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Although the challenge ended a long time ago, I'm kinda bored so I decided to try to factorise some of the numbers.
I initially had an O(n) algorithm, but then, I decided to research big O notation.
Apparently (I could be wrong), O(n) algorithms and O(2n) algorithms basically have the same running time. So do O(n) and O(4n) algorithms. In fact, O(n) and O(cn) algorithms (where c is an integer) essentially have the same running time.
So now, I have an O(8n) algorithm, but it isn't quick enough for 77-bit numbers.
What sort of time complexity would be required to factorise the first few RSA numbers (in under 5-ish minutes)?
My O(8n) algorithm:
import math
num = int(input())
sq = math.sqrt(num)
if num % 2 == 0:
print(2, int(num / 2))
elif sq % 1 == sq:
print(int(sq), int(sq))
else:
sq = round(sq)
a = 3
b = sq + (1 - (sq % 2))
c = ((b + 1) / 2)
d = ((b + 1) / 2)
c -= (1 - (c % 2))
d += (1 - (d % 2))
e = ((c + 1) / 2)
f = ((c + 1) / 2)
e -= (1 - (e % 2))
f += (1 - (f % 2))
g = ((d + 1) / 2) + d
h = ((d + 1) / 2) + d
g -= (1 - (g % 2))
h += (1 - (h % 2))
while a <= sq and num % a != 0 and b > 2 and num % b != 0 and c <= sq and num % c != 0 and d > 2 and num % d != 0 and e <= sq and num % e != 0 and f > 2 and num % f != 0 and g <= sq and num % g != 0 and h > 2 and num % h != 0:
a += 2
b -= 2
c += 2
d -= 2
e += 2
f -= 2
g += 2
h -= 2
if num % a == 0:
print(a, int(num / a))
elif num % b == 0:
print(b, int(num / b))
elif num % c == 0:
print(c, int(num / c))
elif num % d == 0:
print(d, int(num / d))
elif num % e == 0:
print(e, int(num / e))
elif num % f == 0:
print(f, int(num / f))
elif num % g == 0:
print(g, int(num / g))
elif num % h == 0:
print(h, int(num / h))
Your algorithm is poorly-implemented trial division. Throw it away.
Here is my basic prime-number library, using the Sieve of Eratosthenes to enumerate prime numbers, the Miller-Rabin algorithm to recognize primes, and wheel factorization followed by Pollard's rho algorithm to factor composites, which I leave to you to translate to Python:
function primes(n)
i, p, ps, m := 0, 3, [2], n // 2
sieve := makeArray(0..m-1, True)
while i < m
if sieve[i]
ps := p :: ps # insert at head of list
for j from (p*p-3)/2 to m step p
sieve[i] := False
i, p := i+1, p+2
return reverse(ps)
function isPrime(n, k=5)
if n < 2 then return False
for p in [2,3,5,7,11,13,17,19,23,29]
if n % p == 0 then return n == p
s, d = 0, n-1
while d % 2 == 0
s, d = s+1, d/2
for i from 0 to k
x = powerMod(randint(2, n-1), d, n)
if x == 1 or x == n-1 then next i
for r from 1 to s
x = (x * x) % n
if x == 1 then return False
if x == n-1 then next i
return False
return True
function factors(n, limit=10000)
wheel := [1,2,2,4,2,4,2,4,6,2,6]
w, f, fs := 0, 2, []
while f*f <= n and f < limit
while n % f == 0
fs, n := f :: fs, n / f
f, w := f + wheel[w], w+1
if w = 11 then w = 3
if n == 1 return fs
h, t, g, c := 1, 1, 1, 1
while not isPrime(n)
repeat
h := (h*h+c) % n # the hare runs
h := (h*h+c) % n # twice as fast
t := (t*t+c) % n # as the tortoise
g := gcd(t-h, n)
while g == 1
if isPrime(g)
while n % g == 0
fs, n := g :: fs, n / g
h, t, g, c := 1, 1, 1, c+1
return sort(n :: fs)
function powerMod(b, e, m)
x := 1
while e > 0
if e%2 == 1
x, e := (x*b)%m, e-1
else b, e := (b*b)%m, e//2
return x
function gcd(a, b)
if b == 0 then return a
return gcd(b, a % b)
Properly implemented, that algorithm should factor your 79-bit number nearly instantly.
To factor larger numbers, you will have to work harder. Look up "elliptic curve factorization" and "self-initializing quadratic sieve" to find factoring algorithms that you can implement yourself.
I tried to use MillerRabin + PollardP1_rho method to factorize an integer into primes in Python3 for reducing time complexity as much as I could.But it failed some tests,I knew where the problem was.But I am a tyro in algorithm, I didn't know how to fix it.So I will put all relative codes here.
import random
def gcd(a, b):
"""
a, b: integers
returns: a positive integer, the greatest common divisor of a & b.
"""
if a == 0:
return b
if a < 0:
return gcd(-a, b)
while b > 0:
c = a % b
a, b = b, c
return a
def mod_mul(a, b, n):
# Calculate a * b % n iterately.
result = 0
while b > 0:
if (b & 1) > 0:
result = (result + a) % n
a = (a + a) % n
b = (b >> 1)
return result
def mod_exp(a, b, n):
# Calculate (a ** b) % n iterately.
result = 1
while b > 0:
if (b & 1) > 0:
result = mod_mul(result, a, n)
a = mod_mul(a, a, n)
b = (b >> 1)
return result
def MillerRabinPrimeCheck(n):
if n in {2, 3, 5, 7, 11}:
return True
elif (n == 1 or n % 2 == 0 or n % 3 == 0 or n % 5 == 0 or n % 7 == 0 or n % 11 == 0):
return False
k = 0
u = n - 1
while not (u & 1) > 0:
k += 1
u = (u >> 1)
random.seed(0)
s = 5 #If the result isn't right, then add the var s.
for i in range(s):
x = random.randint(2, n - 1)
if x % n == 0:
continue
x = mod_exp(x, u, n)
pre = x
for j in range(k):
x = mod_mul(x, x, n)
if (x == 1 and pre != 1 and pre != n - 1):
return False
pre = x
if x != 1:
return False
return True
def PollardP1_rho(n, c):
'''
Consider c as a constant integer.
'''
i = 1
k = 2
x = random.randrange(1, n - 1) + 1
y = x
while 1:
i += 1
x = (mod_mul(x, x, n) + c) % n
d = gcd(y - x, n)
if 1 < d < n:
return d
elif x == y:
return n
elif i == k:
y = x
k = (k << 1)
result = []
def PrimeFactorsListGenerator(n):
if n <= 1:
pass
elif MillerRabinPrimeCheck(n) == True:
result.append(n)
else:
a = n
while a == n:
a = PollardP1_rho(n, random.randrange(1,n - 1) + 1)
PrimeFactorsListGenerator(a)
PrimeFactorsListGenerator(n // a)
When I tried to test this:
PrimeFactorsListGenerator(4)
It didn't stop and looped this:
PollardP1_rho(4, random.randrange(1,4 - 1) + 1)
I have already tested the functions before PollardP1_rho and they work normally,so I know the function PollardP1_rho cannot deal the number 4 correctly,also the number 5.How can I fix that?
I have solved it myself.
There is 1 mistake in the code.
I should not use a var 'result' outside of the function as a global var,I should define in the function and use result.extend() to ensure the availability of the whole recursive process.So I rewrote PollardP1_rho(n, c) and PrimeFactorsListGenerator(n):
def Pollard_rho(x, c):
'''
Consider c as a constant integer.
'''
i, k = 1, 2
x0 = random.randint(0, x)
y = x0
while 1:
i += 1
x0 = (mod_mul(x0, x0, x) + c) % x
d = gcd(y - x0, x)
if d != 1 and d != x:
return d
if y == x0:
return x
if i == k:
y = x0
k += k
def PrimeFactorsListGenerator(n):
result = []
if n <= 1:
return None
if MillerRabinPrimeCheck(n):
return [n]
p = n
while p >= n:
p = Pollard_rho(p, random.randint(1, n - 1))
result.extend(PrimeFactorsListGenerator(p))
result.extend(PrimeFactorsListGenerator(n // p))
return result
#PrimeFactorsListGenerator(400)
#PrimeFactorsListGenerator(40000)
There is an additional tip: You don't need to write a function mod_mul(a, b, n) at all, using Python built-in pow(a, b, n) will do the trick and it is fully optimized.
How do I check how many numbers from 1 to N (N < 100) have number 3 in it without converting it to a string to check it?
You can use th % mod operator to take the digits of the number one by one, and check them with 3.
Like,
int x;
while(num != 0) // num here goes from 1 to 100
{
x = num % 10;
if(x == 3)
{
//eureka
}
num /= 10;
}
EDIT
Lets check the algorithm for 35.
First iteration
//num = 35
x = num % 10; // x = 5 (35 % 10)
if(x == 3) // is x equal to 3 (NO)
{
//eureka
}
num /= 10; // num = 3 (35 / 10)
While loop check
num != 0 // num = 5
Second Iteration
//num = 35
x = num % 10; // x = 3 (5 % 10)
if(x == 3) // is x equal to 3 (YES)
{
//eureka
}
num /= 10; // num = 0 (5 / 10)
While loop check
num != 0 // num = 0
// While loop exits
I think the simplest way is by remainder and checking if number is between 30 and 39
if((x%10)==3||(x<40&&x>=30))
{
//Thats it
}
You can make use of the modulas operator % such that if(n % 3 == 1){success operation}
A question last week defined the zig zag ordering on an n by m matrix and asked how to list the elements in that order.
My question is how to quickly find the ith item in the zigzag ordering? That is, without traversing the matrix (for large n and m that's much too slow).
For example with n=m=8 as in the picture and (x, y) describing (row, column)
f(0) = (0, 0)
f(1) = (0, 1)
f(2) = (1, 0)
f(3) = (2, 0)
f(4) = (1, 1)
...
f(63) = (7, 7)
Specific question: what is the ten billionth (1e10) item in the zigzag ordering of a million by million matrix?
Let's assume that the desired element is located in the upper half of the matrix. The length of the diagonals are 1, 2, 3 ..., n.
Let's find the desired diagonal. It satisfies the following property:
sum(1, 2 ..., k) >= pos but sum(1, 2, ..., k - 1) < pos. The sum of 1, 2, ..., k is k * (k + 1) / 2. So we just need to find the smallest integer k such that k * (k + 1) / 2 >= pos. We can either use a binary search or solve this quadratic inequality explicitly.
When we know the k, we just need to find the pos - (k - 1) * k / 2 element of this diagonal. We know where it starts and where we should move(up or down, depending on the parity of k), so we can find the desired cell using a simple formula.
This solution has an O(1) or an O(log n) time complexity(it depends on whether we use a binary search or solve the inequation explicitly in step 2).
If the desired element is located in the lower half of the matrix, we can solve this problem for a pos' = n * n - pos + 1 and then use symmetry to get the solution to the original problem.
I used 1-based indexing in this solution, using 0-based indexing might require adding +1 or -1 somewhere, but the idea of the solution is the same.
If the matrix is rectangular, not square, we need to consider the fact the length of diagonals look this way: 1, 2, 3, ..., m, m, m, .., m, m - 1, ..., 1(if m <= n) when we search for the k, so the sum becomes something like k * (k + 1) / 2 if k <= m and k * (k + 1) / 2 + m * (k - m) otherwise.
import math, random
def naive(n, m, ord, swap = False):
dx = 1
dy = -1
if swap:
dx, dy = dy, dx
cur = [0, 0]
for i in range(ord):
cur[0] += dy
cur[1] += dx
if cur[0] < 0 or cur[1] < 0 or cur[0] >= n or cur[1] >= m:
dx, dy = dy, dx
if cur[0] >= n:
cur[0] = n - 1
cur[1] += 2
if cur[1] >= m:
cur[1] = m - 1
cur[0] += 2
if cur[0] < 0: cur[0] = 0
if cur[1] < 0: cur[1] = 0
return cur
def fast(n, m, ord, swap = False):
if n < m:
x, y = fast(m, n, ord, not swap)
return [y, x]
alt = n * m - ord - 1
if alt < ord:
x, y = fast(n, m, alt, swap if (n + m) % 2 == 0 else not swap)
return [n - x - 1, m - y - 1]
if ord < (m * (m + 1) / 2):
diag = int((-1 + math.sqrt(1 + 8 * ord)) / 2)
parity = (diag + (0 if swap else 1)) % 2
within = ord - (diag * (diag + 1) / 2)
if parity: return [diag - within, within]
else: return [within, diag - within]
else:
ord -= (m * (m + 1) / 2)
diag = int(ord / m)
within = ord - diag * m
diag += m
parity = (diag + (0 if swap else 1)) % 2
if not parity:
within = m - within - 1
return [diag - within, within]
if __name__ == "__main__":
for i in range(1000):
n = random.randint(3, 100)
m = random.randint(3, 100)
ord = random.randint(0, n * m - 1)
swap = random.randint(0, 99) < 50
na = naive(n, m, ord, swap)
fa = fast(n, m, ord, swap)
assert na == fa, "(%d, %d, %d, %s) ==> (%s), (%s)" % (n, m, ord, swap, na, fa)
print fast(1000000, 1000000, 9999999999, False)
print fast(1000000, 1000000, 10000000000, False)
So the 10-billionth element (the one with ordinal 9999999999), and the 10-billion-first element (the one with ordinal 10^10) are:
[20331, 121089]
[20330, 121090]
An analytical solution
In the general case, your matrix will be divided in 3 areas:
an initial triangle t1
a skewed part mid where diagonals have a constant length
a final triangle t2
Let's call p the index of your diagonal run.
We want to define two functions x(p) and y(p) that give you the column and row of the pth cell.
Initial triangle
Let's look at the initial triangular part t1, where each new diagonal is one unit longer than the preceding.
Now let's call d the index of the diagonal that holds the cell, and
Sp = sum(di) for i in [0..p-1]
We have p = Sp + k, with 0 <=k <= d and
Sp = d(d+1)/2
if we solve for d, it brings
d²+d-2p = 0, a quadratic equation where we retain only the positive root:
d = (-1+sqrt(1+8*p))/2
Now we want the highest integer value closest to d, which is floor(d).
In the end, we have
p = d + k with d = floor((-1+sqrt(1+8*p))/2) and k = p - d(d+1)/2
Let's call
o(d) the function that equals 1 if d is odd and 0 otherwise, and
e(d) the function that equals 1 if d is even and 0 otherwise.
We can compute x(p) and y(p) like so:
d = floor((-1+sqrt(1+8*p))/2)
k = p - d(d+1)/2
o = d % 2
e = 1 - o
x = e*d + (o-e)*k
y = o*d + (e-o)*k
even and odd functions are used to try to salvage some clarity, but you can replace
e(p) with 1 - o(p) and have slightly more efficient but less symetric formulaes for x and y.
Middle part
let's consider the smallest matrix dimension s, i.e. s = min (m,n).
The previous formulaes hold until x or y (whichever comes first) reaches the value s.
The upper bound of p such as x(i) <= s and y(i) <= s for all i in [0..p]
(i.e. the cell indexed by p is inside the initial triangle t1) is given by
pt1 = s(s+1)/2.
For p >= pt1, diagonal length remains equal to s until we reach the second triangle t2.
when inside mid, we have:
p = s(s+1)/2 + ds + k with k in [0..s[.
which yields:
d = floor ((p - s(s+1)/2)/s)
k = p - ds
We can then use the same even/odd trick to compute x(p) and y(p):
p -= s(s+1)/2
d = floor (p / s)
k = p - d*s
o = (d+s) % 2
e = 1 - o
x = o*s + (e-o)*k
y = e*s + (o-e)*k
if (n > m)
x += d+e
y -= e
else
y += d+o
x -= o
Final triangle
Using symetry, we can calculate pt2 = m*n - s(s+1)/2
We now face nearly the same problem as for t1, except that the diagonal may run in the same direction as for t1 or in the reverse direction (if n+m is odd).
Using symetry tricks, we can compute x(p) and y(p) like so:
p = n*m -1 - p
d = floor((-1+sqrt(1+8*p))/2)
k = p - d*(d+1)/2
o = (d+m+n) % 2
e = 1 - $o;
x = n-1 - (o*d + (e-o)*k)
y = m-1 - (e*d + (o-e)*k)
Putting all together
Here is a sample c++ implementation.
I used 64 bits integers out of sheer lazyness. Most could be replaced by 32 bits values.
The computations could be made more effective by precomputing a few more coefficients.
A good part of the code could be factorized, but I doubt it is worth the effort.
Since this is just a quick and dirty proof of concept, I did not optimize it.
#include <cstdio> // printf
#include <algorithm> // min
using namespace std;
typedef long long tCoord;
void panic(const char * msg)
{
printf("PANIC: %s\n", msg);
exit(-1);
}
struct tPoint {
tCoord x, y;
tPoint(tCoord x = 0, tCoord y = 0) : x(x), y(y) {}
tPoint operator+(const tPoint & p) const { return{ x + p.x, y + p.y }; }
bool operator!=(const tPoint & p) const { return x != p.x || y != p.y; }
};
class tMatrix {
tCoord n, m; // dimensions
tCoord s; // smallest dimension
tCoord pt1, pt2; // t1 / mid / t2 limits for p
public:
tMatrix(tCoord n, tCoord m) : n(n), m(m)
{
s = min(n, m);
pt1 = (s*(s + 1)) / 2;
pt2 = n*m - pt1;
}
tPoint diagonal_cell(tCoord p)
{
tCoord x, y;
if (p < pt1) // inside t1
{
tCoord d = (tCoord)floor((-1 + sqrt(1 + 8 * p)) / 2);
tCoord k = p - (d*(d + 1)) / 2;
tCoord o = d % 2;
tCoord e = 1 - o;
x = o*d + (e - o)*k;
y = e*d + (o - e)*k;
}
else if (p < pt2) // inside mid
{
p -= pt1;
tCoord d = (tCoord)floor(p / s);
tCoord k = p - d*s;
tCoord o = (d + s) % 2;
tCoord e = 1 - o;
x = o*s + (e - o)*k;
y = e*s + (o - e)*k;
if (m > n) // vertical matrix
{
x -= o;
y += d + o;
}
else // horizontal matrix
{
x += d + e;
y -= e;
}
}
else // inside t2
{
p = n * m - 1 - p;
tCoord d = (tCoord)floor((-1 + sqrt(1 + 8 * p)) / 2);
tCoord k = p - (d*(d + 1)) / 2;
tCoord o = (d + m + n) % 2;
tCoord e = 1 - o;
x = n - 1 - (o*d + (e - o)*k);
y = m - 1 - (e*d + (o - e)*k);
}
return{ x, y };
}
void check(void)
{
tPoint move[4] = { { 1, 0 }, { -1, 1 }, { 1, -1 }, { 0, 1 } };
tPoint pos;
tCoord dir = 0;
for (tCoord p = 0; p != n * m ; p++)
{
tPoint dc = diagonal_cell(p);
if (pos != dc) panic("zot!");
pos = pos + move[dir];
if (dir == 0)
{
if (pos.y == m - 1) dir = 2;
else dir = 1;
}
else if (dir == 3)
{
if (pos.x == n - 1) dir = 1;
else dir = 2;
}
else if (dir == 1)
{
if (pos.y == m - 1) dir = 0;
else if (pos.x == 0) dir = 3;
}
else
{
if (pos.x == n - 1) dir = 3;
else if (pos.y == 0) dir = 0;
}
}
}
};
void main(void)
{
const tPoint dim[] = { { 10, 10 }, { 11, 11 }, { 10, 30 }, { 30, 10 }, { 10, 31 }, { 31, 10 }, { 11, 31 }, { 31, 11 } };
for (tPoint d : dim)
{
printf("Checking a %lldx%lld matrix...", d.x, d.y);
tMatrix(d.x, d.y).check();
printf("done\n");
}
tCoord p = 10000000000;
tMatrix matrix(1000000, 1000000);
tPoint cell = matrix.diagonal_cell(p);
printf("Coordinates of %lldth cell: (%lld,%lld)\n", p, cell.x, cell.y);
}
Results are checked against "manual" sweep of the matrix.
This "manual" sweep is a ugly hack that won't work for a one-row or one-column matrix, though diagonal_cell() does work on any matrix (the "diagonal" sweep becomes linear in that case).
The coordinates found for the 10.000.000.000th cell of a 1.000.000x1.000.000 matrix seem consistent, since the diagonal d on which the cell stands is about sqrt(2*1e10), approx. 141421, and the sum of cell coordinates is about equal to d (121090+20330 = 141420). Besides, it is also what the two other posters report.
I would say there is a good chance this lump of obfuscated code actually produces an O(1) solution to your problem.
I don't know how to go about this programming problem.
Given two integers n and m, how many numbers exist such that all numbers have all digits from 0 to n-1 and the difference between two adjacent digits is exactly 1 and the number of digits in the number is atmost 'm'.
What is the best way to solve this problem? Is there a direct mathematical formula?
Edit: The number cannot start with 0.
Example:
for n = 3 and m = 6 there are 18 such numbers (210, 2101, 21012, 210121 ... etc)
Update (some people have encountered an ambiguity):
All digits from 0 to n-1 must be present.
This Python code computes the answer in O(nm) by keeping track of the numbers ending with a particular digit.
Different arrays (A,B,C,D) are used to track numbers that have hit the maximum or minimum of the range.
n=3
m=6
A=[1]*n # Number of ways of being at digit i and never being to min or max
B=[0]*n # number of ways with minimum being observed
C=[0]*n # number of ways with maximum being observed
D=[0]*n # number of ways with both being observed
A[0]=0 # Cannot start with 0
A[n-1]=0 # Have seen max so this 1 moves from A to C
C[n-1]=1 # Have seen max if start with highest digit
t=0
for k in range(m-1):
A2=[0]*n
B2=[0]*n
C2=[0]*n
D2=[0]*n
for i in range(1,n-1):
A2[i]=A[i+1]+A[i-1]
B2[i]=B[i+1]+B[i-1]
C2[i]=C[i+1]+C[i-1]
D2[i]=D[i+1]+D[i-1]
B2[0]=A[1]+B[1]
C2[n-1]=A[n-2]+C[n-2]
D2[0]=C[1]+D[1]
D2[n-1]=B[n-2]+D[n-2]
A=A2
B=B2
C=C2
D=D2
x=sum(d for d in D2)
t+=x
print t
After doing some more research, I think there may actually be a mathematical approach after all, although the math is advanced for me. Douglas S. Stones pointed me in the direction of Joseph Myers' (2008) article, BMO 2008–2009 Round 1 Problem 1—Generalisation, which derives formulas for calculating the number of zig-zag paths across a rectangular board.
As I understand it, in Anirudh's example, our board would have 6 rows of length 3 (I believe this would mean n=3 and r=6 in the article's terms). We can visualize our board so:
0 1 2 example zig-zag path: 0
0 1 2 1
0 1 2 0
0 1 2 1
0 1 2 2
0 1 2 1
Since Myers' formula m(n,r) would generate the number for all the zig-zag paths, that is, the number of all 6-digit numbers where all adjacent digits are consecutive and digits are chosen from (0,1,2), we would still need to determine and subtract those that begin with zero and those that do not include all digits.
If I understand correctly, we may do this in the following way for our example, although generalizing the concept to arbitrary m and n may prove more complicated:
Let m(3,6) equal the number of 6-digit numbers where all adjacent digits
are consecutive and digits are chosen from (0,1,2). According to Myers,
m(3,r) is given by formula and also equals OEIS sequence A029744 at
index r+2, so we have
m(3,6) = 16
How many of these numbers start with zero? Myers describes c(n,r) as the
number of zig-zag paths whose colour is that of the square in the top
right corner of the board. In our case, c(3,6) would include the total
for starting-digit 0 as well as starting-digit 2. He gives c(3,2r) as 2^r,
so we have
c(3,6) = 8. For starting-digit 0 only, we divide by two to get 4.
Now we need to obtain only those numbers that include all the digits in
the range, but how? We can do this be subtracting m(n-1,r) from m(n,r).
In our case, we have all the m(2,6) that would include only 0's and 1's,
and all the m(2,6) that would include 1's and 2's. Myers gives
m(2,anything) as 2, so we have
2*m(2,6) = 2*2 = 4
But we must remember that one of the zero-starting numbers is included
in our total for 2*m(2,6), namely 010101. So all together we have
m(3,6) - c(3,6)/2 - 4 + 1
= 16 - 4 - 4 + 1
= 9
To complete our example, we must follow a similar process for m(3,5),
m(3,4) and m(3,3). Since it's late here, I might follow up tomorrow...
One approach could be to program it recursively, calling the function to add as well as subtract from the last digit.
Haskell code:
import Data.List (sort,nub)
f n m = concatMap (combs n) [n..m]
combs n m = concatMap (\x -> combs' 1 [x]) [1..n - 1] where
combs' count result
| count == m = if test then [concatMap show result] else []
| otherwise = combs' (count + 1) (result ++ [r + 1])
++ combs' (count + 1) (result ++ [r - 1])
where r = last result
test = (nub . sort $ result) == [0..n - 1]
Output:
*Main> f 3 6
["210","1210","1012","2101","12101","10121","21210","21012"
,"21010","121210","121012","121010","101212","101210","101012"
,"212101","210121","210101"]
In response to Anirudh Rayabharam's comment, I hope the following code will be more 'pseudocode' like. When the total number of digits reaches m, the function g outputs 1 if the solution has hashed all [0..n-1], and 0 if not. The function f accumulates the results for g for starting digits [1..n-1] and total number of digits [n..m].
Haskell code:
import qualified Data.Set as S
g :: Int -> Int -> Int -> Int -> (S.Set Int, Int) -> Int
g n m digitCount lastDigit (hash,hashCount)
| digitCount == m = if test then 1 else 0
| otherwise =
if lastDigit == 0
then g n m d' (lastDigit + 1) (hash'',hashCount')
else if lastDigit == n - 1
then g n m d' (lastDigit - 1) (hash'',hashCount')
else g n m d' (lastDigit + 1) (hash'',hashCount')
+ g n m d' (lastDigit - 1) (hash'',hashCount')
where test = hashCount' == n
d' = digitCount + 1
hash'' = if test then S.empty else hash'
(hash',hashCount')
| hashCount == n = (S.empty,hashCount)
| S.member lastDigit hash = (hash,hashCount)
| otherwise = (S.insert lastDigit hash,hashCount + 1)
f n m = foldr forEachNumDigits 0 [n..m] where
forEachNumDigits numDigits accumulator =
accumulator + foldr forEachStartingDigit 0 [1..n - 1] where
forEachStartingDigit startingDigit accumulator' =
accumulator' + g n numDigits 1 startingDigit (S.empty,0)
Output:
*Main> f 3 6
18
(0.01 secs, 571980 bytes)
*Main> f 4 20
62784
(1.23 secs, 97795656 bytes)
*Main> f 4 25
762465
(11.73 secs, 1068373268 bytes)
model your problem as 2 superimposed lattices in 2 dimensions, specifically as pairs (i,j) interconnected with oriented edges ((i0,j0),(i1,j1)) where i1 = i0 + 1, |j1 - j0| = 1, modified as follows:
dropping all pairs (i,j) with j > 9 and its incident edges
dropping all pairs (i,j) with i > m-1 and its incident edges
dropping edge ((0,0), (1,1))
this construction results in a structure like in this diagram:
:
the requested numbers map to paths in the lattice starting at one of the green elements ((0,j), j=1..min(n-1,9)) that contain at least one pink and one red element ((i,0), i=1..m-1, (i,n-1), i=0..m-1 ). to see this, identify the i-th digit j of a given number with point (i,j). including pink and red elements ('extremal digits') guarantee that all available diguts are represented in the number.
Analysis
for convenience, let q1, q2 denote the position-1.
let q1 be the position of a number's first digit being either 0 or min(n-1,9).
let q2 be the position of a number's first 0 if the digit at position q1 is min(n-1,9) and vv.
case 1: first extremal digit is 0
the number of valid prefixes containing no 0 can be expressed as sum_{k=1..min(n-1,9)} (paths_to_0(k,1,q1), the function paths_to_0 being recursively defined as
paths_to_0(0,q1-1,q1) = 0;
paths_to_0(1,q1-1,q1) = 1;
paths_to_0(digit,i,q1) = 0; if q1-i < digit;
paths_to_0(x,_,_) = 0; if x >= min(n-1,9)
// x=min(n-1,9) mustn't occur before position q2,
// x > min(n-1,9) not at all
paths_to_0(x,_,_) = 0; if x <= 0;
// x=0 mustn't occur before position q1,
// x < 0 not at all
and else paths_to_0(digit,i,q1) =
paths_to_0(digit+1,i+1,q1) + paths_to_0(digit-1,i+1,q1);
similarly we have
paths_to_max(min(n-1,9),q2-1,q2) = 0;
paths_to_max(min(n-2,8),q2-1,q2) = 1;
paths_to_max(digit,i,q2) = 0 if q2-i < n-1;
paths_to_max(x,_,_) = 0; if x >= min(n-1,9)
// x=min(n-1,9) mustn't occur before
// position q2,
// x > min(n-1,9) not at all
paths_to_max(x,_,_) = 0; if x < 0;
and else paths_to_max(digit,q1,q2) =
paths_max(digit+1,q1+1,q2) + paths_to_max(digit-1,q1+1,q2);
and finally
paths_suffix(digit,length-1,length) = 2; if digit > 0 and digit < min(n-1,9)
paths_suffix(digit,length-1,length) = 1; if digit = 0 or digit = min(n-1,9)
paths_suffix(digit,k,length) = 0; if length > m-1
or length < q2
or k > length
paths_suffix(digit,k,0) = 1; // the empty path
and else paths_suffix(digit,k,length) =
paths_suffix(digit+1,k+1,length) + paths_suffix(digit-1,k+1,length);
... for a grand total of
number_count_case_1(n, m) =
sum_{first=1..min(n-1,9), q1=1..m-1-(n-1), q2=q1..m-1, l_suffix=0..m-1-q2} (
paths_to_0(first,1,q1)
+ paths_to_max(0,q1,q2)
+ paths_suffix(min(n-1,9),q2,l_suffix+q2)
)
case 2: first extremal digit is min(n-1,9)
case 2.1: initial digit is not min(n-1,9)
this is symmetrical to case 1 with all digits d replaced by min(n,10) - d. as the lattice structure is symmetrical, this means number_count_case_2_1 = number_count_case_1.
case 2.2: initial digit is min(n-1,9)
note that q1 is 1 and the second digit must be min(n-2,8).
thus
number_count_case_2_2 (n, m) =
sum_{q2=1..m-2, l_suffix=0..m-2-q2} (
paths_to_max(1,1,q2)
+ paths_suffix(min(n-1,9),q2,l_suffix+q2)
)
so the grand grand total will be
number_count ( n, m ) = 2 * number_count_case_1 (n, m) + number_count_case_2_2 (n, m);
Code
i don't know whether a closed expression for number_count exists, but the following perl code will compute it (the code is but a proof of concept as it does not use memoization techniques to avoid recomputing results already obtained):
use strict;
use warnings;
my ($n, $m) = ( 5, 7 ); # for example
$n = ($n > 10) ? 10 : $n; # cutoff
sub min
sub paths_to_0 ($$$) {
my (
$d
, $at
, $until
) = #_;
#
if (($d == 0) && ($at == $until - 1)) { return 0; }
if (($d == 1) && ($at == $until - 1)) { return 1; }
if ($until - $at < $d) { return 0; }
if (($d <= 0) || ($d >= $n))) { return 0; }
return paths_to_0($d+1, $at+1, $until) + paths_to_0($d-1, $at+1, $until);
} # paths_to_0
sub paths_to_max ($$$) {
my (
$d
, $at
, $until
) = #_;
#
if (($d == $n-1) && ($at == $until - 1)) { return 0; }
if (($d == $n-2) && ($at == $until - 1)) { return 1; }
if ($until - $at < $n-1) { return 0; }
if (($d < 0) || ($d >= $n-1)) { return 0; }
return paths_to_max($d+1, $at+1, $until) + paths_to_max($d-1, $at+1, $until);
} # paths_to_max
sub paths_suffix ($$$) {
my (
$d
, $at
, $until
) = #_;
#
if (($d < $n-1) && ($d > 0) && ($at == $until - 1)) { return 2; }
if ((($d == $n-1) && ($d == 0)) && ($at == $until - 1)) { return 1; }
if (($until > $m-1) || ($at > $until)) { return 0; }
if ($until == 0) { return 1; }
return paths_suffix($d+1, $at+1, $until) + paths_suffix($d-1, $at+1, $until);
} # paths_suffix
#
# main
#
number_count =
sum_{first=1..min(n-1,9), q1=1..m-1-(n-1), q2=q1..m-1, l_suffix=0..m-1-q2} (
paths_to_0(first,1,q1)
+ paths_to_max(0,q1,q2)
+ paths_suffix(min(n-1,9),q2,l_suffix+q2)
)
my ($number_count, $number_count_2_2) = (0, 0);
my ($first, $q1, i, $l_suffix);
for ($first = 1; $first <= $n-1; $first++) {
for ($q1 = 1; $q1 <= $m-1 - ($n-1); $q1++) {
for ($q2 = $q1; $q2 <= $m-1; $q2++) {
for ($l_suffix = 0; $l_suffix <= $m-1 - $q2; $l_suffix++) {
$number_count =
$number_count
+ paths_to_0($first,1,$q1)
+ paths_to_max(0,$q1,$q2)
+ paths_suffix($n-1,$q2,$l_suffix+$q2)
;
}
}
}
}
#
# case 2.2
#
for ($q2 = 1; $q2 <= $m-2; $q2++) {
for ($l_suffix = 0; $l_suffix <= $m-2 - $q2; $l_suffix++) {
$number_count_2_2 =
$number_count_2_2
+ paths_to_max(1,1,$q2)
+ paths_suffix($n-1,$q2,$l_suffix+$q2)
;
}
}
$number_count = 2 * $number_count + number_count_2_2;