Let's assume that we have the list of loans user has like below:
loan1
loan2
loan3
...
loan10
And we have the function which can accept from 2 to 10 loans:
function(loans).
For ex., the following is possible:
function(loan1, loan2)
function(loan1, loan3)
function(loan1, loan4)
function(loan1, loan2, loan3)
function(loan1, loan2, loan4)
function(loan1, loan2, loan3, loan4, loan5, loan6, loan7, loan8, loan9, loan10)
How to write the code to pass all possible combinations to that function?
On RosettaCode you have implemented generating combinations in many languages, choose yourself.
Here's how we could do it in ruby :
loans= ['loan1','loan2', ... , 'loan10']
def my_function(loans)
array_of_loan_combinations = (0..arr.length).to_a.combination(2).map{|i,j| arr[i...j]}
array_of_loan_combinations.each do |combination|
//do something
end
end
To call :
my_function(loans);
I have written a class to handle common functions for working with the binomial coefficient, which is the type of problem that your problem falls under. It performs the following tasks:
Outputs all the K-indexes in a nice format for any N choose K to a file. The K-indexes can be substituted with more descriptive strings or letters. This method makes solving this type of problem quite trivial.
Converts the K-indexes to the proper index of an entry in the sorted binomial coefficient table. This technique is much faster than older published techniques that rely on iteration. It does this by using a mathematical property inherent in Pascal's Triangle. My paper talks about this. I believe I am the first to discover and publish this technique, but I could be wrong.
Converts the index in a sorted binomial coefficient table to the corresponding K-indexes. I believe it might be faster than the link you have found.
Uses Mark Dominus method to calculate the binomial coefficient, which is much less likely to overflow and works with larger numbers.
The class is written in .NET C# and provides a way to manage the objects related to the problem (if any) by using a generic list. The constructor of this class takes a bool value called InitTable that when true will create a generic list to hold the objects to be managed. If this value is false, then it will not create the table. The table does not need to be created in order to perform the 4 above methods. Accessor methods are provided to access the table.
There is an associated test class which shows how to use the class and its methods. It has been extensively tested with 2 cases and there are no known bugs.
To read about this class and download the code, see Tablizing The Binomial Coeffieicent.
It should not be hard to convert this class to the language of your choice.
To solve your problem, you might want to write a new loans function that takes as input an array of loan objects and works on those objects with the BinCoeff class. In C#, to obtain the array of loans for each unique combination, something like the following example code could be used:
void LoanCombinations(Loan[] Loans)
{
// The Loans array contains all of the loan objects that need
// to be handled.
int LoansCount = Loans.Length;
// Loop though all possible combinations of loan objects.
// Start with 2 loan objects, then 3, 4, and so forth.
for (int N = 2; N <= N; N++)
{
// Loop thru all the possible groups of combinations.
for (int K = N - 1; K < N; K++)
{
// Create the bin coeff object required to get all
// the combos for this N choose K combination.
BinCoeff<int> BC = new BinCoeff<int>(N, K, false);
int NumCombos = BinCoeff<int>.GetBinCoeff(N, K);
int[] KIndexes = new int[K];
// Loop thru all the combinations for this N choose K.
for (int Combo = 0; Combo < NumCombos; Combo++)
{
// Get the k-indexes for this combination, which in this case
// are the indexes to each loan in Loans.
BC.GetKIndexes(Loop, KIndexes);
// Create a new array of Loan objects that correspond to
// this combination group.
Loan[] ComboLoans = new Loan[K];
for (int Loop = 0; Loop < K; Loop++)
ComboLoans[Loop] = Loans[KIndexes[Loop]];
// Call the ProcessLoans function with the loans to be processed.
ProcessLoans(ComboLoans);
}
}
}
}
I have not tested the above code, but in general it should solve your problem.
Related
I have some extremely old legacy procedural code which takes 10 or so enumerated inputs [ i0, i1, i2, ... i9 ] and generates 170 odd enumerated outputs [ r0, r1, ... r168, r169 ]. By enumerated, I mean that each individual input & output has its own set of distinct value sets e.g. [ red, green, yellow ] or [ yes, no ] etc.
I’m putting together the entire state table using the existing code, and instead of puzzling through them by hand, I was wondering if there was an algorithmic way of determining an appropriate function to get to each result from the 10 inputs. Note, not all input columns may be required to determine an individual output column, i.e. r124 might only be dependent on i5, i6 and i9.
These are not continuous functions, and I expect I might end up with some sort of hashing function approach, but I wondered if anyone knew of a more repeatable process I should be using instead? (If only there was some Karnaugh map like approach for multiple value non-binary functions ;-) )
If you are willing to actually enumerate all possible input/output sequences, here is a theoretical approach to tackle this that should be fairly effective.
First, consider the entropy of the output. Suppose that you have n possible input sequences, and x[i] is the number of ways to get i as an output. Let p[i] = float(x[i])/float(n[i]) and then the entropy is - sum(p[i] * log(p[i]) for i in outputs). (Note, since p[i] < 1 the log(p[i]) is a negative number, and therefore the entropy is positive. Also note, if p[i] = 0 then we assume that p[i] * log(p[i]) is also zero.)
The amount of entropy can be thought of as the amount of information needed to predict the outcome.
Now here is the key question. What variable gives us the most information about the output per information about the input?
If a particular variable v has in[v] possible values, the amount of information in specifying v is log(float(in[v])). I already described how to calculate the entropy of the entire set of outputs. For each possible value of v we can calculate the entropy of the entire set of outputs for that value of v. The amount of information given by knowing v is the entropy of the total set minus the average of the entropies for the individual values of v.
Pick the variable v which gives you the best ratio of information_gained_from_v/information_to_specify_v. Your algorithm will start with a switch on the set of values of that variable.
Then for each value, you repeat this process to get cascading nested if conditions.
This will generally lead to a fairly compact set of cascading nested if conditions that will focus on the input variables that tell you as much as possible, as quickly as possible, with as few branches as you can manage.
Now this assumed that you had a comprehensive enumeration. But what if you don't?
The answer to that is that the analysis that I described can be done for a random sample of your possible set of inputs. So if you run your code with, say, 10,000 random inputs, then you'll come up with fairly good entropies for your first level. Repeat with 10,000 each of your branches on your second level, and the same will happen. Continue as long as it is computationally feasible.
If there are good patterns to find, you will quickly find a lot of patterns of the form, "If you put in this that and the other, here is the output you always get." If there is a reasonably short set of nested ifs that give the right output, you're probably going to find it. After that, you have the question of deciding whether to actually verify by hand that each bucket is reliable, or to trust that if you couldn't find any exceptions with 10,000 random inputs, then there are none to be found.
Tricky approach for the validation. If you can find fuzzing software written for your language, run the fuzzing software with the goal of trying to tease out every possible internal execution path for each bucket you find. If the fuzzing software decides that you can't get different answers than the one you think is best from the above approach, then you can probably trust it.
Algorithm is pretty straightforward. Given possible values for each input we can generate all the input vectors possible. Then per each output we can just eliminate these inputs that do no matter for the output. As the result we for each output we can get a matrix showing output values for all the input combinations excluding the inputs that do not matter for given output.
Sample input format (for code snipped below):
var schema = new ConvertionSchema()
{
InputPossibleValues = new object[][]
{
new object[] { 1, 2, 3, }, // input #0
new object[] { 'a', 'b', 'c' }, // input #1
new object[] { "foo", "bar" }, // input #2
},
Converters = new System.Func<object[], object>[]
{
input => input[0], // output #0
input => (int)input[0] + (int)(char)input[1], // output #1
input => (string)input[2] == "foo" ? 1 : 42, // output #2
input => input[2].ToString() + input[1].ToString(), // output #3
input => (int)input[0] % 2, // output #4
}
};
Sample output:
Leaving the heart of the backward conversion below. Full code in a form of Linqpad snippet is there: http://share.linqpad.net/cknrte.linq.
public void Reverse(ConvertionSchema schema)
{
// generate all possible input vectors and record the resul for each case
// then for each output we could figure out which inputs matters
object[][] inputs = schema.GenerateInputVectors();
// reversal path
for (int outputIdx = 0; outputIdx < schema.OutputsCount; outputIdx++)
{
List<int> inputsThatDoNotMatter = new List<int>();
for (int inputIdx = 0; inputIdx < schema.InputsCount; inputIdx++)
{
// find all groups for input vectors where all other inputs (excluding current) are the same
// if across these groups outputs are exactly the same, then it means that current input
// does not matter for given output
bool inputMatters = inputs.GroupBy(input => ExcudeByIndexes(input, new[] { inputIdx }), input => schema.Convert(input)[outputIdx], ObjectsByValuesComparer.Instance)
.Where(x => x.Distinct().Count() > 1)
.Any();
if (!inputMatters)
{
inputsThatDoNotMatter.Add(inputIdx);
Util.Metatext($"Input #{inputIdx} does not matter for output #{outputIdx}").Dump();
}
}
// mapping table (only inputs that matters)
var mapping = new List<dynamic>();
foreach (var inputGroup in inputs.GroupBy(input => ExcudeByIndexes(input, inputsThatDoNotMatter), ObjectsByValuesComparer.Instance))
{
dynamic record = new ExpandoObject();
object[] sampleInput = inputGroup.First();
object output = schema.Convert(sampleInput)[outputIdx];
for (int inputIdx = 0; inputIdx < schema.InputsCount; inputIdx++)
{
if (inputsThatDoNotMatter.Contains(inputIdx))
continue;
AddProperty(record, $"Input #{inputIdx}", sampleInput[inputIdx]);
}
AddProperty(record, $"Output #{outputIdx}", output);
mapping.Add(record);
}
// input x, ..., input y, output z form is needed
mapping.Dump();
}
}
I am trying to solve a simple problem, but at the moment I cannot think of a better solution. I am testing an API that is not documented.
There is an ID used to fetch objects and it has a min and max value with random values missing in-between. I'm trying to test the responses I receive for random objects, but to find objects, I need to have valid IDs.
It would be very inefficient to test random numbers and hope that I get an object back. The best I can do is find a range, get a random number between that range and check if it exists before conducting tests.
A sample list of all of the IDs in the database might look like this:
[1005, 25984, 25986, 29587, 30000, ...]
Assuming the deviation from one value to another will never exceed C, e.g. from the first value to the next value, the difference will never be greater than a pre-defined constant, how would you calculate the min/max of the range given only one value in the range?
Starting from a given value and looping until the last value is found is horrible but that is how it was implemented by previous devs. Below is pseudocode that more or less covers what they do.
// this can be any valid object ID from the database
// assuming the ID's in the database are [1005, 25984, 25986, 29587, 30000]
// "i" could be any one of these values
var i = givenPredefinedObjectId;
var deviation = 100;
// objectWithIdExists() is going to lookup an object with the ID "i" in the database
// if there is no object with the ID "i" , it will return false
// otherwise the object will get tested and return true
while(objectWithIdExists(i)){
i++;
}
for(i; i < i+deviation; i++){
if(objectWithIdExists(i)){
goto while loop;
}
}
endPoint = i - deviation;
Assuming there is no knowledge about the possible values except you can check if they exist and you are given one valid value (there is no array with all possible IDs, that was just an example), how would you find the min/max values?
Unbounded binary search is feasible, with a factor of C slowdown. Given an algorithm for unbounded binary search that, given access to the oracle less_equal(n) for some natural number n, returns n in time O(log n), implement the oracle on input k by querying all of the IDs C*k, C*k+1, ..., C*k+C-1 and reporting that k is less than or equal to n if and only if one ID is found. The running time is O(C*log((max-min)/C)).
I am looking if there is an "easy" or simple way to make an array of something, Lets say Icecreams.. this would be a class of icecream with various Attributes (ID, flavour, Size, scoops), i would like to run an array that gathers every ice cream ordered and then searches through this list for any duplicate values (2+ same size)
First idea i had was a for loop that creates the array than grabs the ice cream ID for the first instance, and checks its "flavour" against the array, if no duplicate is found the ID is increased by 1 (ID++) and then that Ice creams flavour is ran in the array, if a match is found i would set a Boolean to true.
Every approach i seem to take appears to be rather long winded and i haven't got one working as of yet. hoping some fresh/more experienced eyes would help on this.
In answer to below;
The XML would hold something like below
<iceCream id=1>
<flavour>chocolate</flavour>
<scoops>5</scoops>
</iceCream>
<iceCream id=2>
<flavour>banana</flavour>
<scoops>2</scoops>
</iceCream>
I would want to use drools (probably an array list?) to gather each icecream tag and allow me to check if any of the icecreams have the same flavour and output something (set a boolean to true) if a match is found, My understand was to make an array then run each icecream though the array by using its ID to identify it and inside each loop do ID +1 (int ID = 1) then in the lopp ID++. Aswell as search through the flavour childtag.
int ID = 0;
boolean match = false;
ArrayList iceCreams = new ArrayList($cont.getIceCreams());
for(iceCream $Flavour: (ArrayList<iceCream>)iceCreams)
{
ID++
if($Flavour.getFlavour().equals(icecream with id of (ID variable).getFlavour)
{
match = true;
}
}
if(match)
{etc etc etc}
Something along these lines if this helps?
1) If you have control over the first array creation, why dont you make sure that while insertion, you insert only the icecreams that are unique. So, while you are inserting into the array say ID=1, first iterate through the array and check if there is an icecream in the array with ID as 1, if not you put this into the array and do other stuff.
2) Searching part: now while inserting, make sure that you are doing so based on the ascending oder of IDs, so you can perform binary search for the same.
Note: I dont know drools, i have just posted a logic as per my understanding of the problem.
I don't know drools either, but I'll post the some pseudo code for what I think you are trying to accomplish:
for(i = 0; i < len(ice_cream_array); i++)
{
for(j = (i + 1); j < len(ice_cream_array); j++)
{
if (ice_cream_array[i] == ice_cream_array[j])
break from inner loop
else
there is no match
}
}
You may also want to look up bubble sorts and binary searches.
What is the right way to split a string into words ?
(string doesn't contain any spaces or punctuation marks)
For example: "stringintowords" -> "String Into Words"
Could you please advise what algorithm should be used here ?
! Update: For those who think this question is just for curiosity. This algorithm could be used to camеlcase domain names ("sportandfishing .com" -> "SportAndFishing .com") and this algo is currently used by aboutus dot org to do this conversion dynamically.
Let's assume that you have a function isWord(w), which checks if w is a word using a dictionary. Let's for simplicity also assume for now that you only want to know whether for some word w such a splitting is possible. This can be easily done with dynamic programming.
Let S[1..length(w)] be a table with Boolean entries. S[i] is true if the word w[1..i] can be split. Then set S[1] = isWord(w[1]) and for i=2 to length(w) calculate
S[i] = (isWord[w[1..i] or for any j in {2..i}: S[j-1] and isWord[j..i]).
This takes O(length(w)^2) time, if dictionary queries are constant time. To actually find the splitting, just store the winning split in each S[i] that is set to true. This can also be adapted to enumerate all solution by storing all such splits.
As mentioned by many people here, this is a standard, easy dynamic programming problem: the best solution is given by Falk Hüffner. Additional info though:
(a) you should consider implementing isWord with a trie, which will save you a lot of time if you use properly (that is by incrementally testing for words).
(b) typing "segmentation dynamic programming" yields a score of more detail answers, from university level lectures with pseudo-code algorithm, such as this lecture at Duke's (which even goes so far as to provide a simple probabilistic approach to deal with what to do when you have words that won't be contained in any dictionary).
There should be a fair bit in the academic literature on this. The key words you want to search for are word segmentation. This paper looks promising, for example.
In general, you'll probably want to learn about markov models and the viterbi algorithm. The latter is a dynamic programming algorithm that may allow you to find plausible segmentations for a string without exhaustively testing every possible segmentation. The essential insight here is that if you have n possible segmentations for the first m characters, and you only want to find the most likely segmentation, you don't need to evaluate every one of these against subsequent characters - you only need to continue evaluating the most likely one.
If you want to ensure that you get this right, you'll have to use a dictionary based approach and it'll be horrendously inefficient. You'll also have to expect to receive multiple results from your algorithm.
For example: windowsteamblog (of http://windowsteamblog.com/ fame)
windows team blog
window steam blog
Consider the sheer number of possible splittings for a given string. If you have n characters in the string, there are n-1 possible places to split. For example, for the string cat, you can split before the a and you can split before the t. This results in 4 possible splittings.
You could look at this problem as choosing where you need to split the string. You also need to choose how many splits there will be. So there are Sum(i = 0 to n - 1, n - 1 choose i) possible splittings. By the Binomial Coefficient Theorem, with x and y both being 1, this is equal to pow(2, n-1).
Granted, a lot of this computation rests on common subproblems, so Dynamic Programming might speed up your algorithm. Off the top of my head, computing a boolean matrix M such M[i,j] is true if and only if the substring of your given string from i to j is a word would help out quite a bit. You still have an exponential number of possible segmentations, but you would quickly be able to eliminate a segmentation if an early split did not form a word. A solution would then be a sequence of integers (i0, j0, i1, j1, ...) with the condition that j sub k = i sub (k + 1).
If your goal is correctly camel case URL's, I would sidestep the problem and go for something a little more direct: Get the homepage for the URL, remove any spaces and capitalization from the source HTML, and search for your string. If there is a match, find that section in the original HTML and return it. You'd need an array of NumSpaces that declares how much whitespace occurs in the original string like so:
Needle: isashort
Haystack: This is a short phrase
Preprocessed: thisisashortphrase
NumSpaces : 000011233333444444
And your answer would come from:
location = prepocessed.Search(Needle)
locationInOriginal = location + NumSpaces[location]
originalLength = Needle.length() + NumSpaces[location + needle.length()] - NumSpaces[location]
Haystack.substring(locationInOriginal, originalLength)
Of course, this would break if madduckets.com did not have "Mad Duckets" somewhere on the home page. Alas, that is the price you pay for avoiding an exponential problem.
This can be actually done (to a certain degree) without dictionary. Essentially, this is an unsupervised word segmentation problem. You need to collect a large list of domain names, apply an unsupervised segmentation learning algorithm (e.g. Morfessor) and apply the learned model for new domain names. I'm not sure how well it would work, though (but it would be interesting).
This is basically a variation of a knapsack problem, so what you need is a comprehensive list of words and any of the solutions covered in Wiki.
With fairly-sized dictionary this is going to be insanely resource-intensive and lengthy operation, and you cannot even be sure that this problem will be solved.
Create a list of possible words, sort it from long words to short words.
Check if each entry in the list against the first part of the string. If it equals, remove this and append it at your sentence with a space. Repeat this.
A simple Java solution which has O(n^2) running time.
public class Solution {
// should contain the list of all words, or you can use any other data structure (e.g. a Trie)
private HashSet<String> dictionary;
public String parse(String s) {
return parse(s, new HashMap<String, String>());
}
public String parse(String s, HashMap<String, String> map) {
if (map.containsKey(s)) {
return map.get(s);
}
if (dictionary.contains(s)) {
return s;
}
for (int left = 1; left < s.length(); left++) {
String leftSub = s.substring(0, left);
if (!dictionary.contains(leftSub)) {
continue;
}
String rightSub = s.substring(left);
String rightParsed = parse(rightSub, map);
if (rightParsed != null) {
String parsed = leftSub + " " + rightParsed;
map.put(s, parsed);
return parsed;
}
}
map.put(s, null);
return null;
}
}
I was looking at the problem and thought maybe I could share how I did it.
It's a little too hard to explain my algorithm in words so maybe I could share my optimized solution in pseudocode:
string mainword = "stringintowords";
array substrings = get_all_substrings(mainword);
/** this way, one does not check the dictionary to check for word validity
* on every substring; It would only be queried once and for all,
* eliminating multiple travels to the data storage
*/
string query = "select word from dictionary where word in " + substrings;
array validwords = execute(query).getArray();
validwords = validwords.sort(length, desc);
array segments = [];
while(mainword != ""){
for(x = 0; x < validwords.length; x++){
if(mainword.startswith(validwords[x])) {
segments.push(validwords[x]);
mainword = mainword.remove(v);
x = 0;
}
}
/**
* remove the first character if any of valid words do not match, then start again
* you may need to add the first character to the result if you want to
*/
mainword = mainword.substring(1);
}
string result = segments.join(" ");
I want to get the following code to work in the Java ME / J2ME environment. Please help:
Hashtable <Activity, Float>scores = new Hashtable<Activity, Float>();
scores.put(act1, 0.3);
scores.put(act2, 0.5);
scores.put(act3, 0.4);
scores.put(act5, 0.3);
Vector v = new Vector(scores.entrySet());
Collections.sort(v); //error is related to this line
Iterator it = v.iterator();
int cnt = 0;
Activity key;
Float value;
while(it.hasNext()){
cnt++;
Map.Entry e=(Map.Entry)it.next();
key = (Activity)e.getKey();
value = (Float)e.getValue();
System.out.println(key+", "+value);
}
It doesn't work, I get the error:
Exception in thread "main" java.lang.ClassCastException: java.util.Hashtable$Entry cannot be cast to java.lang.Comparable
This points to the line that I've indicated with a comment in the code.
Please help, and bear in mind that I'm using j2me!
The code you've got isn't anywhere near valid J2ME, it's full fat (J2SE) java; J2ME doesn't currently have generics, or a Collections class or a Comparable interface - check for JavaDoc for MIDP 2 and CLDC 1.1, the components of J2ME. Your error mentions those, so definitely didn't come from J2ME, which suggests you might be doing something fundamental wrong in your project setup?
If you do want to do this in J2ME you need to write a sort function yourself, because as far as I can tell no such thing exists. Bubblesort will be easiest to write, since the only way you can easily access sequential members of the hashtable is through Enumerations (via scores.keys() and scores.values()). Assuming you want to sort your activities in ascending order based on the scores (floats) they're associated with, you want something like:
boolean fixedPoint = false;
while (!fixedPoint)
{
fixedPoint = true;
Enumeration e = scores.keys();
if (!e.hasMoreElements()) return;
Object previousKey = e.nextElement();
while (e.hasMoreElements()) {
Object currentKey = e.nextElement();
if ((Float) scores.get(currentKey) > (Float) scores.get(previousKey)) {
swap(currentKey, previousKey);
fixedPoint = false;
}
previousKey = currentKey;
}
}
Also, somewhere you'll need to write a swap function that swaps two elements of the hashtable when given their keys. Worth noting this is NOT the quickest possible implementation -- bubble sort will not be good if you expect to have big big lists. On the other hand, it is very easy with the limited tools that J2ME gives you!
The entrySet method doesn't return the values in the hash table, it returns the key-value pairs. If you want the values you should use the values method instead.
If you want the key-value pairs but sort them only on the value, you have to implement a Comparator for the key-value pairs that compares the values of two pairs, and use the overload of the sort method that takes the Comparator along with the list.