Given a grid of dimensions A*B with values between 1-9, find a sequence of B numbers that maximizes the minimum number of values matched when compared with A rows.
Describe the certain steps you would take to maximize the minimum score.
Example:
Grid Dimension
A = 5 , B = 10
Grid Values
9 3 9 2 9 9 4 5 7 6
6 3 4 2 8 5 7 5 9 2
4 9 5 8 3 7 3 2 7 6
7 5 8 9 9 4 7 3 3 7
2 6 8 3 2 4 5 4 2 2
Possible Answer
6 3 8 2 9 4 7 5 7 4
Score Calculation
This answer scores
5 when compared with Row 1
5 when compared with Row 2
1 when compared with Row 3
4 when compared with Row 4
2 when compared with Row 5
And thus the minimal score for this answer is 1.
I would go for a local hill-climbing approach that you can complement with a randomization to avoid local minima. Something like:
1. Generate a random starting solution S
2. Compute its score score(S, row) for each row. We'll call min_score(S) the minimum score among all rows for S.
3. Attempt to improve the solution with:
For each digit i (1..B) in S:
If i belongs to a row such that score(S, row) > (min_score(S) + 1) then:
Change i to be the digit of a row with min_score(S). If there was only one row with min_score(S), then min_score(S) has improved by 1
Update the scores of all the rows.
If min_score(S) hasn't improved for more than N iterations of 3, go back to 1 and start with a new random solution.
Recently I came across a question in an interview "Print a complete binary tree in concentric circles".
1
2 3
4 5 6 7
8 9 0 1 2 3 4 5
The output should be
1 2 4 8 9 0 1 2 3 4 5 7 3
5 6
Could anyone help me out on how we can solve this problem?
Here is how you can approach the problem. Arrange the tree by levels:
1
2, 3
4, 5, 6, 7
8, 9, 0, 1, 2, 3, 4, 5
So the data you have is k levels L1, L2, ..., Lk. Now answer this questions: After we execute one step, that is when one circle is traversed, how would the tree levels would look like after the traverse elements have been removed from the levels? How should I modify the levels and which elements should I print so that it would seems like I've traversed on circle?
In your example, after the first step the levels would be modified to just:
5, 6
So what was the operation that was executed?
After you've answered the questions just apply the same procedure couple of times until you've printed all elements.
We are studying algorithms and we haven't yet started making any programs with it, so we have a simple exercises. For example:
I have to bubble sort:
4 2 7 16 6
Here is a proccess (we have to look from the right to left in this case):
4 2 7 16 6 (it makes 3 swaps and 4 comparisons in this)
2 4 6 7 16 (it makes 0 swaps and 3 comparison in this)
2 4 6 7 16 (the result)
This how I understand it:
4 2 7 16 6 (6 is swapped with 16)
4 2 7 6 16 (6 is swapped with 7)
4 2 6 7 16 (6 is not swapped with 2)
4 2 6 7 16 (2 is swapped with 4)
2 4 6 7 16 (result after 1st iteration)
now we have numbers which have been arranged properly, we have made 4 comparisons at the beginning and 3 swaps and we know that 2 is the smallest number so we will look at 4 6 7 16 now. However they are arranged properly but still we have to make 3 additional
comparisons.
My question is: do we have to make another 2 comparisons and then the last 1 comparison (which would result in 10 comparisons overall) to finish the bubble sort? Or it stops after 7?
I'm not sure when the Bubble-sort stops.
http://www.cs.pitt.edu/~kirk/cs1501/animations/Sort1.html is this applet counting right? Selection sort for 5 4 3 2 1, I see 2 swaps, but the applet is counting 4 exchanges....
I guess it's a matter of definition. He is doing a swap in the end of every loop, even if he is swapping one element against itself. In his case, the swaps will be:
Original: 5 4 3 2 1
Swap pos 1 and 5: 1 4 3 2 5
Swap pos 2 and 4: 1 2 3 4 5
Swap pos 3 and 3: 1 2 3 4 5
Swap pos 4 and 4: 1 2 3 4 5
(No swap is done for the last element since that will always be in the correct place)
A simple if statement could be used to eliminate the two last swaps.
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.