You are given an infinite matrix whose upper-left square starts with 1. Here are the first five rows of the infinite matrix :
1 2 9 10 25
4 3 8 11 24
5 6 7 12 23
16 15 14 13 22
17 18 19 20 21
Your task is to find out the number in presents at row x and column y after observing a certain kind of patter present in the matrix
Input Format
The first input line contains an integer t: the number of test cases
After this, there are t lines, each containing integer x and y
For each test, print the number present at xth row and yth column.
sample input
3
2 3
1 1
4 2
sample output
8
1
15
Hint: the numbers at the right and bottom border of a left upper square are consecutive (going either down and left, or right and up). First determine in which border your position is, then find out which direction applies, and finally find the correct number at the position (which easy formula gives you the first number in the border?).
Related
What is an efficient algorithm to replace the values in an image while
minimizing the largest value and maintaining order?
Background
I have a 8.5Gb image which is represented as a rows and columns.
Suppose we have a smaller version (there are no duplicates in input):
4, 5, 9,
2, 3, 7,
8, 6, 1
I need to replace the entries at each pixel to the smallest positive value possible (greater than zero) in the entire matrix
while preserving the row-wise and column-wise ordering.
One possible output (duplicates allowed here) is the following and the maximum value is 5 ( I do not believe we can reduce it to 4):
2, 3, 4,
1, 2, 3,
5, 4, 1
The reason it works:
Input: First Row: 4 < 5 < 9 and first Column: 4 > 2 < 8
Output: First Row: 2 < 3 < 4 and First Column 2 > 1 < 5 (column)
The orderings are being maintained. The same for the other rows and columns:
5 > 3 < 6 <=> 3 > 2 < 4
...
...
----------------------------------------- Attempt: My wrong algorithm -----------------------------------------
1. Each row and column will contain unique elements. So start with the first row and assign integers from the range {1, total the number of rows}:
1 2 3
x x x
x x x
The maximum in that row is currently at 3.
2. Go to the next row which is 2,3,7 and again assign numbers in the range {1, total number of rows}. When we assign 1 we look at all the previous rows if there are conflicts. In this case 1 is already present in the previous row. And we need a number which is smaller than 1. So place a zero there (I will offset every entries by on later).
1 2 3
0 1 2
* * *
The maximum in that row is currently 2.
3. Go to the next row and again fill as above. But 1 already occurred before and we need a number larger than the first and second rows:
So, try 2. The next number needs to be larger than 2 and 1 (column) and smaller than 2 (row). That is a huge problem. I need to change too many cells each time.
For severe clarity, I'll add 10 to each of your values.
Input Ordering
14 15 19 - - -
12 13 17 - - -
18 16 11 - - -
Consider each of the values in order, smallest to largest. Each element receives an ordering value that is the smallest integer available at that location. "Available" means that the assigned number is larger than any in the same row or column.
11 and 12 aren't in the same row or column, so we can assign both of those immediately.
Input Ordering
14 15 19 - - -
12 13 17 1 - -
18 16 11 - - 1
When we consider 13, we see that it is in the same row with a 1, so it must have the next larger value:
Input Ordering
14 15 19 - - -
12 13 17 1 2 -
18 16 11 - - 1
14 has the same problem, being above a 1:
Input Ordering
14 15 19 2 - -
12 13 17 1 2 -
18 16 11 - - 1
Continue this process for each number. Take the maximum of the orderings in that number's row and column. Add 1 and assign that ordering.
Input Ordering
14 15 19 2 3 -
12 13 17 1 2 -
18 16 11 - 4 1
Input Ordering
14 15 19 2 3 4
12 13 17 1 2 3
18 16 11 5 4 1
There's a solution. The "dominance" path 18 > 16 > 15 > [14 or 13] > 12 demonstrates that 5 is the lowest max value.
You can also solve this by converting the locations to a directed graph. Nodes in the same row or column have an edge connecting them; the edge is directed from the smaller to the larger. It will be sufficient to order the values and merely connect the adjacent values: given 14->15 and 15->19, we don't need 14->19 as well.
Add a node 0 with label 0 and an edge to each node that has no other input edges.
Now follow a typical labeling iteration: any node with all its inputs labeled receives a label that is one more than the largest of its inputs.
This is the same algorithm as the above, but the correctness and minimalism are much easier to see.
14 -> 15 -> 19
12 -> 13 -> 17
11 -> 16 -> 18
12 -> 14 -> 18
13 -> 15 -> 16
11 -> 17 -> 19
0 -> 11
0 -> 12
Now, if we shake out the topology of this, starting on the left, we get:
0 11 13 17
12 14 15 16 18
19
This makes the numbering obvious: each node is labeled with the length of its longest path from the start node.
Your memory problem should be edited into your question proposal, or given as a new question. You have non-trivial dependencies along rows and columns. If your data do not fit into memory, then you may want to make a disk-hosted data base to store your pre-processed data. For instance, you could store the graph as a list of edges keyed by dependencies:
11 none
12 none
13 12
14 12
15 13, 14
16 11, 15
17 11, 13
18 14, 16
19 15, 17
You haven't described the shape of your data. At the very worst, you should be able to build this graph data base with one pass to do the rows, and then one pass per column -- or multiple columns in each pass, depending on how many you can fit into memory at once.
Then you can apply the algorithm to the items int he data base. You can speed it up if you keep in memory, not only all nodes with no dependencies, but another list with few dependencies -- "few" being dependent on your memory availability.
For instance, make one pass over the data base to grab every cell with 0 or 1 dependencies. Put the independent nodes in your "active" list; as you process those, add nodes only from the "1-dependency" list as they're freed up. Once you've exhausted those sub-graphs, then make a large pass to (1) update the data base; (2) extract the next sets of nodes with 0 or 1 dependency.
Let's look at this with the example you gave. First, we make a couple of lists from the original graph:
0-dep 11, 12
1-dep 13 (on 12), 14 (on 12)
This pass is trivial: we assign 1 to cells 11 and 12; 2 to cells 13 and 14. Now update the graph:
node dep done (assigned values)
15 none 2, 2
16 15 1
17 none 1, 2
18 16 2
19 15, 17
Refresh the in-memory lists:
0-dep 15, 17
1-dep 16 (on 15), 18 (on 16)
On this pass, both 15 and 17 depend on a node with value 2, so they are both assigned 3. Resolving 15 frees node 16, which gets value 4. This, in turn, frees up node 18, which gets the value 5.
In one final pass, we now have node 19 with no outstanding dependencies. it's maximum upstream value is 3, so it gets the value 4.
In the worst case -- you can't even hold all independent nodes in memory at once -- you can still grab as many as you can fit, assign their values in an in-memory pass, and return to the disk for more to process.
Can you handle the data manipulations from here?
I have a 2d game board that expands as tiles are added to the board. Tiles can only be adjacent to existing tiles in the up, down, left and right positions.
So I thought a diamond spiral matrix would be the most efficient way to store the board, but I cannot find a way to convert the x,y coordinates to a 1d array index or the reverse operation.
like this layout
X -3 -2 -1 0 1 2 3
Y 3 13
2 24 5 14
1 23 12 1 6 15
0 22 11 4 0 2 7 16
-1 21 10 3 8 17
-2 20 9 18
-3 19
Tile 1 will always be at position 0, tile 2 will be at 1,2,3 or 4, tile 3 somewhere from 1 to 12 etc.
So I need an algorithm that goes from X,Y to an index and from an index back to the original X and Y.
Anyone know how to do this, or recommend another space filling algorithm that suits my needs. I'm probably going to use Java but would prefer something language neutral.
Thanks
As I can understand form the problem statement, there is no guarantee that the tiles will be filled evenly on the sides. for example:
X -3 -2 -1 0 1 2 3
Y 3 6
2 3 4 5
1 1
0 0 2
-1
So, I think a diamond matrix won't be the best choice.
I would suggest storing them in a hash-map, like implementing a dictionary for 2 letter words.
Also, You need to be more specific to what your requirements are. Like, do you prioritize space complexity over time? Or do you want a fast access time and don't care about memory usage that much.
IMPORTANT :
Also, what is the
Max number of tiles that we have to hold
Max width and height of the board.
I want to device an algorithm to display the following pattern:
1
9 2
10 8 3
14 11 7 4
15 13 12 6 5
Is there a way to convert it into the following array and use the indices of the above matrix and find out the position of the number in the array:
1 9 2 10 8 3 ...
I can't find a pattern to calculate the element using the matrix co-ordinates, that is why I was trying to device the above method of somehow determining the position of the next number in the array.
Just clues:
You need to know formula for sum of arithmetic progression (here sum of natural numbers 1 + 2 + 3 +.. N)
1st step: determine number of diagonal where k-th item of array stands.
2nd step: get direction of this diagonal filling
3rd step: get number of place at this diagonal
4th step: find what number stands here
Can somebody please explain this heuristic function, for example for the following arrangement of 4x4 puzzle, whats the X-Y heuristic cost?
1 2 3 4
5 6 7 8
9 10 11 12
0 13 14 15
(0 indicates blank space)
As from here and here the X-Y heuristic is computed by the sum of the minimum number of column-adjacent blank swaps to get all tiles in their destination column and the minimum number of row adjacent blank swaps to get all tiles in their destination row.
So in this situation:
1 2 3 4
5 6 7 8
9 10 11 12
0 13 14 15
the only misplaced tiles are 13 , 14 and 15, assuming the goal state is
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 0
So in this case the we have to compute at first the number of column swaps the blank has to do to get all the tiles in the correct position. This is equivalent to 3, since the blank has to move three times to the the right column to be in the right position (and to have all the tiles in the right position)
Then we have to compute the number of row swaps the blank has to do. This is 0 thanks to the fact that all the tiles are already on the correct row.
Finally h(n) = 3 + 0 = 3 .
In the FinnAPL Idiom Library, the 19th item is described as “Ascending cardinal numbers (ranking, all different) ,” and the code is as follows:
⍋⍋X
I also found a book review of the same library by R. Peschi, in which he said, “'Ascending cardinal numbers (ranking, all different)' How many of us understand why grading the result of Grade Up has that effect?” That's my question too. I searched extensively on the internet and came up with zilch.
Ascending Cardinal Numbers
For the sake of shorthand, I'll call that little code snippet “rank.” It becomes evident what is happening with rank when you start applying it to binary numbers. For example:
X←0 0 1 0 1
⍋⍋X ⍝ output is 1 2 4 3 5
The output indicates the position of the values after sorting. You can see from the output that the two 1s will end up in the last two slots, 4 and 5, and the 0s will end up at positions 1, 2 and 3. Thus, it is assigning rank to each value of the vector. Compare that to grade up:
X←7 8 9 6
⍋X ⍝ output is 4 1 2 3
⍋⍋X ⍝ output is 2 3 4 1
You can think of grade up as this position gets that number and, you can think of rank as this number gets that position:
7 8 9 6 ⍝ values of X
4 1 2 3 ⍝ position 1 gets the number at 4 (6)
⍝ position 2 gets the number at 1 (7) etc.
2 3 4 1 ⍝ 1st number (7) gets the position 2
⍝ 2nd number (8) gets the position 3 etc.
It's interesting to note that grade up and rank are like two sides of the same coin in that you can alternate between the two. In other words, we have the following identities:
⍋X = ⍋⍋⍋X = ⍋⍋⍋⍋⍋X = ...
⍋⍋X = ⍋⍋⍋⍋X = ⍋⍋⍋⍋⍋⍋X = ...
Why?
So far that doesn't really answer Mr Peschi's question as to why it has this effect. If you think in terms of key-value pairs, the answer lies in the fact that the original keys are a set of ascending cardinal numbers: 1 2 3 4. After applying grade up, a new vector is created, whose values are the original keys rearranged as they would be after a sort: 4 1 2 3. Applying grade up a second time is about restoring the original keys to a sequence of ascending cardinal numbers again. However, the values of this third vector aren't the ascending cardinal numbers themselves. Rather they correspond to the keys of the second vector.
It's kind of hard to understand since it's a reference to a reference, but the values of the third vector are referencing the orginal set of numbers as they occurred in their original positions:
7 8 9 6
2 3 4 1
In the example, 2 is referencing 7 from 7's original position. Since the value 2 also corresponds to the key of the second vector, which in turn is the second position, the final message is that after the sort, 7 will be in position 2. 8 will be in position 3, 9 in 4 and 6 in the 1st position.
Ranking and Shareable
In the FinnAPL Idiom Library, the 2nd item is described as “Ascending cardinal numbers (ranking, shareable) ,” and the code is as follows:
⌊.5×(⍋⍋X)+⌽⍋⍋⌽X
The output of this code is the same as its brother, ascending cardinal numbers (ranking, all different) as long as all the values of the input vector are different. However, the shareable version doesn't assign new values for those that are equal:
X←0 0 1 0 1
⌊.5×(⍋⍋X)+⌽⍋⍋⌽X ⍝ output is 2 2 4 2 4
The values of the output should generally be interpreted as relative, i.e. The 2s have a relatively lower rank than the 4s, so they will appear first in the array.