I'm coding Settlers of Catan for iOS and currently have my data tiles in a matix. Now, I'm trying to attach the numbers to each tile. But I'm stuck as to the correct way to do this algorithmically. There are specific rules for putting down the number tiles on the board. These are:
1) They must go in the pre-defined sequence shown in the image.
2) You must start in a corner of the board.
3) You must work your way around the board counter-clockwise, spiraling toward the center.
4) You must skip the desert tile.
Since there are only 6 places on the board where you can actually start putting tiles down, I realize that it wouldn't be difficult to hand-code the 6 solutions into my data grid. But there is a variation of the game in which the board gets larger, so I figured it would be worth exploring an algorithm.
Any thoughts how this can be achieved?
What you want to do will involve the geometric relationships between tiles (entries in your matrix)
You need a way of storing and querying that information.
There are several methods people have developed for assigning coordinates to hexagons:
http://playtechs.blogspot.com/2007/04/hex-grids.html
http://jemgine.omnisu.com/?page_id=412
http://www-cs-students.stanford.edu/~amitp/Articles/GridToHex.html
One way to do what you want, off the top of my head, would be for each hex to have 6 'pointers' to its neighbors. These pointers could really be indices in your array, of course.
You could detect a 'corner' hex by noting that it will have 3 'null' neighbors. Then you can traverse counter-clockwise from there, remembering which you've already visited.
Update (in response to comment)
Let's suppose that for a given hex, we store the 6 neighbors as I described.
For this discussion, we'll name those neighbors A-F.
We assign these counter-clockwise, because that's convenient for us.
Graphically:
A
_____
B / \ F
/ \
( )
\ /
C \_____/ E
D
For a 'corner' hex, we might have:
A
_____
B / \ NULL
/ \
( )
\ /
C \_____/ NULL
NULL
So if we are looking at this hex, and we list our adjacency information (wrapping around), we have:
A, B, C, NULL, NULL, NULL, A, B, C, ...
A soon as we find 3 NULLs in a row, we know that the next spot should "point" in our starting direction. In this case we should start off in the direction of 'A'.
Another example:
NULL
_____
NULL / \ NULL
/ \
( )
\ /
C \_____/ E
D
Just as before, we make a list, starting from the top.
NULL, NULL, C, D, E, NULL, NULL, NULL, C
As soon as we find 3 NULLs, the next one is our starting direction - 'C' in this case.
Spiraling onward in the same direction from there should be fairly straightforward (well not literally - har har).
PS: thanks to http://ascii.co.uk/art/hexagon for quick ascii art (:
Related
I know, the question was often asked in the past and perhaps the information are given in previous Stack Overflow postings. But learning Forth is a very complicated task and repetition helps to understand the advantages of a concatenative programming language over alternative languages like C.
What I have learned from Forth tutorials is that Forth doesn't provide commands for creating a 2D array, but the user has to realize everything from scratch in the program. I've found two options in occupying memory in Forth. At first by creating a new word:
: namelist s” hello” s” world” ;
or secondly by the CREATE statement:
create temperature 10 allot
temperature 10 cells dump
So far so good; we have created an array of 10 cells in which integer variables can be stored. But what is, if I need to save float numbers? Do I have to convert float always to int or can they saved into the normal cells?
Another open problem is how to store string values in the array. As far as I know, a string contains a pointer plus a size. So in theory I can store only 5 strings in a 10 cell array and additionally I need memory somewhere else which holds the string itself. That doesn't make much sense.
Is there some kind of higher abstraction available to store values in arrays which can be used to write easy to read programs? I mean, if every programmer is using his own Forth method to store something in the memory, other programmers will find it hard to read the code.
create creates a word that returns address of a buffer in the dictionary (data space); it is zero length initially, so you have to reserve required space for it right away.
allot reserves space that is measured in address units (usually bytes), so you have to calculate the required size in bytes.
For example:
create a 10 cells allot
create b 10 floats allot
It is just buffers, and you still need to deal with pointers arithmetic to get or set an item, e.g.:
0.35e 2 floats b + f! \ store the float number into third item (0-based indexing)
Example of a word that creates an array of floats in the dictionary:
: create-floats-array ( length "name" -- ) create floats allot does> swap 1- floats + ;
10 create-floats-array c
0.35e 3 c f! \ store the float number into third item (1-based indexing)
3 c f# f. \ print float number form third item
If you need many arrays and many strings it is better to use appropriate libraries.
For example, see Cell array module and Dynamic text string module from Forth Foundation Library.
A generalised 2darray of elements. Takes the element size as a parameter
\ size is the per element multiplier
\ n size * is the per_row factor
\ m n size * * is the total space required
: 2darray \ create> m n size -- ; does> mix nix -- a
\ size is the number of bytes for one element
\
create 2dup * , \ multiplier to apply to m index
dup , \ multiplier to apply to n index
* * allot \ calculate total bytes and allot them
does> \ mix nix -- a
>r r# cell+ # * \ offset from n index
swap r# # * + \ offset with m index
r> + 2 cells+ \ 2 cells offset for the 'admin' cells
;
Examples
50 40 /float 2darray 50x40floats
50 40 2 cells 2darray 50x40stringpairs
even
20 constant /record
10 200 /record 2darray 2000records
You're confused about strings. The string just goes into memory and memory at that address is allocated for that string, and it's there forever (unless you change that).
So if you wanted to store 5 ( c-addr u) strings in an allocated block of memory (calling it an array is a bit of a stretch), you can just store the c-addr in cell n and the length u in cell n+1.
If you're worried about 10 cells being a lot of space (it's really nothing to worry about) and only want to use 5 cells, you can store your strings as counted strings, using words like C" - counted strings store the length in the first byte, every subsequent byte is a character.
Also, you can store things into the dictionary at the current dp using the word , (comma).
I have a set S. It contains N subsets (which in turn contain some sub-subsets of various lengths):
1. [[a,b],[c,d],[*]]
2. [[c],[d],[e,f],[*]]
3. [[d,e],[f],[f,*]]
N. ...
I also have a list L of 'unique' elements that are contained in the set S:
a, b, c, d, e, f, *
I need to find all possible combinations between each sub-subset from each subset so, that each resulting combination has exactly one element from the list L, but any number of occurrences of the element [*] (it is a wildcard element).
So, the result of the needed function working with the above mentioned set S should be (not 100% accurate):
- [a,b],[c],[d,e],[f];
- [a,b],[c],[*],[d,e],[f];
- [a,b],[c],[d,e],[f],[*];
- [a,b],[c],[d,e],[f,*],[*];
So, basically I need an algorithm that does the following:
take a sub-subset from the subset 1,
add one more sub-subset from the subset 2 maintaining the list of 'unique' elements acquired so far (the check on the 'unique' list is skipped if the sub-subset contains the * element);
Repeat 2 until N is reached.
In other words, I need to generate all possible 'chains' (it is pairs, if N == 2, and triples if N==3), but each 'chain' should contain exactly one element from the list L except the wildcard element * that can occur many times in each generated chain.
I know how to do this with N == 2 (it is a simple pair generation), but I do not know how to enhance the algorithm to work with arbitrary values for N.
Maybe Stirling numbers of the second kind could help here, but I do not know how to apply them to get the desired result.
Note: The type of data structure to be used here is not important for me.
Note: This question has grown out from my previous similar question.
These are some pointers (not a complete code) that can take you to right direction probably:
I don't think you will need some advanced data structures here (make use of erlang list comprehensions). You must also explore erlang sets and lists module. Since you are dealing with sets and list of sub-sets, they seems like an ideal fit.
Here is how things with list comprehensions will get solved easily for you: [{X,Y} || X <- [[c],[d],[e,f]], Y <- [[a,b],[c,d]]]. Here i am simply generating a list of {X,Y} 2-tuples but for your use case you will have to put real logic here (including your star case)
Further note that with list comprehensions, you can use output of one generator as input of a later generator e.g. [{X,Y} || X1 <- [[c],[d],[e,f]], X <- X1, Y1 <- [[a,b],[c,d]], Y <- Y1].
Also for removing duplicates from a list of things L = ["a", "b", "a"]., you can anytime simply do sets:to_list(sets:from_list(L)).
With above tools you can easily generate all possible chains and also enforce your logic as these chains get generated.
Please take a look at the screenshot below and see if you can tell me why this won't work. The examples in on the reference page for TextRecognize look pretty impressive, I don't think recognizing single letters like this should be a problem. I've tried resizing the letters as well as having the image sharpened.
For convenience in case you want to try this yourself I have included the image that I use at the bottom of this post. You can also find plenty more like this by searching for "Wordfeud" in Google Image Search.
Very cool question!
TextRecognize uses heuristics to recognize whole words from the English language. This is
the gotcha that makes recognizing single letters very hard
Consider the following line of thought:
s = Import["http://i.stack.imgur.com/JHYuh.png"];
p = ImagePartition[s, 32]
Now pick letters to form the English word 'EXIT':
x = {p[[1, 13]], p[[6, 6]], p[[3, 13]], p[[1, 12]]}
Now clean up these images a bit, like so:
d = ImageAssemble[ Map[ImageTake[#, {3, 27}, {2, 20}] &, x ]];
Then this returns the string "EXIT":
TextRecognize[d]
This is an approach completely different from using TextRecognize, so I am posting this as a separate answer. It uses the same image recognition technique from the How do I find Waldo with Mathematica.
First get the puzzle:
wordfeud = Import["http://i.stack.imgur.com/JHYuh.png"]
And then get the pieces of the puzzle:
Grid[pieces = ImagePartition[s, 32]]
Let's be interested in the letter E:
LetterE = pieces[[4, 3]]
Get the correlation image:
correlation =
ImageCorrelate[wordfeud, Binarize[LetterE],
NormalizedSquaredEuclideanDistance]
And highlight the matches:
positions = Dilation[ColorNegate[Binarize[correlation, .1]], DiskMatrix[20]];
found = ImageMultiply[wordfeud, ImageAdd[ColorConvert[positions, "GrayLevel"], .5]]
As before, this requires a bit of tuning on binarizing the correlation image, but other than
that this should help to identify bits and pieces of this puzzle.
I thought the quality of your image might be interfering. Binarizing your image did not help : recognition was zilch. I also tried a very sharp black and white image of a crossword puzzle solution. (see below) Again, nothing was recognized whether in regular or binarized format.
So I removed the black background leaving only the letters and their thin black frames. Again, recognition was about 0%.
When I removed the frames from around some of the letters AND binarized the image the only parts that were recognizable were those regions in which there was nothing but letters. (see below)
Notice in the output below, ANTS, TIRES, and TEXAS are correctly identified (as well as VECTORS), but just about nothing else.
Notice also that, even though the strings were widely spaced, mma interpreted them as words, rather than separate letters. Note "TEXAS" instead of "T E X A S".
TextRecognize[Binarize#img]
(* output *)
ANTS FFWWW FEEWF
E R o If IU I?
E A FI5F WWWFF 5
5552? L E F F
T s E NTT BT|
H0RWW#0WVlWF;EE F
5 W E ; OCS
FOFT W W R AL%AE
A TT I T ? _
i iE#W'NF WG%S W
A A EW F I i
SWWTW W ALTFCWD N
H A V 5 A F F
PLATT EWWLIGHT
W N E T
HE TIRES C
TEXAS VECTORS
I didn't have the patience to completely clean up the image. It would have been much faster to retype the text by hand.
Conclusion: Don't use text recognition in mma unless you have absolutely clear text against an even-colored, bright, preferrably white, background.
The results also varied depending on the file format used. Avoid .pdf altogether.
Edit
acl captured and tried to recognize the last 5 lines (above Edit). His results (in a comment below): mostly gibberish.
I decided to do the same. But since Prashant warned that text size makes a difference, I zoomed in first so that the text appear (to my eyes) to be about 20 pica. Below is the picture of the text I scanned and TextRecognized.
Here's the result of an unbinarized TextRecognize (at that large size):
Gliii. Q lk-ii`t`*¥ if EY £\[CloseCurlyDoubleQuote]1\[Euro]'EE \
Di'¥C~E\"P ITF SKI' T»f}!E'!',IL:?E\[CloseCurlyDoubleQuote] I 2 VEEE5\
\[CloseCurlyQuote] LEP \"- \"VE
1. ur e=\\..r.1.»».»\\\\ rw r 1»»\\|a'*r | r .fm -»'-an \
\[OpenCurlyQuote] -.-rr -_.»~|-.'i~-.w~,.-- nv n.w~»-\
\[OpenCurlyDoubleQuote]~"
Now, here's the result for the TextRecognize of the binarized image. The original image was a .png from Jing.
I didn't have the patience to completely clean up the image. It would \
have been much faster to retype the
text by hand.
Conclusion: Don't use text recognition in mma unless you have \
absolutely clear text against an even-
colored, bright, preferrably white, background.
The results also varied depending on the file format used. Avoid .pdf \
altogether.
I am coding a survey that outputs a .csv file. Within this csv I have some entries that are space delimited, which represent multi-select questions (e.g. questions with more than one response). In the end I want to parse these space delimited entries into their own columns and create headers for them so i know where they came from.
For example I may start with this (note that the multiselect columns have an _M after them):
Q1, Q2_M, Q3, Q4_M
6, 1 2 88, 3, 3 5 99
6, , 3, 1 2
and I want to go to this:
Q1, Q2_M_1, Q2_M_2, Q2_M_88, Q3, Q4_M_1, Q4_M_2, Q4_M_3, Q4_M_5, Q4_M_99
6, 1, 1, 1, 3, 0, 0, 1, 1, 1
6,,,,3,1,1,0,0,0
I imagine this is a relatively common issue to deal with but I have not been able to find it in the R section. Any ideas how to do this in R after importing the .csv ? My general thoughts (which often lead to inefficient programs) are that I can:
(1) pull column numbers that have the special suffix with grep()
(2) loop through (or use an apply) each of the entries in these columns and determine the levels of responses and then create columns accordingly
(3) loop through (or use an apply) and place indicators in appropriate columns to indicate presence of selection
I appreciate any help and please let me know if this is not clear.
I agree with ran2 and aL3Xa that you probably want to change the format of your data to have a different column for each possible reponse. However, if you munging your dataset to a better format proves problematic, it is possible to do what you asked.
process_multichoice <- function(x) lapply(strsplit(x, " "), as.numeric)
q2 <- c("1 2 3 NA 4", "2 5")
processed_q2 <- process_multichoice(q2)
[[1]]
[1] 1 2 3 NA 4
[[2]]
[1] 2 5
The reason different columns for different responses are suggested is because it is still quite unpleasant trying to retrieve any statistics from the data in this form. Although you can do things like
# Number of reponses given
sapply(processed_q2, length)
#Frequency of each response
table(unlist(processed_q2), useNA = "ifany")
EDIT: One more piece of advice. Keep the code that processes your data separate from the code that analyses it. If you create any graphs, keep the code for creating them separate again. I've been down the road of mixing things together, and it isn't pretty. (Especially when you come back to the code six months later.)
I am not entirely sure what you trying to do respectively what your reasons are for coding like this. Thus my advice is more general – so just feel to clarify and I will try to give a more concrete response.
1) I say that you are coding the survey on your own, which is great because it means you have influence on your .csv file. I would NEVER use different kinds of separation in the same .csv file. Just do the naming from the very beginning, just like you suggested in the second block.
Otherwise you might geht into trouble with checkboxes for example. Let's say someone checks 3 out of 5 possible answers, the next only checks 1 (i.e. "don't know") . Now it will be much harder to create a spreadsheet (data.frame) type of results view as opposed to having an empty field (which turns out to be an NA in R) that only needs to be recoded.
2) Another important question is whether you intend to do a panel survey(i.e longitudinal study asking the same participants over and over again) . That (among many others) would be a good reason to think about saving your data to a MySQL database instead of .csv . RMySQL can connect directly to the database and access its tables and more important its VIEWS.
Views really help with survey data since you can rearrange the data in different views, conditional on many different needs.
3) Besides all the personal / opinion and experience, here's some (less biased) literature to get started:
Complex Surveys: A Guide to Analysis Using R (Wiley Series in Survey Methodology
The book is comparatively simple and leaves out panel surveys but gives a lot of R Code and examples which should be a practical start.
To prevent re-inventing the wheel you might want to check LimeSurvey, a pretty decent (not speaking of the templates :) ) tool for survey conductors. Besides I TYPO3 CMS extensions pbsurvey and ke_questionnaire (should) work well too (only tested pbsurvey).
Multiple choice items should always be coded as separate variables. That is, if you have 5 alternatives and multiple choice, you should code them as i1, i2, i3, i4, i5, i.e. each one is a binary variable (0-1). I see that you have values 3 5 99 for Q4_M variable in the first example. Does that mean that you have 99 alternatives in an item? Ouch...
First you should go on and create separate variables for each alternative in a multiple choice item. That is, do:
# note that I follow your example with Q4_M variable
dtf_ins <- as.data.frame(matrix(0, nrow = nrow(<initial dataframe>), ncol = 99))
# name vars appropriately
names(dtf_ins) <- paste("Q4_M_", 1:99, sep = "")
now you have a data.frame with 0s, so what you need to do is to get 1s in an appropriate position (this is a bit cumbersome), a function will do the job...
# first you gotta change spaces to commas and convert character variable to a numeric one
y <- paste("c(", gsub(" ", ", ", x), ")", sep = "")
z <- eval(parse(text = y))
# now you assing 1 according to indexes in z variable
dtf_ins[1, z] <- 1
And that's pretty much it... basically, you would like to reconsider creating a data.frame with _M variables, so you can write a function that does this insertion automatically. Avoid for loops!
Or, even better, create a matrix with logicals, and just do dtf[m] <- 1, where dtf is your multiple-choice data.frame, and m is matrix with logicals.
I would like to help you more on this one, but I'm recuperating after a looong night! =) Hope that I've helped a bit! =)
Thanks for all the responses. I agree with most of you that this format is kind of silly but it is what I have to work with (survey is coded and going into use next week). This is what I came up with from all the responses. I am sure this is not the most elegant or efficient way to do it but I think it should work.
colnums <- grep("_M",colnames(dat))
responses <- nrow(dat)
for (i in colnums) {
vec <- as.vector(dat[,i]) #turn into vector
b <- lapply(strsplit(vec," "),as.numeric) #split up and turn into numeric
c <- sort(unique(unlist(b))) #which values were used
newcolnames <- paste(colnames(dat[i]),"_",c,sep="") #column names
e <- matrix(nrow=responses,ncol=length(c)) #create new matrix for indicators
colnames(e) <- newcolnames
#next loop looks for responses and puts indicators in the correct places
for (i in 1:responses) {
e[i,] <- ifelse(c %in% b[[i]],1,0)
}
dat <- cbind(dat,e)
}
Suggestions for improvement are welcome.
Is there any input that SHA-1 will compute to a hex value of fourty-zeros, i.e. "0000000000000000000000000000000000000000"?
Yes, it's just incredibly unlikely. I.e. one in 2^160, or 0.00000000000000000000000000000000000000000000006842277657836021%.
Also, becuase SHA1 is cryptographically strong, it would also be computationally unfeasible (at least with current computer technology -- all bets are off for emergent technologies such as quantum computing) to find out what data would result in an all-zero hash until it occurred in practice. If you really must use the "0" hash as a sentinel be sure to include an appropriate assertion (that you did not just hash input data to your "zero" hash sentinel) that survives into production. It is a failure condition your code will permanently need to check for. WARNING: Your code will permanently be broken if it does.
Depending on your situation (if your logic can cope with handling the empty string as a special case in order to forbid it from input) you could use the SHA1 hash ('da39a3ee5e6b4b0d3255bfef95601890afd80709') of the empty string. Also possible is using the hash for any string not in your input domain such as sha1('a') if your input has numeric-only as an invariant. If the input is preprocessed to add any regular decoration then a hash of something without the decoration would work as well (eg: sha1('abc') if your inputs like 'foo' are decorated with quotes to something like '"foo"').
I don't think so.
There is no easy way to show why it's not possible. If there was, then this would itself be the basis of an algorithm to find collisions.
Longer analysis:
The preprocessing makes sure that there is always at least one 1 bit in the input.
The loop over w[i] will leave the original stream alone, so there is at least one 1 bit in the input (words 0 to 15). Even with clever design of the bit patterns, at least some of the values from 0 to 15 must be non-zero since the loop doesn't affect them.
Note: leftrotate is circular, so no 1 bits will get lost.
In the main loop, it's easy to see that the factor k is never zero, so temp can't be zero for the reason that all operands on the right hand side are zero (k never is).
This leaves us with the question whether you can create a bit pattern for which (a leftrotate 5) + f + e + k + w[i] returns 0 by overflowing the sum. For this, we need to find values for w[i] such that w[i] = 0 - ((a leftrotate 5) + f + e + k)
This is possible for the first 16 values of w[i] since you have full control over them. But the words 16 to 79 are again created by xoring the first 16 values.
So the next step could be to unroll the loops and create a system of linear equations. I'll leave that as an exercise to the reader ;-) The system is interesting since we have a loop that creates additional equations until we end up with a stable result.
Basically, the algorithm was chosen in such a way that you can create individual 0 words by selecting input patterns but these effects are countered by xoring the input patterns to create the 64 other inputs.
Just an example: To make temp 0, we have
a = h0 = 0x67452301
f = (b and c) or ((not b) and d)
= (h1 and h2) or ((not h1) and h3)
= (0xEFCDAB89 & 0x98BADCFE) | (~0x98BADCFE & 0x10325476)
= 0x98badcfe
e = 0xC3D2E1F0
k = 0x5A827999
which gives us w[0] = 0x9fb498b3, etc. This value is then used in the words 16, 19, 22, 24-25, 27-28, 30-79.
Word 1, similarly, is used in words 1, 17, 20, 23, 25-26, 28-29, 31-79.
As you can see, there is a lot of overlap. If you calculate the input value that would give you a 0 result, that value influences at last 32 other input values.
The post by Aaron is incorrect. It is getting hung up on the internals of the SHA1 computation while ignoring what happens at the end of the round function.
Specifically, see the pseudo-code from Wikipedia. At the end of the round, the following computation is done:
h0 = h0 + a
h1 = h1 + b
h2 = h2 + c
h3 = h3 + d
h4 = h4 + e
So an all 0 output can happen if h0 == -a, h1 == -b, h2 == -c, h3 == -d, and h4 == -e going into this last section, where the computations are mod 2^32.
To answer your question: nobody knows whether there exists an input that produces all zero outputs, but cryptographers expect that there are based upon the simple argument provided by daf.
Without any knowledge of SHA-1 internals, I don't see why any particular value should be impossible (unless explicitly stated in the description of the algorithm). An all-zero value is no more or less probable than any other specific value.
Contrary to all of the current answers here, nobody knows that. There's a big difference between a probability estimation and a proof.
But you can safely assume it won't happen. In fact, you can safely assume that just about ANY value won't be the result (assuming it wasn't obtained through some SHA-1-like procedures). You can assume this as long as SHA-1 is secure (it actually isn't anymore, at least theoretically).
People doesn't seem realize just how improbable it is (if all humanity focused all of it's current resources on finding a zero hash by bruteforcing, it would take about xxx... ages of the current universe to crack it).
If you know the function is safe, it's not wrong to assume it won't happen. That may change in the future, so assume some malicious inputs could give that value (e.g. don't erase user's HDD if you find a zero hash).
If anyone still thinks it's not "clean" or something, I can tell you that nothing is guaranteed in the real world, because of quantum mechanics. You assume you can't walk through a solid wall just because of an insanely low probability.
[I'm done with this site... My first answer here, I tried to write a nice answer, but all I see is a bunch of downvoting morons who are wrong and can't even tell the reason why are they doing it. Your community really disappointed me. I'll still use this site, but only passively]
Contrary to all answers here, the answer is simply No.
The hash value always contains bits set to 1.