I found this question online and I really have no idea what the question is even asking. I would really appreciate some help in first understanding the question, and a solution if possible. Thanks!
To see if a number is divisible by 3, you need to add up the digits of its decimal notation, and check if the sum is divisible by 3.
To see if a number is divisible by 11, you need to split its decimal notation into pairs of digits (starting from the right end), add up corresponding numbers and check if the sum is divisible by 11.
For any prime p (except for 2 and 5) there exists an integer r such that a similar divisibility test exists: to check if a number is divisible by p, you need to split its decimal notation into r-tuples of digits (starting from the right end), add up these r-tuples and check whether their sum is divisible by p.
Given a prime int p, find the minimal r for which such divisibility test is valid and output it.
The input consists of a single integer p - a prime between 3 and 999983, inclusive, not equal to 5.
Example
input
3
output
1
input
11
output
2
This is a very cool problem! It uses modular arithmetic and some basic number theory to devise the solution.
Let's say we have p = 11. What divisibility rule applies here? How many digits at once do we need to take, to have a divisibility rule?
Well, let's try a single digit at a time. That would mean, that if we have 121 and we sum its digits 1 + 2 + 1, then we get 4. However we see, that although 121 is divisible by 11, 4 isn't and so the rule doesn't work.
What if we take two digits at a time? With 121 we get 1 + 21 = 22. We see that 22 IS divisible by 11, so the rule might work here. And in fact, it does. For p = 11, we have r = 2.
This requires a bit of intuition which I am unable to convey in text (I really have tried) but it can be proven that for a given prime p other than 2 and 5, the divisibility rule works for tuples of digits of length r if and only if the number 99...9 (with r nines) is divisible by p. And indeed, for p = 3 we have 9 % 3 = 0, while for p = 11 we have 9 % 11 = 9 (this is bad) and 99 % 11 = 0 (this is what we want).
If we want to find such an r, we start with r = 1. We check if 9 is divisible by p. If it is, then we found the r. Otherwise, we go further and we check if 99 is divisible by p. If it is, then we return r = 2. Then, we check if 999 is divisible by p and if so, return r = 3 and so on. However, the 99...9 numbers can get very large. Thankfully, to check divisibility by p we only need to store the remainder modulo p, which we know is small (at least smaller than 999983). So the code in C++ would look something like this:
int r(int p) {
int result = 1;
int remainder = 9 % p;
while (remainder != 0) {
remainder = (remainder * 10 + 9) % p;
result++;
}
return result;
}
I have no idea how they expect a random programmer with no background to figure out the answer from this.
But here is the brief introduction to modulo arithmetic that should make this doable.
In programming, n % k is the modulo operator. It refers to taking the remainder of n / k. It satisfies the following two important properties:
(n + m) % k = ((n % k) + (m % k)) % k
(n * m) % k = ((n % k) * (m % k)) % k
Because of this, for any k we can think of all numbers with the same remainder as somehow being the same. The result is something called "the integers modulo k". And it satisfies most of the rules of algebra that you're used to. You have the associative property, the commutative property, distributive law, addition by 0, and multiplication by 1.
However if k is a composite number like 10, you have the unfortunate fact that 2 * 5 = 10 which means that modulo 10, 2 * 5 = 0. That's kind of a problem for division.
BUT if k = p, a prime, then things become massively easier. If (a*m) % p = (b*m) % p then ((a-b) * m) % p = 0 so (a-b) * m is divisible by p. And therefore either (a-b) or m is divisible by p.
For any non-zero remainder m, let's look at the sequence m % p, m^2 % p, m^3 % p, .... This sequence is infinitely long and can only take on p values. So we must have a repeat where, a < b and m^a % p = m^b %p. So (1 * m^a) % p = (m^(b-a) * m^a) % p. Since m doesn't divide p, m^a doesn't either, and therefore m^(b-a) % p = 1. Furthermore m^(b-a-1) % p acts just like m^(-1) = 1/m. (If you take enough math, you'll find that the non-zero remainders under multiplication is a finite group, and all the remainders forms a field. But let's ignore that.)
(I'm going to drop the % p everywhere. Just assume it is there in any calculation.)
Now let's let a be the smallest positive number such that m^a = 1. Then 1, m, m^2, ..., m^(a-1) forms a cycle of length a. For any n in 1, ..., p-1 we can form a cycle (possibly the same, possibly different) n, n*m, n*m^2, ..., n*m^(a-1). It can be shown that these cycles partition 1, 2, ..., p-1 where every number is in a cycle, and each cycle has length a. THEREFORE, a divides p-1. As a side note, since a divides p-1, we easily get Fermat's little theorem that m^(p-1) has remainder 1 and therefore m^p = m.
OK, enough theory. Now to your problem. Suppose we have a base b = 10^i. The primality test that they are discussing is that a_0 + a_1 * b + a_2 * b^2 + a_k * b^k is divisible by a prime p if and only if a_0 + a_1 + ... + a_k is divisible by p. Looking at (p-1) + b, this can only happen if b % p is 1. And if b % p is 1, then in modulo arithmetic b to any power is 1, and the test works.
So we're looking for the smallest i such that 10^i % p is 1. From what I showed above, i always exists, and divides p-1. So you just need to factor p-1, and try 10 to each power until you find the smallest i that works.
Note that you should % p at every step you can to keep those powers from getting too big. And with repeated squaring you can speed up the calculation. So, for example, calculating 10^20 % p could be done by calculating each of the following in turn.
10 % p
10^2 % p
10^4 % p
10^5 % p
10^10 % p
10^20 % p
This is an almost direct application of Fermat's little theorem.
First, you have to reformulate the "split decimal notation into tuples [...]"-condition into something you can work with:
to check if a number is divisible by p, you need to split its decimal notation into r-tuples of digits (starting from the right end), add up these r-tuples and check whether their sum is divisible by p
When you translate it from prose into a formula, what it essentially says is that you want
for any choice of "r-tuples of digits" b_i from { 0, ..., 10^r - 1 } (with only finitely many b_i being non-zero).
Taking b_1 = 1 and all other b_i = 0, it's easy to see that it is necessary that
It's even easier to see that this is also sufficient (all 10^ri on the left hand side simply transform into factor 1 that does nothing).
Now, if p is neither 2 nor 5, then 10 will not be divisible by p, so that Fermat's little theorem guarantees us that
, that is, at least the solution r = p - 1 exists. This might not be the smallest such r though, and computing the smallest one is hard if you don't have a quantum computer handy.
Despite it being hard in general, for very small p, you can simply use an algorithm that is linear in p (you simply look at the sequence
10 mod p
100 mod p
1000 mod p
10000 mod p
...
and stop as soon as you find something that equals 1 mod p).
Written out as code, for example, in Scala:
def blockSize(p: Int, n: Int = 10, r: Int = 1): Int =
if n % p == 1 then r else blockSize(p, n * 10 % p, r + 1)
println(blockSize(3)) // 1
println(blockSize(11)) // 2
println(blockSize(19)) // 18
or in Python:
def blockSize(p: int, n: int = 10, r: int = 1) -> int:
return r if n % p == 1 else blockSize(p, n * 10 % p, r + 1)
print(blockSize(3)) # 1
print(blockSize(11)) # 2
print(blockSize(19)) # 18
A wall of numbers, just in case someone else wants to sanity-check alternative approaches:
11 -> 2
13 -> 6
17 -> 16
19 -> 18
23 -> 22
29 -> 28
31 -> 15
37 -> 3
41 -> 5
43 -> 21
47 -> 46
53 -> 13
59 -> 58
61 -> 60
67 -> 33
71 -> 35
73 -> 8
79 -> 13
83 -> 41
89 -> 44
97 -> 96
101 -> 4
103 -> 34
107 -> 53
109 -> 108
113 -> 112
127 -> 42
131 -> 130
137 -> 8
139 -> 46
149 -> 148
151 -> 75
157 -> 78
163 -> 81
167 -> 166
173 -> 43
179 -> 178
181 -> 180
191 -> 95
193 -> 192
197 -> 98
199 -> 99
Thank you andrey tyukin.
Simple terms to remember:
When x%y =z then (x%y)%y again =z
(X+y)%z == (x%z + y%z)%z
keep this in mind.
So you break any number into some r digits at a time together. I.e. break 3456733 when r=6 into 3 * 10 power(6 * 1) + 446733 * 10 power(6 * 0).
And you can break 12536382626373 into 12 * 10 power (6 * 2). + 536382 * 10 power (6 * 1) + 626373 * 10 power (6 * 0)
Observe that here r is 6.
So when we say we combine the r digits and sum them together and apply modulo. We are saying we apply modulo to coefficients of above breakdown.
So how come coefficients sum represents whole number’s sum?
When the “10 power (6* anything)” modulo in the above break down becomes 1 then that particular term’s modulo will be equal to the coefficient’s modulo. That means the 10 power (r* anything) is of no effect. You can check why it will have no effect by using the formulas 1&2.
And the other similar terms 10 power (r * anything) also will have modulo as 1. I.e. if you can prove that (10 power r)modulo is 1. Then (10 power r * anything) is also 1.
But the important thing is we should have 10 power (r) equal to 1. Then every 10 power (r * anything) is 1 that leads to modulo of number equal to sum of r digits divided modulo.
Conclusion: find r in (10 power r) such that the given prime number will leave 1 as reminder.
That also mean the smallest 9…..9 which is divisible by given prime number decides r.
I was wondering if there is an algorithm that checks wether a given number is factorable into a set of prime numbers and if no give out the nearest number.
The problem comes always up when I use the FFT.
Thanks a lot for your help guys.
In general this looks like a hard problem, particularly finding the next largest integer that factors into your set of primes. However, if your set of primes isn't too big, one approach would be to turn this into an integer optimization problem by taking the logs. Here is how to find the smallest number > n that factors into a set of primes p_1...p_k
choose integers x_1,...,x_k to minimize (x_1 log p_1 + ... + x_k log p_k - log n)
Subject to:
x_1 log p_1 + ... + x_k log p_k >= log n
x_i >= 0 for all i
The x_i will give you the exponents for the primes. Here is an implementation in R using lpSolve:
minfact<-function(x,p){
sol<-lp("min",log(p),t(log(p)),">=",log(x),all.int=T)
prod(p^sol$solution)
}
> p<-c(2,3,13,31)
> x<-124363183
> y<-minfact(x,p)
> y
[1] 124730112
> factorize(y)
Big Integer ('bigz') object of length 13:
[1] 2 2 2 2 2 2 2 2 3 13 13 31 31
> y-x
[1] 366929
>
Using big integers, this works pretty well even for large numbers:
> p<-c(2,3,13,31,53,79)
> x<-as.bigz("1243631831278461278641361")
> y<-minfact(x,p)
y
>
Big Integer ('bigz') :
[1] 1243634072805560436129792
> factorize(y)
Big Integer ('bigz') object of length 45:
[1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
[26] 2 2 2 2 2 2 2 2 3 3 3 3 13 31 31 31 31 53 53 53
>
Your question is about well-known factorization problem - which could not be resolved with 'fast' (polynomial) time. Lenstra's elliptic algorithm is the most efficient (known) way in common case, but it requires strong knowledge of numbers theory - and it's also sub-exponential (but not polynomial).
Other algorithms are listed in the page by first link in my post, but such things as direct try (brute force) are much more slower, of cause.
Please, note, that under "could not be resolved with polynomial time" - I mean that there's no way to do this now - but not that such way does not exist (at least now, number theory can not provide such solution for this problem)
Here is a brute force method in C++. It returns the factorization of the nearest factorable number. If N has two equidistant factorable neighbours, it returns the smallest one.
GCC 4.7.3: g++ -Wall -Wextra -std=c++0x factorable-neighbour.cpp
#include <iostream>
#include <vector>
using ints = std::vector<int>;
ints factor(int n, const ints& primes) {
ints f(primes.size(), 0);
for (int i = 0; i < primes.size(); ++i) {
while (0< n && !(n % primes[i])) {
n /= primes[i];
++f[i]; } }
// append the "remainder"
f.push_back(n);
return f;
}
ints closest_factorable(int n, const ints& primes) {
int d = 0;
ints r;
while (true) {
r = factor(n + d, primes);
if (r[r.size() - 1] == 1) { break; }
++d;
r = factor(n - d, primes);
if (r[r.size() - 1] == 1) { break; }
}
r.pop_back();
return r; }
int main() {
for (int i = 0; i < 30; ++i) {
for (const auto& f : closest_factorable(i, {2, 3, 5, 7, 11})) {
std::cout << f << " "; }
std::cout << "\n"; }
}
I suppose that you have a (small) set of prime numbers S and an integer n and you want to know is n factors only using number in S. The easiest way seems to be the following:
P <- product of s in S
while P != 1 do
P <- GCD(P, n)
n <- n/P
return n == 1
You compute the GCD using Euclid's algorithm.
The idea is the following: Suppose that S = {p1, p2, ... ,pk}. You can write n uniquely as
n = p1^n1 p2^n2 ... pk^nk * R
where R is coprime wrt the pi. You want to know whether R=1.
Then
GCD(n, P) = prod ( pi such that ni <> 0 ).
Therefore n/p decrease all those non zeros ni by 1 so that they eventually become 0. At the end only R remains.
For example: S = {2,3,5}, n = 5600 = 2^5*5^2*7. Then P = 2*3*5 = 30. One gets GCD(n, p)=10=2*5. And therefore n/GCD(n,p) = 560 = 2^4*5*7.
You are now back to the same problem: You want to know if 560 can be factored using S = {2,5} hence the loop. So the next steps are
GCD(560, 10) = 10. 560/GCD = 56 = 2^3 * 7.
GCD(56, 10) = 2. 56/2 = 28 = 2^2 * 7
GCD(28, 2) = 2. 28/2 = 14 = 2 * 7
GCD(14, 2) = 2. 14/2 = 7
GCD(7, 2) = 1 so that R = 7 ! Your answer if FALSE.
kissfft has a function
int kiss_fft_next_fast_size(int n)
that returns the next largest N that is an aggregate of 2,3,5.
Also related is a kf_factor function that factorizes a number n, pulling out the "nice" FFT primes first (e.g. 4's are pulled out before 2's)
You are given N blocks of height 1…N. In how many ways can you arrange these blocks in a row such that when viewed from left you see only L blocks (rest are hidden by taller blocks) and when seen from right you see only R blocks? Example given N=3, L=2, R=1 there is only one arrangement {2, 1, 3} while for N=3, L=2, R=2 there are two ways {1, 3, 2} and {2, 3, 1}.
How should we solve this problem by programming? Any efficient ways?
This is a counting problem, not a construction problem, so we can approach it using recursion. Since the problem has two natural parts, looking from the left and looking from the right, break it up and solve for just one part first.
Let b(N, L, R) be the number of solutions, and let f(N, L) be the number of arrangements of N blocks so that L are visible from the left. First think about f because it's easier.
APPROACH 1
Let's get the initial conditions and then go for recursion. If all are to be visible, then they must be ordered increasingly, so
f(N, N) = 1
If there are suppose to be more visible blocks than available blocks, then nothing we can do, so
f(N, M) = 0 if N < M
If only one block should be visible, then put the largest first and then the others can follow in any order, so
f(N,1) = (N-1)!
Finally, for the recursion, think about the position of the tallest block, say N is in the kth spot from the left. Then choose the blocks to come before it in (N-1 choose k-1) ways, arrange those blocks so that exactly L-1 are visible from the left, and order the N-k blocks behind N it in any you like, giving:
f(N, L) = sum_{1<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * (N-k)!
In fact, since f(x-1,L-1) = 0 for x<L, we may as well start k at L instead of 1:
f(N, L) = sum_{L<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * (N-k)!
Right, so now that the easier bit is understood, let's use f to solve for the harder bit b. Again, use recursion based on the position of the tallest block, again say N is in position k from the left. As before, choose the blocks before it in N-1 choose k-1 ways, but now think about each side of that block separately. For the k-1 blocks left of N, make sure that exactly L-1 of them are visible. For the N-k blocks right of N, make sure that R-1 are visible and then reverse the order you would get from f. Therefore the answer is:
b(N,L,R) = sum_{1<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * f(N-k, R-1)
where f is completely worked out above. Again, many terms will be zero, so we only want to take k such that k-1 >= L-1 and N-k >= R-1 to get
b(N,L,R) = sum_{L <= k <= N-R+1} (N-1 choose k-1) * f(k-1, L-1) * f(N-k, R-1)
APPROACH 2
I thought about this problem again and found a somewhat nicer approach that avoids the summation.
If you work the problem the opposite way, that is think of adding the smallest block instead of the largest block, then the recurrence for f becomes much simpler. In this case, with the same initial conditions, the recurrence is
f(N,L) = f(N-1,L-1) + (N-1) * f(N-1,L)
where the first term, f(N-1,L-1), comes from placing the smallest block in the leftmost position, thereby adding one more visible block (hence L decreases to L-1), and the second term, (N-1) * f(N-1,L), accounts for putting the smallest block in any of the N-1 non-front positions, in which case it is not visible (hence L stays fixed).
This recursion has the advantage of always decreasing N, though it makes it more difficult to see some formulas, for example f(N,N-1) = (N choose 2). This formula is fairly easy to show from the previous formula, though I'm not certain how to derive it nicely from this simpler recurrence.
Now, to get back to the original problem and solve for b, we can also take a different approach. Instead of the summation before, think of the visible blocks as coming in packets, so that if a block is visible from the left, then its packet consists of all blocks right of it and in front of the next block visible from the left, and similarly if a block is visible from the right then its packet contains all blocks left of it until the next block visible from the right. Do this for all but the tallest block. This makes for L+R packets. Given the packets, you can move one from the left side to the right side simply by reversing the order of the blocks. Therefore the general case b(N,L,R) actually reduces to solving the case b(N,L,1) = f(N,L) and then choosing which of the packets to put on the left and which on the right. Therefore we have
b(N,L,R) = (L+R choose L) * f(N,L+R)
Again, this reformulation has some advantages over the previous version. Putting these latter two formulas together, it's much easier to see the complexity of the overall problem. However, I still prefer the first approach for constructing solutions, though perhaps others will disagree. All in all it just goes to show there's more than one good way to approach the problem.
What's with the Stirling numbers?
As Jason points out, the f(N,L) numbers are precisely the (unsigned) Stirling numbers of the first kind. One can see this immediately from the recursive formulas for each. However, it's always nice to be able to see it directly, so here goes.
The (unsigned) Stirling numbers of the First Kind, denoted S(N,L) count the number of permutations of N into L cycles. Given a permutation written in cycle notation, we write the permutation in canonical form by beginning the cycle with the largest number in that cycle and then ordering the cycles increasingly by the first number of the cycle. For example, the permutation
(2 6) (5 1 4) (3 7)
would be written in canonical form as
(5 1 4) (6 2) (7 3)
Now drop the parentheses and notice that if these are the heights of the blocks, then the number of visible blocks from the left is exactly the number of cycles! This is because the first number of each cycle blocks all other numbers in the cycle, and the first number of each successive cycle is visible behind the previous cycle. Hence this problem is really just a sneaky way to ask you to find a formula for Stirling numbers.
well, just as an empirical solution for small N:
blocks.py:
import itertools
from collections import defaultdict
def countPermutation(p):
n = 0
max = 0
for block in p:
if block > max:
n += 1
max = block
return n
def countBlocks(n):
count = defaultdict(int)
for p in itertools.permutations(range(1,n+1)):
fwd = countPermutation(p)
rev = countPermutation(reversed(p))
count[(fwd,rev)] += 1
return count
def printCount(count, n, places):
for i in range(1,n+1):
for j in range(1,n+1):
c = count[(i,j)]
if c > 0:
print "%*d" % (places, count[(i,j)]),
else:
print " " * places ,
print
def countAndPrint(nmax, places):
for n in range(1,nmax+1):
printCount(countBlocks(n), n, places)
print
and sample output:
blocks.countAndPrint(10)
1
1
1
1 1
1 2
1
2 3 1
2 6 3
3 3
1
6 11 6 1
6 22 18 4
11 18 6
6 4
1
24 50 35 10 1
24 100 105 40 5
50 105 60 10
35 40 10
10 5
1
120 274 225 85 15 1
120 548 675 340 75 6
274 675 510 150 15
225 340 150 20
85 75 15
15 6
1
720 1764 1624 735 175 21 1
720 3528 4872 2940 875 126 7
1764 4872 4410 1750 315 21
1624 2940 1750 420 35
735 875 315 35
175 126 21
21 7
1
5040 13068 13132 6769 1960 322 28 1
5040 26136 39396 27076 9800 1932 196 8
13068 39396 40614 19600 4830 588 28
13132 27076 19600 6440 980 56
6769 9800 4830 980 70
1960 1932 588 56
322 196 28
28 8
1
40320 109584 118124 67284 22449 4536 546 36 1
40320 219168 354372 269136 112245 27216 3822 288 9
109584 354372 403704 224490 68040 11466 1008 36
118124 269136 224490 90720 19110 2016 84
67284 112245 68040 19110 2520 126
22449 27216 11466 2016 126
4536 3822 1008 84
546 288 36
36 9
1
You'll note a few obvious (well, mostly obvious) things from the problem statement:
the total # of permutations is always N!
with the exception of N=1, there is no solution for L,R = (1,1): if a count in one direction is 1, then it implies the tallest block is on that end of the stack, so the count in the other direction has to be >= 2
the situation is symmetric (reverse each permutation and you reverse the L,R count)
if p is a permutation of N-1 blocks and has count (Lp,Rp), then the N permutations of block N inserted in each possible spot can have a count ranging from L = 1 to Lp+1, and R = 1 to Rp + 1.
From the empirical output:
the leftmost column or topmost row (where L = 1 or R = 1) with N blocks is the sum of the
rows/columns with N-1 blocks: i.e. in #PengOne's notation,
b(N,1,R) = sum(b(N-1,k,R-1) for k = 1 to N-R+1
Each diagonal is a row of Pascal's triangle, times a constant factor K for that diagonal -- I can't prove this, but I'm sure someone can -- i.e.:
b(N,L,R) = K * (L+R-2 choose L-1) where K = b(N,1,L+R-1)
So the computational complexity of computing b(N,L,R) is the same as the computational complexity of computing b(N,1,L+R-1) which is the first column (or row) in each triangle.
This observation is probably 95% of the way towards an explicit solution (the other 5% I'm sure involves standard combinatoric identities, I'm not too familiar with those).
A quick check with the Online Encyclopedia of Integer Sequences shows that b(N,1,R) appears to be OEIS sequence A094638:
A094638 Triangle read by rows: T(n,k) =|s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind (1<=k<=n; in other words, the unsigned Stirling numbers of the first kind in reverse order).
1, 1, 1, 1, 3, 2, 1, 6, 11, 6, 1, 10, 35, 50, 24, 1, 15, 85, 225, 274, 120, 1, 21, 175, 735, 1624, 1764, 720, 1, 28, 322, 1960, 6769, 13132, 13068, 5040, 1, 36, 546, 4536, 22449, 67284, 118124, 109584, 40320, 1, 45, 870, 9450, 63273, 269325, 723680, 1172700
As far as how to efficiently compute the Stirling numbers of the first kind, I'm not sure; Wikipedia gives an explicit formula but it looks like a nasty sum. This question (computing Stirling #s of the first kind) shows up on MathOverflow and it looks like O(n^2), as PengOne hypothesizes.
Based on #PengOne answer, here is my Javascript implementation:
function g(N, L, R) {
var acc = 0;
for (var k=1; k<=N; k++) {
acc += comb(N-1, k-1) * f(k-1, L-1) * f(N-k, R-1);
}
return acc;
}
function f(N, L) {
if (N==L) return 1;
else if (N<L) return 0;
else {
var acc = 0;
for (var k=1; k<=N; k++) {
acc += comb(N-1, k-1) * f(k-1, L-1) * fact(N-k);
}
return acc;
}
}
function comb(n, k) {
return fact(n) / (fact(k) * fact(n-k));
}
function fact(n) {
var acc = 1;
for (var i=2; i<=n; i++) {
acc *= i;
}
return acc;
}
$("#go").click(function () {
alert(g($("#N").val(), $("#L").val(), $("#R").val()));
});
Here is my construction solution inspired by #PengOne's ideas.
import itertools
def f(blocks, m):
n = len(blocks)
if m > n:
return []
if m < 0:
return []
if n == m:
return [sorted(blocks)]
maximum = max(blocks)
blocks = list(set(blocks) - set([maximum]))
results = []
for k in range(0, n):
for left_set in itertools.combinations(blocks, k):
for left in f(left_set, m - 1):
rights = itertools.permutations(list(set(blocks) - set(left)))
for right in rights:
results.append(list(left) + [maximum] + list(right))
return results
def b(n, l, r):
blocks = range(1, n + 1)
results = []
maximum = max(blocks)
blocks = list(set(blocks) - set([maximum]))
for k in range(0, n):
for left_set in itertools.combinations(blocks, k):
for left in f(left_set, l - 1):
other = list(set(blocks) - set(left))
rights = f(other, r - 1)
for right in rights:
results.append(list(left) + [maximum] + list(right))
return results
# Sample
print b(4, 3, 2) # -> [[1, 2, 4, 3], [1, 3, 4, 2], [2, 3, 4, 1]]
We derive a general solution F(N, L, R) by examining a specific testcase: F(10, 4, 3).
We first consider 10 in the leftmost possible position, the 4th ( _ _ _ 10 _ _ _ _ _ _ ).
Then we find the product of the number of valid sequences in the left and in the right of 10.
Next, we'll consider 10 in the 5th slot, calculate another product and add it to the previous one.
This process will go on until 10 is in the last possible slot, the 8th.
We'll use the variable named pos to keep track of N's position.
Now suppose pos = 6 ( _ _ _ _ _ 10 _ _ _ _ ). In the left of 10, there are 9C5 = (N-1)C(pos-1) sets of numbers to be arranged.
Since only the order of these numbers matters, we could look at 1, 2, 3, 4, 5.
To construct a sequence with these numbers so that 3 = L-1 of them are visible from the left, we can begin by placing 5 in the leftmost possible slot ( _ _ 5 _ _ ) and follow similar steps to what we did before.
So if F were defined recursively, it could be used here.
The only difference now is that the order of numbers in the right of 5 is immaterial.
To resolve this issue, we'll use a signal, INF (infinity), for R to indicate its unimportance.
Turning to the right of 10, there will be 4 = N-pos numbers left.
We first consider 4 in the last possible slot, position 2 = R-1 from the right ( _ _ 4 _ ).
Here what appears in the left of 4 is immaterial.
But counting arrangements of 4 blocks with the mere condition that 2 of them should be visible from the right is no different than counting arrangements of the same blocks with the mere condition that 2 of them should be visible from the left.
ie. instead of counting sequences like 3 1 4 2, one can count sequences like 2 4 1 3
So the number of valid arrangements in the right of 10 is F(4, 2, INF).
Thus the number of arrangements when pos == 6 is 9C5 * F(5, 3, INF) * F(4, 2, INF) = (N-1)C(pos-1) * F(pos-1, L-1, INF)* F(N-pos, R-1, INF).
Similarly, in F(5, 3, INF), 5 will be considered in a succession of slots with L = 2 and so on.
Since the function calls itself with L or R reduced, it must return a value when L = 1, that is F(N, 1, INF) must be a base case.
Now consider the arrangement _ _ _ _ _ 6 7 10 _ _.
The only slot 5 can take is the first, and the following 4 slots may be filled in any manner; thus F(5, 1, INF) = 4!.
Then clearly F(N, 1, INF) = (N-1)!.
Other (trivial) base cases and details could be seen in the C implementation below.
Here is a link for testing the code
#define INF UINT_MAX
long long unsigned fact(unsigned n) { return n ? n * fact(n-1) : 1; }
unsigned C(unsigned n, unsigned k) { return fact(n) / (fact(k) * fact(n-k)); }
unsigned F(unsigned N, unsigned L, unsigned R)
{
unsigned pos, sum = 0;
if(R != INF)
{
if(L == 0 || R == 0 || N < L || N < R) return 0;
if(L == 1) return F(N-1, R-1, INF);
if(R == 1) return F(N-1, L-1, INF);
for(pos = L; pos <= N-R+1; ++pos)
sum += C(N-1, pos-1) * F(pos-1, L-1, INF) * F(N-pos, R-1, INF);
}
else
{
if(L == 1) return fact(N-1);
for(pos = L; pos <= N; ++pos)
sum += C(N-1, pos-1) * F(pos-1, L-1, INF) * fact(N-pos);
}
return sum;
}