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I'm trying to write a Haskell program that calculates multiples. Basically, when given two integers a and b, I want to find how many integers 1 ≤ bi ≤ b are multiple of any integer 2 ≤ ai ≤ a. For example, if a = 3 and b = 30, I want to know how many integers in the range of 1-30 are a multiple of 2 or 3; there are 20 such integers: 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30.
I have a C program that does this. I'm trying to get this translated into Haskell, but part of the difficulty is getting around the loops that I've used since Haskell doesn't use loops. I appreciate any and all help in translating this!
My C program for reference (sorry if formatting is off):
#define PRIME_RANGE 130
#define PRIME_CNT 32
#define UPPER_LIMIT (1000000000000000ull) //10^15
#define MAX_BASE_MULTIPLES_COUNT 25000000
typedef struct
{
char primeFactorFlag;
long long multiple;
}multipleInfo;
unsigned char primeFlag[PRIME_RANGE + 1];
int primes[PRIME_CNT];
int primeCnt = 0;
int maxPrimeStart[PRIME_CNT];
multipleInfo baseMultiples[MAX_BASE_MULTIPLES_COUNT];
multipleInfo mergedMultiples[MAX_BASE_MULTIPLES_COUNT];
int baseMultiplesCount, mergedMultiplesCount;
void findOddMultiples(int a, long long b, long long *count);
void generateBaseMultiples(void);
void mergeLists(multipleInfo listSource[], int countS, multipleInfo
listDest[], int *countD);
void sieve(void);
int main(void)
{
int i, j, a, n, startInd, endInd;
long long b, multiples;
//Generate primes
sieve();
primes[primeCnt] = PRIME_RANGE + 1;
generateBaseMultiples();
baseMultiples[baseMultiplesCount].multiple = UPPER_LIMIT + 1;
//Input and Output
scanf("%d", &n);
for(i = 1; i <= n; i++)
{
scanf("%d%lld", &a, &b);
//If b <= a, all are multiple except 1
if(b <= a)
printf("%lld\n",b-1);
else
{
//Add all even multiples
multiples = b / 2;
//Add all odd multiples
findOddMultiples(a, b, &multiples);-
printf("%lld\n", multiples);
}
}
return 0;
}
void findOddMultiples(int a, long long b, long long *count)
{
int i, k;
long long currentNum;
for(k = 1; k < primeCnt && primes[k] <= a; k++)
{
for(i = maxPrimeStart[k]; i < maxPrimeStart[k + 1] &&
baseMultiples[i].multiple <= b; i++)
{
currentNum = b/baseMultiples[i].multiple;
currentNum = (currentNum + 1) >> 1; // remove even multiples
if(baseMultiples[i].primeFactorFlag) //odd number of factors
(*count) += currentNum;
else
(*count) -= currentNum;
}
}
}
void addTheMultiple(long long value, int primeFactorFlag)
{
baseMultiples[baseMultiplesCount].multiple = value;
baseMultiples[baseMultiplesCount].primeFactorFlag = primeFactorFlag;
baseMultiplesCount++;
}
void generateBaseMultiples(void)
{
int i, j, t, prevCount;
long long curValue;
addTheMultiple(3, 1);
mergedMultiples[0] = baseMultiples[0];
mergedMultiplesCount = 1;
maxPrimeStart[1] = 0;
prevCount = mergedMultiplesCount;
for(i = 2; i < primeCnt; i++)
{
maxPrimeStart[i] = baseMultiplesCount;
addTheMultiple(primes[i], 1);
for(j = 0; j < prevCount; j++)
{
curValue = mergedMultiples[j].multiple * primes[i];
if(curValue > UPPER_LIMIT)
break;
addTheMultiple(curValue, 1 - mergedMultiples[j].primeFactorFlag);
}
if(i < primeCnt - 1)
mergeLists(&baseMultiples[prevCount], baseMultiplesCount - prevCount, mergedMultiples, &mergedMultiplesCount);
prevCount = mergedMultiplesCount;
}
maxPrimeStart[primeCnt] = baseMultiplesCount;
}
void mergeLists(multipleInfo listSource[], int countS, multipleInfo listDest[], int *countD)
{
int limit = countS + *countD;
int i1, i2, j, k;
//Copy one list in unused safe memory
for(j = limit - 1, k = *countD - 1; k >= 0; j--, k--)
listDest[j] = listDest[k];
//Merge the lists
for(i1 = 0, i2 = countS, k = 0; i1 < countS && i2 < limit; k++)
{
if(listSource[i1].multiple <= listDest[i2].multiple)
listDest[k] = listSource[i1++];
else
listDest[k] = listDest[i2++];
}
while(i1 < countS)
listDest[k++] = listSource[i1++];
while(i2 < limit)
listDest[k++] = listDest[i2++];
*countD = k;
}
void sieve(void)
{
int i, j, root = sqrt(PRIME_RANGE);
primes[primeCnt++] = 2;
for(i = 3; i <= PRIME_RANGE; i+= 2)
{
if(!primeFlag[i])
{
primes[primeCnt++] = i;
if(root >= i)
{
for(j = i * i; j <= PRIME_RANGE; j += i << 1)
primeFlag[j] = 1;
}
}
}
}
First, unless I'm grossly misunderstanding, the number of multiples you have there is wrong. The number of multiples of 2 between 1 and 30 is 15, and the number of multiples of 3 between 1 and 30 is 10, so there should be 25 numbers there.
EDIT: I did misunderstand; you want unique multiples.
To get unique multiples, you can use Data.Set, which has the invariant that the elements of the Set are unique and ordered ascendingly.
If you know you aren't going to exceed x = maxBound :: Int, you can get even better speedups using Data.IntSet. I've also included some test cases and annotated with comments what they run at on my machine.
{-# LANGUAGE BangPatterns #-}
{-# OPTIONS_GHC -O2 #-}
module Main (main) where
import System.CPUTime (getCPUTime)
import Data.IntSet (IntSet)
import qualified Data.IntSet as IntSet
main :: IO ()
main = do
test 3 30 -- 0.12 ms
test 131 132 -- 0.14 ms
test 500 300000 -- 117.63 ms
test :: Int -> Int -> IO ()
test !a !b = do
start <- getCPUTime
print (numMultiples a b)
end <- getCPUTime
print $ "Needed " ++ show ((fromIntegral (end - start)) / 10^9) ++ " ms.\n"
numMultiples :: Int -> Int -> Int
numMultiples !a !b = IntSet.size (foldMap go [2..a])
where
go :: Int -> IntSet
go !x = IntSet.fromAscList [x, x+x .. b]
I'm not really into understanding your C, so I implemented a solution afresh using the algorithm discussed here. The N in the linked algorithm is the product of the primes up to a in your problem description.
So first we'll need a list of primes. There's a standardish trick for getting a list of primes that is at once very idiomatic and relatively efficient:
primes :: [Integer]
primes = 2:filter isPrime [3..]
-- Doesn't work right for n<2, but we never call it there, so who cares?
isPrime :: Integer -> Bool
isPrime n = go primes n where
go (p:ps) n | p*p>n = True
| otherwise = n `rem` p /= 0 && go ps n
Next up: we want a way to iterate over the positive square-free divisors of N. This can be achieved by iterating over the subsets of the primes less than a. There's a standard idiomatic way to get a powerset, namely:
-- import Control.Monad
-- powerSet :: [a] -> [[a]]
-- powerSet = filterM (const [False, True])
That would be a fine component to use, but since at the end of the day we only care about the product of each powerset element and the value of the Mobius function of that product, we would end up duplicating a lot of multiplications and counting problems. It's cheaper to compute those two things directly while producing the powerset. So:
-- Given the prime factorization of a square-free number, produce a list of
-- its divisors d together with mu(d).
divisorsWithMu :: Num a => [a] -> [(a, a)]
divisorsWithMu [] = [(1, 1)]
divisorsWithMu (p:ps) = rec ++ [(p*d, -mu) | (d, mu) <- rec] where
rec = divisorsWithMu ps
With that in hand, we can just iterate and do a little arithmetic.
f :: Integer -> Integer -> Integer
f a b = b - sum
[ mu * (b `div` d)
| (d, mu) <- divisorsWithMu (takeWhile (<=a) primes)
]
And that's all the code. Crunched 137 lines of C down to 15 lines of Haskell -- not bad! Try it out in ghci:
> f 3 30
20
As an additional optimization, one could consider modifying divisorsWithMu to short-circuit when its divisor is bigger than b, as we know such terms will not contribute to the final sum. This makes a noticeable difference for large a, as without it there are exponentially many elements in the powerset. Here's how that modification looks:
-- Given an upper bound and the prime factorization of a square-free number,
-- produce a list of its divisors d that are no larger than the upper bound
-- together with mu(d).
divisorsWithMuUnder :: (Ord a, Num a) => a -> [a] -> [(a, a)]
divisorsWithMuUnder n [] = [(1, 1)]
divisorsWithMuUnder n (p:ps) = rec ++ [(p*d, -mu) | (d, mu) <- rec, p*d<=n]
where rec = divisorsWithMuUnder n ps
f' :: Integer -> Integer -> Integer
f' a b = b - sum
[ mu * (b `div` d)
| (d, mu) <- divisorsWithMuUnder b (takeWhile (<=a) primes)
]
Not much more complicated; the only really interesting difference is that there's now a condition in the list comprehension. Here's an example of f' finishing quickly for inputs that would take infeasibly long with f:
> f' 100 100000
88169
With data-ordlist package mentioned by Daniel Wagner in the comments, it is just
f a b = length $ unionAll [ [p,p+p..b] | p <- takeWhile (<= a) primes]
That is all. Some timings, for non-compiled code run inside GHCi:
~> f 100 (10^5)
88169
(0.05 secs, 48855072 bytes)
~> f 131 (3*10^6)
2659571
(0.55 secs, 1493586480 bytes)
~> f 131 132
131
(0.00 secs, 0 bytes)
~> f 500 300000
274055
(0.11 secs, 192704760 bytes)
Compiling will surely make the memory consumption a non-issue, by converting the length to a counting loop.
You'll have to use recursion in place of loops.
In (most) procedural or object-orientated languages, you should hardly ever (never?) be using recursion. It is horribly inefficient, as a new stack frame must be created each time the recursive function is called.
However, in a functional language, like Haskell, the compiler is often able to optimize the recursion away into a loop, which makes it much faster then its procedural counterparts.
I've converted your sieve function into a set of recursive functions in C. I'll leave it to you to convert it into Haskell:
int main(void) {
//...
int root = sqrt(PRIME_RANGE);
primes[primeCnt++] = 2;
sieve(3, PRIME_RANGE, root);
//...
}
void sieve(int i, int end, int root) {
if(i > end) {
return;
}
if(!primeFlag[i]) {
primes[primeCnt++] = i;
if(root >= i) {
markMultiples(i * i, PRIME_RANGE, i);
}
}
i += 2;
sieve(i, end, root);
}
void markMultiples(int j, int end, int prime) {
if(j > end) {
return;
}
primeFlag[j] = 1;
j += i << 1;
markMultiples(j, end, prime);
}
The point of recursion is that the same function is called repeatedly, until a condition is met. The results of one recursive call are passed onto the next call, until the condition is met.
Also, why are you bit-fiddling instead of just multiplying or dividing by 2? Any half-decent compiler these days can convert most multiplications and divisions by 2 into a bit-shift.
I know the Miller–Rabin primality test is probabilistic. However I want to use it for a programming task that leaves no room for error.
Can we assume that it is correct with very high probability if the input numbers are 64-bit integers (i.e. long long in C)?
Miller–Rabin is indeed probabilistic, but you can trade accuracy for computation time arbitrarily. If the number you test is prime, it will always give the correct answer. The problematic case is when a number is composite, but is reported to be prime. We can bound the probability of this error using the formula on Wikipedia: If you select k different bases randomly and test them, the error probability is less than 4-k. So even with k = 9, you only get a 3 in a million chance of being wrong. And with k = 40 or so it becomes ridiculously unlikely.
That said, there is a deterministic version of Miller–Rabin, relying on the correctness of the generalized Riemann hypothesis. For the range u
up to 264, it is enough to check a = 2, 3, 5, 7, 11, 13, 17, 19, 23. I have a C++ implementation online which was field-tested in lots of programming contests. Here's an instantiation of the template for unsigned 64-bit ints:
bool isprime(uint64_t n) { //determines if n is a prime number
const int pn = 9, p[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23 };
for (int i = 0; i < pn; ++i)
if (n % p[i] == 0) return n == p[i];
if (n < p[pn - 1]) return 0;
uint64_t s = 0, t = n - 1;
while (~t & 1)
t >>= 1, ++s;
for (int i = 0; i < pn; ++i) {
uint64_t pt = PowerMod(p[i], t, n);
if (pt == 1) continue;
bool ok = 0;
for (int j = 0; j < s && !ok; ++j) {
if (pt == n - 1) ok = 1;
pt = MultiplyMod(pt, pt, n);
}
if (!ok) return 0;
}
return 1;
}
PowerMod and MultiplyMod are just primitives to multiply and exponentiate under a given modulus, using square-and-{multiply,add}.
For n < 2^64, it is possible to perform strong-pseudoprime tests to the seven bases 2, 325, 9375, 28178, 450775, 9780504, and 1795265022 and completely determine the primality of n; see http://miller-rabin.appspot.com/.
A faster primality test performs a strong-pseudoprime test to base 2 followed by a Lucas pseudoprime test. It takes about 3 times as long as a single strong-pseudoprime test, so is more than twice as fast as the 7-base Miller-Rabin test. The code is more complex, but not dauntingly so.
I can post code if you're interested; let me know in the comments.
In each iteration of Miller-Rabin you need to choose a random number. If you are unlucky this random number doesn't reveal certain composites. A small example of this is that 2^341 mod 341 = 2, passing the test
But the test guarantees that it only lets a composite pass with probability <1/4. So if you run the test 64 times with different random values, the probability drops below 2^(-128) which is enough in practice.
You should take a look at the Baillie–PSW primality test. While it may have false positives, there are no known examples for this and according to wikipedia has been verified that no composite number below 2^64 passes the test. So it should fit your requirements.
There are efficient deterministic variants of the MR test for 64-bit values - which do not rely on the GRH - having been exhaustively tested by exploiting GPUs and other known results.
I've listed the pertinent sections of a C program I wrote that tests the primality of any 64-bit value: (n > 1), using Jaeschke's and Sinclair's bases for the deterministic MR variant. It makes use of gcc and clang's __int128 extended type for exponentiation. If not available, an explicit routine is required. Maybe others will find this useful...
#include <inttypes.h>
/******************************************************************************/
static int sprp (uint64_t n, uint64_t a)
{
uint64_t m = n - 1, r, y;
unsigned int s = 1, j;
/* assert(n > 2 && (n & 0x1) != 0); */
while ((m & (UINT64_C(1) << s)) == 0) s++;
r = m >> s; /* r, s s.t. 2^s * r = n - 1, r in odd. */
if ((a %= n) == 0) /* else (0 < a < n) */
return (1);
{
unsigned __int128 u = 1, w = a;
while (r != 0)
{
if ((r & 0x1) != 0)
u = (u * w) % n; /* (mul-rdx) */
if ((r >>= 1) != 0)
w = (w * w) % n; /* (sqr-rdx) */
}
if ((y = (uint64_t) u) == 1)
return (1);
}
for (j = 1; j < s && y != m; j++)
{
unsigned __int128 u = y;
u = (u * u) % n; /* (sqr-rdx) */
if ((y = (uint64_t) u) <= 1) /* (n) is composite: */
return (0);
}
return (y == m);
}
/******************************************************************************/
static int is_prime (uint64_t n)
{
const uint32_t sprp32_base[] = /* (Jaeschke) */ {
2, 7, 61, 0};
const uint32_t sprp64_base[] = /* (Sinclair) */ {
2, 325, 9375, 28178, 450775, 9780504, 1795265022, 0};
const uint32_t *sprp_base;
/* assert(n > 1); */
if ((n & 0x1) == 0) /* even: */
return (n == 2);
sprp_base = (n <= UINT32_MAX) ? sprp32_base : sprp64_base;
for (; *sprp_base != 0; sprp_base++)
if (!sprp(n, *sprp_base)) return (0);
return (1); /* prime. */
}
/******************************************************************************/
Note that the MR (sprp) test is slightly modified to pass values on an iteration where the base is a multiple of the candidate, as mentioned in the 'remarks' section of the website
Update: while this has fewer base tests than Niklas' answer, it's important to note that the bases: {3, 5, 7, 11, 13, 17, 19, 23, 29} provide a cheap test that allows us to eliminate candidates exceeding: 29 * 29 = 841 -
simply using the GCD.
For (n > 29 * 29), we can clearly eliminate any even value as prime. The product of the small primes: (3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29} = 3234846615, fits nicely in a 32-bit unsigned value. A gcd(n, 3234846615) is a lot cheaper than a MR test! If the result is not (1), then (n) > 841 has a small factor.
Merten's (?) theorem suggests that this simple gcd(u64, u64) test eliminates ~ 68% of all odd candidates (as composites). If you're using M-R to search for primes (randomly or incrementally), rather than just a 'one-off' test, this is certainly worth while!
Your computer is not perfect; it has a finite probability of failing in such a way as to produce an incorrect result to a calculation. Providing the probability of the M-R test giving a false result is greatly less than the probability of some other computer failure, then you are fine. There is no reason to run the M-R test for less than 64 iterations (a 1 in 2^128 chance of error). Most examples will fail in the first few iterations, so only the actual primes will be thoroughly tested. Use 128 iterations for a 1 in 2^256 chance of error.
I came across this question.A number is called lucky if the sum of its digits, as well as the sum of the squares of its digits is a prime number. How many numbers between A and B are lucky? 1 <= A <= B <= 1018. I tried this.
First I generated all possible primes between 1 and the number that could be resulted by summing the squares (81 *18 = 1458).
I read in A and B find out maximum number that could be generated by summing the digits If B is a 2 digit number ( the max number is 18 generated by 99).
For each prime number between 1 an max number. I applied integer partition algorithm.
For each possible partition I checked whether their sum of squares of their digits form prime. If so the possible permutations of that partition are generated and if they lie with in range they are lucky numbers.
This is the implementation:
#include<stdio.h>
#include<malloc.h>
#include<math.h>
#include <stdlib.h>
#include<string.h>
long long luckynumbers;
int primelist[1500];
int checklucky(long long possible,long long a,long long b){
int prime =0;
while(possible>0){
prime+=pow((possible%10),(float)2);
possible/=10;
}
if(primelist[prime]) return 1;
else return 0;
}
long long getmax(int numdigits){
if(numdigits == 0) return 1;
long long maxnum =10;
while(numdigits>1){
maxnum = maxnum *10;
numdigits-=1;
}
return maxnum;
}
void permuteandcheck(char *topermute,int d,long long a,long long b,int digits){
if(d == strlen(topermute)){
long long possible=atoll(topermute);
if(possible >= getmax(strlen(topermute)-1)){ // to skip the case of getting already read numbers like 21 and 021(permuted-210
if(possible >= a && possible <= b){
luckynumbers++;
}
}
}
else{
char lastswap ='\0';
int i;
char temp;
for(i=d;i<strlen(topermute);i++){
if(lastswap == topermute[i])
continue;
else
lastswap = topermute[i];
temp = topermute[d];
topermute[d] = topermute[i];
topermute[i] = temp;
permuteandcheck(topermute,d+1,a,b,digits);
temp = topermute[d];
topermute[d] = topermute[i];
topermute[i] = temp;
}
}
}
void findlucky(long long possible,long long a,long long b,int digits){
int i =0;
if(checklucky(possible,a,b)){
char topermute[18];
sprintf(topermute,"%lld",possible);
permuteandcheck(topermute,0,a,b,digits);
}
}
void partitiongenerator(int k,int n,int numdigits,long long possible,long long a,long long b,int digits){
if(k > n || numdigits > digits-1 || k > 9) return;
if(k == n){
possible+=(k*getmax(numdigits));
findlucky(possible,a,b,digits);
return;
}
partitiongenerator(k,n-k,numdigits+1,(possible + k*getmax(numdigits)),a,b,digits);
partitiongenerator(k+1,n,numdigits,possible,a,b,digits);
}
void calcluckynumbers(long long a,long long b){
int i;
int numdigits = 0;
long long temp = b;
while(temp > 0){
numdigits++;
temp/=10;
}
long long maxnum =getmax(numdigits)-1;
int maxprime=0,minprime =0;
temp = maxnum;
while(temp>0){
maxprime+=(temp%10);
temp/=10;
}
int start = 2;
for(;start <= maxprime ;start++){
if(primelist[start]) {
partitiongenerator(0,start,0,0,a,b,numdigits);
}
}
}
void generateprime(){
int i = 0;
for(i=0;i<1500;i++)
primelist[i] = 1;
primelist[0] = 0;
primelist[1] = 0;
int candidate = 2;
int topCandidate = 1499;
int thisFactor = 2;
while(thisFactor * thisFactor <= topCandidate){
int mark = thisFactor + thisFactor;
while(mark <= topCandidate){
*(primelist + mark) = 0;
mark += thisFactor;
}
thisFactor++;
while(thisFactor <= topCandidate && *(primelist+thisFactor) == 0) thisFactor++;
}
}
int main(){
char input[100];
int cases=0,casedone=0;
long long a,b;
generateprime();
fscanf(stdin,"%d",&cases);
while(casedone < cases){
luckynumbers = 0;
fscanf(stdin,"%lld %lld",&a,&b);
int i =0;
calcluckynumbers(a,b);
casedone++;
}
}
The algorithm is too slow. I think the answer can be found based on the property of numbers.Kindly share your thoughts. Thank you.
Excellent solution OleGG, But your code is not optimized. I have made following changes to your code,
It does not require to go through 9*9*i for k in count_lucky function, because for 10000 cases it would run that many times, instead i have reduced this value through start and end.
i have used ans array to store intermediate results. It might not look like much but over 10000 cases this is the major factor that reduces the time.
I have tested this code and it passed all the test cases. Here is the modified code:
#include <stdio.h>
const int MAX_LENGTH = 18;
const int MAX_SUM = 162;
const int MAX_SQUARE_SUM = 1458;
int primes[1460];
unsigned long long dyn_table[20][164][1460];
//changed here.......1
unsigned long long ans[19][10][164][1460]; //about 45 MB
int start[19][163];
int end[19][163];
//upto here.........1
void gen_primes() {
for (int i = 0; i <= MAX_SQUARE_SUM; ++i) {
primes[i] = 1;
}
primes[0] = primes[1] = 0;
for (int i = 2; i * i <= MAX_SQUARE_SUM; ++i) {
if (!primes[i]) {
continue;
}
for (int j = 2; i * j <= MAX_SQUARE_SUM; ++j) {
primes[i*j] = 0;
}
}
}
void gen_table() {
for (int i = 0; i <= MAX_LENGTH; ++i) {
for (int j = 0; j <= MAX_SUM; ++j) {
for (int k = 0; k <= MAX_SQUARE_SUM; ++k) {
dyn_table[i][j][k] = 0;
}
}
}
dyn_table[0][0][0] = 1;
for (int i = 0; i < MAX_LENGTH; ++i) {
for (int j = 0; j <= 9 * i; ++j) {
for (int k = 0; k <= 9 * 9 * i; ++k) {
for (int l = 0; l < 10; ++l) {
dyn_table[i + 1][j + l][k + l*l] += dyn_table[i][j][k];
}
}
}
}
}
unsigned long long count_lucky (unsigned long long maxp) {
unsigned long long result = 0;
int len = 0;
int split_max[MAX_LENGTH];
while (maxp) {
split_max[len] = maxp % 10;
maxp /= 10;
++len;
}
int sum = 0;
int sq_sum = 0;
unsigned long long step_result;
unsigned long long step_;
for (int i = len-1; i >= 0; --i) {
step_result = 0;
int x1 = 9*i;
for (int l = 0; l < split_max[i]; ++l) {
//changed here........2
step_ = 0;
if(ans[i][l][sum][sq_sum]!=0)
{
step_result +=ans[i][l][sum][sq_sum];
continue;
}
int y = l + sum;
int x = l*l + sq_sum;
for (int j = 0; j <= x1; ++j) {
if(primes[j + y])
for (int k=start[i][j]; k<=end[i][j]; ++k) {
if (primes[k + x]) {
step_result += dyn_table[i][j][k];
step_+=dyn_table[i][j][k];
}
}
}
ans[i][l][sum][sq_sum] = step_;
//upto here...............2
}
result += step_result;
sum += split_max[i];
sq_sum += split_max[i] * split_max[i];
}
if (primes[sum] && primes[sq_sum]) {
++result;
}
return result;
}
int main(int argc, char** argv) {
gen_primes();
gen_table();
//changed here..........3
for(int i=0;i<=18;i++)
for(int j=0;j<=163;j++)
{
for(int k=0;k<=1458;k++)
if(dyn_table[i][j][k]!=0ll)
{
start[i][j] = k;
break;
}
for(int k=1460;k>=0;k--)
if(dyn_table[i][j][k]!=0ll)
{
end[i][j]=k;
break;
}
}
//upto here..........3
int cases = 0;
scanf("%d",&cases);
for (int i = 0; i < cases; ++i) {
unsigned long long a, b;
scanf("%lld %lld", &a, &b);
//changed here......4
if(b == 1000000000000000000ll)
b--;
//upto here.........4
printf("%lld\n", count_lucky(b) - count_lucky(a-1));
}
return 0;
}
Explanation:
gen_primes() and gen_table() are pretty much self explanatory.
count_lucky() works as follows:
split the number in split_max[], just storing single digit number for ones, tens, hundreds etc. positions.
The idea is: suppose split_map[2] = 7, so we need to calculate result for
1 in hundreds position and all 00 to 99.
2 in hundreds position and all 00 to 99.
.
.
7 in hundreds position and all 00 to 99.
this is actually done(in l loop) in terms of sum of digits and sum of square of digits which has been precalcutaled.
for this example: sum will vary from 0 to 9*i & sum of square will vary from 0 to 9*9*i...this is done in j and k loops.
This is repeated for all lengths in i loop
This was the idea of OleGG.
For optimization following is considered:
its useless to run sum of squares from 0 to 9*9*i as for particular sums of digits it would not go upto the full range. Like if i = 3 and sum equals 5 then sum of square would not vary from 0 to 9*9*3.This part is stored in start[] and end[] arrays using precomputed values.
value for particular number of digits and particular digit at most significant position of number and upto particular sum and upto particular sum of square isstored for memorization. Its too long but still its about 45 MB.
I believe this could be further optimized.
You should use DP for this task. Here is my solution:
#include <stdio.h>
const int MAX_LENGTH = 18;
const int MAX_SUM = 162;
const int MAX_SQUARE_SUM = 1458;
int primes[1459];
long long dyn_table[19][163][1459];
void gen_primes() {
for (int i = 0; i <= MAX_SQUARE_SUM; ++i) {
primes[i] = 1;
}
primes[0] = primes[1] = 0;
for (int i = 2; i * i <= MAX_SQUARE_SUM; ++i) {
if (!primes[i]) {
continue;
}
for (int j = 2; i * j <= MAX_SQUARE_SUM; ++j) {
primes[i*j] = 0;
}
}
}
void gen_table() {
for (int i = 0; i <= MAX_LENGTH; ++i) {
for (int j = 0; j <= MAX_SUM; ++j) {
for (int k = 0; k <= MAX_SQUARE_SUM; ++k) {
dyn_table[i][j][k] = 0;
}
}
}
dyn_table[0][0][0] = 1;
for (int i = 0; i < MAX_LENGTH; ++i) {
for (int j = 0; j <= 9 * i; ++j) {
for (int k = 0; k <= 9 * 9 * i; ++k) {
for (int l = 0; l < 10; ++l) {
dyn_table[i + 1][j + l][k + l*l] += dyn_table[i][j][k];
}
}
}
}
}
long long count_lucky (long long max) {
long long result = 0;
int len = 0;
int split_max[MAX_LENGTH];
while (max) {
split_max[len] = max % 10;
max /= 10;
++len;
}
int sum = 0;
int sq_sum = 0;
for (int i = len-1; i >= 0; --i) {
long long step_result = 0;
for (int l = 0; l < split_max[i]; ++l) {
for (int j = 0; j <= 9 * i; ++j) {
for (int k = 0; k <= 9 * 9 * i; ++k) {
if (primes[j + l + sum] && primes[k + l*l + sq_sum]) {
step_result += dyn_table[i][j][k];
}
}
}
}
result += step_result;
sum += split_max[i];
sq_sum += split_max[i] * split_max[i];
}
if (primes[sum] && primes[sq_sum]) {
++result;
}
return result;
}
int main(int argc, char** argv) {
gen_primes();
gen_table();
int cases = 0;
scanf("%d", &cases);
for (int i = 0; i < cases; ++i) {
long long a, b;
scanf("%lld %lld", &a, &b);
printf("%lld\n", count_lucky(b) - count_lucky(a-1));
}
return 0;
}
Brief explanation:
I'm calculating all primes up to 9 * 9 * MAX_LENGTH using Eratosthenes method;
Later, using DP, I'm building table dyn_table where value X in dyn_table[i][j][k] means that we have exactly X numbers of length i with sum of digits equal to j and sum of its squares equal to k
Then we can easily count amount of lucky numbers from 1 to 999..999(len times of 9). For this we just sum up all dyn_table[len][j][k] where both j and k are primes.
To calculate amount of lucky number from 1 to random X we split interval from 1 to X into intervals with length equal to 10^K (see *count_lucky* function).
And our last step is subtract count_lucky(a-1) (cause we are including a in our interval) from count_lucky(b).
That's all. Precalculation work for O(log(MAX_NUMBER)^3), each step have also this complexity.
I've tested my solution against linear straightforward one and results were equal
Instead of enumerating the space of numbers, enumerate the different "signatures" of numbers that are lucky. and then print all the differnet combination of those.
This can be done with trivial backtracking:
#define _GNU_SOURCE
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#define bitsizeof(e) (CHAR_BIT * sizeof(e))
#define countof(e) (sizeof(e) / sizeof((e)[0]))
#define BITMASK_NTH(type_t, n) ( ((type_t)1) << ((n) & (bitsizeof(type_t) - 1)))
#define OP_BIT(bits, n, shift, op) \
((bits)[(unsigned)(n) / (shift)] op BITMASK_NTH(typeof(*(bits)), n))
#define TST_BIT(bits, n) OP_BIT(bits, n, bitsizeof(*(bits)), & )
#define SET_BIT(bits, n) (void)OP_BIT(bits, n, bitsizeof(*(bits)), |= )
/* fast is_prime {{{ */
static uint32_t primes_below_1M[(1U << 20) / bitsizeof(uint32_t)];
static void compute_primes_below_1M(void)
{
SET_BIT(primes_below_1M, 0);
SET_BIT(primes_below_1M, 1);
for (uint32_t i = 2; i < bitsizeof(primes_below_1M); i++) {
if (TST_BIT(primes_below_1M, i))
continue;
for (uint32_t j = i * 2; j < bitsizeof(primes_below_1M); j += i) {
SET_BIT(primes_below_1M, j);
}
}
}
static bool is_prime(uint64_t n)
{
assert (n < bitsizeof(primes_below_1M));
return !TST_BIT(primes_below_1M, n);
}
/* }}} */
static uint32_t prime_checks, found;
static char sig[10];
static uint32_t sum, square_sum;
static void backtrack(int startdigit, int ndigits, int maxdigit)
{
ndigits++;
for (int i = startdigit; i <= maxdigit; i++) {
sig[i]++;
sum += i;
square_sum += i * i;
prime_checks++;
if (is_prime(sum) && is_prime(square_sum)) {
found++;
}
if (ndigits < 18)
backtrack(0, ndigits, i);
sig[i]--;
sum -= i;
square_sum -= i * i;
}
}
int main(void)
{
compute_primes_below_1M();
backtrack(1, 0, 9);
printf("did %d signature checks, found %d lucky signatures\n",
prime_checks, found);
return 0;
}
When I run it it does:
$ time ./lucky
did 13123091 signature checks, found 933553 lucky signatures
./lucky 0.20s user 0.00s system 99% cpu 0.201 total
Instead of found++ you want to generate all the distinct permutations of digits that you can build with that number. I also precompute the first 1M of primes ever.
I've not checked if the code is 100% correct, you may have to debug it a bit. But the rought idea is here, and I'm able to generate all the lucky permutation below 0.2s (even without bugs it should not be more than twice as slow).
And of course you want to generate the permutations that verify A <= B. You may want to ignore generating partitions that have more digits than B or less than A too. Anyway you can improve on my general idea from here.
(Note: The blurb at the start is because I cut and paste code I wrote for project euler, hence the very fast is_prime that works for N <= 1M ;) )
For those who weren't aware already, this is a problem on the website InterviewStreet.com (and in my opinion, the most difficult one there). My approach started off similar to (and was inspired by) OleGG's below. However, after creating the first [19][163][1459] table that he did (which I'll call table1), I went in a slightly different direction. I created a second table of ragged length [19][x][3] (table2), where x is the number of unique sum pairs for the corresponding number of digits. And for the third dimension of the table, with length 3, the 1st element is the quantity of unique "sum pairs" with the sum and squareSum values held by the 2nd and 3rd elements.
For example:
//pseudocode
table2[1] = new long[10][3]
table2[1] = {{1, 0, 0}, {1, 1, 1}, {1, 2, 4},
{1, 3, 9}, {1, 4, 16}, {1, 5, 25},
{1, 6, 36}, {1, 7, 49}, {1, 8, 64}, {1, 9, 81}}
table2[2] = new long[55][3]
table2[3] = new long[204][3]
table2[4] = new long[518][3]
.
.
.
.
table2[17] = new long[15552][3]
table2[18] = new long[17547][3]
The numbers I have for the second dimension length of the array (10, 55, 204, 518, ..., 15552, 17547) can be verified by querying table1, and in a similar fashion table2 can be populated. Now using table2 we can solve large "lucky" queries much faster than OleGG's posted method, although still employing a similar "split" process as he did. For example, if you need to find lucky(00000-54321) (i.e. the lucky numbers between 0 and 54321), it breaks down to the sum of the following 5 lines:
lucky(00000-54321) = {
lucky(00000-49999) +
lucky(50000-53999) +
lucky(54000-54299) +
lucky(54300-53319) +
lucky(54320-54321)
}
Which breaks down further:
lucky(00000-49999) = {
lucky(00000-09999) +
lucky(10000-19999) +
lucky(20000-29999) +
lucky(30000-39999) +
lucky(40000-49999)
}
.
.
lucky(54000-54299) = {
lucky(54000-54099) +
lucky(54100-54199) +
lucky(54200-54299)
}
.
.
.
etc
Each of these values can be obtained easily by querying table2. For example, lucky(40000-49999) is found by adding 4 and 16 to the 2nd and 3rd elements of the third dimension table2:
sum = 0
for (i = 0; i < 518; i++)
if (isPrime[table2[4][i][1] + 4] && isPrime[table2[4][i][2] + 4*4])
sum += table2[4][i][0]
return sum
Or for lucky(54200-54299):
sum = 0
for (i = 0; i < 55; i++)
if (isPrime[table2[2][i][1] + (5+4+2)]
&& isPrime[table2[2][i][2] + (5*5+4*4+2*2)])
sum += table2[2][i][0]
return sum
Now, OleGG's solution performed significantly faster than anything else I'd tried up until then, but with my modifications described above, it performs even better than before (by a factor of roughly 100x for a large test set). However, it is still not nearly fast enough for the blind test cases given on InterviewStreet. Through some clever hack I was able to determine I am currently running about 20x too slow to complete their test set in the allotted time. However, I can find no further optimizations. The biggest time sink here is obviously iterating through the second dimension of table2, and the only way to avoid that would be to tabulate the results of those sums. However, there are too many possibilities to compute them all in the time given (5 seconds) or to store them all in the space given (256MB). For example, the lucky(54200-54299) loop above could be pre-computed and stored as a single value, but if it was, we'd also need to pre-compute lucky(123000200-123000299) and lucky(99999200-99999299), etc etc. I've done the math and it is way too many calculations to pre-compute.
I have just solved this problem.
It's just a Dynamic Programming problem. Take DP[n](sum-square_sum) as the DP function, and DP[n](sum-square_sum) is the count of all of the numbers whose digits is less than or equal to n, with the sum and square_sum of digits of the number is respectively represented by sum and square_sum. For example:
DP[1](1-1) = 1 # only 1 satisfies the condition
DP[2](1-1) = 2 # both 1 and 10 satisfies the condition
DP[3](1-1） = 3 # 1 10 100
DP[3](2-4） = 3 # 11 110 101
Since we can easily figure out the first DP state DP[1][..][..], it is:
(0-0) => 1 (1-1) => 1 (2-4) => 1 (3-9) => 1 (4-16) => 1
(5-25) => 1 (6-36) => 1 (7-49) => 1 (8-64) => 1 (9-81) => 1
then we can deduce DP[1] from DP[1], and then DP[3] ... DP[18]
the deduce above is made by the fact that every time when n increase by 1, for example from DP[1] to DP[2], we got a new digit (0..9), and the set of (sum, square_sum) pair (i.e. DP[n]) must be updated.
Finally, we can traverse the DP[18] set and count of the numbers that are lucky.
Well, how about the time and space complexity of the algorithm above?
As we know sum <= 18*9=162, square_sum <= 18*9*9 = 1458, so the set of (sum, square_sum) pair
(i.e. DP[n]) is very small, less than 162*1458=236196, in fact it's much smaller than 236196;
The fact is: my ruby program counting all the lucky numbers between 0 and 10^18 finishes in less than 1s.
ruby lucky_numbers.rb 0.55s user 0.00s system 99% cpu 0.556 total
and I test my program by writing a test function using brute force algorithm, and it's right for numbers less than 10^7 .
Based on the requirements, you can do it in different ways. If I was doing it, I would calculate the prime numbers using 'Sieve of Eratosthenes' in the required range (A to (9*2)*B.length), cache them (again, depending on your setup, you can use in-memory or disk cache) and use it for the next run.
I just coded a fast solution (Java), as below (NOTE: Integer overflow is not checked. Just a fast example. Also, my code is not optimized.):
import java.util.ArrayList;
import java.util.Arrays;
public class LuckyNumbers {
public static void main(String[] args) {
int a = 0, b = 1000;
LuckyNumbers luckyNums = new LuckyNumbers();
ArrayList<Integer> luckyList = luckyNums.findLuckyNums(a, b);
System.out.println(luckyList);
}
private ArrayList<Integer> findLuckyNums(int a, int b) {
ArrayList<Integer> luckyList = new ArrayList<Integer>();
int size = ("" + b).length();
int maxNum = 81 * 4; //9*2*b.length() - 9 is used, coz it's the max digit
System.out.println("Size : " + size + " MaxNum : " + maxNum);
boolean[] primeArray = sieve(maxNum);
for(int i=a;i<=b;i++) {
String num = "" + i;
int sumDigits = 0;
int sumSquareDigits = 0;
for(int j=0;j<num.length();j++) {
int digit = Integer.valueOf("" + num.charAt(j));
sumDigits += digit;
sumSquareDigits += Math.pow(digit, 2);
}
if(primeArray[sumDigits] && primeArray[sumSquareDigits]) {
luckyList.add(i);
}
}
return luckyList;
}
private boolean[] sieve(int n) {
boolean[] prime = new boolean[n + 1];
Arrays.fill(prime, true);
prime[0] = false;
prime[1] = false;
int m = (int) Math.sqrt(n);
for (int i = 2; i <= m; i++) {
if (prime[i]) {
for (int k = i * i; k <= n; k += i) {
prime[k] = false;
}
}
}
return prime;
}
}
And the output was:
[11, 12, 14, 16, 21, 23, 25, 32, 38, 41, 49, 52, 56, 58, 61, 65, 83, 85, 94, 101, 102, 104, 106, 110, 111, 113, 119, 120, 131, 133, 137, 140, 146, 160, 164, 166, 173, 179, 191, 197, 199, 201, 203, 205, 210, 223, 229, 230, 232, 250, 289, 292, 298, 302, 308, 311, 313, 317, 320, 322, 331, 335, 337, 344, 346, 353, 355, 364, 368, 371, 373, 377, 379, 380, 386, 388, 397, 401, 409, 410, 416, 434, 436, 443, 449, 461, 463, 467, 476, 490, 494, 502, 506, 508, 520, 533, 535, 553, 559, 560, 566, 580, 595, 601, 605, 610, 614, 616, 634, 638, 641, 643, 647, 650, 656, 661, 665, 674, 683, 689, 698, 713, 719, 731, 733, 737, 739, 746, 764, 773, 779, 791, 793, 797, 803, 805, 829, 830, 836, 838, 850, 863, 869, 883, 892, 896, 904, 911, 917, 919, 922, 928, 937, 940, 944, 955, 968, 971, 973, 977, 982, 986, 991]
I haven't carefully analyzed your current solution but this might improve on it:
Since the order of digits doesn't matter, you should go through all possible combinations of digits 0-9 of length 1 to 18, keeping track of the sum of digits and their squares and adding one digit at a time, using result of previous calculation.
So if you know that for 12 sum of digits is 3 and of squares is 5, look at numbers
120, 121, 122... etc and calculate sums for them trivially from the 3 and 5 for 12.
Sometimes the fastest solution is incredibly simple:
uint8_t precomputedBitField[] = {
...
};
bool is_lucky(int number) {
return precomputedBitField[number >> 8] & (1 << (number & 7));
}
Just modify your existing code to generate "precomputedBitField".
If you're worried about size, to cover all numbers from 0 to 999 it will only cost you 125 bytes, so this method will probably be smaller (and a lot faster) than any other alternative.
I was trying to come up with a solution using Pierre's enumeration method, but never came up with a sufficiently fast way to count the permutations. OleGG's counting method is very clever, and pirate's optimizations are necessary to make it fast enough. I came up with one minor improvement, and one workaround to a serious problem.
First, the improvement: you don't have to step through all the sums and squaresums one by one checking for primes in pirate's j and k loops. You have (or can easily generate) a list of primes. If you use the other variables to figure out which primes are in range, you can just step through the list of suitable primes for sum and squaresum. An array of primes and a lookup table to quickly determine at which index the prime >= a number is at is helpful. However, this is probably only a fairly minor improvement.
The big issue is with pirate's ans cache array. It is not 45MB as claimed; with 64 bit entries, it is something like 364MB. This is outside the (current) allowed memory limits for C and Java. It can be reduced to 37MB by getting rid of the "l" dimension, which is unnecessary and hurts cache performance anyway. You're really interested in caching counts for l + sum and l*l + squaresum, not l, sum, and squaresum individually.
First I would like to add that a lucky number can be calculated by a sieve, the explanation of the sieve can be found here: http://en.wikipedia.org/wiki/Lucky_number
so you can improve your solution speed using a sieve to determine the numbers,
This was an interview question:
Given a sequence of n numbers (n can be any number, assume n <= 100 for this question), say for eg. 11, 23, 9, 17, 20, 8, 5, 6 . Problem is to write a recursive function in C to add each number in the sequence to get the sum. If this sum is of more than one digit then sum the digits again and again if the sum is of more than one digit then sum the digits again. Follow this process until the sum is reduced to one digit no. Now add all the sums obtained in the process to output the final sum.
For illustration take above sequence: 11, 23, 9, 17, 20, 8, 5, 6
SUM(11, 23, 9, 17, 20, 8, 5, 6) = 99 => SUM(9, 9) = 18 => SUM(1, 8) = 9
Now add all the sums obtained, i.e. SUM(99, 18, 9) = 126 <== should be the output.
Please note that the function should be a recursive function in C.
Not sure about C, but the algorithm would look similar to this.
SUM(1, 2, ... n) = 1 + SUM(2, ... n) and so on to get the total, then repeat once the final number is found to be more than one digit.
Here's an Erlang implementation you could use as a guide
-module('summation').
-export([start/1]).
sumlist(List)->
lists:foldl(fun(X, Sum) -> X + Sum end, 0, List). << Inherently recursive
num_to_list(Value) ->
Str = integer_to_list(Value),
lists:map(fun(X) -> X - 48 end, Str). << Inherently recursive
accumulate([_H], List) ->
io:fwrite("~w~n", [List]),
List;
accumulate(Value, List) ->
Tmp = sumlist(Value),
accumulate(num_to_list(Tmp), [Tmp|List]). % << Recurse here
start(List)->
Value = accumulate(List, []),
sumlist(Value).
testing
25> c(summation).
{ok,summation}
26> summation:start([11, 23, 9, 17, 20, 8, 5, 6]).
[9,18,99]
126
27>
#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
const int n = 8;
int sumDigits(int x)
{
int d = 0;
while (x != 0)
{
d += x % 10;
x /= 10;
}
return d;
}
int sumArr(int* a, int start)
{
return (start == n)? 0: a[start] + sumArr(a, start + 1);
}
int sum(int x)
{
return (x < 10)? x: x + sum(sumDigits(x));
}
int main(int argc, _TCHAR* argv[])
{
int* a = new int[n];
a[0] = 11; a[1] = 23; a[2] = 9; a[3] = 17; a[4] = 20; a[5] = 8; a[6] = 5; a[7] = 6;
//for (int i = 0; i < n; i++) a[i] = rand() % 100;
//for (int i = 0; i < n; i++) printf("a[%d] = %d\n", i, a[i]);
printf("sum = %d\n", sum(sumArr(a, 0)));
return 0;
}
This outputs:
sum = 126
Here's a Scala implementation:
def sum(lst: List[Int]): Int = {
val sum1 = lst.reduceLeft(_+_)
println(sum1)
sum1 match {
case nb if nb < 10 => sum1
case _ => {
val lst2 = sum1.toString.toList.map(_.toString).map(Integer.parseInt(_))
sum1 + sum(lst2)
}
}
}
val lst = List(11, 23, 9, 17, 20, 8, 5, 6)
val totalSum = sum(lst)
println(totalSum)
Result:
99
18
9
126
I'm really beginning to love, how concise Scala is.
As others had said : the point here is to understand the recursion.
There are 3 place we can use recursion :
sum all the digits in a Integral number:
sum_digital :: (Integral a) => a -> a
sum_digital d
| d < 10 = d
| otherwise = d `mod` 10 + sum_digital (d `div` 10)
chain all the sums from a start value and the rules
chain :: (Integral a) => a -> [a]
chain a
| a < 10 = [a]
| otherwise = a : chain (sum_digital a)
final one. sum of a list
mySum :: (Integral a) => [a]-> a
mySum [] = 0
mySum (x:xs) = x + mySum xs
Put all these together:
*Main> mySum $ chain $ mySum [11, 23, 9, 17, 20, 8, 5, 6]
126
The C version is left for you as the exercise:)
I just want to add this one to 77v's answer in order to make everything hardcore recursive as possible. I know this is a year ago already, and his C++ solution works quite nice already. But I really had no fun that I though I can make that one last function called sumDigits in to recursion. So to rid myself of boredom, here it is:
long sumDigits(long x, long d = 0)
{
if (x != 0)
{
d = x % 10;
return d + sumDigits(x / 10, d);
}
else
return 0;
}
It's the same, 7 lines long and accepts one argument. Note that the second one is defaulted to 0. It's used as a memory for the recursion itself. The user may ignore that second argument entirely. The function is also used the same way as 77v's implementation. You can in fact directly replace his function with this one. Hence making all the function in his solution recursion based. Which makes an already awesome work more awesome! Lol! :D
#include <iostream>
using namespace std;
class RecursiveSum {
public:
int nDigits(int a) {
int d = 0;
while (a > 0) {
d++;
a /= 10;
}
return d;
}
long sum(int *arr, int b, int e, int s) {
if (!arr)
return 0;
if (b < e) {
s += arr[b++];
return sum(arr, b, e, s);
} else { // b >= e
if (s < 10)
return s;
int nd = nDigits(s);
int* narr = new int[nd];
long n = s, itr = 0;
while (n > 0) {
narr[itr++] = n % 10;
n /= 10;
}
s += sum(narr, 0, nd, 0);
delete[] narr;
return s;
}
}
};