Related
After asking this question on math.stackexchange.com I figured this might be a better place after all...
I have a small list of positive numbers rounded to (say) two decimals:
1.15 (can be 1.145 - 1.154999...)
1.92 (can be 1.915 - 1.924999...)
2.36 (can be 2.355 - 2.364999...)
2.63 (can be 2.625 - 2.634999...)
2.78 (can be 2.775 - 2.784999...)
3.14 (can be 3.135 - 3.144999...)
24.04 (can be 24.035 - 24.044999...)
I suspect that these numbers are fractions of integers and that all numerators or all denominators are equal. Choosing 100 as a common denominator would work in this case, that would leave the last value as 2404/100. But there could be a 'simpler' solution with much smaller integers.
How do I efficiently find the smallest common numerator and/or denominator? Or (if that is different) the one that would result in the smallest maximum denominator resp. numerator?
Of course I could brute force for small lists/numbers and few decimals. That would find 83/72, 138/72, 170/72, 189/72, 200/72, 226/72 and 1731/72 for this example.
Assuming the numbers don't have too many significant digits and aren't too big you can try increasing the denominator until you find a valid solution. It is not just brute-forcing. Additionally the following script is staying at the number violating the constraints as long as there is nothing found, in the hope of getting the denominator higher faster, without having to calculate for the non-problematic numbers.
It works based on the following formula:
x / y < a / b if x * b < a * y
This means a denominator d is valid if:
ceil(loNum * d / loDen) * hiDen < hiNum * d
The ceil(...) part calculates the smallest possible numerator satisfying the constraint of the low boundary and the rest is checking if it also satysfies the high boundary.
Better would be to work with real integer calculations, e.g. just longs in Java, then the ceil part becomes:
(loNum * d + loDen - 1) / loDen
function findRatios(arr) {
let lo = [], hi = [], consecutive = 0, d = 1
for (let i = 0; i < arr.length; i++) {
let x = '' + arr[i], len = x.length, dot = x.indexOf('.'),
num = parseInt(x.substr(0, dot) + x.substr(dot + 1)) * 10,
den = Math.pow(10, len - dot),
loGcd = gcd(num - 5, den), hiGcd = gcd(num + 5, den)
lo[i] = {num: (num - 5) / loGcd, den: den / loGcd}
hi[i] = {num: (num + 5) / hiGcd, den: den / hiGcd}
}
for (let index = 0; consecutive < arr.length; index = (index + 1) % arr.length) {
if (!valid(d, lo[index], hi[index])) {
consecutive = 1
d++
while (!valid(d, lo[index], hi[index]))
d++
} else {
consecutive++
}
}
for (let i = 0; i < arr.length; i++)
console.log(Math.ceil(lo[i].num * d / lo[i].den) + ' / ' + d)
}
function gcd(x, y) {
while(y) {
let t = y
y = x % y
x = t
}
return x
}
function valid(d, lo, hi) {
let n = Math.ceil(lo.num * d / lo.den)
return n * hi.den < hi.num * d
}
findRatios([1.15, 1.92, 2.36, 2.63, 2.78, 3.14, 24.04])
I am giving a Matrix of N x M. For a Submatrix of Length X which starts at position (a, b) i have to find the largest element present in a Submatrix.
My Approach:
Do as the question says:
Simple 2 loops
for(i in range(a, a + x))
for(j in range(b, b + x)) max = max(max,A[i][j]) // N * M
A little Advance:
1. Make a segment tree for every i in range(0, N)
2. for i in range(a, a + x) query(b, b + x) // N * logM
Is there any better solution having O(log n) complexity only ?
A Sparse Table Algorithm Approach
:- <O( N x M x log(N) x log(M)) , O(1)>.
Precomputation Time - O( N x M x log(N) x log(M))
Query Time - O(1)
For understanding this method you should have knowledge of finding RMQ using sparse Table Algorithm for one dimension.
We can use 2D Sparse Table Algorithm for finding Range Minimum Query.
What we do in One Dimension:-
we preprocess RMQ for sub arrays of length 2^k using dynamic programming. We will keep an array M[0, N-1][0, logN] where M[i][j] is the index of the minimum value in the sub array starting at i.
For calculating M[i][j] we must search for the minimum value in the first and second half of the interval. It’s obvious that the small pieces have 2^(j – 1) length, so the pseudo code for calculation this is:-
if (A[M[i][j-1]] < A[M[i + 2^(j-1) -1][j-1]])
M[i][j] = M[i][j-1]
else
M[i][j] = M[i + 2^(j-1) -1][j-1]
Here A is actual array which stores values.Once we have these values preprocessed, let’s show how we can use them to calculate RMQ(i, j). The idea is to select two blocks that entirely cover the interval [i..j] and find the minimum between them. Let k = [log(j - i + 1)]. For computing RMQ(i, j) we can use the following formula:-
if (A[M[i][k]] <= A[M[j - 2^k + 1][k]])
RMQ(i, j) = A[M[i][k]]
else
RMQ(i , j) = A[M[j - 2^k + 1][k]]
For 2 Dimension :-
Similarly We can extend above rule for 2 Dimension also , here we preprocess RMQ for sub matrix of length 2^K, 2^L using dynamic programming & keep an array M[0,N-1][0, M-1][0, logN][0, logM]. Where M[x][y][k][l] is the index of the minimum value in the sub matrix starting at [x , y] and having length 2^K, 2^L respectively.
pseudo code for calculation M[x][y][k][l] is:-
M[x][y][i][j] = GetMinimum(M[x][y][i-1][j-1], M[x + (2^(i-1))][y][i-1][j-1], M[x][y+(2^(j-1))][i-1][j-1], M[x + (2^(i-1))][y+(2^(j-1))][i-1][j-1])
Here GetMinimum function will return the index of minimum element from provided elements. Now we have preprocessed, let's see how to calculate RMQ(x, y, x1, y1). Here [x, y] starting point of sub matrix and [x1, y1] represent end point of sub matrix means bottom right point of sub matrix. Here we have to select four sub matrices blocks that entirely cover [x, y, x1, y1] and find minimum of them. Let k = [log(x1 - x + 1)] & l = [log(y1 - y + 1)]. For computing RMQ(x, y, x1, y1) we can use following formula:-
RMQ(x, y, x1, y1) = GetMinimum(M[x][y][k][l], M[x1 - (2^k) + 1][y][k][l], M[x][y1 - (2^l) + 1][k][l], M[x1 - (2^k) + 1][y1 - (2^l) + 1][k][l]);
pseudo code for above logic:-
// remember Array 'M' store index of actual matrix 'P' so for comparing values in GetMinimum function compare the values of array 'P' not of array 'M'
SparseMatrix(n , m){ // n , m is dimension of matrix.
for i = 0 to 2^i <= n:
for j = 0 to 2^j <= m:
for x = 0 to x + 2^i -1 < n :
for y = 0 to y + (2^j) -1 < m:
if i == 0 and j == 0:
M[x][y][i][j] = Pair(x , y) // store x, y
else if i == 0:
M[x][y][i][j] = GetMinimum(M[x][y][i][j-1], M[x][y+(2^(j-1))][i][j-1])
else if j == 0:
M[x][y][i][j] = GetMinimum(M[x][y][i-1][j], M[x+ (2^(i-1))][y][i-1][j])
else
M[x][y][i][j] = GetMinimum(M[x][y][i-1][j-1], M[x + (2^(i-1))][y][i-1][j-1], M[x][y+(2^(j-1))][i-1][j-1], M[x + (2^(i-1))][y+(2^(j-1))][i-1][j-1]);
}
RMQ(x, y, x1, y1){
k = log(x1 - x + 1)
l = log(y1 - y + 1)
ans = GetMinimum(M[x][y][k][l], M[x1 - (2^k) + 1][y][k][l], M[x][y1 - (2^l) + 1][k][l], M[x1 - (2^k) + 1][y1 - (2^l) + 1][k][l]);
return P[ans->x][ans->y] // ans->x represent Row number stored in ans and similarly ans->y represent column stored in ans
}
Here is the sample code in c++, for the pseudo code given by #Chapta, as was requested by some user.
int M[1000][1000][10][10];
int **matrix;
void precompute_max(){
for (int i = 0 ; (1<<i) <= n; i += 1){
for(int j = 0 ; (1<<j) <= m ; j += 1){
for (int x = 0 ; x + (1<<i) -1 < n; x+= 1){
for (int y = 0 ; y + (1<<j) -1 < m; y+= 1){
if (i == 0 and j == 0)
M[x][y][i][j] = matrix[x][y]; // store x, y
else if (i == 0)
M[x][y][i][j] = max(M[x][y][i][j-1], M[x][y+(1<<(j-1))][i][j-1]);
else if (j == 0)
M[x][y][i][j] = max(M[x][y][i-1][j], M[x+ (1<<(i-1))][y][i-1][j]);
else
M[x][y][i][j] = max(M[x][y][i-1][j-1], M[x + (1<<(i-1))][y][i-1][j-1], M[x][y+(1<<(j-1))][i-1][j-1], M[x + (1<<(i-1))][y+(1<<(j-1))][i-1][j-1]);
// cout << "from i="<<x<<" j="<<y<<" of length="<<(1<<i)<<" and length="<<(1<<j) <<"max is: " << M[x][y][i][j] << endl;
}
}
}
}
}
int compute_max(int x, int y, int x1, int y1){
int k = log2(x1 - x + 1);
int l = log2(y1 - y + 1);
// cout << "Value of k="<<k<<" l="<<l<<endl;
int ans = max(M[x][y][k][l], M[x1 - (1<<k) + 1][y][k][l], M[x][y1 - (1<<l) + 1][k][l], M[x1 - (1<<k) + 1][y1 - (1<<l) + 1][k][l]);
return ans;
}
This code first precomputes, the 2 dimensional sparse table, and then queries it in constant time.
Additional info: the sparse table stores the maximum element and not the indices to the maximum element.
AFAIK, there can be no O(logn approach) as the matrix follows no order. However, if you have an order such that every row is sorted in ascending from left to right and every column is sorted ascending from up to down, then you know that A[a+x][b+x] (bottom-right cell of the submatrix) is the largest element in that submatrix. Thus, finding the maximum takes O(1) time once the matrix is sorted. However, sorting the matrix, if not already sorted, will cost O(NxM log{NxM})
I have written the below code outline, basically to sum an array (a) where each element is multiplied by a value x^i:
y = a(0)
i = 0
{y = sum from i=0 to (n-1) a(i) * x^i AND 0 <= n <= a.length} //Invariant
while (i < (n-1))
{y = sum from i=0 to (n-1) a(i) * x^i AND 0 <= n <= a.length AND i < (n-1)}
y = y + a(i)*x^i
i = i + 1
end while
{y = sum from i=0 to (n-1) a(i) * x^i} //Postcondition
Note that I do not expect the code to compile - it's just a sensible outline of how the code should work. I need to improve the efficiency of the code by using a tracking variable, and thus, a linking invariant to bridge said variable with the rest of the code. This is where I am stuck. What would be useful to track in this case? I have thought about retaining sum values at each iteration, but I'm not sure if that does the trick. If I could figure out what to track, I'm pretty sure it would be trivial to link it to the space. Can anyone see how my algorithm might be improved via a tracking variable?
Your invariant logic has off-by-1 problems. Here is a corrected version that tracks partial power operations.
// Precondition: 1 <= n <= a.length
// Invariant:
{ 0 <= i < n AND xi = x^i AND y = sum(j = 0..i) . a(j) * x^j }
// Establish invariant at i = 0:
// xi = x^0 = 1 AND y = sum(j=0..0) . a(j) * x^j = a(0) * x^0 = a(0)
i = 0;
xi = 1;
y = a(0);
while (i < n - 1) {
i = i + 1; // Break the invariant
xi = xi * x; // Re-establish it
y = y + a(i) * xi
}
// Invariant was last established at i = n-1, so we have post condition:
{ y = sum(j = 0..n-1) . a(j) * x^j }
The more common and numerically stable way to calculate polynomials is with Horner's Rule
y = 0
for i = n-1 downto 0 do y = y * x + a(i)
So it seems like you're trying to end up with this:
(a(0)*x^0) + (a(1)*x^1) + ... + (a(n-1)*x^(n-1))
Is that right?
The only way I can see to improve performance would be if the ^ operation is more costly than the * operation. In that case, you could keep track of the x^n variable as you go, multiplying x by the value through each iteration.
In fact, in that case you could probably start at the end of the array and work your way backwards, multiplying by x each time, to produce:
(((...((a(n-1)*x+a(n-2))*x+...)+a(2))*x+a(1))*x)+a(0)
That would theoretically be slightly faster than recalculating x^i each time, but it's not going to be algorithmically faster. It probably wouldn't be an order of magnitude faster.
I have three numbers x, y , z.
For a range between numbers x and y.
How can i find the total numbers whose % with z is 0 i.e. how many numbers between x and y are divisible by z ?
It can be done in O(1): find the first one, find the last one, find the count of all other.
I'm assuming the range is inclusive. If your ranges are exclusive, adjust the bounds by one:
find the first value after x that is divisible by z. You can discard x:
x_mod = x % z;
if(x_mod != 0)
x += (z - x_mod);
find the last value before y that is divisible by y. You can discard y:
y -= y % z;
find the size of this range:
if(x > y)
return 0;
else
return (y - x) / z + 1;
If mathematical floor and ceil functions are available, the first two parts can be written more readably. Also the last part can be compressed using math functions:
x = ceil (x, z);
y = floor (y, z);
return max((y - x) / z + 1, 0);
if the input is guaranteed to be a valid range (x >= y), the last test or max is unneccessary:
x = ceil (x, z);
y = floor (y, z);
return (y - x) / z + 1;
(2017, answer rewritten thanks to comments)
The number of multiples of z in a number n is simply n / z
/ being the integer division, meaning decimals that could result from the division are simply ignored (for instance 17/5 => 3 and not 3.4).
Now, in a range from x to y, how many multiples of z are there?
Let see how many multiples m we have up to y
0----------------------------------x------------------------y
-m---m---m---m---m---m---m---m---m---m---m---m---m---m---m---
You see where I'm going... to get the number of multiples in the range [ x, y ], get the number of multiples of y then subtract the number of multiples before x, (x-1) / z
Solution: ( y / z ) - (( x - 1 ) / z )
Programmatically, you could make a function numberOfMultiples
function numberOfMultiples(n, z) {
return n / z;
}
to get the number of multiples in a range [x, y]
numberOfMultiples(y) - numberOfMultiples(x-1)
The function is O(1), there is no need of a loop to get the number of multiples.
Examples of results you should find
[30, 90] ÷ 13 => 4
[1, 1000] ÷ 6 => 166
[100, 1000000] ÷ 7 => 142843
[777, 777777777] ÷ 7 => 111111001
For the first example, 90 / 13 = 6, (30-1) / 13 = 2, and 6-2 = 4
---26---39---52---65---78---91--
^ ^
30<---(4 multiples)-->90
I also encountered this on Codility. It took me much longer than I'd like to admit to come up with a good solution, so I figured I would share what I think is an elegant solution!
Straightforward Approach 1/2:
O(N) time solution with a loop and counter, unrealistic when N = 2 billion.
Awesome Approach 3:
We want the number of digits in some range that are divisible by K.
Simple case: assume range [0 .. n*K], N = n*K
N/K represents the number of digits in [0,N) that are divisible by K, given N%K = 0 (aka. N is divisible by K)
ex. N = 9, K = 3, Num digits = |{0 3 6}| = 3 = 9/3
Similarly,
N/K + 1 represents the number of digits in [0,N] divisible by K
ex. N = 9, K = 3, Num digits = |{0 3 6 9}| = 4 = 9/3 + 1
I think really understanding the above fact is the trickiest part of this question, I cannot explain exactly why it works.
The rest boils down to prefix sums and handling special cases.
Now we don't always have a range that begins with 0, and we cannot assume the two bounds will be divisible by K.
But wait! We can fix this by calculating our own nice upper and lower bounds and using some subtraction magic :)
First find the closest upper and lower in the range [A,B] that are divisible by K.
Upper bound (easier): ex. B = 10, K = 3, new_B = 9... the pattern is B - B%K
Lower bound: ex. A = 10, K = 3, new_A = 12... try a few more and you will see the pattern is A - A%K + K
Then calculate the following using the above technique:
Determine the total number of digits X between [0,B] that are divisible by K
Determine the total number of digits Y between [0,A) that are divisible by K
Calculate the number of digits between [A,B] that are divisible by K in constant time by the expression X - Y
Website: https://codility.com/demo/take-sample-test/count_div/
class CountDiv {
public int solution(int A, int B, int K) {
int firstDivisible = A%K == 0 ? A : A + (K - A%K);
int lastDivisible = B%K == 0 ? B : B - B%K; //B/K behaves this way by default.
return (lastDivisible - firstDivisible)/K + 1;
}
}
This is my first time explaining an approach like this. Feedback is very much appreciated :)
This is one of the Codility Lesson 3 questions. For this question, the input is guaranteed to be in a valid range. I answered it using Javascript:
function solution(x, y, z) {
var totalDivisibles = Math.floor(y / z),
excludeDivisibles = Math.floor((x - 1) / z),
divisiblesInArray = totalDivisibles - excludeDivisibles;
return divisiblesInArray;
}
https://codility.com/demo/results/demoQX3MJC-8AP/
(I actually wanted to ask about some of the other comments on this page but I don't have enough rep points yet).
Divide y-x by z, rounding down. Add one if y%z < x%z or if x%z == 0.
No mathematical proof, unless someone cares to provide one, but test cases, in Perl:
#!perl
use strict;
use warnings;
use Test::More;
sub multiples_in_range {
my ($x, $y, $z) = #_;
return 0 if $x > $y;
my $ret = int( ($y - $x) / $z);
$ret++ if $y%$z < $x%$z or $x%$z == 0;
return $ret;
}
for my $z (2 .. 10) {
for my $x (0 .. 2*$z) {
for my $y (0 .. 4*$z) {
is multiples_in_range($x, $y, $z),
scalar(grep { $_ % $z == 0 } $x..$y),
"[$x..$y] mod $z";
}
}
}
done_testing;
Output:
$ prove divrange.pl
divrange.pl .. ok
All tests successful.
Files=1, Tests=3405, 0 wallclock secs ( 0.20 usr 0.02 sys + 0.26 cusr 0.01 csys = 0.49 CPU)
Result: PASS
Let [A;B] be an interval of positive integers including A and B such that 0 <= A <= B, K be the divisor.
It is easy to see that there are N(A) = ⌊A / K⌋ = floor(A / K) factors of K in interval [0;A]:
1K 2K 3K 4K 5K
●········x········x··●·····x········x········x···>
0 A
Similarly, there are N(B) = ⌊B / K⌋ = floor(B / K) factors of K in interval [0;B]:
1K 2K 3K 4K 5K
●········x········x········x········x···●····x···>
0 B
Then N = N(B) - N(A) equals to the number of K's (the number of integers divisible by K) in range (A;B]. The point A is not included, because the subtracted N(A) includes this point. Therefore, the result should be incremented by one, if A mod K is zero:
N := N(B) - N(A)
if (A mod K = 0)
N := N + 1
Implementation in PHP
function solution($A, $B, $K) {
if ($K < 1)
return 0;
$c = floor($B / $K) - floor($A / $K);
if ($A % $K == 0)
$c++;
return (int)$c;
}
In PHP, the effect of the floor function can be achieved by casting to the integer type:
$c = (int)($B / $K) - (int)($A / $K);
which, I think, is faster.
Here is my short and simple solution in C++ which got 100/100 on codility. :)
Runs in O(1) time. I hope its not difficult to understand.
int solution(int A, int B, int K) {
// write your code in C++11
int cnt=0;
if( A%K==0 or B%K==0)
cnt++;
if(A>=K)
cnt+= (B - A)/K;
else
cnt+=B/K;
return cnt;
}
(floor)(high/d) - (floor)(low/d) - (high%d==0)
Explanation:
There are a/d numbers divisible by d from 0.0 to a. (d!=0)
Therefore (floor)(high/d) - (floor)(low/d) will give numbers divisible in the range (low,high] (Note that low is excluded and high is included in this range)
Now to remove high from the range just subtract (high%d==0)
Works for integers, floats or whatever (Use fmodf function for floats)
Won't strive for an o(1) solution, this leave for more clever person:) Just feel this is a perfect usage scenario for function programming. Simple and straightforward.
> x,y,z=1,1000,6
=> [1, 1000, 6]
> (x..y).select {|n| n%z==0}.size
=> 166
EDIT: after reading other's O(1) solution. I feel shamed. Programming made people lazy to think...
Division (a/b=c) by definition - taking a set of size a and forming groups of size b. The number of groups of this size that can be formed, c, is the quotient of a and b. - is nothing more than the number of integers within range/interval ]0..a] (not including zero, but including a) that are divisible by b.
so by definition:
Y/Z - number of integers within ]0..Y] that are divisible by Z
and
X/Z - number of integers within ]0..X] that are divisible by Z
thus:
result = [Y/Z] - [X/Z] + x (where x = 1 if and only if X is divisible by Y otherwise 0 - assuming the given range [X..Y] includes X)
example :
for (6, 12, 2) we have 12/2 - 6/2 + 1 (as 6%2 == 0) = 6 - 3 + 1 = 4 // {6, 8, 10, 12}
for (5, 12, 2) we have 12/2 - 5/2 + 0 (as 5%2 != 0) = 6 - 2 + 0 = 4 // {6, 8, 10, 12}
The time complexity of the solution will be linear.
Code Snippet :
int countDiv(int a, int b, int m)
{
int mod = (min(a, b)%m==0);
int cnt = abs(floor(b/m) - floor(a/m)) + mod;
return cnt;
}
here n will give you count of number and will print sum of all numbers that are divisible by k
int a = sc.nextInt();
int b = sc.nextInt();
int k = sc.nextInt();
int first = 0;
if (a > k) {
first = a + a/k;
} else {
first = k;
}
int last = b - b%k;
if (first > last) {
System.out.println(0);
} else {
int n = (last - first)/k+1;
System.out.println(n * (first + last)/2);
}
Here is the solution to the problem written in Swift Programming Language.
Step 1: Find the first number in the range divisible by z.
Step 2: Find the last number in the range divisible by z.
Step 3: Use a mathematical formula to find the number of divisible numbers by z in the range.
func solution(_ x : Int, _ y : Int, _ z : Int) -> Int {
var numberOfDivisible = 0
var firstNumber: Int
var lastNumber: Int
if y == x {
return x % z == 0 ? 1 : 0
}
//Find first number divisible by z
let moduloX = x % z
if moduloX == 0 {
firstNumber = x
} else {
firstNumber = x + (z - moduloX)
}
//Fist last number divisible by z
let moduloY = y % z
if moduloY == 0 {
lastNumber = y
} else {
lastNumber = y - moduloY
}
//Math formula
numberOfDivisible = Int(floor(Double((lastNumber - firstNumber) / z))) + 1
return numberOfDivisible
}
public static int Solution(int A, int B, int K)
{
int count = 0;
//If A is divisible by K
if(A % K == 0)
{
count = (B / K) - (A / K) + 1;
}
//If A is not divisible by K
else if(A % K != 0)
{
count = (B / K) - (A / K);
}
return count;
}
This can be done in O(1).
Here you are a solution in C++.
auto first{ x % z == 0 ? x : x + z - x % z };
auto last{ y % z == 0 ? y : y - y % z };
auto ans{ (last - first) / z + 1 };
Where first is the first number that ∈ [x; y] and is divisible by z, last is the last number that ∈ [x; y] and is divisible by z and ans is the answer that you are looking for.
I know that there is an algorithm that permits, given a combination of number (no repetitions, no order), calculates the index of the lexicographic order.
It would be very useful for my application to speedup things...
For example:
combination(10, 5)
1 - 1 2 3 4 5
2 - 1 2 3 4 6
3 - 1 2 3 4 7
....
251 - 5 7 8 9 10
252 - 6 7 8 9 10
I need that the algorithm returns the index of the given combination.
es: index( 2, 5, 7, 8, 10 ) --> index
EDIT: actually I'm using a java application that generates all combinations C(53, 5) and inserts them into a TreeMap.
My idea is to create an array that contains all combinations (and related data) that I can index with this algorithm.
Everything is to speedup combination searching.
However I tried some (not all) of your solutions and the algorithms that you proposed are slower that a get() from TreeMap.
If it helps: my needs are for a combination of 5 from 53 starting from 0 to 52.
Thank you again to all :-)
Here is a snippet that will do the work.
#include <iostream>
int main()
{
const int n = 10;
const int k = 5;
int combination[k] = {2, 5, 7, 8, 10};
int index = 0;
int j = 0;
for (int i = 0; i != k; ++i)
{
for (++j; j != combination[i]; ++j)
{
index += c(n - j, k - i - 1);
}
}
std::cout << index + 1 << std::endl;
return 0;
}
It assumes you have a function
int c(int n, int k);
that will return the number of combinations of choosing k elements out of n elements.
The loop calculates the number of combinations preceding the given combination.
By adding one at the end we get the actual index.
For the given combination there are
c(9, 4) = 126 combinations containing 1 and hence preceding it in lexicographic order.
Of the combinations containing 2 as the smallest number there are
c(7, 3) = 35 combinations having 3 as the second smallest number
c(6, 3) = 20 combinations having 4 as the second smallest number
All of these are preceding the given combination.
Of the combinations containing 2 and 5 as the two smallest numbers there are
c(4, 2) = 6 combinations having 6 as the third smallest number.
All of these are preceding the given combination.
Etc.
If you put a print statement in the inner loop you will get the numbers
126, 35, 20, 6, 1.
Hope that explains the code.
Convert your number selections to a factorial base number. This number will be the index you want. Technically this calculates the lexicographical index of all permutations, but if you only give it combinations, the indexes will still be well ordered, just with some large gaps for all the permutations that come in between each combination.
Edit: pseudocode removed, it was incorrect, but the method above should work. Too tired to come up with correct pseudocode at the moment.
Edit 2: Here's an example. Say we were choosing a combination of 5 elements from a set of 10 elements, like in your example above. If the combination was 2 3 4 6 8, you would get the related factorial base number like so:
Take the unselected elements and count how many you have to pass by to get to the one you are selecting.
1 2 3 4 5 6 7 8 9 10
2 -> 1
1 3 4 5 6 7 8 9 10
3 -> 1
1 4 5 6 7 8 9 10
4 -> 1
1 5 6 7 8 9 10
6 -> 2
1 5 7 8 9 10
8 -> 3
So the index in factorial base is 1112300000
In decimal base, it's
1*9! + 1*8! + 1*7! + 2*6! + 3*5! = 410040
This is Algorithm 2.7 kSubsetLexRank on page 44 of Combinatorial Algorithms by Kreher and Stinson.
r = 0
t[0] = 0
for i from 1 to k
if t[i - 1] + 1 <= t[i] - 1
for j from t[i - 1] to t[i] - 1
r = r + choose(n - j, k - i)
return r
The array t holds your values, for example [5 7 8 9 10]. The function choose(n, k) calculates the number "n choose k". The result value r will be the index, 251 for the example. Other inputs are n and k, for the example they would be 10 and 5.
zero-base,
# v: array of length k consisting of numbers between 0 and n-1 (ascending)
def index_of_combination(n,k,v):
idx = 0
for p in range(k-1):
if p == 0: arrg = range(1,v[p]+1)
else: arrg = range(v[p-1]+2, v[p]+1)
for a in arrg:
idx += combi[n-a, k-1-p]
idx += v[k-1] - v[k-2] - 1
return idx
Null Set has the right approach. The index corresponds to the factorial-base number of the sequence. You build a factorial-base number just like any other base number, except that the base decreases for each digit.
Now, the value of each digit in the factorial-base number is the number of elements less than it that have not yet been used. So, for combination(10, 5):
(1 2 3 4 5) == 0*9!/5! + 0*8!/5! + 0*7!/5! + 0*6!/5! + 0*5!/5!
== 0*3024 + 0*336 + 0*42 + 0*6 + 0*1
== 0
(10 9 8 7 6) == 9*3024 + 8*336 + 7*42 + 6*6 + 5*1
== 30239
It should be pretty easy to calculate the index incrementally.
If you have a set of positive integers 0<=x_1 < x_2< ... < x_k , then you could use something called the squashed order:
I = sum(j=1..k) Choose(x_j,j)
The beauty of the squashed order is that it works independent of the largest value in the parent set.
The squashed order is not the order you are looking for, but it is related.
To use the squashed order to get the lexicographic order in the set of k-subsets of {1,...,n) is by taking
1 <= x1 < ... < x_k <=n
compute
0 <= n-x_k < n-x_(k-1) ... < n-x_1
Then compute the squashed order index of (n-x_k,...,n-k_1)
Then subtract the squashed order index from Choose(n,k) to get your result, which is the lexicographic index.
If you have relatively small values of n and k, you can cache all the values Choose(a,b) with a
See Anderson, Combinatorics on Finite Sets, pp 112-119
I needed also the same for a project of mine and the fastest solution I found was (Python):
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
def index(comb,n,k):
r=nCr(n,k)
for i in range(k):
if n-comb[i]<k-i:continue
r=r-nCr(n-comb[i],k-i)
return r
My input "comb" contained elements in increasing order You can test the code with for example:
import itertools
k=3
t=[1,2,3,4,5]
for x in itertools.combinations(t, k):
print x,index(x,len(t),k)
It is not hard to prove that if comb=(a1,a2,a3...,ak) (in increasing order) then:
index=[nCk-(n-a1+1)Ck] + [(n-a1)C(k-1)-(n-a2+1)C(k-1)] + ... =
nCk -(n-a1)Ck -(n-a2)C(k-1) - .... -(n-ak)C1
There's another way to do all this. You could generate all possible combinations and write them into a binary file where each comb is represented by it's index starting from zero. Then, when you need to find an index, and the combination is given, you apply a binary search on the file. Here's the function. It's written in VB.NET 2010 for my lotto program, it works with Israel lottery system so there's a bonus (7th) number; just ignore it.
Public Function Comb2Index( _
ByVal gAr() As Byte) As UInt32
Dim mxPntr As UInt32 = WHL.AMT.WHL_SYS_00 '(16.273.488)
Dim mdPntr As UInt32 = mxPntr \ 2
Dim eqCntr As Byte
Dim rdAr() As Byte
modBinary.OpenFile(WHL.WHL_SYS_00, _
FileMode.Open, FileAccess.Read)
Do
modBinary.ReadBlock(mdPntr, rdAr)
RP: If eqCntr = 7 Then GoTo EX
If gAr(eqCntr) = rdAr(eqCntr) Then
eqCntr += 1
GoTo RP
ElseIf gAr(eqCntr) < rdAr(eqCntr) Then
If eqCntr > 0 Then eqCntr = 0
mxPntr = mdPntr
mdPntr \= 2
ElseIf gAr(eqCntr) > rdAr(eqCntr) Then
If eqCntr > 0 Then eqCntr = 0
mdPntr += (mxPntr - mdPntr) \ 2
End If
Loop Until eqCntr = 7
EX: modBinary.CloseFile()
Return mdPntr
End Function
P.S. It takes 5 to 10 mins to generate 16 million combs on a Core 2 Duo. To find the index using binary search on file takes 397 milliseconds on a SATA drive.
Assuming the maximum setSize is not too large, you can simply generate a lookup table, where the inputs are encoded this way:
int index(a,b,c,...)
{
int key = 0;
key |= 1<<a;
key |= 1<<b;
key |= 1<<c;
//repeat for all arguments
return Lookup[key];
}
To generate the lookup table, look at this "banker's order" algorithm. Generate all the combinations, and also store the base index for each nItems. (For the example on p6, this would be [0,1,5,11,15]). Note that by you storing the answers in the opposite order from the example (LSBs set first) you will only need one table, sized for the largest possible set.
Populate the lookup table by walking through the combinations doing Lookup[combination[i]]=i-baseIdx[nItems]
EDIT: Never mind. This is completely wrong.
Let your combination be (a1, a2, ..., ak-1, ak) where a1 < a2 < ... < ak. Let choose(a,b) = a!/(b!*(a-b)!) if a >= b and 0 otherwise. Then, the index you are looking for is
choose(ak-1, k) + choose(ak-1-1, k-1) + choose(ak-2-1, k-2) + ... + choose (a2-1, 2) + choose (a1-1, 1) + 1
The first term counts the number of k-element combinations such that the largest element is less than ak. The second term counts the number of (k-1)-element combinations such that the largest element is less than ak-1. And, so on.
Notice that the size of the universe of elements to be chosen from (10 in your example) does not play a role in the computation of the index. Can you see why?
Sample solution:
class Program
{
static void Main(string[] args)
{
// The input
var n = 5;
var t = new[] { 2, 4, 5 };
// Helping transformations
ComputeDistances(t);
CorrectDistances(t);
// The algorithm
var r = CalculateRank(t, n);
Console.WriteLine("n = 5");
Console.WriteLine("t = {2, 4, 5}");
Console.WriteLine("r = {0}", r);
Console.ReadKey();
}
static void ComputeDistances(int[] t)
{
var k = t.Length;
while (--k >= 0)
t[k] -= (k + 1);
}
static void CorrectDistances(int[] t)
{
var k = t.Length;
while (--k > 0)
t[k] -= t[k - 1];
}
static int CalculateRank(int[] t, int n)
{
int k = t.Length - 1, r = 0;
for (var i = 0; i < t.Length; i++)
{
if (t[i] == 0)
{
n--;
k--;
continue;
}
for (var j = 0; j < t[i]; j++)
{
n--;
r += CalculateBinomialCoefficient(n, k);
}
n--;
k--;
}
return r;
}
static int CalculateBinomialCoefficient(int n, int k)
{
int i, l = 1, m, x, y;
if (n - k < k)
{
x = k;
y = n - k;
}
else
{
x = n - k;
y = k;
}
for (i = x + 1; i <= n; i++)
l *= i;
m = CalculateFactorial(y);
return l/m;
}
static int CalculateFactorial(int n)
{
int i, w = 1;
for (i = 1; i <= n; i++)
w *= i;
return w;
}
}
The idea behind the scenes is to associate a k-subset with an operation of drawing k-elements from the n-size set. It is a combination, so the overall count of possible items will be (n k). It is a clue that we could seek the solution in Pascal Triangle. After a while of comparing manually written examples with the appropriate numbers from the Pascal Triangle, we will find the pattern and hence the algorithm.
I used user515430's answer and converted to python3. Also this supports non-continuous values so you could pass in [1,3,5,7,9] as your pool instead of range(1,11)
from itertools import combinations
from scipy.special import comb
from pandas import Index
debugcombinations = False
class IndexedCombination:
def __init__(self, _setsize, _poolvalues):
self.setsize = _setsize
self.poolvals = Index(_poolvalues)
self.poolsize = len(self.poolvals)
self.totalcombinations = 1
fast_k = min(self.setsize, self.poolsize - self.setsize)
for i in range(1, fast_k + 1):
self.totalcombinations = self.totalcombinations * (self.poolsize - fast_k + i) // i
#fill the nCr cache
self.choose_cache = {}
n = self.poolsize
k = self.setsize
for i in range(k + 1):
for j in range(n + 1):
if n - j >= k - i:
self.choose_cache[n - j,k - i] = comb(n - j,k - i, exact=True)
if debugcombinations:
print('testnth = ' + str(self.testnth()))
def get_nth_combination(self,index):
n = self.poolsize
r = self.setsize
c = self.totalcombinations
#if index < 0 or index >= c:
# raise IndexError
result = []
while r:
c, n, r = c*r//n, n-1, r-1
while index >= c:
index -= c
c, n = c*(n-r)//n, n-1
result.append(self.poolvals[-1 - n])
return tuple(result)
def get_n_from_combination(self,someset):
n = self.poolsize
k = self.setsize
index = 0
j = 0
for i in range(k):
setidx = self.poolvals.get_loc(someset[i])
for j in range(j + 1, setidx + 1):
index += self.choose_cache[n - j, k - i - 1]
j += 1
return index
#just used to test whether nth_combination from the internet actually works
def testnth(self):
n = 0
_setsize = self.setsize
mainset = self.poolvals
for someset in combinations(mainset, _setsize):
nthset = self.get_nth_combination(n)
n2 = self.get_n_from_combination(nthset)
if debugcombinations:
print(str(n) + ': ' + str(someset) + ' vs ' + str(n2) + ': ' + str(nthset))
if n != n2:
return False
for x in range(_setsize):
if someset[x] != nthset[x]:
return False
n += 1
return True
setcombination = IndexedCombination(5, list(range(1,10+1)))
print( str(setcombination.get_n_from_combination([2,5,7,8,10])))
returns 188