I'm learning a spread of programming languages in a class, and we're working on an APLX project at the moment. A restriction we have to work around is we cannot use If, For, While, etc. No loops or conditionals. I have to be able to take a plane of numbers, ranging 0-7, and replace each number 2 or greater into the depth of that number, and, ideally, change the 1's to 0's. For example:
0100230 => 0000560
I have no idea how I'm supposed to do the replacement with depth aspect, though the change from ones to zeros is quite simple. I'm able to produce the set of integers in a table and I understand how to replace specific values, but only with other specific values, not values that would have to be determined during the function. The depth should be the row depth, rather than the multi-dimensional depth.
For the record this is not the whole of the program, the program itself is a poker dealing and scoring program. This is a specific aspect of the scoring methodology that my professor recommended I use.
TOTALS„SCORE PHAND;TYPECOUNT;DEPTH;ISCOUNT;TEMPS;REPLACE
:If (½½PHAND) = 0
PHAND„DEAL PHAND
:EndIf
TYPECOUNT„CHARS°.¹PHAND
DEPTH„2Þ(½TYPECOUNT)
REPLACE „ 2 3 4 5 6 7
ISCOUNT „ +/ TYPECOUNT
ISCOUNT „ ³ISCOUNT
((1=,ISCOUNT)/,ISCOUNT)„0
©((2=,ISCOUNT)/,ISCOUNT)„1
©TEMPS „ ISCOUNT
Œ„ISCOUNT
Œ„PHAND
You may have missed the first lessons of your prof and it might help to look at at again to learn about vectors and how easy you can work with them - once you unlearned the ideas of other programming languages ;-)
Assume you have a vector A with numbers from 1 to 7:
A←⍳7
A
1 2 3 4 5 6 7
Now, if you wanted to search for values > 3, you'd do:
A>3
0 0 0 1 1 1 1
The result is a vector, too, and you can easily combine the two in lots of operations:
multiplication to only keep values > 0 and replace others with 0:
A×A>3
0 0 0 4 5 6 7
or add 500 to values >3
A+500×A>3
1 2 3 504 505 506 507
or, find the indices of values > 3:
(A>3)×⍳⍴A
0 0 0 4 5 6 7
Now, looking at your q again, the word 'depth' has a specific meaning in APL and I guess you meant something different. Do I understand correctly that you want to replace values > 2 with the ' indices' of these values?
Well, with what I've shown before, this is easy:
A←0 1 0 0 2 3 0
(A≥2)×⍳⍴A
0 0 0 0 5 6 0
edit: looking at multi-dimensional arrays:
let's look into this example:
A←(⍳5)∘.×⍳10
A
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
Now, which numbers are > 20 and < 30?
z←(A>20)∧A<30
z
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 1 1 0
0 0 0 0 0 1 1 0 0 0
0 0 0 0 1 0 0 0 0 0
Then, you can multiply the values with that boolean result to filter out only the ones satisfying the condition:
A×z
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 21 24 27 0
0 0 0 0 0 24 28 0 0 0
0 0 0 0 25 0 0 0 0 0
Or, perhaps you're interested in the column-index of the values?
z×[2]⍳¯1↑⍴z
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 7 8 9 0
0 0 0 0 0 6 7 0 0 0
0 0 0 0 5 0 0 0 0 0
NB: this statement might not work in all APL-dialects. Here's another way to formulate this:
z×((1↑⍴z)⍴0)∘.+⍳¯1↑⍴z
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 7 8 9 0
0 0 0 0 0 6 7 0 0 0
0 0 0 0 5 0 0 0 0 0
I hope this gives you some ideas to play with. In general, using booleans to manipulate arrays in mathematical operations is an extremely powerful idea in APL which will take you loooooong ways ;-)
Also, if you'd like to see more of the same, have a look at the FinnAPL Idioms - some useful shorties grown over the years ;-)
edit re. "maintaining untouched values":
going back to example array A:
A←(⍳5)∘.×⍳10
A
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
Replacing values between 20 and 30 with the power 2 of these values, keeping all others unchanged:
touch←(A>20)∧A<30
(touch×A*2)+A×~touch
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 441 576 729 30
4 8 12 16 20 576 784 32 36 40
5 10 15 20 625 30 35 40 45 50
I hope you get the idea...
Or better: ask a new q, as otherwise this would truly take epic dimensions, whereas the idea of stackoverflow is more like "one issue - one question"...
Related
So I have a game board represented by a 1d array of 64 squares and I wish to know when a piece is trying to escape the borders.
So for example:
Assume a king in the game of chess is on the X:
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 X 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
it is easy to calculate the attacked squares with a direction vector:
-9, -8, -7,
-1, 1,
7, 8, 9
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1 1 1 0 0
0 0 0 1 X 1 0 0
0 0 0 1 1 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
also if he is on the upper or lower edges of the board:
I just add an if that checks if the square + directionvector[i] > 63 or
square + directionvector[i] < 0
so:
0 0 0 1 X 1 0 0
0 0 0 1 1 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
The problem then comes when it sits on the side edges:
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 1
1 0 0 0 0 0 1 X
1 0 0 0 0 0 1 1
1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
This is what will happen it will "jump" to the wrong place as you see above.
Do you know of any way of checking this?
thanks
At the beginning of your program, you can precompute the legal moves for every piece and every square and store them somewhere.
Assuming that your squares are numbered from 0 to 63, then KING_MOVES[i] will store the legal moves for a king on square i (where i ranges from 0 to 63). For example, KING_MOVES[0] = { 1, 8, 9 } and KING_MOVES[31] = { 22, 23, 30, 38, 39 }. Similarly, for a knight in the corner, KNIGHT_MOVES[0] = { 10, 17 }.
0 1 2 3 4 5 6 7
8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23
24 25 26 27 28 29 30 31
32 33 34 35 36 37 38 39
40 41 42 43 44 45 46 47
48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63
The number of legal moves is different from every square (a king has 8 moves in the center, but only 3 in the corner). You can store the number of legal moves separately (such as NUM_KING_MOVES[31] = 5) or you can terminate each list with a special value like -1 or 64.
To precompute these lists, you can convert each 1-D number to a pair of (row, column) coordinates and generate moves naively, with boundary checks. For a king on square 31, you convert 31 to (row 3, column 7). When moving to the right, you end up on (row 3, column 8), which is illegal, so you discard this move. When moving to the top-left, you end up on (row 2, column 6), which is legal, so you convert that back to a 1-D representation: 2 * 8 + 6 = 22. You append this move to KING_MOVES[31] and so forth.
This gives you the best of both worlds: faster and shorter code for move generation with no wasted memory on every board due to padding. The downside is a few kilobytes of extra memory.
If you're going down this path, you could also take a look at bitboards for even faster move generation.
You could get the row number from the cell index as i / 8 (floored), and use that to check if e.g. a step to the right or left passed to another row.
E.g. here, on a four-column board, stepping to the "right" from position 7, we go to position 8. The row numbers for these are 7 / 4 = 1 and 8 / 4 = 2, so we changed to another row.
....
...7
8...
....
But that's harder to do for diagonal moves, because you expect them to change the row. In the end, this is pretty much just turning the board into a 2D array, and the step offsets into 2D values, where you can check for overflow in the obvious way.
IMO, in principle what you're asking is impossible for a 1D board. With the board defined as a 1D vector (and similarly, the offsets as 1D offsets), then there are no edges within the board, and nothing to escape. Instead of this:
.ab.
....
...c
d...
you really just have this:
.ab........cd...
And it's obvious that c and d are just as adjacent to each other as a and b are.
actual problem is like this which I got from an Online competition. I solved it but my solution, which is in C, couldn't produce answer in time for large numbers. I need to solve it in C.
Given below is a word from the English dictionary arranged as a matrix:
MATHE
ATHEM
THEMA
HEMAT
EMATI
MATIC
ATICS
Tracing the matrix is starting from the top left position and at each step move either RIGHT or DOWN, to reach the bottom right of the matrix. It is assured that any such tracing generates the same word. How many such tracings can be possible for a given word of length m+n-1 written as a matrix of size m * n?
1 ≤ m,n ≤ 10^6
I have to print the number of ways S the word can be traced as explained in the problem statement. If the number is larger than 10^9+7, I have to print S mod (10^9 + 7).
In the testcases, m and n can be very large.
Imagine traversing the matrix, whatever path you choose you need to take exatcly n+m-2 steps to make the word, among of which n-1 are down and m-1 are to the right, their order may change but the numbers n-1 and m-1 remain same. So the problem got reduced to only select n-1 positions out of n+m-2, so the answer is
C(n+m-2,n-1)=C(n+m-2,m-1)
How to calculate C(n,r) for this problem:
You must be knowing how to multiply two numbers in modular arithmetics, i.e.
(a*b)%mod=(a%mod*b%mod)%mod,
now to calculate C(n,r) you also need to divide, but division in modular arithmetic can be performed by using modular multiplicative inverse of the number i.e.
((a)*(a^-1))%mod=1
Ofcourse a^-1 in modular arithmetic need not equal to 1/a, and can be computed using Extended Euclidean Algorithm, as in your case mod is a prime number therefore
(a^(-1))=a^(mod-2)%mod
a^(mod-2) can be computed efficiently using repetitive squaring method.
I would suggest a dynamic programming approach for this problem since calculation of factorials of large numbers shall involve a lot of time, especially since you have multiple queries.
Starting from a small matrix (say 2x1), keep finding solutions for bigger matrices. Note that this solution works since in finding the solution for bigger matrix, you can use the value calculated for smaller matrices and speed up your calculation.
The complexity of the above soltion IMO is polynomial in M and N for an MxN matrix.
Use Laplace's triangle, incorrectly named also "binomial"
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 1 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 1 1 0 0
1 2 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 1 1 1 0
1 2 3 0 0
1 3 0 0 0
1 0 0 0 0
0 0 0 0 0
1 1 1 1 1
1 2 3 4 0
1 3 6 0 0
1 4 0 0 0
1 0 0 0 0
1 1 1 1 1
1 2 3 4 5
1 3 6 10 0
1 4 10 0 0
1 5 0 0 0
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 0
1 5 15 0 0
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 35
1 5 15 35 0
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 35
1 5 15 35 70
Got it? Notice, that elements could be counted as binomial members. The diag members are here: C^1_2, C^2_4,C^3_6,C^4_8, and so on. Choose which you need.
I'm working on J.
How can I convert this matrix:
(i.10)*/(i.10)
0 0 0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7 8 9
0 2 4 6 8 10 12 14 16 18
0 3 6 9 12 15 18 21 24 27
0 4 8 12 16 20 24 28 32 36
0 5 10 15 20 25 30 35 40 45
0 6 12 18 24 30 36 42 48 54
0 7 14 21 28 35 42 49 56 63
0 8 16 24 32 40 48 56 64 72
0 9 18 27 36 45 54 63 72 81
in array?
0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 . . .
I tried
(i.10)*/(i.10)"0
and then I've added
~.(i.10)*/(i.10)"0
to eliminate doubles, but it doesn't work.
If you want to turn a 2-dimensional table (matrix) into a 1-dimensional list (vector or "array", though in the J world "array" usually means "rectangle with any number [N] of dimensions"), you can use ravel (,):
matrix =: (i.10)*/(i.10)
list =: , matrix
list
0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 ...
Now using nub (~.) to remove duplicates should work:
~. list
0 1 2 3 4 5 6 7 8 9 10 12 ...
Note that, in J, the shape of an array usually carries important information, so flattening a matrix like this would be fairly unusual. Still, nothing stopping you.
BTW, you can save yourself some keystrokes by using the adverb ~, which will copy the left argument of a dyad to the right side as well, so you could just say:
matrix =: */~ i. 10
and get the same result as (i.10) */ (i.10).
I am working on a minigame called 'Pogo Painter', and I need some mathematical solutions. Below is an image (made with Paint) to illustrate a bit what it's all about.
Four players, each of different color, must claim squares to gain points. The minigame will be similar to this: http://www.youtube.com/watch?v=rKCQfAlaRrc, but slightly different. The players will be allowed to run around the playground and claim any of the squares, and points are gathered when a pattern is closed. For example, claiming blue square on A3 will create a closed blue pattern.
What kind of variables should I declare and how do I check if the pattern is closed?
Please answer if you have a solution :)
Here’s another (Discrete Optimization) way to model your problem.
Notation
View your grid as a ‘graph’ with n^2 nodes, and edges of length 1 (Edges connect two neighboring nodes.) Let the nodes be numbered 1:n^2. (For ease of notation, you can use a double array (x,y) to denote each node if you prefer.)
Decision Variables
There are k colors, one for each player (1 through 4). 0 is an unclaimed cell (white)
X_ik = 1 if player k has claimed node i. 0 otherwise.
To start out
X_i0 = 1 for all nodes i.
All nodes start out as white (0).
Neighboring sets: Two nodes i and j are ‘neighbors’ if they are adjacent to each other. (Any given node i can have at most 4 neighbors: Up down right and left.)
Edge variables:
We can now define a new set of edge variables Y_ijk that connect two adjacent nodes (i and j) with a common color k.
Y_ijk = 1 if neighboring nodes i and j are both of color k. 0 Otherwise.
(That is, X_ik = X_jk) for non-zero k.
We now have an undirected graph. Checking for ‘closed patterns’ is the same as detecting cycles.
Detecting Cycles:
A simple DFS search will do, since we have undirected cycles. Start with each colored node i, and check for cycles. If a path leads you back to a visited node, cycles exist. You can award points accordingly.
Finally, one suggestion as you design the game. You can reward points according to the “longest cycle” you detect. The shortest cycle gets 4 points, one point for each edge (or one point for each node in the cycle) whichever works best for you.
1 1
1 1 scores 4 points
1 1 1
1 1 1 scores 6 points
1 1 1
1 1 1
1 1 scores 8 points
Hope that helps.
Okay,
This is plenty of text, but it's simple.
An N-by-N square will satisfy as the game-board.
Each time a player claims a square,
If the square is not attached to any square of that player, then you must give that square a unique ID.
If the square is attached,
Count how many neighbours of each ID it has.
( See the demos I put below, to see what this means)
For each group
patterns_count += group_size - 1
If the number of groups is more than 1
Change the ID of that group as well as every other square connected to it so they all share the same ID
You must remember which IDs belong to which players.
This is what you have in your example
1 1 1 0 0 0 0 2 2
1 0 0 0 1 3 3 0 0
1 1 0 0 3 3 0 0 0
0 1 0 0 4 5 0 0 0
0 0 0 6 4 0 0 0 0
7 7 0 0 0 0 8 8 8
0 7 7 0 9 8 8 0 8
A A 7 0 9 8 0 0 8
A 0 7 0 0 0 8 8 8
And this is what it would turn out like after blue grabs A-3
1 1 1 0 0 0 0 2 2
1 0 0 0 1 3 3 0 0
1 1 0 0 3 3 0 0 0
0 1 0 0 4 5 0 0 0
0 0 0 6 4 0 0 0 0
7 7 0 0 0 0 8 8 8
0 7 7 0 9 8 8 0 8
A A 7 0 9 8 0 0 8
A 0 7 0 0 8 8 8 8
More examples of the algorithm in use
1 1 1 0
1 0 1 0
1 1 0
0 0 0 0
2 neighbours. 2x'1'
1x closed pattern.
1 1 1 0
1 0 1 0
1 1 1 0
0 0 0 0
--
1 1 1 0 0
1 0 1 0 0
1 1 0 0
1 0 1 0 0
1 1 1 0 0
3 neighbours: 3x'1'
2x closed patterns
1 1 1 0 0
1 0 1 0 0
1 1 1 0 0
1 0 1 0 0
1 1 1 0 0
--
1 1 1 0 0
1 0 1 0 0
1 1 2 2
0 0 2 0 2
0 0 2 2 2
4 neighbours: 2x'1', 2x'2'
2 Closed patterns
1 1 1 0 0
1 0 1 0 0
1 1 1 1 1
0 0 1 0 1
0 0 1 1 1
But I also consider these a closed pattern. You haven't given any description as to what should be considered one and what shouldn't be.
1 1 0
1 1 0
0 0 0
1 1 1
1 1 1
0 0 0
1 1 1
1 1 1
1 1
I'm analysing a dataset with some data-mining tools.The response variable has ten levels and I'm trying to create a classifier.
Here comes the problem.When using nnet and bagging function,the result is not that good and the 5th level is even not in the prediction.
I want to use a confusion matrix to analyse the classifier.but as the 5th level is not shown in the prediction I can't get a well-formed matrix.So how can I get a well-formed matrix?i.e. I want a 10*10 matrix.
The confusion matrix:
library("mda")#This is where **confusion** comes from
> confusion(pre.bag$class,CLASS)#here confusion acts like table
true
predicted 1 2 3 4 6 7 8 9 10 5
1 338 9 6 0 5 12 10 1 15 46
2 9 549 1 59 18 0 3 0 0 6
3 18 1 44 0 0 0 2 0 0 4
4 0 1 0 21 0 0 0 0 0 0
6 2 13 0 1 299 2 9 0 0 0
7 5 2 1 0 10 231 6 0 1 0
8 0 0 0 0 0 5 76 0 0 0
9 5 1 0 0 0 0 0 62 0 0
10 7 3 1 0 0 2 1 6 181 16
attr(,"error")
[1] 0.1231743
attr(,"mismatch")
[1] 0.03386642
Try this:
pred <- factor(pre.bag$class, levels=levels(CLASS) )
confusion(pre.bag$class, CLASS)
(Tested with an fda-object.)