i'm currently solving a problem that states:
A company filed for bankruptcy and decided to pay the employees with the last remaining valuable items in the company only if it can be distributed evenly among them so that all of them have at least received 1 item and that the difference between the employee carrying the most valuable items and the employee carrying the least valuable items can not exceed a certain value x;
Input:
First row contains number of employee;
Second row contains the x value so that the the difference between the employee carrying the most valuable items and the employee carrying the least valuable items can not exceed;
Third row contains all the items with their value;
Output:
First number is the least valuable basket of items value and the second is the most valuable basket;
Example:
Input:
5
4
2 5 3 11 4 3 1 15 7 8 10
Output:
13 15
Input:
5
4
1 1 1 11 1 3 1 2 7 8
Output:
NO (It's impossible to distribute evenly)
Input:
5
10
1 1 1 1
Output:
NO (It's impossible to distribute evenly)
My solution to resolve this problem taking the first input is to, sort the items in ascending or descending order so from
2 5 3 11 4 3 1 15 7 8 10 --> 1 2 3 3 4 5 7 8 10 11 15
then create an adjacency list or just store it in simple variables where we add the biggest number to the lowest basket while iterating the item values array
Element 0: 15
Element 1: 11 <- 3 (sum 14)
Element 2: 10 <- 3 (sum 13)
Element 3: 8 <- 4 <- 1 (sum 13)
Element 4: 7 <- 5 <- 2 (sum 14)
So that my solution will have O(nlogN + 2n), first part using merge sort and then finding max e min value, what do you guys think about this solution?
I have a vector of length k where each element is a 2-by-m matrix that is a mapping between indices. All elements are integers from the set {1,2,...,mn}, there are no duplicates within a given 2-by-m matrix, and I know what m and n are. I want an efficient way of finding n-1 of these 2-by-m matrices such that combining these mappings gives me all of the elements in {1,2,...,mn}. To make things more concrete, assume that m=n=3 and my vector is
4 3 6
2 7 5
4 6 3
1 9 8
9 1 8
6 2 7
7 8 3
2 1 4
I want the algorithm to output
4 3 6
2 7 5
1 8 9
which is found by combining the first two mappings.
I've got a circular buffer with positive natural values, e.g.
1 5
4 2
11 7
2 9
We're going to partition it into exactly two continuous parts, while keeping this order. These two parts in this example could be:
(4 1 5) and (2 7 9 2 11),
(7 9 2 11 4) and (1 5 2),
etc.
The idea is to keep order and take two continuous subsequences.
And the problem now is to partition it so that the sums of these subsequences are closes to each other, i.e. the difference between the sums must be closest to zero.
In this case, I believe the solution would be: (2 7 9 2) and (11 4 1 5) with sums, respectively, 20 and 21.
How to do this optimally?
Algorithm:
Calculate the total sum.
Let the current sum = 0.
Start off with 2 pointers at any point (both starting off at the same point).
Increase the second pointer, adding the number it passed, until the current sum is more than half of the total sum.
Increase the first pointer, subtracting the number it passed, until the current sum is less than half of the total sum.
Stop if either:
The first pointer is back where it started, or
The best sum is 0.5 or 0 from half the total sum (in which case the difference will be 1 or 0).
The difference can be 1 only if the total sum is odd, in which case the difference can never be 0. (Thanks Artur!)
Otherwise repeat from step 3.
Check all the current sums we got in this process and keep the one that's closest to half, along with indices of the partition that got that sum.
Running time:
The running time will be O(n), since we only ever increase the pointers and the first one only goes around once, and the second one can't go around more than twice.
Example:
Input:
1 5
4 2
11 7
2 9
Total sum = 41.
Half of sum = 20.5.
So, let's say we start off at 1. (I just put it on a straight line to make it easier to draw)
p1, p2
V
1 5 2 7 9 2 11 4
sum = 0
p1 p2
V V
1 5 2 7 9 2 11 4
sum = 1
p1 p2
V V
1 5 2 7 9 2 11 4
sum = 6
p1 p2
V V
1 5 2 7 9 2 11 4
sum = 8
p1 p2
V V
1 5 2 7 9 2 11 4
sum = 15
p1 p2
V V
1 5 2 7 9 2 11 4
sum = 24
p1 p2
V V
1 5 2 7 9 2 11 4
sum = 23
p1 p2
V V
1 5 2 7 9 2 11 4
sum = 18
p1 p2
V V
1 5 2 7 9 2 11 4
sum = 20
Here the sum (20) is 0.5 from half the total sum (20.5), so we can stop.
The above corresponds to (11 4 1 5) (2 7 9 2), with a difference in sums of 1.
I have a sqaure matrix and a smaller square which moves inside the matrix at all possible positions (does not go out of the matrix). I need to find the smallest number in all such possible overlappings.
The problem is that the sizes of both can go upto thousands. Any fast way to do that?
I know one way - if there's an array instead of a matrix and a window instead of a square, we can do that in linear time using a deque.
Thanks in advance.
EDIT: Examples
Matrix:
1 3 6 2 5
8 2 3 4 5
3 8 6 1 5
7 4 8 2 1
8 0 9 0 5
For a square of size 3, total 9 overlappings are possible. For each overlapping the minimum numbers in matrix form are:
1 1 1
2 1 1
0 0 0
It is possible in O(k * n^2) with your deque idea:
If your smaller square is k x k, iterate the first row of elements from 1 to k in your matrix and treat it as an array by precomputing the minimum of the elements from 1 to k, from 2 to k + 1 etc in each column of the matrix (this precomputation will take O(k * n^2)). This is what your first row will be:
*********
1 3 6 2 5
8 2 3 4 5
3 8 6 1 5
*********
7 4 8 2 1
8 0 9 0 5
The precomputation I mentioned will give you the minimum in each of its columns, so you will have reduced the problem to your 1d array problem.
Then continue with the row of elements from 2 to k + 1:
1 3 6 2 5
*********
8 2 3 4 5
3 8 6 1 5
7 4 8 2 1
*********
8 0 9 0 5
There will be O(n) rows and you will be able to solve each one in O(n) because our precomputation allows us to reduce them to basic arrays.
I was asked to find a 11x11-grid containing the digits such that one can read the squares of 1,...,100. Here read means that you fix the starting position and direction (8 possibilities) and if you can find for example the digits 1,0,0,0,0,4 consecutively, you have found the squares of 1, 2, 10, 100 and 20. I made a program (the algorithm is not my own. I modified slightly a program which uses best-first search to find a solution but it is too slow. Does anyone know a better algorithm to solve the problem?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <vector>
#include <algorithm>
using namespace std;
int val[21][21];//number which is present on position
int vnum[21][21];//number of times the position is used - useful if you want to backtrack
//5 unit borders
int mx[4]={-1,0,1,0};//movement arrays
int my[4]={0,-1,0,1};
int check(int x,int y,int v,int m)//check if you can place number - if you can, return number of overlaps
{
int c=1;
while(v)//extract digits one by one
{
if(vnum[x][y] && (v%10)!=val[x][y])
return 0;
if(vnum[x][y])
c++;
v/=10;
x+=mx[m];
y+=my[m];
}
return c;
}
void apply(int x,int y,int v,int m)//place number - no sanity checks
{
while(v)//extract digits one by one
{
val[x][y]=v%10;
vnum[x][y]++;
v/=10;
x+=mx[m];
y+=my[m];
}
}
void deapply(int x,int y,int v,int m)//remove number - no sanity checks
{
while(v)
{
vnum[x][y]--;
v/=10;
x+=mx[m];
y+=my[m];
}
}
int best=100;
void recur(int num)//go down a semi-random path
{
if(num<best)
{
best=num;
if(best)
printf("FAILED AT %d\n",best);
else
printf("SUCCESS\n");
for(int x=5;x<16;x++) // 16 and 16
{
for(int y=5;y<16;y++)
{
if(vnum[x][y]==0)
putchar('.');
else
putchar(val[x][y]+'0');
}
putchar('\n');
}
fflush(stdout);
}
if(num==0)
return;
int s=num*num,t;
vector<int> poss;
for(int x=5;x<16;x++)
for(int y=5;y<16;y++)
for(int m=0;m<4;m++)
if(t=check(x,y,s,m))
poss.push_back((x)|(y<<8)|(m<<16)|(t<<24));//compress four numbers into an int
if(poss.size()==0)
return;
sort(poss.begin(),poss.end());//essentially sorting by t
t=poss.size()-1;
while(t>=0 && (poss[t]>>24)==(poss.back()>>24))
t--;
t++;
//t is now equal to the smallest index which has the maximal overlap
t=poss[rand()%(poss.size()-t)+t];//select random index>=t
apply(t%256,(t>>8)%256,s,(t>>16)%256);//extract random number
recur(num-1);//continue down path
}
int main()
{
srand((unsigned)time(0));//seed
while(true)
{
for(int i=0;i<21;i++)//reset board
{
memset(val[i],-1,21*sizeof(int));
memset(vnum[i],-1,21*sizeof(int));
}
for(int i=5;i<16;i++)
{
memset(val[i]+5,0,11*sizeof(int));
memset(vnum[i]+5,0,11*sizeof(int));
}
recur(100);
}
}
Using a random search so far I only got to 92 squares with one unused spot (8 missing numbers: 5041 9025 289 10000 4356 8464 3364 3249)
1 5 2 1 2 9 7 5 6 9 5
6 1 0 8 9 3 8 4 4 1 2
9 7 2 2 5 0 0 4 8 8 2
1 6 5 9 6 0 4 4 7 7 4
4 4 2 7 6 1 2 9 0 2 2
2 9 6 1 7 8 4 4 0 9 3
6 5 5 3 2 6 0 1 4 0 6
4 7 6 1 8 1 1 8 2 8 1
8 0 1 3 4 8 1 5 3 2 9
0 5 9 6 9 8 8 6 7 4 5
6 6 2 9 1 7 3 9 6 9
The algorithm basically uses as solution encoding a permutation on the input (search space is 100!) and then places each number in the "topmost" legal position. The solution value is measured as the sum of the squares of the lengths of the numbers placed (to give more importance to long numbers) and the number of "holes" remaining (IMO increasing the number of holes should raise the likehood that another number will fit in).
The code has not been optimized at all and is only able to decode a few hundred solutions per second. Current solution has been found after 196k attempts.
UPDATE
Current best solution with this approach is 93 without free holes (7 missing numbers: 676 7225 3481 10000 3364 7744 5776):
9 6 0 4 8 1 0 0 9 3 6
6 4 0 0 2 2 5 6 8 8 9
1 7 2 9 4 1 5 4 7 6 3
5 8 2 3 8 6 4 9 6 5 7
2 4 4 4 1 8 2 8 2 7 2
1 0 8 9 9 1 3 4 4 9 1
2 1 2 9 6 1 0 6 2 4 1
2 3 5 5 3 9 9 4 0 9 6
5 0 0 6 1 0 3 5 2 0 3
2 7 0 4 2 2 5 2 8 0 9
9 8 2 2 6 5 3 4 7 6 1
This is a solution (all 100 numbers placed) however using a 12x12 grid (MUCH easier)
9 4 6 8 7 7 4 4 5 5 1 7
8 3 0 5 5 9 2 9 6 7 6 4
4 4 8 3 6 2 6 0 1 7 8 4
4 8 4 2 9 1 4 0 5 6 1 4
9 1 6 9 4 8 1 5 4 2 0 1
9 4 4 7 2 2 5 2 2 5 0 0
4 6 2 2 5 8 4 2 7 4 0 2
0 3 3 3 6 4 0 0 6 3 0 9
9 8 0 1 2 1 7 9 5 5 9 1
6 8 4 2 3 5 2 6 3 2 0 6
9 9 8 2 5 2 9 9 4 2 2 7
1 1 5 6 6 1 9 3 6 1 5 4
It has been found using a truly "brute force" approach, starting from a random matrix and keeping randomly changing digits when that improved the coverage.
This solution has been found by an highly unrolled C++ program automatically generated by a Python script.
Update 2
Using an incremental approach (i.e. keeping a more complex data structure so that when changing a matrix element the number of targets covered can be updated instead than recomputed) I got a much faster search (about 15k matrices/second investigated with a Python implementation running with PyPy).
In a few minutes this version was able to find a 99 quasi-solution (a number is still missing):
7 0 5 6 5 1 1 5 7 1 6
4 6 3 3 9 8 8 6 7 6 1
3 9 0 8 2 6 1 1 4 7 8
1 1 0 8 9 9 0 0 4 4 6
3 4 9 0 4 9 0 4 6 7 1
6 4 4 6 8 6 3 2 5 2 9
9 7 8 4 1 1 4 0 5 4 2
6 2 4 1 5 2 2 1 2 9 7
9 8 2 5 2 2 7 3 6 5 0
3 1 2 5 0 0 6 3 0 5 4
7 5 6 9 2 1 6 5 3 4 6
UPDATE 3
Ok. After a some time (no idea how much) the same Python program actually found a complete solution (several ones indeed)... here is one
6 4 6 9 4 1 2 9 7 3 6
9 2 7 7 4 4 8 1 2 1 7
1 0 6 2 7 0 4 4 8 3 4
2 1 2 2 5 5 9 2 9 6 5
9 2 5 5 2 0 2 6 3 9 1
1 6 3 6 0 0 9 3 7 0 6
6 0 0 4 9 0 1 6 0 0 4
9 8 4 4 8 0 1 4 5 2 3
2 4 8 2 8 1 6 8 6 7 5
1 7 6 9 2 4 5 4 2 7 6
6 6 3 8 8 5 6 1 5 2 1
The searching program can be found here...
You've got 100 numbers and 121 cells to work with, so you'll need to be very efficient. We should try to build up the grid, so that each time we fill a cell, we attain a new number in our list.
For now, let's only worry about 68 4-digit numbers. I think a good chunk of the shorter numbers will be in our grid without any effort.
Start with a 3x3 or 4x4 set of numbers in the top-left of your grid. It can be arbitrary, or fine-tune for slightly better results. Now let's fill in the rest of the grid one square at a time.
Repeat these steps:
Fill an empty cell with a digit
Check which numbers that knocked off the list
If it didn't knock off any 4-digit numbers, try a different digit or cell
Eventually you may need to fill 2 cells or even 3 cells to achieve a new 4-digit number, but this should be uncommon, except at the end (at which point, hopefully there's a lot of empty space). Continue the process for the (few?) remaining 3-digit numbers.
There's a lot room for optimizations and tweaks, but I think this technique is fast and promising and a good starting point. If you get an answer, share it with us! :)
Update
I tried my approach and only got 87 out of the 100:
10894688943
60213136008
56252211674
61444925224
59409675697
02180334817
73260193640
.5476685202
0052034645.
...4.948156
......4671.
My guess is that both algorithms are too slow. Some optimization algorithm might work like best-first search or simulated annealing but my I don't have much experience on programming those.
Have you tried any primary research on Two-Dimensional Bin Packing (2DBP) algorithms? Google Scholars is a good start. I did this a while ago when building an application to generate mosaics.
All rectangular bin packing algorithms can be divided into 4 groups based on their support for the following constraints:
Must the resulting bin be guillotine cuttable? I.e. do you have to later slice the bin(s) in half until all the pieces are unpacked?
Can the pieces be rotated to fit into the bin? Not an issue with square pieces, so this makes more algorithms available to you.
Out of all the algorithms I looked into, the most efficient solution is an Alternate Directions (AD) algorithm with a Tabu Search optimization layer. There are dissertations which prove this. I may be able to dig-up some links if this helps.
Some ideas off the top of my head, without investing much time into thinking about details.
I would start by counting the number of occurrences of each digit in all squares 1..100. The total number of digits will be obviously larger than 121, but by analyzing individual frequencies you can deduce which digits must be grouped on a single line to form as many different squares as possible. For example, if 0 has the highest frequency, you have to try to put as many squares containing a 0 on the same line.
You could maintain a count of digits for each line, and each time you place a digit, you update the count. This lets you easily compute which square numbers have been covered by that particular line.
So, the program will still be brute-force, but it will exploit the problem structure much better.
PS: Counting digit frequencies is the easiest way to decide whether a certain permutation of digits constitutes a square.