What is an efficient algorithm for finding the digit in nth position in the following string
112123123412345123456 ... 123456789101112 ...
Storing the entire string in memory is not feasible for very large n, so I am looking for an algorithm that can find the nth digit in the above string which works if n is very large (i.e. an alternative to just generating the first n digits of the string).
There are several levels here: the digit is part of a number x, the number x is part of a sequence 1,2,3...x...y and that sequence is part of a block of sequences that lead up to numbers like y that have z digits. We'll tackle these levels one by one.
There are 9 numbers with 1 digit:
first: 1 (sequence length: 1 * 1)
last: 9 (sequence length: 9 * 1)
average sequence length: (1 + 9) / 2 = 5
1-digit block length: 9 * 5 = 45
There are 90 numbers with 2 digits:
first: 10 (sequence length: 9 * 1 + 1 * 2)
last: 99 (sequence length: 9 * 1 + 90 * 2)
average sequence length: 9 + (2 + 180) / 2 = 100
2-digit block length: 90 * 100 = 9000
There are 900 numbers with 3 digits:
first: 100 (sequence length: 9 * 1 + 90 * 2 + 1 * 3)
last: 999 (sequence length: 9 * 1 + 90 * 2 + 900 * 3)
average sequence length: 9 + 180 + (3 + 2,700) / 2 = 1,540.5
3-digit block length: 900 * 1,540.5 = 1,386,450
If you continue to calculate these values, you'll find which block (of sequences up to how many digits) the digit you're looking for is in, and you'll know the start and end point of this block.
Say you want the millionth digit. You find that it's in the 3-digit block, and that this block is located in the total sequence at:
start of 3-digit block: 45 + 9,000 + = 9,045
start of 4-digit block: 45 + 9,000 + 1,386,450 = 1,395,495
So in this block we're looking for digit number:
1,000,000 - 9,045 = 990,955
Now you can use e.g. a binary search to find which sequence the 990,955th digit is in; you start with the 3-digit number halfway in the 3-digit block:
first: 100 (sequence length: 9 + 180 + 1 * 3)
number: 550 (sequence length: 9 + 180 + 550 * 3)
average sequence length: 9 + 180 + (3 + 1650) / 2 = 1,015.5
total sequence length: 550 * 1,015.5 = 558,525
Which is too small; so we try 550 * 3/4 = 825, see if that is too small or large, and go up or down in increasingly smaller steps until we know which sequence the 990,995th digit is in.
Say it's in the sequence for the number n; then we calculate the total length of all 3-digit sequences up to n-1, and this will give us the location of the digit we're looking for in the sequence for the number n. Then we can use the numbers 9*1, 90*2, 900*3 ... to find which number the digit is in, and then what the digit is.
We have three types of structures that we would like to be able to search on, (1) the sequence of concatenating d-digit numbers, for example, single digit:
123456...
or 3-digit:
100101102103
(2) the rows in a section,
where each section builds on the previous section added to a prefix. For example, section 1:
1
12
123
...
or section 3:
1234...10111213...100
1234...10111213...100102
1234...10111213...100102103
<----- prefix ----->
and (3) the full sections, although the latter we can just enumerate since they grow exponentially and help build our section prefixes. For (1), we can use simple division if we know the digit count; for (2), we can binary search.
Here's Python code that also answers the big ones:
def getGreatest(n, d, prefix):
rows = 9 * 10**(d - 1)
triangle = rows * (d + rows * d) // 2
l = 0
r = triangle
while l < r:
mid = l + ((r - l) >> 1)
triangle = mid * prefix + mid * (d + mid * d) // 2
prevTriangle = (mid-1) * prefix + (mid-1) * (d + (mid-1) * d) // 2
nextTriangle = (mid+1) * prefix + (mid+1) * (d + (mid+1) * d) // 2
if triangle >= n:
if prevTriangle < n:
return prevTriangle
else:
r = mid - 1
else:
if nextTriangle >= n:
return triangle
else:
l = mid
return l * prefix + l * (d + l * d) // 2
def solve(n):
debug = 1
d = 0
p = 0.1
prefixes = [0]
sections = [0]
while sections[d] < n:
d += 1
p *= 10
rows = int(9 * p)
triangle = rows * (d + rows * d) // 2
section = rows * prefixes[d-1] + triangle
sections.append(sections[d-1] + section)
prefixes.append(prefixes[d-1] + rows * d)
section = sections[d - 1]
if debug:
print("section: %s" % section)
n = n - section
rows = getGreatest(n, d, prefixes[d - 1])
if debug:
print("rows: %s" % rows)
n = n - rows
d = 1
while prefixes[d] < n:
d += 1;
if prefixes[d] == n:
return 9;
prefix = prefixes[d - 1]
if debug:
print("prefix: %s" % prefix)
n -= prefix
if debug:
print((n, d, prefixes, sections))
countDDigitNums = n // d
remainder = n % d
prev = 10**(d - 1) - 1
num = prev + countDDigitNums
if debug:
print("num: %s" % num)
if remainder:
return int(str(num + 1)[remainder - 1])
else:
s = str(num);
return int(s[len(s) - 1])
ns = [
1, # 1
2, # 1
3, # 2
100, # 1
2100, # 2
31000, # 2
999999999999999999, # 4
1000000000000000000, # 1
999999999999999993, # 7
]
for n in ns:
print(n)
print(solve(n))
print('')
Well, you have a series of sequences each increasing by a single number.
If you have "x" of them, then the sequences up to that point occupy x * (x + 1) / 2 character positions. Or, another way of saying this is that the "x"s sequence starts at x * (x - 1) / 2 (assuming zero-based indexing). These are called triangular numbers.
So, all you need to do is to find the "x" value where the cumulative amount is closest to a given "n". Here are three ways:
Search for a closed from solution. This exists, but the formula is rather complicated. (Here is one reference for the sum of triangular numbers.)
Pre-calculate a table in memory with values up to, say, 1,000,000. that will get you to 10^10 sizes.
Use a "binary" search and the formula. So, generate the sequence of values for 1, 2, 4, 8, and so on and then do a binary search to find the exact sequence.
Once you know the sequence where the value lies, determining the value is simply a matter of arithmetic.
I am trying to do this using recursion with memoization ,I have identified the following base cases .
I) when n==k there is only one group with all the balls.
II) when k>n then no groups can have atleast k balls,hence zero.
I am unable to move forward from here.How can this be done?
As an illustration when n=6 ,k=2
(2,2,2)
(4,2)
(3,3)
(6)
That is 4 different groupings can be formed.
This can be represented by the two dimensional recursive formula described below:
T(0, k) = 1
T(n, k) = 0 n < k, n != 0
T(n, k) = T(n-k, k) + T(n, k + 1)
^ ^
There is a box with k balls, No box with k balls, advance to next k
put them
In the above, T(n,k) is the number of distributions of n balls such that each box gets at least k.
And the trick is to think of k as the lowest possible number of balls, and seperate the problem to two scenarios: Is there a box with exactly k balls (if so, place them and recurse with n-k balls), or not (and then, recurse with minimal value of k+1, and same number of balls).
Example, to calculate your example: T(6,2) (6 balls, minimum 2 per box):
T(6,2) = T(4,2) + T(6,3)
T(4,2) = T(2,2) + T(4,3) = T(0,2) + T(2,3) + T(1,3) + T(4,4) =
= T(0,2) + T(2,3) + T(1,3) + T(0,4) + T(4,5) =
= 1 + 0 + 0 + 1 + 0
= 2
T(6,3) = T(3,3) + T(6,4) = T(0,3) + T(3,4) + T(2,4) + T(6,5)
= T(0,3) + T(3,4) + T(2,4) + T(1,5) + T(6,6) =
= T(0,3) + T(3,4) + T(2,4) + T(1,5) + T(0,6) + T(6,7) =
= 1 + 0 + 0 + 0 + 1 + 0
= 2
T(6,2) = T(4,2) + T(6,3) = 2 + 2 = 4
Using Dynamic Programming, it can be calculated in O(n^2) time.
This case can be solved pretty simple:
Number of buckets
The maximum-number of buckets b can be determined as follows:
b = roundDown(n / k)
Each valid distribution can use at most b buckets.
Number of distributions with x buckets
For a given number of buckets the number of distribution can be found pretty simple:
Distribute k balls to each bucket. Find the number of ways to distribute the remaining balls (r = n - k * x) to x buckets:
total_distributions(x) = bincoefficient(x , n - k * x)
EDIT: this will onyl work, if order matters. Since it doesn't for the question, we can use a few tricks here:
Each distribution can be mapped to a sequence of numbers. E.g.: d = {d1 , d2 , ... , dx}. We can easily generate all of these sequences starting with the "first" sequence {r , 0 , ... , 0} and subsequently moving 1s from the left to the right. So the next sequence would look like this: {r - 1 , 1 , ... , 0}. If only sequences matching d1 >= d2 >= ... >= dx are generated, no duplicates will be generated. This constraint can easily be used to optimize this search a bit: We can only move a 1 from da to db (with a = b - 1), if da - 1 >= db + 1 is given, since otherwise the constraint that the array is sorted is violated. The 1s to move are always the rightmost that can be moved. Another way to think of this would be to view r as a unary number and simply split that string into groups such that each group is atleast as long as it's successor.
countSequences(x)
sequence[]
sequence[0] = r
sequenceCount = 1
while true
int i = findRightmostMoveable(sequence)
if i == -1
return sequenceCount
sequence[i] -= 1
sequence[i + 1] -= 1
sequenceCount
findRightmostMoveable(sequence)
for i in [length(sequence) - 1 , 0)
if sequence[i - 1] > sequence[i] + 1
return i - 1
return -1
Actually findRightmostMoveable could be optimized a bit, if we look at the structure-transitions of the sequence (to be more precise the difference between two elements of the sequence). But to be honest I'm by far too lazy to optimize this further.
Putting the pieces together
range(1 , roundDown(n / k)).map(b -> countSequences(b)).sum()
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;
}
Given an integer N I want to find two integers A and B that satisfy A × B ≥ N with the following conditions:
The difference between A × B and N is as low as possible.
The difference between A and B is as low as possible (to approach a square).
Example: 23. Possible solutions 3 × 8, 6 × 4, 5 × 5. 6 × 4 is the best since it leaves just one empty space in the grid and is "less" rectangular than 3 × 8.
Another example: 21. Solutions 3 × 7 and 4 × 6. 3 × 7 is the desired one.
A brute force solution is easy. I would like to see if a clever solution is possible.
Easy.
In pseudocode
a = b = floor(sqrt(N))
if (a * b >= N) return (a, b)
a += 1
if (a * b >= N) return (a, b)
return (a, b+1)
and it will always terminate, the distance between a and b at most only 1.
It will be much harder if you relax second constraint, but that's another question.
Edit: as it seems that the first condition is more important, you have to attack the problem
a bit differently. You have to specify some method to measure the badness of not being square enough = 2nd condition, because even prime numbers can be factorized as 1*number, and we fulfill the first condition. Assume we have a badness function (say a >= b && a <= 2 * b), then factorize N and try different combinations to find best one. If there aren't any good enough, try with N+1 and so on.
Edit2: after thinking a bit more I come with this solution, in Python:
from math import sqrt
def isok(a, b):
"""accept difference of five - 2nd rule"""
return a <= b + 5
def improve(a, b, N):
"""improve result:
if a == b:
(a+1)*(b-1) = a^2 - 1 < a*a
otherwise (a - 1 >= b as a is always larger)
(a+1)*(b-1) = a*b - a + b - 1 =< a*b
On each iteration new a*b will be less,
continue until we can, or 2nd condition is still met
"""
while (a+1) * (b-1) >= N and isok(a+1, b-1):
a, b = a + 1, b - 1
return (a, b)
def decomposite(N):
a = int(sqrt(N))
b = a
# N is square, result is ok
if a * b >= N:
return (a, b)
a += 1
if a * b >= N:
return improve(a, b, N)
return improve(a, b+1, N)
def test(N):
(a, b) = decomposite(N)
print "%d decomposed as %d * %d = %d" % (N, a, b, a*b)
[test(x) for x in [99, 100, 101, 20, 21, 22, 23]]
which outputs
99 decomposed as 11 * 9 = 99
100 decomposed as 10 * 10 = 100
101 decomposed as 13 * 8 = 104
20 decomposed as 5 * 4 = 20
21 decomposed as 7 * 3 = 21
22 decomposed as 6 * 4 = 24
23 decomposed as 6 * 4 = 24
I think this may work (your conditions are somewhat ambiguous). this solution is somewhat similar to other one, in basically produces rectangular matrix which is almost square.
you may need to prove that A+2 is not optimal condition
A0 = B0 = ceil (sqrt N)
A1 = A0+1
B1 = B0-1
if A0*B0-N > A1*B1-N: return (A1,B1)
return (A0,B0)
this is solution if first condition is dominant (and second condition is not used)
A0 = B0 = ceil (sqrt N)
if A0*B0==N: return (A0,B0)
return (N,1)
Other conditions variations will be in between
A = B = ceil (sqrt N)