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Farey sequence of order n is the sequence of completely reduced fractions, between 0 and 1 which when in lowest terms have denominators less than or equal to n, arranged in order of increasing size. Detailed explanation here.
Problem
The problem is, given n and k, where n = order of seq and k = element index, can we find the particular element from the sequence. For examples answer for (n=5, k =6) is 1/2.
Lead
There are many less than optimal solution available, but am looking for a near-optimal one. One such algorithm is discussed here, for which I am unable to understand the logic hence unable to apply the examples.
Question
Can some please explain the solution with more detail, preferably with an example.
Thank you.
I've read the method provided in your link, and the accepted C++ solution to it. Let me post them, for reference:
Editorial Explanation
Several less-than-optimal solutions exist. Using a priority queue, one
can iterate through the fractions (generating them one by one) in O(K
log N) time. Using a fancier math relation, this can be reduced to
O(K). However, neither of these solution obtains many points, because
the number of fractions (and thus K) is quadratic in N.
The “good” solution is based on meta-binary search. To construct this
solution, we need the following subroutine: given a fraction A/B
(which is not necessarily irreducible), find how many fractions from
the Farey sequence are less than this fraction. Suppose we had this
subroutine; then the algorithm works as follows:
Determine a number X such that the answer is between X/N and (X+1)/N; such a number can be determined by binary searching the range
1...N, thus calling the subroutine O(log N) times.
Make a list of all fractions A/B in the range X/N...(X+1)/N. For any given B, there is at most one A in this range, and it can be
determined trivially in O(1).
Determine the appropriate order statistic in this list (doing this in O(N log N) by sorting is good enough).
It remains to show how we can construct the desired subroutine. We
will show how it can be implemented in O(N log N), thus giving a O(N
log^2 N) algorithm overall. Let us denote by C[j] the number of
irreducible fractions i/j which are less than X/N. The algorithm is
based on the following observation: C[j] = floor(X*B/N) – Sum(C[D],
where D divides j). A direct implementation, which tests whether any D
is a divisor, yields a quadratic algorithm. A better approach,
inspired by Eratosthene’s sieve, is the following: at step j, we know
C[j], and we subtract it from all multiples of j. The running time of
the subroutine becomes O(N log N).
Relevant Code
#include <cassert>
#include <algorithm>
#include <fstream>
#include <iostream>
#include <vector>
using namespace std;
const int kMaxN = 2e5;
typedef int int32;
typedef long long int64_x;
// #define int __int128_t
// #define int64 __int128_t
typedef long long int64;
int64 count_less(int a, int n) {
vector<int> counter(n + 1, 0);
for (int i = 2; i <= n; i += 1) {
counter[i] = min(1LL * (i - 1), 1LL * i * a / n);
}
int64 result = 0;
for (int i = 2; i <= n; i += 1) {
for (int j = 2 * i; j <= n; j += i) {
counter[j] -= counter[i];
}
result += counter[i];
}
return result;
}
int32 main() {
// ifstream cin("farey.in");
// ofstream cout("farey.out");
int64_x n, k; cin >> n >> k;
assert(1 <= n);
assert(n <= kMaxN);
assert(1 <= k);
assert(k <= count_less(n, n));
int up = 0;
for (int p = 29; p >= 0; p -= 1) {
if ((1 << p) + up > n)
continue;
if (count_less((1 << p) + up, n) < k) {
up += (1 << p);
}
}
k -= count_less(up, n);
vector<pair<int, int>> elements;
for (int i = 1; i <= n; i += 1) {
int b = i;
// find a such that up/n < a / b and a / b <= (up+1) / n
int a = 1LL * (up + 1) * b / n;
if (1LL * up * b < 1LL * a * n) {
} else {
continue;
}
if (1LL * a * n <= 1LL * (up + 1) * b) {
} else {
continue;
}
if (__gcd(a, b) != 1) {
continue;
}
elements.push_back({a, b});
}
sort(elements.begin(), elements.end(),
[](const pair<int, int>& lhs, const pair<int, int>& rhs) -> bool {
return 1LL * lhs.first * rhs.second < 1LL * rhs.first * lhs.second;
});
cout << (int64_x)elements[k - 1].first << ' ' << (int64_x)elements[k - 1].second << '\n';
return 0;
}
Basic Methodology
The above editorial explanation results in the following simplified version. Let me start with an example.
Let's say, we want to find 7th element of Farey Sequence with N = 5.
We start with writing a subroutine, as said in the explanation, that gives us the "k" value (how many Farey Sequence reduced fractions there exist before a given fraction - the given number may or may not be reduced)
So, take your F5 sequence:
k = 0, 0/1
k = 1, 1/5
k = 2, 1/4
k = 3, 1/3
k = 4, 2/5
k = 5, 1/2
k = 6, 3/5
k = 7, 2/3
k = 8, 3/4
k = 9, 4/5
k = 10, 1/1
If we can find a function that finds the count of the previous reduced fractions in Farey Sequence, we can do the following:
int64 k_count_2 = count_less(2, 5); // result = 4
int64 k_count_3 = count_less(3, 5); // result = 6
int64 k_count_4 = count_less(4, 5); // result = 9
This function is written in the accepted solution. It uses the exact methodology explained in the last paragraph of the editorial.
As you can see, the count_less() function generates the same k values as in our hand written list.
We know the values of the reduced fractions for k = 4, 6, 9 using that function. What about k = 7? As explained in the editorial, we will list all the reduced fractions in range X/N and (X+1)/N, here X = 3 and N = 5.
Using the function in the accepted solution (its near bottom), we list and sort the reduced fractions.
After that we will rearrange our k values, as in to fit in our new array as such:
k = -, 0/1
k = -, 1/5
k = -, 1/4
k = -, 1/3
k = -, 2/5
k = -, 1/2
k = -, 3/5 <-|
k = 0, 2/3 | We list and sort the possible reduced fractions
k = 1, 3/4 | in between these numbers
k = -, 4/5 <-|
k = -, 1/1
(That's why there is this piece of code: k -= count_less(up, n);, it basically remaps the k values)
(And we also subtract one more during indexing, i.e.: cout << (int64_x)elements[k - 1].first << ' ' << (int64_x)elements[k - 1].second << '\n';. This is just to basically call the right position in the generated array.)
So, for our new re-mapped k values, for N = 5 and k = 7 (original k), our result is 2/3.
(We select the value k = 0, in our new map)
If you compile and run the accepted solution, it will give you this:
Input: 5 7 (Enter)
Output: 2 3
I believe this is the basic point of the editorial and accepted solution.
when I submit to leetcode, it run case 500/502 but failed, reason: 1808548329. But when I run it on my own mac, it gave the same answer as the accepted one.
my code :
int trailingZeroes(int n) {
int count = 0;
int tmp = 0; //check every number in [1, i]
for (int i = 1; i <= n; i++) {
tmp = i;
while (tmp % 5 == 0) {
count++;
tmp /= 5;
}
}
return count;
}
and the ac answer:
int trailingZeroes2(int n) {
return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
}
they run the same result, on my mac:
std::cout << trailingZeroes(1808548329) << std::endl; //452137076
std::cout << trailingZeroes2(1808548329) << std::endl; //452137076
Is the reason that first solution not accepted because of time complexity?
(cus' i am running it on my own mac, but it gives the same answer that the ac gave)
how can i calculate the time complexity of the first solution,
is it O(NlogN) ? I am not sure. can u do me a favor？ : -)
edited, remove pics.
Your solution is O(n).
The inner loop repeats at least once every 5 items
The inner loop repeats at least twice every 25 items
...
The inner loop repeats at least k times every 5^k items.
Summing it together gives you that the inner loop runs:
n/5 + n/25 + n/125 + ... + 1 =
n (1/5 + 1/25 + 1/125 + ... + 1/n)
This is sum of geometric series, which is in O(n)
In addition, the outer loop itself has O(n) iterations, with each constant cost, if ignoring the inner loops, so this remains O(n).
The alternative solution, however runs in O(logn), which is significantly more efficient.
So, this problem I dont have any clue how to solve it the problem statement is :
Given a set S of N integers the task is decide if it is possible to
divide them into K non-empty subsets such that the sum of elements in
every of the K subsets is equal.
N can be at max 20. K can be at max 8
The problem is to be solved specifically using DP+Bitmasks!
I cannot understand where to start ! As there are K sets to be maintained , I cannot take K states each representing some or the other!!
If I try taking the whole set as a state and K as the other, I have issues in creating a recurrent relation!
Can you help??
The link to original problem Problem
You can solve the problem in O(N * 2^N), so the K is meaningless for the complexity.
First let me warn you about the corner case N < K with all the numbers being zero, in which the answer is "no".
The idea of my algorithm is the following. Assume we have computed the sum of each of the masks (that can be done in O(2^N)). We know that for each of the groups, the sum should be the total sum divided by K.
We can do a DP with masks in which the state is just a binary mask telling which numbers have been used. The key idea in removing the K from the algorithm complexity is noticing that if we know which numbers have been used, we know the sum so far, so we also know which group we are filling now (current sum / group sum). Then just try to select the next number for the group: it will be valid if we do not exceed the group expected sum.
You can check my C++ code:
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
typedef long long ll;
ll v[21 + 5];
ll sum[(1 << 21) + 5];
ll group_sum;
int n, k;
void compute_sums(int position, ll current_sum, int mask)
{
if (position == -1)
{
sum[mask] = current_sum;
return;
}
compute_sums(position - 1, current_sum, mask << 1);
compute_sums(position - 1, current_sum + v[position], (mask << 1) + 1);
}
void solve_case()
{
cin >> n >> k;
for (int i = 0; i < n; ++i)
cin >> v[i];
memset(sum, 0, sizeof(sum));
compute_sums(n - 1, 0, 0);
group_sum = sum[(1 << n) - 1];
if (group_sum % k != 0)
{
cout << "no" << endl;
return;
}
if (group_sum == 0)
{
if (n >= k)
cout << "yes" << endl;
else
cout << "no" << endl;
return;
}
group_sum /= k;
vector<int> M(1 << n, 0);
M[0] = 1;
for (int mask = 0; mask < (1 << n); ++mask)
{
if (M[mask])
{
int current_group = sum[mask] / group_sum;
for (int i = 0; i < n; ++i)
{
if ((mask >> i) & 1)
continue;
if (sum[mask | (1 << i)] <= group_sum * (current_group + 1))
M[mask | (1 << i)] = 1;
}
}
}
if (M[(1 << n) - 1])
cout << "yes" << endl;
else
cout << "no" << endl;
}
int main()
{
int cases;
cin >> cases;
for (int z = 1; z <= cases; ++z)
solve_case();
}
Here's the working O(K*2^N*N) implementation in JavaScript. From the pseudo code https://discuss.codechef.com/questions/58420/sanskar-editorial
http://jsfiddle.net/d7q4o0nj/
function equality(set, size, count) {
if(size < count) { return false; }
var total = set.reduce(function(p, c) { return p + c; }, 0);
if((total % count) !== 0) { return false }
var subsetTotal = total / count;
var search = {0: true};
var nextSearch = {};
for(var i=0; i<count; i++) {
for(var bits=0; bits < (1 << size); bits++){
if(search[bits] !== true) { continue; }
var sum = 0;
for(var j=0; j < size; j++) {
if((bits & (1 << j)) !== 0) { sum += set[j]; }
}
sum -= i * subsetTotal;
for(var j=0; j < size; j++) {
if((bits & (1 << j)) !== 0) { continue; }
var testBits = bits | (1 << j);
var tmpTotal = sum + set[j];
if(tmpTotal == subsetTotal) { nextSearch[testBits] = true; }
else if(tmpTotal < subsetTotal) { search[testBits] = true; }
}
}
search = nextSearch;
nextSearch = {};
}
if(search[(1 << size) - 1] === true) {
return true;
}
return false;
}
console.log(true, equality([1,2,3,1,2,3], 6, 2));
console.log(true, equality([1, 2, 4, 5, 6], 5, 3));
console.log(true, equality([10,20,10,20,10,20,10,20,10,20], 10, 5));
console.log(false, equality([1,2,4,5,7], 5, 3));
EDIT The algorithm finds all of the bitmasks (which represent subsets bits) that meet the criteria (having a sum tmpTotal less than or equal to the ideal subset sum subsetTotal). Repeating this process by the amount of subsets required count, you either have a bitmask where all size bits are set which means success or the test fails.
EXAMPLE
set = [1, 2, 1, 2]
size = 4
count = 2, we want to try to partition the set into 2 subsets
subsetTotal = (1+2+1+2) / 2 = 3
Iteration 1:
search = {0b: true, 1b: true, 10b: true, 100b: true, 1000b: true, 101b: true}
nextSearch = {11b: true, 1100b: true, 110b: true, 1001b: true }
Iteration 2:
search = {11b: true, 1100b: true, 110b: true, 1001b: true, 111b: true, 1101b: true }
nextSearch = {1111b: true}
Final Check
(1 << size) == 10000b, (1 << size) - 1 == 1111b
Since nextSearch[ 1111b ] exists we return success.
UPD: I confused N and K with each other and my idea is true but not efficient.Efficient idea added at the end
Assume that so far you've created k-1 subsets, and now you want to create the k-th subset. For creating the k-th subset, you need to be able to answer these two questions:
1- What should be the sum of elements of k-th subset?
2- Which elements have been used so far ?
Answering the first question is easy, the sum should be equal to sum of all elements divided by K, let's name it subSum.
For second question, we need to have the state of each element, used or not. Here we need to use bitmask idea.
Here's the dp recurrence:
dp[i][mask] = means is it possible to create i subsets with sum of each equals to subSum, using the elements which are 1(not used) in mask (in its bit representation), So dp[i][mask] is a boolean type.
dp[i][mask] = OR(dp[i-1][mask2]) for all possible mask2 states. mask2 will be produced by converting some 1's of mask to 0's, i.e. those 1's that we want to be the elements of i-th subset.
For checking all possible mask2, you need to check all 2^n possible subsets of available 1's bits.Therefore, totaly, the time complexity will be O(N*(2^n)*(2^n)). In your problem is 20*2^8*2^8= 10*2^17 < 10^7 which can pass the time limit.
Obviously, for base case you have to handle dp[0][mask] on your own, without using the recurrence.Final answer is whether dp[K][2^N-1] is true or not.
__UPD__: For getting a better performance,before get into DP, you could preprocess all subsets with sum of subSum. Then, for calculating mask2, you just need to iterate over the preprocessed list, and see whether the AND operation of them with mask would result in the subset in the list or not.
UPD2:
For having an efficient solution, instead of finding proper mask2, we could use the fact that at each step, we know the sum of elements till that point. So we could add elements one by one into the mask, and whenever we had a sum which is divisible by K we could go to the next step for creating next subset.
if (sum of used elements of mask is divisible by K)
dp[i][mask]= dp[i+1][mask];
else
dp[i][mask]|=dp[i][mask ^(1<<i)] provided that i-th item is not used and can not exceed the current sum more than i*subSum.
First, define two integers N and K, where N >= K, both known at compile time. For example: N = 8 and K = 3.
Next, define a set of integers [0, N) (or [1, N] if that makes the answer simpler) and call it S. For example: {0, 1, 2, 3, 4, 5, 6, 7}
The number of subsets of S with K elements is given by the formula C(N, K). Example
My problem is this: Create a perfect minimal hash for those subsets. The size of the example hash table will be C(8, 3) or 56.
I don't care about ordering, only that there be 56 entries in the hash table, and that I can determine the hash quickly from a set of K integers. I also don't care about reversibility.
Example hash: hash({5, 2, 3}) = 42. (The number 42 isn't important, at least not here)
Is there a generic algorithm for this that will work with any values of N and K? I wasn't able to find one by searching Google, or my own naive efforts.
There is an algorithm to code and decode a combination into its number in the lexicographical order of all combinations with a given fixed K. The algorithm is linear to N for both code and decode of the combination. What language are you interested in?
EDIT: here is example code in c++(it founds the lexicographical number of a combination in the sequence of all combinations of n elements as opposed to the ones with k elements but is really good starting point):
typedef long long ll;
// Returns the number in the lexicographical order of all combinations of n numbers
// of the provided combination.
ll code(vector<int> a,int n)
{
sort(a.begin(),a.end());
int cur = 0;
int m = a.size();
ll res =0;
for(int i=0;i<a.size();i++)
{
if(a[i] == cur+1)
{
res++;
cur = a[i];
continue;
}
else
{
res++;
int number_of_greater_nums = n - a[i];
for(int j = a[i]-1,increment=1;j>cur;j--,increment++)
res += 1LL << (number_of_greater_nums+increment);
cur = a[i];
}
}
return res;
}
// Takes the lexicographical code of a combination of n numbers and returns the
// combination
vector<int> decode(ll kod, int n)
{
vector<int> res;
int cur = 0;
int left = n; // Out of how many numbers are we left to choose.
while(kod)
{
ll all = 1LL << left;// how many are the total combinations
for(int i=n;i>=0;i--)
{
if(all - (1LL << (n-i+1)) +1 <= kod)
{
res.push_back(i);
left = n-i;
kod -= all - (1LL << (n-i+1)) +1;
break;
}
}
}
return res;
}
I am sorry I have an algorithm for the problem you are asking for right now, but I believe it will be a good exercise to try to understand what I do above. Truth is this is one of the algorithms I teach in the course "Design and analysis of algorithms" and that is why I had it pre-written.
This is what you (and I) need:
hash() maps k-tuples from [1..n] onto the set 1..C(n,k)\subset N.
The effort is k subtractions (and O(k) is a lower bound anyway, see Strandjev's remark above):
// bino[n][k] is (n "over" k) = C(n,k) = {n \choose k}
// these are assumed to be precomputed globals
int hash(V a,int n, int k) {// V is assumed to be ordered, a_k<...<a_1
// hash(a_k,..,a_2,a_1) = (n k) - sum_(i=1)^k (n-a_i i)
// ii is "inverse i", runs from left to right
int res = bino[n][k];
int i;
for(unsigned int ii = 0; ii < a.size(); ++ii) {
i = a.size() - ii;
res = res - bino[n-a[ii]][i];
}
return res;
}
A taxicab number is an integer that can be expressed as the sum of two cubes of integers in two different ways: a^3+b^3 = c^3+d^3. Design an algorithm to find all taxicab numbers with a, b, c, and d less than N.
Please give both the space and time complexity in terms of N.
I could do it in o(N^2.logN) time with O(N^2) space.
Best algorithm I've found so far:
Form all pairs: N^2
Sort the sum: N^2 logN
Find duplicates less than N
But this takes N^2 space. Can we do better?
But this takes N^2 space. Can we do better?
There exists an O(N) space solution based on a priority queue. Time complexity is O(N^2 logN). To sketch out the idea of the algorithm, here is the matrix M such that M[i][j] = i^3 + j^3 (of course, the matrix is never created in memory):
0 1 8 27 64 125
1 2 9 28 65 126
8 9 16 35 72 133
27 28 35 54 91 152
64 65 72 91 128 189
125 126 133 152 189 250
Observe that every line and every row is sorted in ascending order. Let PQ be the priority queue. First we put the biggest element in the priority queue. Then perform the following, as long as the PQ is not empty:
Pop the biggest element from PQ
add adjacent element above if the PQ doesn't have any element from that row
add adjacent element on the left if the PQ doesn't have any element from that column, and if it is not under the diagonal of the matrix (to avoid redundant elements)
Note that
You don't need to create the matrix in memory to implement the algorithm
The elements will be popped from the PQ in descending order, from the biggest element of the matrix to its smallest one (avoiding elements from the redundant half part of the matrix).
Everytime the PQ issues the same value twice then we have found a taxicab number.
As an illustration, here is an implementation in C++. The time complexity is O(N^2 logN) and space complexity O(N).
#include <iostream>
#include <cassert>
#include <queue>
using namespace std;
typedef unsigned int value_type;
struct Square
{
value_type i;
value_type j;
value_type sum_of_cubes;
Square(value_type i, value_type j) : i(i), j(j), sum_of_cubes(i*i*i+j*j*j) {}
friend class SquareCompare;
bool taxicab(const Square& sq) const
{
return sum_of_cubes == sq.sum_of_cubes && i != sq.i && i != sq.j;
}
friend ostream& operator<<(ostream& os, const Square& sq);
};
class SquareCompare
{
public:
bool operator()(const Square& a, const Square& b)
{
return a.sum_of_cubes < b.sum_of_cubes;
}
};
ostream& operator<<(ostream& os, const Square& sq)
{
return os << sq.i << "^3 + " << sq.j << "^3 = " << sq.sum_of_cubes;
}
int main()
{
const value_type N=2001;
value_type count = 0;
bool in_i [N];
bool in_j [N];
for (value_type i=0; i<N; i++) {
in_i[i] = false;
in_j[i] = false;
}
priority_queue<Square, vector<Square>, SquareCompare> p_queue;
p_queue.push(Square(N-1, N-1));
in_i[N-1] = true;
in_j[N-1] = true;
while(!p_queue.empty()) {
Square sq = p_queue.top();
p_queue.pop();
in_i[sq.i] = false;
in_j[sq.j] = false;
// cout << "pop " << sq.i << " " << sq.j << endl;
if (sq.i > 0 && !in_i[sq.i - 1] && sq.i-1 >= sq.j) {
p_queue.push(Square(sq.i-1, sq.j));
in_i[sq.i-1] = true;
in_j[sq.j] = true;
// cout << "push " << sq.i-1 << " " << sq.j << endl;
}
if (sq.j > 0 && !in_j[sq.j-1] && sq.i >= sq.j - 1) {
p_queue.push(Square(sq.i, sq.j-1));
in_i[sq.i] = true;
in_j[sq.j - 1] = true;
// cout << "push " << sq.i << " " << sq.j-1 << endl;
}
if (sq.taxicab(p_queue.top())) {
/* taxicab number */
cout << sq << " " << p_queue.top() << endl;
count++;
}
}
cout << endl;
cout << "there are " << count << " taxicab numbers with a, b, c, d < " << N << endl;
return 0;
}
The answers given by Novneet Nov and user3017842 are both correct ideas for finding the taxicab numbers with storage O(N) using minHeap.
Just a little bit more explanation why the minHeap of size N works.
First, if you had all the sums (O(N^2)) and could sort them (O(N^2lgN)) you would just pick the duplicates as you traverse the sorted array. Well, in our case using a minHeap we can traverse in-order all the sums: we just need to ensure that the minHeap always contains the minimum unprocessed sum.
Now, we have a huge number of sums (O(N^2)). But, notice that this number can be split into N groups each of which has an easily defined minimum!
(fix a, change b from 0 to N-1 => here are your N groups. The sum in one group with a smaller b is smaller than one with a bigger b in the same group - because a is the same).
The minimum of union of these groups is in the union of mins of these
groups. Therefore, if you keep all minimums of these groups in the
minHeap you are guaranteed to have the total minimum in the minHeap.
Now, when you extract Min from the heap, you just add next smallest element from the group of this extracted min (so if you extracted (a, b) you add (a, b+1)) and you are guaranteed that your minHeap still contains the next unprocessed min of all the sums.
I found the solution/code here : Time complexity O(N^2 logN), space complexity O(N)
The solution is implemented by help of priority queues.
Reverse thinking can be easily done by looking at the code. It can be done in an array of size N because the min sums are deleted from the array after comparing to the next minimum and then the array is made to size N by adding a new sum - (i^3 + (j+1)^3).
A intuitive proof is here :
Initially, we have added (1,1),(2,2),(3,3),...,(N,N) in the min-priority queue.
Suppose a^+b^3=c^3+d^3, and (a,b) is the minimum that will be taken out of the priority queue next. To be able to detect this taxicab number, (c,d) must also be in the priority queue which would be taken out after (a,b).
Note: We would be adding (a,b+1) after extracting (a,b) so there is no way that extraction of (a,b) would result in addition of (c,d) to the priority queue, so it must already exist in the priority queue.
Now lets assume that (c,d) is not in the priority queue, because we haven't gotten to it yet. Instead, there is some (c,d−k) in the priority queue where k>0.
Since (a,b) is being taken out,
a^3+b^3≤c^3+(d−k)^3
However, a^3+b^3=c^3+d^3
Therefore,
c^3+d^3≤c^3+(d−k)^3
d≤d−k
k≤0
Since k>0, this is impossible. Thus our assumption can never come to pass.
Thus for every (a,b) which is being removed from the min-PQ, (c,d) is already in the min-PQ (or was just removed) if a^3+b^3=c^3+d^3
The time complexity of the algorithm can't be less than O(N2) in any case, since you might print up to O(N2) taxicab numbers.
To reduce space usage you could, in theory, use the suggestion mentioned here: little link. Basically, the idea is that first you try all possible pairs a, b and find the solution to this:
a = 1 − (p − 3 * q)(p2 + 3 * q2)
b = −1 + (p + 3 * q)(p2 + 3q2)
Then you can find the appropriate c, d pair using:
c = (p + 3 * q) - (p2 + 3 * q2)
d = -(p - 3 * q) + (p2 + 3 * q2)
and check whether they are both less than N. The issue here is that solving that system of equations might get a bit messy (by 'a bit' I mean very tedious).
The O(N2) space solution is much simpler, and it'd probably be efficient enough since anything of quadratic time complexity that can run in reasonable time limits will probably be fine with quadratic space usage.
I hope that helped!
version1 uses List and sorting
O(n^2*logn) time and O(n^2) space
public static void Taxicab1(int n)
{
// O(n^2) time and O(n^2) space
var list = new List<int>();
for (int i = 1; i <= n; i++)
{
for (int j = i; j <= n; j++)
{
list.Add(i * i * i + j * j * j);
}
}
// O(n^2*log(n^2)) time
list.Sort();
// O(n^2) time
int prev = -1;
foreach (var next in list)
{
if (prev == next)
{
Console.WriteLine(prev);
}
prev = next;
}
}
version2 uses HashSet
O(n^2) time and O(n^2) space
public static void Taxicab2(int n)
{
// O(n^2) time and O(n^2) space
var set = new HashSet<int>();
for (int i = 1; i <= n; i++)
{
for (int j = i; j <= n; j++)
{
int x = i * i * i + j * j * j;
if (!set.Add(x))
{
Console.WriteLine(x);
}
}
}
}
version3 uses min oriented Priority Queue
O(n^2*logn) time and O(n) space
public static void Taxicab3(int n)
{
// O(n) time and O(n) space
var pq = new MinPQ<SumOfCubes>();
for (int i = 1; i <= n; i++)
{
pq.Push(new SumOfCubes(i, i));
}
// O(n^2*logn) time
var sentinel = new SumOfCubes(0, 0);
while (pq.Count > 0)
{
var current = pq.Pop();
if (current.Result == sentinel.Result)
Console.WriteLine($"{sentinel.A}^3+{sentinel.B}^3 = {current.A}^3+{current.B}^3 = {current.Result}");
if (current.B <= n)
pq.Push(new SumOfCubes(current.A, current.B + 1));
sentinel = current;
}
}
where SummOfCubes
public class SumOfCubes : IComparable<SumOfCubes>
{
public int A { get; private set; }
public int B { get; private set; }
public int Result { get; private set; }
public SumOfCubes(int a, int b)
{
A = a;
B = b;
Result = a * a * a + b * b * b;
}
public int CompareTo(SumOfCubes other)
{
return Result.CompareTo(other.Result);
}
}
github
create an array: 1^3, 2^3, 3^3, 4^3, ....... k^3. such that k^3 < N and (k+1)^3 > N. the array size would be ~ (N)^(1/3). the array is sorted order.
use 2sum technique (link) in lineal time proportional to the array size. if we find 2 pairs of numbers, that is a hit.
looping through step 2 by decreasing N by 1 each time.
This will use O(N^(1/3)) extra space and ~ O(N^(4/3)) time.
A easy way of understanding Time complexity O(N^2 logN), space complexity O(N) is to think it as a merge of N sorted arrays plus a bookkeeping of the previously merged element.
It seems like a simple brute-force algorithm with proper bounds solves it in time proportional to n^1.33 and space proportional to n. Or could anyone point me to the place where I'm mistaken?
Consider 4 nested loops, each running from 1 to cubic root of n. Using these loops we can go over all possible combinations of 4 values and find the pairs forming taxicab numbers. It means each loop takes time proportional to cubic root of n, or n^(1/3). Multiply this value 4 times and get:
(n^(1/3)^4 = n^(4/3) = n^1.33
I wrote a solution in JavaScript and benchmarked it, and it seems to be working. One caveat is that the result is only partially sorted.
Here is my JavaScript code (it's not optimal yet, could be optimized even more):
function taxicab(n) {
let a = 1, b = 1, c = 1, d = 1,
cubeA = a**3 + b**3,
cubeB = c**3 + d**3,
results = [];
while (cubeA < n) { // loop over a
while (cubeA < n) { // loop over b
// avoid running nested loops if this number is already in results
if (results.indexOf(cubeA) === -1) {
while (cubeB <= cubeA) { // loop over c
while (cubeB <= cubeA) { // loop over d
if (cubeB === cubeA && a!=c && a!=d) { // found a taxicab number!
results.push(cubeA);
}
d++;
cubeB = c**3 + d**3;
} // end loop over d
c++;
d = c;
cubeB = c**3 + d**3;
} // end loop over c
}
b++;
cubeA = a**3 + b**3;
c = d = 1;
cubeB = c**3 + d**3;
} // end loop over d
a++;
b = a;
cubeA = a**3 + b**3;
} // end loop over a
return results;
}
Running taxicab(1E8) takes around 30 seconds in a browser console and yields 485 numbers as a result. Ten times smaller value taxicab(1E7) (10 millions) takes almost 1.4 seconds and yields 150 numbers. 10^1.33 * 1.4 = 29.9, i.e. multiplying n by 10 leads to the running time increased by 10^1.33 times. The result array is unsorted, but after quickly sorting it we get correct result, as it seems:
[1729, 4104, 13832, 20683, 32832, 39312, 40033, 46683, 64232, 65728,
110656, 110808, 134379, 149389, 165464, 171288, 195841, 216027, 216125,
262656, 314496, 320264, 327763, 373464, 402597, 439101, 443889, 513000,
513856, 515375, 525824, 558441, 593047, 684019, 704977, 805688, 842751,
885248, 886464, 920673, 955016, 984067, 994688, 1009736, 1016496, 1061424,
1073375, 1075032, 1080891, 1092728, 1195112, 1260441, 1323712, 1331064,
1370304, 1407672, 1533357, 1566728, 1609272, 1728216, 1729000, 1734264,
1774656, 1845649, 2048391, 2101248, 2301299, 2418271, 2515968, 2562112,
2585375, 2622104, 2691451, 2864288, 2987712, 2991816, 3220776, 3242197,
3375001, 3375008, 3511872, 3512808, 3551112, 3587409, 3628233, 3798613,
3813992, 4033503, 4104000, 4110848, 4123000, 4174281, 4206592, 4342914,
4467528, 4505949, 4511808, 4607064, 4624776, 4673088, …]
Here is a code for benchmarking:
// run taxicab(n) for k trials and return the average running time
function benchmark(n, k) {
let t = 0;
k = k || 1; // how many times to repeat the trial to get an averaged result
for(let i = 0; i < k; i++) {
let t1 = new Date();
taxicab(n);
let t2 = new Date();
t += t2 - t1;
}
return Math.round(t/k);
}
Finally, I tested it:
let T = benchmark(1E7, 3); // 1376 - running time for n = 10 million
let T2 = benchmark(2E7, 3);// 4821 - running time for n = 20 million
let powerLaw = Math.log2(T2/T); // 1.3206693816701993
So it means time is proportional to n^1.32 in this test. Repeating this many times with different values always yields around the same result: from 1.3 to 1.4.
First of all, we will construct the taxicab numbers instead of searching for them. The range we will use to construct a taxicab number i.e Ta(2) will go up to n^1/3 not n. Because if you cube a number bigger than n^1/3 it will be bigger than n and also we can't cube negative numbers to prevent that case by definition. We will use a HashSet to remember the sums of two cubed numbers in the algorithm. This will help us to lookup previous cubed sums in O(1) time while we are iterating over every possible pair of numbers in the range I mentioned earlier.
Time complexity: O(n^2/3)
Space complexity: O(n^1/3)
def taxicab_numbers(n: int) -> list[int]:
taxicab_numbers = []
max_num = math.floor(n ** (1. / 3.))
seen_sums = set()
for i in range(1, max_num + 1):
for j in range(i, max_num + 1):
cube_sum = i ** 3 + j ** 3
if cube_sum in seen_sums:
taxicab_numbers.append(cube_sum)
else:
seen_sums.add(cube_sum)
return taxicab_numbers
import java.util.*;
public class A5Q24 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter number:");
int n = sc.nextInt();
// start checking every int less than the input
for (int a = 2;a <= n;a++) {
int count = 0;
// number of ways that number be expressed in sum of two number cubes
for (int i = 1; Math.pow(i, 3) < a; i++) {
// if the cube of number smaller is greater than the number than it goes out
for (int j = 1; j <= i; j++) {
if (Math.pow(i, 3) + Math.pow(j, 3) == a)
count++;
}
}
if (count == 2)
System.out.println(a);
}
sc.close();
}
}
I think we can also do better on time (O (N ^ 2)) with O(N ^ 2) memory, using a hashmap to check if a pair of cubes has already be seen. In Python:
def find_taxicab_numbers(n: int) -> List[Tuple[int, int, int, int, int]]:
"""
find all taxicab numbers smaller than n, i.e. integers that can be expressed as the sum of two cubes of positive
integers in two different ways so that a^3 + b^3 = c^3 + d^3.
Time: O(n ^ 2) (two loops, one dict lookup). Space: O(n ^ 2)) (all possible cubes)
:param n: upper bound for a, b, c, d
:return: list of tuples of int: a, b, c, d, and taxicab numbers
"""
cubes = [i ** 3 for i in range(n)]
seen_sum_cubes = dict() # mapping sum cubes -> a, b
taxicabs = list() # list of a, b, c, d, taxicab
# check all possible sums of cubes
for i in range(n):
for j in range(i):
sum_cubes = cubes[i] + cubes[j]
if sum_cubes in seen_sum_cubes:
prev_i, prev_j = seen_sum_cubes[sum_cubes]
taxicabs.append((i, j, prev_i, prev_j, sum_cubes))
else:
seen_sum_cubes[sum_cubes] = (i, j)
return taxicabs