITS_PER_WORD) | big[i] instead of using operator + because that is precisely what is needed here; no 64-bit addition with carry propagation needs to take place. This means that the two operands could be written directly to their separate registers in a register pair, or to fields in an equivalent struct like LARGE_INTEGER.

On 32-bit systems the 64-bit division cannot be inlined as a few CPU instructions, because the compiler cannot know that the algorithm guarantees quotient and remainder to fit into 32-bit registers. Hence the compiler calls a helper function that can handle all eventualities.

These systems may profit from using a smaller limb, i.e. RADIX = 10000 and uint instead of ulong for holding intermediate (double-limb) results. An alternative for languages like C/C++ would be to call a suitable compiler intrinsic that wraps the raw 32-bit by 32-bit to 64-bit multiply (assuming that division by the constant radix is to be implemented by multiplication with the inverse). Conversely, on 64-bit systems the limb size can be increased to 19 digits if the compiler offers a suitable 64-by-64-to-128 bit multiply primitive or allows inline assembler.

Decimal Doubling

Repeated doubling seems to be everyone's favourite, so let's do that next. Variables for intermediate results need to hold one 'digit' plus one carry bit, which gives 18 digits per limb for long. Going to ulong cannot improve things (there's 0.04 bit missing to 19 digits plus carry), and so we might as well stick with long.

On a binary computer, decimal limbs do not coincide with computer word boundaries. That makes it necessary to perform a modulo operation on the limbs during each step of the calculation. Here, this modulo op can be reduced to a subtraction of the modulus in the event of carry, which is faster than performing a division. The branching in the inner loop can be eliminated by bit twiddling but that would be needlessly obscure for a demonstration of the basic algorithm.

class E16_DecimalDoubling
{
    const int DIGITS_PER_LIMB = 18;                  // == floor(log10(2) * (63 - 1)), b/o carry
    const long LIMB_MODULUS = 1000000000000000000L;  // == 10^18

    public static int digit_sum_for_power_of_2 (int power_of_2) 
    {
        Trace.Assert(power_of_2 > 0);

        int total_digits = (int)Math.Ceiling(Math.Log10(2) * power_of_2);
        int total_limbs = (total_digits + DIGITS_PER_LIMB - 1) / DIGITS_PER_LIMB;
        var a = new long[total_limbs];
        int limbs = 1;

        a[0] = 2;

        for (int i = 1; i < power_of_2; ++i)
        {
            int carry = 0;

            for (int j = 0; j < limbs; ++j)
            {
                long new_limb = (a[j] << 1) | carry;

                carry = 0;

                if (new_limb >= LIMB_MODULUS)
                {
                    new_limb -= LIMB_MODULUS;
                    carry = 1;
                }

                a[j] = new_limb;
            }

            if (carry != 0)
            {
                a[limbs++] = carry;
            }
        }

        return E16_Common.digit_sum(a);
    }
}

This is just as simple as radix conversion, but except for very small exponents it does not perform anywhere near as well (despite its huge meta-digits of 18 decimal places). The reason is that the code must perform (exponent - 1) doublings, and the work done in each pass corresponds to about half the total number of digits (limbs).

Repeated Squaring

The idea behind powering by repeated squaring is to replace a large number of doublings with a small number of multiplications.

1000 = 2^3 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9

x^1000 = x^(2^3 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9)

x^1000 = x^2^3 * x^2^5 * x^2^6 * x^2^7 * x^2*8 * x^2^9

x^2^3 can be obtained by squaring x three times, x^2^5 by squaring five times, and so on. On a binary computer the decomposition of the exponent into powers of two is readily available because it is the bit pattern representing that number. However, even non-binary computers should be able to test whether a number is odd or even, or to divide a number by two.

The multiplication can be done by coding the pencil-and-paper method; here I'm using a helper function that computes one row of a product and adds it into the result at a suitably shifted position, so that the rows of partial products do not need to be stored for a separate addition step later. Intermediate values during computation can be up to two 'digits' in size, so that the limbs can be only half as wide as for repeated doubling (where only one extra bit had to fit in addition to a 'digit').

Note: the radix of the computations is not a power of 2, and so the squarings of 2 cannot be computed by simple shifting here. On the positive side, the code can be used for computing powers of bases other than 2.

class E16_DecimalSquaring
{
    const int DIGITS_PER_LIMB = 9;  // language limit 18, half needed for holding the carry
    const int LIMB_MODULUS = 1000000000;

    public static int digit_sum_for_power_of_2 (int e)
    {
        Trace.Assert(e > 0);

        int total_digits = (int)Math.Ceiling(Math.Log10(2) * e);
        int total_limbs = (total_digits + DIGITS_PER_LIMB - 1) / DIGITS_PER_LIMB;
        var squared_power = new List<int>(total_limbs) { 2 };
        var result = new List<int>(total_limbs);

        result.Add((e & 1) == 0 ? 1 : 2);

        while ((e >>= 1) != 0)
        {
            squared_power = multiply(squared_power, squared_power);

            if ((e & 1) == 1)
                result = multiply(result, squared_power);
        }

        return E16_Common.digit_sum(result);
    }

    static List<int> multiply (List<int> lhs, List<int> rhs)
    {
        var result = new List<int>(lhs.Count + rhs.Count);

        resize_to_capacity(result);

        for (int i = 0; i < rhs.Count; ++i)
            addmul_1(result, i, lhs, rhs[i]);

        trim_leading_zero_limbs(result);

        return result;
    }

    static void addmul_1 (List<int> result, int offset, List<int> multiplicand, int multiplier)
    {
        // it is assumed that the caller has sized `result` appropriately before calling this primitive
        Trace.Assert(result.Count >= offset + multiplicand.Count + 1);

        long carry = 0;

        foreach (long limb in multiplicand)
        {
            long temp = result[offset] + limb * multiplier + carry;
            carry = temp / LIMB_MODULUS;
            result[offset++] = (int)(temp - carry * LIMB_MODULUS);
        }

        while (carry != 0)
        {
            long final_temp = result[offset] + carry;
            carry = final_temp / LIMB_MODULUS;
            result[offset++] = (int)(final_temp - carry * LIMB_MODULUS);
        }
    }

    static void resize_to_capacity (List<int> operand)
    {
        operand.AddRange(Enumerable.Repeat(0, operand.Capacity - operand.Count));
    }

    static void trim_leading_zero_limbs (List<int> operand)
    {
        int i = operand.Count;

        while (i > 1 && operand[i - 1] == 0)
            --i;

        operand.RemoveRange(i, operand.Count - i);
    }   
}

The efficiency of this approach is roughly on par with radix conversion but there are specific improvements that apply here. Efficiency of the squaring can be doubled by writing a special squaring routine that utilises the fact that ai*bj == aj*bi if a == b, which cuts the number of multiplications in half.

Also, there are methods for computing addition chains that involve fewer operations overall than using the exponent bits for determining the squaring/multiplication schedule.

Helper Code and Benchmarks

The helper code for summing decimal digits in the meta-digits (decimal limbs) produced by the sample code is trivial, but I'm posting it here anyway for your convenience:

internal class E16_Common
{
    internal static int digit_sum (int limb)
    {
        int sum = 0;

        for ( ; limb > 0; limb /= 10)
            sum += limb % 10;

        return sum;
    }

    internal static int digit_sum (long limb)
    {
        const int M1E9 = 1000000000;

        return digit_sum((int)(limb / M1E9)) + digit_sum((int)(limb % M1E9));
    }

    internal static int digit_sum (IEnumerable<int> limbs)
    {
        return limbs.Aggregate(0, (sum, limb) => sum + digit_sum(limb));
    }

    internal static int digit_sum (IEnumerable<long> limbs)
    {
        return limbs.Select((limb) => digit_sum(limb)).Sum();
    }
}

This can be made more efficient in various ways but overall it is not critical.

All three solutions take O(n^2) time where n is the exponent. In other words, they will take a hundred times as long when the exponent grows by a factor of ten. Radix conversion and repeated squaring can both be improved to roughly O(n log n) by employing divide-and-conquer strategies; I doubt whether the doubling scheme can be improved in a similar fastion but then it was never competitive to begin with.

All three solutions presented here can be used to print the actual results, by stringifying the meta-digits with suitable padding and concatenating them. I've coded the functions as returning the digit sum instead of the arrays/lists with decimal limbs only in order to keep the sample code simple and to ensure that all functions have the same signature, for benchmarking.

In these benchmarks, the .Net BigInteger type was wrapped like this:

static int digit_sum_via_BigInteger (int power_of_2)
{
    return System.Numerics.BigInteger.Pow(2, power_of_2)
        .ToString()
        .ToCharArray()
        .Select((c) => (int)c - '0')
        .Sum();
}

Finally, the benchmarks for the C# code:

# testing decimal doubling ...

   1000:    1366 in      0,052 ms
  10000:   13561 in      3,485 ms
 100000:  135178 in    339,530 ms
1000000: 1351546 in 33.505,348 ms

# testing decimal squaring ...

   1000:    1366 in      0,023 ms
  10000:   13561 in      0,299 ms
 100000:  135178 in     24,610 ms
1000000: 1351546 in  2.612,480 ms

# testing radix conversion ...

   1000:    1366 in      0,018 ms
  10000:   13561 in      0,619 ms
 100000:  135178 in     60,618 ms
1000000: 1351546 in  5.944,242 ms

# testing BigInteger + LINQ ...

   1000:    1366 in      0,021 ms
  10000:   13561 in      0,737 ms
 100000:  135178 in     69,331 ms
1000000: 1351546 in  6.723,880 ms

As you can see, the radix conversion is almost as slow as the solution using the builtin BigInteger class. The reason is that the runtime is of the newer type that does performs certain standard optimisations only for signed integer types but not for unsigned ones (here: implementing division by a constant as multiplication with the inverse).

I haven't found an easy means of inspecting the native code for existing .Net assemblies, so I decided on a different path of investigation: I coded a variant of E16_RadixConversion for comparison where ulong and uint were replaced by long and int respectively, and BITS_PER_WORD decreased by 1 accordingly. Here are the timings:

# testing radix conv Int63 ...

   1000:    1366 in      0,004 ms
  10000:   13561 in      0,202 ms
 100000:  135178 in     18,414 ms
1000000: 1351546 in  1.834,305 ms

More than three times as fast as the version that uses unsigned types! Clear evidence of numbskullery in the compiler...

In order to showcase the effect of different limb sizes I templated the solutions in C++ on the unsigned integer types used as limbs. The timings are prefixed with the byte size of a limb and the number of decimal digits in a limb, separated by a colon. There is no timing for the often-seen case of manipulating digit characters in strings, but it is safe to say that such code will take at least twice as long as the code that uses double digits in byte-sized limbs.

# E16_DecimalDoubling

[1:02] e =    1000 ->    1366      0.308 ms
[2:04] e =    1000 ->    1366      0.152 ms
[4:09] e =    1000 ->    1366      0.070 ms
[8:18] e =    1000 ->    1366      0.071 ms

[1:02] e =   10000 ->   13561     30.533 ms
[2:04] e =   10000 ->   13561     13.791 ms
[4:09] e =   10000 ->   13561      6.436 ms
[8:18] e =   10000 ->   13561      2.996 ms

[1:02] e =  100000 ->  135178   2719.600 ms
[2:04] e =  100000 ->  135178   1340.050 ms
[4:09] e =  100000 ->  135178    588.878 ms
[8:18] e =  100000 ->  135178    290.721 ms

[8:18] e = 1000000 -> 1351546  28823.330 ms

For the exponent of 10^6 there is only the timing with 64-bit limbs, since I didn't have the patience to wait many minutes for full results. The picture is similar for radix conversion, except that there is no row for 64-bit limbs because my compiler does not have a native 128-bit integral type.

# E16_RadixConversion

[1:02] e =    1000 ->    1366      0.080 ms
[2:04] e =    1000 ->    1366      0.026 ms
[4:09] e =    1000 ->    1366      0.048 ms

[1:02] e =   10000 ->   13561      4.537 ms
[2:04] e =   10000 ->   13561      0.746 ms
[4:09] e =   10000 ->   13561      0.243 ms

[1:02] e =  100000 ->  135178    445.092 ms
[2:04] e =  100000 ->  135178     68.600 ms
[4:09] e =  100000 ->  135178     19.344 ms

[4:09] e = 1000000 -> 1351546   1925.564 ms

The interesting thing is that simply compiling the code as C++ doesn't make it any faster - i.e., the optimiser couldn't find any low-hanging fruit that the C# jitter missed, apart from not toeing the line with regard to penalising unsigned integers. That's the reason why I like prototyping in C# - performance in the same ballpark as (unoptimised) C++ and none of the hassle.

Here's the meat of the C++ version (sans reams of boring stuff like helper templates and so on) so that you can see that I didn't cheat to make C# look better:

template<typename W>
struct E16_RadixConversion
{
   typedef W limb_t;
   typedef typename detail::E16_traits<W>::long_t long_t;

   static unsigned const BITS_PER_WORD = sizeof(limb_t) * CHAR_BIT;
   static unsigned const RADIX_DIGITS = std::numeric_limits<limb_t>::digits10;
   static limb_t const RADIX = detail::pow10_t<limb_t, RADIX_DIGITS>::RESULT;

   static unsigned digit_sum_for_power_of_2 (unsigned e)
   {
      std::vector<limb_t> digits;

      compute_digits_for_power_of_2(e, digits);

      return digit_sum(digits);
   }

   static void compute_digits_for_power_of_2 (unsigned e, std::vector<limb_t> &result)
   {
      assert(e > 0);

      unsigned total_digits = unsigned(std::ceil(std::log10(2) * e));
      unsigned total_limbs = (total_digits + RADIX_DIGITS - 1) / RADIX_DIGITS;

      result.resize(0);
      result.reserve(total_limbs);

      std::vector<limb_t> bin((e + BITS_PER_WORD) / BITS_PER_WORD);

      bin.back() = limb_t(limb_t(1) << (e % BITS_PER_WORD));

      while (!bin.empty())
      {
         long_t rest = 0;

         for (std::size_t i = bin.size(); i-- > 0; )
         {
            long_t temp = (rest << BITS_PER_WORD) | bin[i];
            long_t quot = temp / RADIX;

            rest = temp - quot * RADIX;
            bin[i] = limb_t(quot);
         }

         result.push_back(limb_t(rest));

         if (bin.back() == 0)
            bin.pop_back();
      }
   }
};

Conclusion

These benchmarks also show that this Euler task - like many others - seems designed to be solved on a ZX81 or an Apple ][, not on our modern toys that are a million times as powerful. There's no challenge involved here unless the limits are increased drastically (an exponent of 10^5 or 10^6 would be much more adequate).

A good overview of the practical state of the art can be got from GMP's overview of algorithms. Another excellent overview of the algorithms is chapter 1 of "Modern Computer Arithmetic" by Richard Brent and Paul Zimmermann. It contains exactly what one needs to know for coding challenges and competitions, but unfortunately the depth is not equal to that of Donald Knuth's treatment in "The Art of Computer Programming".

The radix conversion solution adds a useful technique to one's code challenge toolchest, since the given code can be trivially extended for converting any old big integer instead of only the bit pattern 1 << exponent. The repeated squaring solutiono can be similarly useful since changing the sample code to power something other than 2 is again trivial.

The approach of performing computations directly in powers of 10 can be useful for challenges where decimal results are required, because performance is in the same ballpark as native computation but there is no need for a separate conversion step (which can require similar amounts of time as the actual computation).

Answer 10

One alternative to representing the digits as a sequence of integers is to represent the number base 2^32 as a list of 32 bit integers, which is what many big integer libraries do. You then have to convert the number to base 10 for output. This doesn't gain you very much for this particular problem - you can write 2^1000 straight away, then have to divide by 10 many times instead of multiplying 2 by itself 1000 times ( or, as 1000 is 0b1111101000. calculating the product of 2^8,32,64,128,256,512 using repeated squaring 2^8 = (((2^2)^2)^2))) which requires more space and a multiplication method, but is far fewer operations ) - is closer to normal big integer use, so you may find it more useful in later problems ( if you try to calculate the last ten digits of 28433×2^(7830457)+1 using the digit-per int method and repeated addition, it may take some time (though in that case you could use modulo arthimetic, rather than adding strings of millions of digits) ).