how to generate Narcissistic numbers faster?

Question

The “Narcissistic numbers”, are n digit numbers where the sum of all the nth power of their digits is equal to the number.

So, 153 is a narcissistic number because 1^3 + 5^3 + 3^3 = 153 .

Now given N, find all Narcissistic numbers that are N digit length ?

My Approach : was to iterate over all numbers doing sum of powers of digits

and check if its the same number or not, and I per calculated the powers.

but that's not good enough, so is there any faster way ?!

Update: In nature there is just 88 narcissistic numbers, and the largest is 39 digits long, But I just need the numbers with length 12 or less.

My Code :

long long int powers[11][12];
// powers[x][y] is x^y. and its already calculated

bool isNarcissistic(long long int x,int n){
    long long int r = x;
    long long int sum = 0;

    for(int i=0; i<n ; ++i){
        sum += powers[x%10][n];
        if(sum > r)
            return false;
        x /= 10;
    }
    return (sum == r);
}

void find(int n,vector<long long int> &vv){
    long long int start = powers[10][n-1];
    long long int end = powers[10][n];

    for(long long int i=start ; i<end ; ++i){
        if(isNarcissistic(i,n))
            vv.push_back(i);
    }
}

Answer 1

由于总共只有88个narcisstic数字，你可以将它们存储在一个查找表中并迭代它： http ： //mathworld.wolfram.com/NarcissisticNumber.html

Answer 2

Start from the other end. Iterate over the set of all nondecreasing sequences of d digits, compute the sum of the d -th powers, and check whether that produces (after sorting) the sequence you started with.

Since there are

9×10^(d-1)

d -digit numbers, but only

(10+d-1) `choose` d

nondecreasing sequences of d digits, that reduces the search space by a factor close to d! .

Answer 3

The code below implements the idea of @Daniel Fischer. It duplicates the table referenced at Mathworld and then prints a few more 11-digit numbers and verifies that there are none with 12 digits as stated here .

It would actually be simplier and probably a little faster to generate all possible histograms of non-increasing digit strings rather than the strings themselves. By a histogram I mean a table indexed 0-9 of frequencies of the respective digit. These can be compared directly without sorting. But the code below runs in < 1 sec, so I'm not going to implement the histogram idea.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_DIGITS 12

// pwr[d][n] is d^n
long long pwr[10][MAX_DIGITS + 1];

// Digits and final index of number being considered.
int digits[MAX_DIGITS];
int m;

// Fill pwr.
void fill_tbls(void)
{
  for (int d = 0; d < 10; d++) {
    pwr[d][0] = 1;
    for (int p = 1; p <= MAX_DIGITS; p++) 
      pwr[d][p] = pwr[d][p-1] * d;
  }
}

// qsort comparison for integers descending
int cmp_ints_desc(const void *vpa, const void *vpb)
{
  const int *pa = vpa, *pb = vpb;
  return *pb - *pa;
}

// Test current number and print if narcissistic.
void test(void)
{
  long long sum = 0;
  for (int i = 0; i <= m; i++)
    sum += pwr[digits[i]][m + 1];
  int sum_digits[MAX_DIGITS * 2];
  int n = 0;
  for (long long s = sum; s; s /= 10)
    sum_digits[n++] = s % 10;
  if (n == m + 1) {
    qsort(sum_digits, n, sizeof(int), cmp_ints_desc);
    if (memcmp(sum_digits, digits, n * sizeof(int)) == 0) 
      printf("%lld\n", sum);
  }
}

// Recursive generator of non-increasing digit strings.
// Calls test when a string is complete.
void gen(int i, int min, int max)
{
  if (i > m) 
    test();
  else {
    for (int d = min; d <= max; d++) {
      digits[i] = d;
      gen(i + 1, 0, d); 
    }
  }
}

// Fill tables and generate.
int main(void)
{
  fill_tbls();
  for (m = 0; m < MAX_DIGITS; m++)
    gen(0, 1, 9);
  return 0;
}

Answer 4

I wrote a program in Lua which found all the narcissistic numbers in 30829.642 seconds. The basis of the program is a recursive digit-value count array generator function which calls a checking function when it's generated the digit-value count for all the digit-values. Each nested loop iterates:

FROM i= The larger of 0 and the solution to a+x*d^o+(sx)*(d-1)^o >= 10^(o-1) for x where - 'a' is the accumulative sum of powers of digits so far, - 'd' is the current digit-value (0-9 for base 10), - 'o' is the total number of digits (which the sum of the digit-value count array must add up to), - 's' represents the remaining slots available until the array adds to 'o'

UP TO i<= The smaller of 's' and the solution to a+x*d^o < 10^o for x with the same variables.

This ensures that the numbers checked will ALWAYS have the same number of digits as 'o', and therefore be more likely to be narcissistic while avoiding unnecessary computation.

In the loop, it does the recursive call for which it decrements the digit-value 'd' adds the current digit-value's contribution (a=a+i*d^o) and takes the i digit-slots used up away from 's'.

The gist of what I wrote is:

local function search(o,d,s,a,...) --Original number of digits, digit-value, remaining digits, accumulative sum, number of each digit-value in descending order (such as 5 nines)
    if d>0 then
        local d0,d1=d^o,(d-1)^o
        local dd=d0-d1
        --a+x*d^o+(s-x)*(d-1)^o >= 10^(o-1) , a+x*d^o < 10^o
        for i=max(0,floor((10^(o-1)-s*d1-a)/dd)),min(s,ceil((10^o-a)/dd)-1) do
            search(o,d-1,s-i,a+i*d0,i,...) --The digit counts are passed down as extra arguments.
        end
    else
        --Check, with the count of zeroes set to 's', if the sum 'a' has the same count of each digit-value as the list specifies, and if so, add it to a list of narcissists.
    end
end

local digits=1 --Skip the trivial single digit narcissistic numbers.
while #found<89 do
    digits=digits+1
    search(digits,9,digits,0)
end

EDIT: I forgot to mention that my program finds 89 narcissistic numbers! These are what it finds:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748, 92727, 93084, 548834, 1741725, 4210818, 9800817, 9926315, 24678050, 24678051, 88593477, 146511208, 472335975, 534494836, 912985153, 4679307774, 32164049650, 32164049651, 40028394225, 42678290603, 44708635679, 49388550606, 82693916578, 94204591914, 28116440335967, 4338281769391370, 4338281769391371, 21897142587612075, 35641594208964132, 35875699062250035, 1517841543307505039, 3289582984443187032, 4498128791164624869, 4929273885928088826, 63105425988599693916, 128468643043731391252,449177399146038697307, 21887696841122916288858, 27879694893054074471405, 27907865009977052567814, 28361281321319229463398, 35452590104031691935943, 174088005938065293023722, 188451485447897896036875, 239313664430041569350093, 1550475334214501539088894, 1553242162893771850669378, 3706907995955475988644380, 3706907995955475988644381, 4422095118095899619457938, 121204998563613372405438066, 121270696006801314328439376, 128851796696487777842012787, 174650464499531377631639254, 177265453171792792366489765, 14607640612971980372614873089, 19008174136254279995012734740, 19008174136254279995012734741, 23866716435523975980390369295, 1145037275765491025924292050346, 1927890457142960697580636236639, 2309092682616190307509695338915, 17333509997782249308725103962772, 186709961001538790100634132976990, 186709961001538790100634132976991, 1122763285329372541592822900204593, 12639369517103790328947807201478392, 12679937780272278566303885594196922, 1219167219625434121569735803609966019, 12815792078366059955099770545296129367, 115132219018763992565095597973971522400, 115132219018763992565095597973971522401

Answer 5

For posterity ;-) This is most similar to @Krakow10's approach, generating bags of digits recursively, starting with 9, then 8, then 7 ... to 0.

It's Python3 code and finds all base-10 solutions with 1 through 61 digits (the first "obviously impossible" width) in less than 10 minutes (on my box). It's by far the fastest code I've ever heard of for this problem. What's the trick? No trick - just tedium ;-) As we go along, the partial sum so far yields a world of constraints on feasible continuations. The code just pays attention to those, and so is able to cut off vast masses of searches early.

Note: this doesn't find 0. I don't care. While all the references say there are 88 solutions, their tables all have 89 entries. Some eager editor must have added "0" later, and then everyone else mindlessly copied it ;-)

EDIT New version is over twice as fast, by exploiting some partial-sum constraints earlier in the search - now finishes in a little over 4 minutes on my box.

def nar(width):
    from decimal import Decimal as D
    import decimal
    decimal.getcontext().prec = width + 10
    if width * 9**width < 10**(width - 1):
        raise ValueError("impossible at %d" % width)
    pows = [D(i) ** width for i in range(10)]
    mintotal, maxtotal = D(10)**(width - 1), D(10)**width - 1

    def extend(d, todo, total):
        # assert d > 0
        powd = pows[d]
        d1 = d-1
        powd1 = pows[d1]
        L = total + powd1 * todo # largest possible taking no d's
        dL = powd - powd1  # the change to L when i goes up 1
        for i in range(todo + 1):
            if i:
                total += powd
                todo -= 1
                L += dL
                digitcount[d] += 1
            if total > maxtotal:
                break
            if L < mintotal:
                continue
            if total < mintotal or L > maxtotal:
                yield from extend(d1, todo, total)
                continue
            # assert mintotal <= total <= L <= maxtotal
            t1 = total.as_tuple().digits
            t2 = L.as_tuple().digits
            # assert len(t1) == len(t2) == width
            # Every possible continuation has sum between total and
            # L, and has a full-width sum.  So if total and L have
            # some identical leading digits, a solution must include
            # all such leading digits.  Count them.
            c = [0] * 10
            for a, b in zip(t1, t2):
                if a == b:
                    c[a] += 1
                else:
                    break
            else:  # the tuples are identical
                # assert d == 1 or todo == 0
                # assert total == L
                # This is the only sum we can get - no point to
                # recursing.  It's a solution iff each digit has been
                # picked exactly as many times as it appears in the
                # sum.
                # If todo is 0, we've picked all the digits.
                # Else todo > 0, and d must be 1:  all remaining
                # digits must be 0.
                digitcount[0] += todo
                # assert sum(c) == sum(digitcount) == width
                if digitcount == c:
                    yield total
                digitcount[0] -= todo
                continue
            # The tuples aren't identical, but may have leading digits
            # in common.  If, e.g., "9892" is a common prefix, then to
            # get a solution we must pick at least one 8, at least two
            # 9s, and at least one 2.
            if any(digitcount[j] < c[j] for j in range(d, 10)):
                # we're done picking digits >= d, but don't have
                # enough of them
                continue
            # for digits < d, force as many as we need for the prefix
            newtodo, newtotal = todo, total
            added = []
            for j in range(d):
                need = c[j] - digitcount[j]
                # assert need >= 0
                if need:
                    newtodo -= need
                    added.append((j, need))
            if newtodo < 0:
                continue
            for j, need in added:
                newtotal += pows[j] * need
                digitcount[j] += need
            yield from extend(d1, newtodo, newtotal)
            for j, need in added:
                digitcount[j] -= need
        digitcount[d] -= i

    digitcount = [0] * 10
    yield from extend(9, width, D(0))
    assert all(i == 0 for i in digitcount)

if __name__ == "__main__":
    from datetime import datetime
    start_t = datetime.now()
    width = total = 0
    while True:
        this_t = datetime.now()
        width += 1
        print("\nwidth", width)
        for t in nar(width):
            print("   ", t)
            total += 1
        end_t = datetime.now()
        print(end_t - this_t, end_t - start_t, total)

Answer 6

I think the idea is to generate similar numbers. For example, 61 is similar to 16 as you are just summing

6^n +1^n

so

6^n+1^n=1^n+6^n

In this way you can reduce significant amount of numbers. For example in 3 digits scenario,

121==112==211,

you get the point. You need to generate those numbers first. And you need to generate those numbers without actually iterating from 0-n.

Answer 7

Python version is:

def generate_power_list(power):
return [i**power for i in range(0,10)]


def find_narcissistic_numbers_naive(min_length, max_length):
for number_length in range(min_length, max_length):

    power_dict = generate_power_dictionary(number_length)

    max_number = 10 ** number_length
    number = 10** (number_length -1)
    while number < max_number:

        value = 0
        for digit in str(number):
            value += power_dict[digit]

        if value == number:
            logging.debug('narcissistic %s ' % number)

        number += 1

Recursive solution:

In this solution each recursion handles a single digit of the array of digits being used, and tries all appropriate combinations of that digit

def execute_recursive(digits, number_length):
index = len(digits)
if digits:
    number = digits[-1]
else:
    number = 0
results = []
digits.append(number)    

if len(digits) < number_length:
    while number < 10:
        results += execute_recursive(digits[:], number_length)
        number += 1
        digits[index] = number

else:
    while number < 10:
        digit_value = sum_digits(digits)
        if check_numbers_match_group(digit_value, digits):
            results.append(digit_value)
            logging.debug(digit_value)

        number += 1
        digits[index] = number

return results


def find_narcissistic_numbers(min_length, max_length):
for number_length in range(min_length, max_length):
    digits = []
    t_start = time.clock()
    results = execute_recursive(digits, number_length)
    print 'duration: %s for number length: %s' %(time.clock() - t_start, number_length)

Narcissistic number check In the base version, when checking that a number matched the digits, we iterated through each digit type, to ensure that there were the same number of each type. In this version we have added the optimisation of checking the digit length is correct before doing the full check.

I expected that this would have more of an effect on small number lengths, because as number length increases, there will tend to be more numbers in the middle of the distribution. This was somewhat upheld by the results:

1. n=16: 11.5% improvement
2. n=19: 9.8% improvement

def check_numbers_match_group(number, digits):
number_search = str(number)

# new in v1.3
if len(number_search) != len(digits):
    return False

for digit in digit_list:
    if number_search.count(digit[0]) != digits.count(digit[1]):
        return False

return True

Answer 8

I think you could use Multinomial theorem for some optimisation of cheacking if it is Narcissistic number.
you can calculate (a+b+c+..)^n- sum of non n-th powers values
for example for n=2 you should compare x and (a+b)^2-2*a*b where a and b is digits of number x

Answer 9

'''We can use Nar() function to calculate the Narcissitic Number.'''

import math
def Nar(num):
   sum=0
   n=len(str(num))
   while n>0:
     temp=num%10
     sum=sum+math.pow(temp,n)
     num=num/10
   return sum
x=input()
y=Nar(x)
if x==y:
  print x,' is a Narcissistic number'
else:
 print x,' is not a Narcissistic number'