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Counting such number whose digit product multiple to itself smaller than given number

Revision en19, by SPyofgame, 2020-11-07 10:59:37

The problem

Lets $$$f(x) = $$$ product of digits of $$$x$$$. Example: $$$f(123) = 1 * 2 * 3 = 6$$$, $$$f(990) = 9 * 9 * 0 = 0$$$, $$$f(1) = 1$$$

The statement is, given such limitation $$$N$$$, count the number of positive $$$x$$$ that $$$1 \leq x * f(x) \leq N$$$

Example: For $$$N = 20$$$, there are 5 valid numbers $$$1, 2, 3, 4, 11$$$



The limitation

  • Subtask 1: $$$N \leq 10^6$$$
  • Subtask 2: $$$N \leq 10^{10}$$$
  • Subtask 3: $$$N \leq 10^{14}$$$
  • Subtask 4: $$$N \leq 10^{18}$$$


My approach for subtask 1

  • If $$$(x > N)$$$ or $$$(f(x) > N)$$$ then $$$(x * f(x) > N)$$$. So we will only care about $$$x \leq N$$$ that $$$x * f(x) \leq N$$$
Calculating x * f(x) - O(log10(x))
Counting - O(n log10(n))


My approach for subtask 2

  • If $$$x$$$ contains $$$0$$$ then $$$f(x) = 0 \Rightarrow x \times f(x) < 1$$$. We only care about such $$$x$$$ without $$$0$$$ digit
Building function - O(result + log10(n))


Here is the solution:

Let takes some $$$x$$$ satisfy $$$1 \leq x * f(x) \leq N$$$

We can easily prove that $$$f(x) \leq x$$$, and because $$$x * f(x) \leq N$$$, we have $$$f(x) \leq \sqrt{N}$$$ (notice that $$$x$$$ might bigger than $$$\sqrt{N}$$$)

Since $$$f(x)$$$ is product of digits of $$$x$$$, which can be obtain by such digits {$$$1, 2, \dots, 9$$$}. So $$$f(x) = 2^a \times 3^b \times 5^c \times 7^d$$$

So we can bruteforces all possible tuple of such $$$(a, b, c, d)$$$ satisfy ($$$P = 2^a \times 3^b \times 5^c \times 7^d \leq \sqrt{N}$$$). There are small amount of such tuples (493 tuples for $$$N = 10^9$$$ and 5914 tuples for $$$N = 10^{18}$$$)

Find all possible tuples - O(quartic_root(N))

For each tuples, we need to counting the numbers of such $$$x$$$ that $$$1 \leq x \times f(x) \leq N$$$ and $$$f(x) = P$$$.

  • We have the value $$$P$$$, so $$$\lceil \frac{1}{P} \rceil \leq x \leq \lfloor \frac{N}{P} \rfloor$$$.
  • We have the value $$$f(x) = P$$$, so $$$x$$$ can be made by digits having the product exactly $$$P$$$, so we can do some DP-digit

So now we have to solve this DP-digit problem: Calculate numbers of such $$$x$$$ ($$$L \leq x \leq R$$$) whose $$$f(x) = P$$$



Solving Subproblem

We try to build each digits by digits for $$$X$$$. Because $$$X \leq N$$$, so we have to build about $$$18$$$ digits.

Lets make a recursive function $$$magic(X, N, p2, p3, p5, p7)$$$

Lets make some definition
Notice that
Ugly precalculation code - O(1)
magic function - O(18*p2*p3*p5*p7)


About the proving stuff

1) $$$\forall x \in \mathbb{N}, f(x) \leq x$$$

  • Proof: $$$x = \overline{\dots dcba} = \dots + d \times 10^3 + c \times 10^2 + b \times 10^1 + a \times 10^0 \geq \dots \times d \times c \times b \times a = f(x)$$$

2) If $$$x$$$ satisfy then $$$f(x) \leq \sqrt{R}$$$ must be satisfied

  • Proof: $$$x \times f(x) \leq R \Rightarrow f(x) \times f(x) \leq R \Rightarrow f(x) \leq \sqrt{R}$$$

3) $$$\exists\ a, b, c, d \in \mathbb{N} \rightarrow f(x) = 2^a \times 3^b \times 5^c \times 7^d$$$

  • Since $$$x = \overline{\dots dcba} \Rightarrow (0 \leq \dots, d, c, b, a \leq 9)$$$ and $$$f(x) = \dots \times d \times c \times b \times a$$$

  • And we also have $$$\forall$$$ digit $$$v$$$ ($$$v \in \mathbb{N}, 0 \leq v \leq 9$$$) $$$\rightarrow \exists\ a, b, c, d \in \mathbb{N} \rightarrow v = 2^a \times 3^b \times 5^c \times 7^d$$$

  • And because $$$f(x)$$$ is the product of digits of $$$x$$$, hence the statement is correct

4) If we know $$$f(x)$$$ we can find such $$$x$$$ satisfy $$$x \in [L, R]$$$

  • Proof: Since $$$f(x)$$$ is created from factors of digits of $$$x$$$, so $$$x$$$ can also be generated using the factors

5) Number of tuples $$$(a, b, c, d)$$$ satisfy $$$P = 2^a \times 3^b \times 5^c \times 7^d \leq \sqrt{R}$$$ is very small

  • Lets $$$O(k(x)) = O(log_2(x) \times log_3(x) \times log_5(x) \times log_7(x))$$$

  • Since each value $$$x$$$ have the upper bound of $$$log_x(\sqrt{R})$$$. So the complexity is about $$$O(log_2(\sqrt{R}) \times log_3(\sqrt{R}) \times log_5(\sqrt{R}) \times log_7(\sqrt{R})) = O(k(\sqrt{R})) \leq O(log_2(\sqrt{R})^4)$$$

  • But actually for $$$R \leq 10^{18}$$$, the complexity is just about $$$O(k(\sqrt[4]{R}))$$$

Weak proof - N = 10^k
Weak proof - N = 2^k
Code


About the complexity:

  • $$$O(h(x)) = O(log_{10}(R))$$$ is number of digits we have to build
  • $$$O(k(x)) = O(log_2(R) \times log_3(R) \times log_5(R) \times log_7(R)) = O(log(R)^4)$$$ is product of all prime digits $$$p$$$ with its maximum power $$$k$$$ satisfy $$$p^k \leq N$$$
  • $$$O(g(x)) = O(k(\sqrt{R}))$$$ is number of such valid tuples, but for $$$1 \leq N \leq 10^{18}$$$ it is about $$$\approx O(k(\sqrt[4]{R})) \leq O(log_2^4{\sqrt[4]{R}})$$$
  • The space complexity is $$$O(SPACE) = O(h(x) \times k(x)) = O(log_2(R) \times log_3(R) \times log_5(R) \times log_7(R) \times log_{10}(R)) = O(log(R)^5)$$$
  • The time complexity is $$$O(TIME) = O(SPACE) + O(g(x) \times k(x)) \approx O(log(R)^4 \times log_2^4{\sqrt[4]{R}})$$$


Other

  • Does anyone know the real complexity for $$$O(k(x))$$$ ? — Counting such tuples of $$$(a, b, c, d \in \mathbb{N})$$$ that $$$P = 2^a \times 3^b \times 5^c \times 7^d \leq N$$$ ?

Since my $$$O(log_2(N))$$$ complexity seem very far than the real counting and $$$O(log_2(\sqrt{N}))$$$ is closer but just correct for $$$N \leq 10^{18}$$$

  • Thanks for reading and sorry for updating this blog publicly so many times

History

 
 
 
 
Revisions
 
 
  Rev. Lang. By When Δ Comment
en21 English SPyofgame 2020-11-08 18:08:04 60
en20 English SPyofgame 2020-11-07 11:02:44 132 Tiny change: '\leq N$ ? \n> Since my ' -> '\leq N$ ? Since my '
en19 English SPyofgame 2020-11-07 10:59:37 1409 Tiny change: '4, 11$\n\n-------\n\n#### T' -> '4, 11$\n\n========\n\n#### T'
en18 English SPyofgame 2020-11-06 06:29:46 3554 (published)
en17 English SPyofgame 2020-11-06 06:02:44 2761 Tiny change: 'sqrt{R})^4 \times log_2(R)^4)$\n\n-' -> 'sqrt{R})^4)$\n\n-' (saved to drafts)
en16 English SPyofgame 2020-11-05 18:40:17 1012
en15 English SPyofgame 2020-11-05 18:39:30 3010 Tiny change: 'leq x$\n\n> $x = \ove' -> 'leq x$\n\n Proof: $x = \ove'
en14 English SPyofgame 2020-11-05 18:11:26 3 Tiny change: '$p^k \leq MAXN$, which ' -> '$p^k \leq N$, which '
en13 English SPyofgame 2020-11-05 17:56:06 40
en12 English SPyofgame 2020-11-05 17:53:26 43
en11 English SPyofgame 2020-11-05 17:42:05 84
en10 English SPyofgame 2020-11-05 16:44:33 16 Tiny change: ' summary="Function f(x) - O(' -> ' summary="Calculating x * f(x) - O('
en9 English SPyofgame 2020-11-05 15:53:37 88 Tiny change: 'uch those numbers\n- $pw10[' -> 'uch those variables we use\n- $pw10['
en8 English SPyofgame 2020-11-05 15:52:01 107
en7 English SPyofgame 2020-11-05 15:51:08 61
en6 English SPyofgame 2020-11-05 15:50:12 6163 Tiny change: ' build yet)\n\n> Wher' -> ' build yet\n\n> Wher' (published)
en5 English SPyofgame 2020-11-05 15:05:42 552 Tiny change: 'y $O(\sqrt{\sqrt{N}}' -> 'y $O(\sqrt[4]{\sqrt{N}}' (saved to drafts)
en4 English SPyofgame 2020-11-05 05:03:48 23
en3 English SPyofgame 2020-11-05 05:03:05 12
en2 English SPyofgame 2020-11-05 05:02:48 170 Tiny change: 'unction - O(result)' -> 'unction - approx O(result)' (published)
en1 English SPyofgame 2020-11-05 04:54:10 1886 Initial revision (saved to drafts)