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  • How would I find all sets of N single-digit, non-repeating numbers that add up to a given sum in PHP

    - by TerranRich
    Let's say I want to find all sets of 5 single-digit, non-repeating numbers that add up to 30... I'd end up with [9,8,7,5,1], [9,8,7,4,2], [9,8,6,4,3], [9,8,6,5,2], [9,7,6,5,3], and [8,7,6,5,4]. Each of those sets contains 5 non-repeating digits that add up to 30, the given sum. Any help would be greatly appreciated. Even just a starting point for me to use would be awesome. I came up with one method, which seems like a long way of going about it: get all unique 5-digit numbers (12345, 12346, 12347, etc.), add up the digits, and see if it equals the given sum (e.g. 30). If it does, add it to the list of possible matching sets. I'm doing this for a personal project, which will help me in solving Kakuro puzzles without actually solving the whole thing at once. Yeah, it may be cheating, but it's... it's not THAT bad... :P

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  • need help with Java solution /newbie

    - by Racket
    Hi, I'm new to programming in general so i'm trying to be as specific as possible in this question. There's this book that i'm doing some exercises on. I managed to do more than half of what they say, but it's just one input that I have been struggling to find out. I'll write the question and thereafter my code, "Write an application that creates and prints a random phone number of the form XXX-XXX-XXXX. Include the dashes in the output. Do not let the first three digits contain an 8 or 9 (but don't be more restrictive than that), and make sure that the second set of three digits is not greater than 742. Hint: Think through the easiest way to construct the phone number. Each diigit does not have to be determined separately." OK, the highlighted sentence is what i'm looking at. Here's my code: import java.util.Random; public class PP33 { public static void main (String[] args) { Random rand = new Random(); int num1, num2, num3; num1 = rand.nextInt(900) + 100; num2 = rand.nextInt(643) + 100; num3 = rand.nextInt(9000) + 1000; System.out.println(num1+"-"+num2+"-"+num3); } } How am I suppose to do this? I'm on chapter 3 so we have not yet discussed if statements etcetera, but Aliases, String class, Packages, Import declaration, Random Class, Math Class, Formatting output (decimal- & numberFormat), Printf, Enumeration & Wrapper classes + autoboxing. So consider answer the question based only on these assumptions, please. The code doesn't have any errors. Thank you!

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  • jquery calculation sum two different type of item

    - by st4n
    I'm writing a script like the example shown in the demo where All of the "Total" values (Including the "Grand Total") are Automatically Calculated using the calc () method. at this link: But I have some fields in which to apply the equation qty * price, and others where I want to do other operations .. you can tell me how? thank you very much i try with this, but it is a very stupid code .. and the grandTotal .. not sum the two different fields: function recalc() { $("[id^=total_item]").calc("qty * price", { qty: $("input[name^=qty_item_]"), price: $("[id^=price_item_]") }, function (s){ // return the number as a dollar amount return "$" + s.toFixed(2); }, function ($this){ // sum the total of the $("[id^=total_item]") selector var sum = $this.sum(); $("#grandTotal").text( // round the results to 2 digits "$" + sum.toFixed(2) ); }); $("[id^=total_otheritem]").calc("qty1 / price1", { qty1: $("input[name^=qty_other_]"), price1: $("[id^=price_other_]") }, function (s){ // return the number as a dollar amount return "$" + s.toFixed(2); }, function ($this){ var sum = $this.sum(); $("#grandTotal").text( // round the results to 2 digits "$" + sum.toFixed(2) ); }); }

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  • How do I make this nested for loop, testing sums of cubes, more efficient?

    - by Brian J. Fink
    I'm trying to iterate through all the combinations of pairs of positive long integers in Java and testing the sum of their cubes to discover if it's a Fibonacci number. I'm currently doing this by using the value of the outer loop variable as the inner loop's upper limit, with the effect being that the outer loop runs a little slower each time. Initially it appeared to run very quickly--I was up to 10 digits within minutes. But now after 2 full days of continuous execution, I'm only somewhere in the middle range of 15 digits. At this rate it may end up taking a whole year just to finish running this program. The code for the program is below: import java.lang.*; import java.math.*; public class FindFib { public static void main(String args[]) { long uLimit=9223372036854775807L; //long maximum value BigDecimal PHI=new BigDecimal(1D+Math.sqrt(5D)/2D); //Golden Ratio for(long a=1;a<=uLimit;a++) //Outer Loop, 1 to maximum for(long b=1;b<=a;b++) //Inner Loop, 1 to current outer { //Cube the numbers and add BigDecimal c=BigDecimal.valueOf(a).pow(3).add(BigDecimal.valueOf(b).pow(3)); System.out.print(c+" "); //Output result //Upper and lower limits of interval for Mobius test: [c*PHI-1/c,c*PHI+1/c] BigDecimal d=c.multiply(PHI).subtract(BigDecimal.ONE.divide(c,BigDecimal.ROUND_HALF_UP)), e=c.multiply(PHI).add(BigDecimal.ONE.divide(c,BigDecimal.ROUND_HALF_UP)); //Mobius test: if integer in interval (floor values unequal) Fibonacci number! if (d.toBigInteger().compareTo(e.toBigInteger())!=0) System.out.println(); //Line feed else System.out.print("\r"); //Carriage return instead } //Display final message System.out.println("\rDone. "); } } Now the use of BigDecimal and BigInteger was delibrate; I need them to get the necessary precision. Is there anything other than my variable types that I could change to gain better efficiency?

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  • jQuery Validate on form submitted by JavaScript

    - by Daniel
    My form is submitted by a link using JavaScript, but I am also trying to validate the from justing jQuery validate. The validation doesn't work when submitted by the link, but it does if I change the link to a submit button. What am I doing wrong? My form: <form id="findmatch" method="post" action="search"> <div> <label class="formlabel">Match Type <input type="text" name="matchtype" id="matchtype" class="forminput" /> </label> <label class="formlabel">Location (postcode) <input type="text" name="location" id="location" class="forminput" /> </label> <label class="formlabel">Radius (miles) <input type="text" name="Radius" id="Radius" class="forminput" /> </label> <label class="formlabel">Keywords <input type="text" onblur="javascript:usePointFromPostcode(document.getElementById('location').value, showCompleteLatLng)" onchange="javascript:usePointFromPostcode(document.getElementById('location').value, showCompleteLatLng)" name="keywords" id="keywords" class="forminput" /> </label> <input id="lat" class="hidden" name="lat" type="text" value="" /> <input id="lon" class="hidden" name="lon" type="text" value="" /> <a href="javascript:document.getElementById('findmatch').submit();" onmouseover="javascript:usePointFromPostcode(document.getElementById('location').value, showCompleteLatLng)" class="submit">Search</a> </div> </form> And my jQuery is <script type="text/javascript"> $(document).ready(function () { $("#findmatch").validate({ rules: { location: "required", Radius: { required: true, digits: true }, keywords: "required" }, messages: { location: "Please enter your postcode", Radius: { required: "Please enter a radius", digits: "Please only enter numbers" }, keywords: "Please enter the keywords you wish to search for" } }); }); </script>

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  • Find telephonenumbers - finding number with and without an phone extension

    - by nWorx
    Hello there I've a table with about 130 000 records with telephonenumbers. The numbers are all formated like this +4311234567. The numbers always include international country code, local area code and then the phonenumber and sometimes an extension. There is a webservice which checks for the caller's number in the table. That service works already. But now the client wants that also if someone calls from a company which number is already in the database but not his extension, that the service will return some result. Example for table. **id** | **telephonenumber** | **name** | 1 | +431234567 | company A | 2 | +431234567890 | employee in company A | 3 | +4398765432 | company b now if somebody from company A calls with a different extension for example +43123456777, than it should return id1. But the problem is, that I don't know how many digits the extensions have. It could have 3,4 or more digits. Are there any patterns for string kind of matchings? The data is stored in a sql2005 database. Thanks

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  • Why is my Android app force closing when I try to check if an EditText has a double

    - by user336861
    Scanner scanner = new Scanner(lapsPerMile_st); if (!scanner.hasNextDouble()) { Context context = getApplicationContext(); String msg = "Please Enter Digits and Decmials Only"; int duration = Toast.LENGTH_LONG; Toast.makeText(context, msg, duration).show(); lapsPerMileEditText.setText(""); return; } else { //Edit box has only digits, Set data and display stats data.setLapsPerMile(Integer.parseInt(lapsPerMile_st)); lapsRunLabel.setVisibility(0); lapsRunTextView.setText(Integer.toString(data.getLapsRun())); milesRunLabel.setVisibility(0); milesRunTextView.setText(Double.toString(data.getLapsRun()/data.getLapsPerMile())); } <EditText android:id="@+id/mileCount" android:layout_width="100dp" android:layout_height="wrap_content" android:layout_marginTop="110dp" android:inputType="numberDecimal" android:maxLength="4" /> For some reason if I enter a non decimal number such as 3, or 5, it works fine but when I enter a floating point such as 3.4 or 5.8 it force closes. I cant seem to figure out whats going on. Any ideas? Thanks

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  • jQuery - Compatibility Problem with Internet Explorer 7 and Opera

    - by Marius
    Hello there, I have this counter which counts + 1 every time somebody shares content from the site. When it happens, the social icon that was clicked will bounce. It works in Firefox,Chrome, IE8, and Opera, however the bouncing animation is wrong in opera. $.fn.countExternal = function(animSpeed, num) { // for each counter this.each(function(){ // select all the digit containers var span = $(this).children(); // count the num of digit containers var len = $(span).length; // get the current count u = $(span).text(); // copy variable and add increment(s) v = num + ''; // foreach digit container... for (i=v.length - 1; i >= 0; i--) { // ...check which digits are not affected by the increment(s) if (v.charAt(i) == u.charAt(i)) { break; } } // slice from the total number of digit containers the digits containers which needs updating. slce = len - (v.length - (i + 1)) var updates = $(span).slice(slce); // loop through each digit container and fade out ... $(updates).fadeTo(animSpeed, 0,function(){ $(updates).each(function(index){ f = i + 1 + index; // ...then pick the right digit and update the digit... $(this).text(v.charAt(f)); // ...before fading back in. Cycle complete. $(this).fadeTo(animSpeed, 1); }); }); }); }; }) (jQuery); Demo (NSFW) is here (look underneath the social sharing icons). Any idea how I can solve the IE, and possibly the Opera compatibility problem? Thank you for your time.

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  • Finding the digit root of a number

    - by Jessica M.
    Study question is to find the digit root of a already provided number. The teacher provides us with the number 2638. In order to find the digit root you have to add each digit separately 2 + 6 + 3 + 8 = 19. Then you take the result 19 and add those two digits together 1 + 9 = 10. Do the same thing again 1 + 0 = 1. The digit root is 1. My first step was to use the variable total to add up the number 2638 to find the total of 19. Then I tried to use the second while loop to separate the two digits by using the % I have to try and solve the problem by using basic integer arithmetic (+, -, *, /). 1.Is it necessary and or possible to solve the problem using nested while loops? 2.Is my math correct? 3. As I wrote it here it does not run in Eclipse. Am I using the while loops correctly? import acm.program.*; public class Ch4Q7 extends ConsoleProgram { public void run(){ println("This program attempts to find the digit root of your number: "); int n = readInt("Please enter your number: "); int total = 0; int root = total; while (n > 0 ){ total = total + (n %10); n = (n / 10); } while ( total > 0 ){ root = total; total = ((total % 10) + total / 10); } println("your root should be " + root); } }

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  • How can I improve this SQL to avoid several problems with its results?

    - by Josh Curren
    I am having some problems with trying to search. Currently this will only return results that have at least 1 row in the maintenance_parts table. I would like it to return results even if there are 0 parts rows. My second problem is that when you search for a vehicle and it should return multiple results (multiple maintenance rows) it will only return 1 result for that vehicle. Some Background Info: The user has 2 fields to fill out. The fields are vehicle and keywords. The vehicle field is meant to allow searching based on the make, model, VIN, truck number (often is 2 - 3 digits or a letter prefix followed by 2 digits), and a few other fields that belong to the truck table. The keywords are meant to search most fields in the maintenance and maintenance_parts tables (things like the description of the work, parts name, parts number). The maintenance_parts table can contain 0, 1, or more rows for each maintenance row. The truck table contains exactly 1 row for every maintenance row. A truck can have multiple maintenance records. "SELECT M.maintenance_id, M.some_id, M.type_code, M.service_date, M.mileage, M.mg_id, M.mg_type, M.comments, M.work_done, MATCH( M.comments, M.work_done) AGAINST( '$keywords' ) + MATCH( P.part_num, P.part_desc, P.part_ref) AGAINST( '$keywords' ) + MATCH( T.truck_number, T.make, T.model, T.engine, T.vin_number, T.transmission_number, T.comments) AGAINST( '$vehicle' ) AS score FROM maintenance M, maintenance_parts P, truck T WHERE M.maintenance_id = P.maintenance_id AND M.some_id = T.truck_id AND M.type_code = 'truck' AND ( (MATCH( T.truck_number, T.make, T.model, T.engine, T.vin_number, T.transmission_number, T.comments) AGAINST( '$vehicle' ) OR T.truck_number LIKE '%$vehicle%') OR MATCH( P.part_num, P.part_desc, P.part_ref) AGAINST( '$keywords' ) OR MATCH( M.comments, M.work_done) AGAINST( '$keywords' ) ) AND M.status = 'A' GROUP BY maintenance_id ORDER BY score DESC, maintenance_id DESC LIMIT 0, $limit"

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  • Project Euler #119 Make Faster

    - by gangqinlaohu
    Trying to solve Project Euler problem 119: The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a number with this property is 614656 = 28^4. We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum. You are given that a2 = 512 and a10 = 614656. Find a30. Question: Is there a more efficient way to find the answer than just checking every number until a30 is found? My Code int currentNum = 0; long value = 0; for (long a = 11; currentNum != 30; a++){ //maybe a++ is inefficient int test = Util.sumDigits(a); if (isPower(a, test)) { currentNum++; value = a; System.out.println(value + ":" + currentNum); } } System.out.println(value); isPower checks if a is a power of test. Util.sumDigits: public static int sumDigits(long n){ int sum = 0; String s = "" + n; while (!s.equals("")){ sum += Integer.parseInt("" + s.charAt(0)); s = s.substring(1); } return sum; } program has been running for about 30 minutes (might be overflow on the long). Output (so far): 81:1 512:2 2401:3 4913:4 5832:5 17576:6 19683:7 234256:8 390625:9 614656:10 1679616:11 17210368:12 34012224:13 52521875:14 60466176:15 205962976:16 612220032:17

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  • matching certain numbers at the end of a string

    - by user697473
    I have a vector of strings: s <- c('abc1', 'abc2', 'abc3', 'abc11', 'abc12', 'abcde1', 'abcde2', 'abcde3', 'abcde11', 'abcde12', 'nonsense') I would like a regular expression to match only the strings that begin with abc and end with 3, 11, or 12. In other words, the regex has to exclude abc1 but not abc11, abc2 but not abc12, and so on. I thought that this would be easy to do with lookahead assertions, but I haven't found a way. Is there one? EDIT: Thanks to posters below for pointing out a serious ambiguity in the original post. In reality, I have many strings. They all end in digits: some in 0, some in 9, some in the digits in between. I am looking for a regex that will match all strings except those that end with a letter followed by a 1 or a 2. (The regex should also match only those strings that start with abc, but that's an easy problem.) I tried to use negative lookahead assertions to create such a regex. But I didn't have any success.

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  • mpasdlta files -- what are they?

    - by Tmdean
    I noticed a bunch of folders in the root of my hard drive named with a string of hex digits that contain files named with a GUID ending with "mpasdlta.vdm" and "mpavdlta.vdm". From some Googling, I've determined that these files are spyware and virus definition files used by Microsoft Security Essentials. Are these files safe to delete? (Why doesn't Microsoft follow their own guidelines and store application data in the folders intended for that purpose? grumble grumble)

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  • OAS log files filling up hard drive

    - by Andrew Hampton
    We've had issues with log files for Oracle Application Server filling up the hard drive on our server. The files are in the /network/admin folder and are named server.log_XXXXX.trc and client.log_XXXXX.trc where XXXXX are 5 digits. The files are typically anywhere from 1-2MB in size but can be up to 100MB and thousands of them are created at a rate of about 5-10 per minute. Does anyone know how to disable these logs? Thanks!

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  • How to prevent a Windows 7 PC from sleeping when CPU usage is over X%?

    - by MaxVT
    I often leave the PC running into the night to process video files, so it shouldn't sleep while it's working but it would be nice if it went into sleep when it's done. During the export the CPU is always above a set %, and when idle it's typically in the single digits. Is there some tool or setting that would prevent the PC from going to sleep as long as the CPU usage (let's say averaged over one minute) stays above a specified limit?

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  • flex and bison: wrong output

    - by user2972227
    I am doing a homework using flex and bison to make a complex number calculator. But my program cannot give a correct output. .lex file: %option noyywrap %{ #include<stdio.h> #include<stdlib.h> #include "complex_cal.h" #define YYSTYPE complex #include "complex_cal.tab.h" void RmWs(char* str); %} /* Add your Flex definitions here */ /* Some definitions are already provided to you*/ ws [ \t]+ digits [0-9] number (0|[1-9]+{digits}*)\.?{digits}* im [i] complexnum {ws}*[-]*{ws}*{number}{ws}*[+|-]{ws}*{number}{ws}*{im}{ws}* op [-+*/()] %% {complexnum} {RmWs(yytext); sscanf(yytext,"%lf %lf",&(yylval.real),&(yylval.img)); return CNUMBER;} {ws} /**/ {op} return *yytext; %% /* function provided to student to remove */ /* all the whitespaces from a string. */ void RmWs(char* str){ int i=0,j=0; char temp[strlen(str)+1]; strcpy(temp,str); while (temp[i]!='\0'){ while (temp[i]==' '){i++;} str[j]=temp[i]; i++; j++; } str[j]='\0'; } .y file: %{ #include <stdio.h> #include <stdlib.h> #include "complex_cal.h" /* prototypes of the provided functions */ complex complex_add (complex, complex); complex complex_sub (complex, complex); complex complex_mul (complex, complex); complex complex_div (complex, complex); /* prototypes of the provided functions */ int yylex(void); int yyerror(const char*); %} %token CNUMBER %left '+' '-' %left '*' '/' %nonassoc '(' ')' %% /* start: Add your grammar rules and actions here */ complexexp: complexexp '+' complexexpmultidiv {$$=complex_add($1, $3);} | complexexp '-' complexexpmultidiv {$$=complex_sub($1, $3);} | complexexpmultidiv {$$.real=$1.real;$$.img=$1.img;} ; complexexpmultidiv: complexexpmultidiv '*' complexsimple {$$=complex_mul($1, $3);} | complexexpmultidiv '/' complexsimple {$$=complex_div($1, $3);} | complexsimple {$$.real=$1.real;$$.img=$1.img;} ; complexsimple: '(' complexexp ')' {$$.real=$2.real;$$.img=$2.img;} | '(' CNUMBER ')' {$$.real=$2.real;$$.img=$2.img;} ; /* end: Add your grammar rules and actions here */ %% int main(){ return yyparse(); } int yyerror(const char* s){ printf("%s\n", s); return 0; } /* function provided to do complex addition */ /* input : complex numbers c1, c2 */ /* output: nothing */ /* side effect : none */ /* return value: result of addition in c3 */ complex complex_add (complex c1, complex c2){ /* c1 + c2 */ complex c3; c3.real = c1.real + c2.real; c3.img = c1.img + c2.img; return c3; } /* function provided to do complex subtraction */ /* input : complex numbers c1, c2 */ /* output: nothing */ /* side effect : none */ /* return value: result of subtraction in c3 */ complex complex_sub (complex c1, complex c2){ /* c1 - c2 */ complex c3; c3.real = c1.real - c2.real; c3.img = c1.img - c2.img; return c3; } /* function provided to do complex multiplication */ /* input : complex numbers c1, c2 */ /* output: nothing */ /* side effect : none */ /* return value: result of multiplication in c3 */ complex complex_mul (complex c1, complex c2){ /* c1 * c2 */ complex c3; c3.real = c1.real*c2.real - c1.img*c2.img; c3.img = c1.img*c2.real + c1.real*c2.img; return c3; } /* function provided to do complex division */ /* input : complex numbers c1, c2 */ /* output: nothing */ /* side effect : none */ /* return value: result of c1/c2 in c3 */ complex complex_div (complex c1, complex c2){ /* c1 / c2 (i.e. c1 divided by c2 ) */ complex c3; double d; /*divisor calculation using the conjugate of c2*/ d = c2.real*c2.real + c2.img*c2.img; c3.real = (c1.real*c2.real + c1.img*c2.img)/d; c3.img = (c1.img*c2.real - c1.real*c2.img)/d; return c3; } .h file: #include <string.h> /* struct for holding a complex number */ typedef struct { double real; double img; } complex; /* define the return type of FLEX */ #define YYSTYPE complex Script for compiling the file: bison -d -v complex_cal.y flex -ocomplex_cal.lex.yy.c complex_cal.lex gcc -o complex_cal complex_cal.lex.yy.c complex_cal.tab.c ./complex_cal Some correct sample run of the program: input:(5+6i)*(6+1i) output:24.000000+41.000000i input:(7+8i)/(-3-4i)*(5+7i) output:-11.720000-14.040000i input:(7+8i)/((-3-4i)*(5+7i)) output:-0.128108+0.211351i But when I run this program, the program only give an output which is identical to my input. For example, when I input (5+6i)(6+1i), it just gives (5+6i)(6+1i). Even if I input any other things, for example, input "abc" it just gives "abc" and is not syntax error. I don't know where the problem is and I hope to know how to solve it.

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  • Threading extra state through a parser in Scala

    - by Travis Brown
    I'll give you the tl;dr up front I'm trying to use the state monad transformer in Scalaz 7 to thread extra state through a parser, and I'm having trouble doing anything useful without writing a lot of t m a -> t m b versions of m a -> m b methods. An example parsing problem Suppose I have a string containing nested parentheses with digits inside them: val input = "((617)((0)(32)))" I also have a stream of fresh variable names (characters, in this case): val names = Stream('a' to 'z': _*) I want to pull a name off the top of the stream and assign it to each parenthetical expression as I parse it, and then map that name to a string representing the contents of the parentheses, with the nested parenthetical expressions (if any) replaced by their names. To make this more concrete, here's what I'd want the output to look like for the example input above: val target = Map( 'a' -> "617", 'b' -> "0", 'c' -> "32", 'd' -> "bc", 'e' -> "ad" ) There may be either a string of digits or arbitrarily many sub-expressions at a given level, but these two kinds of content won't be mixed in a single parenthetical expression. To keep things simple, we'll assume that the stream of names will never contain either duplicates or digits, and that it will always contain enough names for our input. Using parser combinators with a bit of mutable state The example above is a slightly simplified version of the parsing problem in this Stack Overflow question. I answered that question with a solution that looked roughly like this: import scala.util.parsing.combinator._ class ParenParser(names: Iterator[Char]) extends RegexParsers { def paren: Parser[List[(Char, String)]] = "(" ~> contents <~ ")" ^^ { case (s, m) => (names.next -> s) :: m } def contents: Parser[(String, List[(Char, String)])] = "\\d+".r ^^ (_ -> Nil) | rep1(paren) ^^ ( ps => ps.map(_.head._1).mkString -> ps.flatten ) def parse(s: String) = parseAll(paren, s).map(_.toMap) } It's not too bad, but I'd prefer to avoid the mutable state. What I want Haskell's Parsec library makes adding user state to a parser trivially easy: import Control.Applicative ((*>), (<$>), (<*)) import Data.Map (fromList) import Text.Parsec paren = do (s, m) <- char '(' *> contents <* char ')' h : t <- getState putState t return $ (h, s) : m where contents = flip (,) [] <$> many1 digit <|> (\ps -> (map (fst . head) ps, concat ps)) <$> many1 paren main = print $ runParser (fromList <$> paren) ['a'..'z'] "example" "((617)((0)(32)))" This is a fairly straightforward translation of my Scala parser above, but without mutable state. What I've tried I'm trying to get as close to the Parsec solution as I can using Scalaz's state monad transformer, so instead of Parser[A] I'm working with StateT[Parser, Stream[Char], A]. I have a "solution" that allows me to write the following: import scala.util.parsing.combinator._ import scalaz._, Scalaz._ object ParenParser extends ExtraStateParsers[Stream[Char]] with RegexParsers { protected implicit def monadInstance = parserMonad(this) def paren: ESP[List[(Char, String)]] = (lift("(" ) ~> contents <~ lift(")")).flatMap { case (s, m) => get.flatMap( names => put(names.tail).map(_ => (names.head -> s) :: m) ) } def contents: ESP[(String, List[(Char, String)])] = lift("\\d+".r ^^ (_ -> Nil)) | rep1(paren).map( ps => ps.map(_.head._1).mkString -> ps.flatten ) def parse(s: String, names: Stream[Char]) = parseAll(paren.eval(names), s).map(_.toMap) } This works, and it's not that much less concise than either the mutable state version or the Parsec version. But my ExtraStateParsers is ugly as sin—I don't want to try your patience more than I already have, so I won't include it here (although here's a link, if you really want it). I've had to write new versions of every Parser and Parsers method I use above for my ExtraStateParsers and ESP types (rep1, ~>, <~, and |, in case you're counting). If I had needed to use other combinators, I'd have had to write new state transformer-level versions of them as well. Is there a cleaner way to do this? I'd love to see an example of a Scalaz 7's state monad transformer being used to thread state through a parser, but Scala 6 or Haskell examples would also be useful.

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  • How would you go about tackling this problem? [SOLVED in C++]

    - by incrediman
    Intro: EDIT: See solution at the bottom of this question (c++) I have a programming contest coming up in about half a week, and I've been prepping :) I found a bunch of questions from this canadian competition, they're great practice: http://cemc.math.uwaterloo.ca/contests/computing/2009/stage2/day1.pdf I'm looking at problem B ("Dinner"). Any idea where to start? I can't really think of anything besides the naive approach (ie. trying all permutations) which would take too long to be a valid answer. Btw, the language there says c++ and pascal I think, but i don't care what language you use - I mean really all I want is a hint as to the direction I should proceed in, and perhpas a short explanation to go along with it. It feels like I'm missing something obvious... Of course extended speculation is more than welcome, but I just wanted to clarify that I'm not looking for a full solution here :) Short version of the question: You have a binary string N of length 1-100 (in the question they use H's and G's instead of one's and 0's). You must remove all of the digits from it, in the least number of steps possible. In each step you may remove any number of adjacent digits so long as they are the same. That is, in each step you can remove any number of adjacent G's, or any number of adjacent H's, but you can't remove H's and G's in one step. Example: HHHGHHGHH Solution to the example: 1. HHGGHH (remove middle Hs) 2. HHHH (remove middle Gs) 3. Done (remove Hs) -->Would return '3' as the answer. Note that there can also be a limit placed on how large adjacent groups have to be when you remove them. For example it might say '2', and then you can't remove single digits (you'd have to remove pairs or larger groups at a time). Solution I took Mark Harrison's main algorithm, and Paradigm's grouping idea and used them to create the solution below. You can try it out on the official test cases if you want. //B.cpp //include debug messages? #define DEBUG false #include <iostream> #include <stdio.h> #include <vector> using namespace std; #define FOR(i,n) for (int i=0;i<n;i++) #define FROM(i,s,n) for (int i=s;i<n;i++) #define H 'H' #define G 'G' class String{ public: int num; char type; String(){ type=H; num=0; } String(char type){ this->type=type; num=1; } }; //n is the number of bits originally in the line //k is the minimum number of people you can remove at a time //moves is the counter used to determine how many moves we've made so far int n, k, moves; int main(){ /*Input from File*/ scanf("%d %d",&n,&k); char * buffer = new char[200]; scanf("%s",buffer); /*Process input into a vector*/ //the 'line' is a vector of 'String's (essentially contigious groups of identical 'bits') vector<String> line; line.push_back(String()); FOR(i,n){ //if the last String is of the correct type, simply increment its count if (line.back().type==buffer[i]) line.back().num++; //if the last String is of the wrong type but has a 0 count, correct its type and set its count to 1 else if (line.back().num==0){ line.back().type=buffer[i]; line.back().num=1; } //otherwise this is the beginning of a new group, so create the new group at the back with the correct type, and a count of 1 else{ line.push_back(String(buffer[i])); } } /*Geedily remove groups until there are at most two groups left*/ moves=0; int I;//the position of the best group to remove int bestNum;//the size of the newly connected group the removal of group I will create while (line.size()>2){ /*START DEBUG*/ if (DEBUG){ cout<<"\n"<<moves<<"\n----\n"; FOR(i,line.size()) printf("%d %c \n",line[i].num,line[i].type); cout<<"----\n"; } /*END DEBUG*/ I=1; bestNum=-1; FROM(i,1,line.size()-1){ if (line[i-1].num+line[i+1].num>bestNum && line[i].num>=k){ bestNum=line[i-1].num+line[i+1].num; I=i; } } //remove the chosen group, thus merging the two adjacent groups line[I-1].num+=line[I+1].num; line.erase(line.begin()+I);line.erase(line.begin()+I); moves++; } /*START DEBUG*/ if (DEBUG){ cout<<"\n"<<moves<<"\n----\n"; FOR(i,line.size()) printf("%d %c \n",line[i].num,line[i].type); cout<<"----\n"; cout<<"\n\nFinal Answer: "; } /*END DEBUG*/ /*Attempt the removal of the last two groups, and output the final result*/ if (line.size()==2 && line[0].num>=k && line[1].num>=k) cout<<moves+2;//success else if (line.size()==1 && line[0].num>=k) cout<<moves+1;//success else cout<<-1;//not everyone could dine. /*START DEBUG*/ if (DEBUG){ cout<<" moves."; } /*END DEBUG*/ }

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  • Python 3.4 adds re.fullmatch()

    - by Jan Goyvaerts
    Python 3.4 does not bring any changes to its regular expression syntax compared to previous 3.x releases. It does add one new function to the re module called fullmatch(). This function takes a regular expression and a subject string as its parameters. It returns True if the regular expression can match the string entirely. It returns False if the string cannot be matched or if it can only be matched partially. This is useful when using a regular expression to validate user input. Do note that fullmatch() will return True if the subject string is the empty string and the regular expression can find zero-length matches. A zero-length match of a zero-length string is a complete match. So if you want to check whether the user entered a sequence of digits, use \d+ rather than \d* as the regex.

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  • Project Euler 8: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 8.  As always, any feedback is welcome. # Euler 8 # http://projecteuler.net/index.php?section=problems&id=8 # Find the greatest product of five consecutive digits # in the following 1000-digit number import time start = time.time() number = '\ 73167176531330624919225119674426574742355349194934\ 96983520312774506326239578318016984801869478851843\ 85861560789112949495459501737958331952853208805511\ 12540698747158523863050715693290963295227443043557\ 66896648950445244523161731856403098711121722383113\ 62229893423380308135336276614282806444486645238749\ 30358907296290491560440772390713810515859307960866\ 70172427121883998797908792274921901699720888093776\ 65727333001053367881220235421809751254540594752243\ 52584907711670556013604839586446706324415722155397\ 53697817977846174064955149290862569321978468622482\ 83972241375657056057490261407972968652414535100474\ 82166370484403199890008895243450658541227588666881\ 16427171479924442928230863465674813919123162824586\ 17866458359124566529476545682848912883142607690042\ 24219022671055626321111109370544217506941658960408\ 07198403850962455444362981230987879927244284909188\ 84580156166097919133875499200524063689912560717606\ 05886116467109405077541002256983155200055935729725\ 71636269561882670428252483600823257530420752963450' max = 0 for i in xrange(0, len(number) - 5): nums = [int(x) for x in number[i:i+5]] val = reduce(lambda agg, x: agg*x, nums) if val > max: max = val print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Why is it impossible to produce truly random numbers?

    - by Vinoth Kumar
    I was trying to solve a hobby problem that required generating a million random numbers. But I quickly realized, it is becoming difficult to make them unique. I picked up Algorithm Design Manual to read about random number generation. It has the following paragraph that I am fully not able to understand. Unfortunately, generating random numbers looks a lot easier than it really is. Indeed, it is fundamentally impossible to produce truly random numbers on any deterministic device. Von Neumann [Neu63] said it best: “Anyone who considers arithmetical methods of producing random digits is, of course, in a state of sin.” The best we can hope for are pseudo-random numbers, a stream of numbers that appear as if they were generated randomly. Why is it impossible to produce truly random numbers in any deterministic device? What does this sentence mean?

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  • Using a Predicate as a key to a Dictionary

    - by Tom Hines
    I really love Linq and Lambda Expressions in C#.  I also love certain community forums and programming websites like DaniWeb. A user on DaniWeb posted a question about comparing the results of a game that is like poker (5-card stud), but is played with dice. The question stemmed around determining what was the winning hand.  I looked at the question and issued some comments and suggestions toward a potential answer, but I thought it was a neat homework exercise. [A little explanation] I eventually realized not only could I compare the results of the hands (by name) with a certain construct – I could also compare the values of the individual dice with the same construct. That piece of code eventually became a Dictionary with the KEY as a Predicate<int> and the Value a Func<T> that returns a string from the another structure that contains the mapping of an ENUM to a string.  In one instance, that string is the name of the hand and in another instance, it is a string (CSV) representation of of the digits in the hand. An added benefit is that the digits re returned in the order they would be for a proper poker hand.  For instance the hand 1,2,5,3,1 would be returned as ONE_PAIR (1,1,5,3,2). [Getting to the point] 1: using System; 2: using System.Collections.Generic; 3:   4: namespace DicePoker 5: { 6: using KVP_E2S = KeyValuePair<CDicePoker.E_DICE_POKER_HAND_VAL, string>; 7: public partial class CDicePoker 8: { 9: /// <summary> 10: /// Magical construction to determine the winner of given hand Key/Value. 11: /// </summary> 12: private static Dictionary<Predicate<int>, Func<List<KVP_E2S>, string>> 13: map_prd2fn = new Dictionary<Predicate<int>, Func<List<KVP_E2S>, string>> 14: { 15: {new Predicate<int>(i => i.Equals(0)), PlayerTie},//first tie 16:   17: {new Predicate<int>(i => i > 0), 18: (m => string.Format("Player One wins\n1={0}({1})\n2={2}({3})", 19: m[0].Key, m[0].Value, m[1].Key, m[1].Value))}, 20:   21: {new Predicate<int>(i => i < 0), 22: (m => string.Format("Player Two wins\n2={2}({3})\n1={0}({1})", 23: m[0].Key, m[0].Value, m[1].Key, m[1].Value))}, 24:   25: {new Predicate<int>(i => i.Equals(0)), 26: (m => string.Format("Tie({0}) \n1={1}\n2={2}", 27: m[0].Key, m[0].Value, m[1].Value))} 28: }; 29: } 30: } When this is called, the code calls the Invoke method of the predicate to return a bool.  The first on matching true will have its value invoked. 1: private static Func<DICE_HAND, E_DICE_POKER_HAND_VAL> GetHandEval = dh => 2: map_dph2fn[map_dph2fn.Keys.Where(enm2fn => enm2fn(dh)).First()]; After coming up with this process, I realized (with a little modification) it could be called to evaluate the individual values in the dice hand in the event of a tie. 1: private static Func<List<KVP_E2S>, string> PlayerTie = lst_kvp => 2: map_prd2fn.Skip(1) 3: .Where(x => x.Key.Invoke(RenderDigits(dhPlayerOne).CompareTo(RenderDigits(dhPlayerTwo)))) 4: .Select(s => s.Value) 5: .First().Invoke(lst_kvp); After that, I realized I could now create a program completely without “if” statements or “for” loops! 1: static void Main(string[] args) 2: { 3: Dictionary<Predicate<int>, Action<Action<string>>> main = new Dictionary<Predicate<int>, Action<Action<string>>> 4: { 5: {(i => i.Equals(0)), PlayGame}, 6: {(i => true), Usage} 7: }; 8:   9: main[main.Keys.Where(m => m.Invoke(args.Length)).First()].Invoke(Display); 10: } …and there you have it. :) ZIPPED Project

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  • How do I get a Netgear WNDA3100V2 working?

    - by Michal
    I have Ubuntu 11.10 on my desktop. A month ago I bought Linksys AE1000 adapter,I did not check that it's not working on Ubuntu and because I've lost receipt I'm stuck with it. Last week I bought Netgear adapter and this time I did check and it meant to be plug an play but it was not. I have checked many forums and managed to install software, system does sees adapter but it's not connecting to network. I have found that it may not like WPA so I have created my own password-letters and digits,no spaces-still nothing.I don't understand why. This is my next attempt with Linux and I'm not with IT background so it takes time and research before I can resolve something but I really want to learn. I so wish to learn on Ubuntu.One day, I've checked Fedora16 and my old Linksys AE1000 worked without any instalations.

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  • Project Euler 13: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 13.  As always, any feedback is welcome. # Euler 13 # http://projecteuler.net/index.php?section=problems&id=13 # Work out the first ten digits of the sum of the # following one-hundred 50-digit numbers. import time start = time.time() number_string = '\ 37107287533902102798797998220837590246510135740250\ 46376937677490009712648124896970078050417018260538\ 74324986199524741059474233309513058123726617309629\ 91942213363574161572522430563301811072406154908250\ 23067588207539346171171980310421047513778063246676\ 89261670696623633820136378418383684178734361726757\ 28112879812849979408065481931592621691275889832738\ 44274228917432520321923589422876796487670272189318\ 47451445736001306439091167216856844588711603153276\ 70386486105843025439939619828917593665686757934951\ 62176457141856560629502157223196586755079324193331\ 64906352462741904929101432445813822663347944758178\ 92575867718337217661963751590579239728245598838407\ 58203565325359399008402633568948830189458628227828\ 80181199384826282014278194139940567587151170094390\ 35398664372827112653829987240784473053190104293586\ 86515506006295864861532075273371959191420517255829\ 71693888707715466499115593487603532921714970056938\ 54370070576826684624621495650076471787294438377604\ 53282654108756828443191190634694037855217779295145\ 36123272525000296071075082563815656710885258350721\ 45876576172410976447339110607218265236877223636045\ 17423706905851860660448207621209813287860733969412\ 81142660418086830619328460811191061556940512689692\ 51934325451728388641918047049293215058642563049483\ 62467221648435076201727918039944693004732956340691\ 15732444386908125794514089057706229429197107928209\ 55037687525678773091862540744969844508330393682126\ 18336384825330154686196124348767681297534375946515\ 80386287592878490201521685554828717201219257766954\ 78182833757993103614740356856449095527097864797581\ 16726320100436897842553539920931837441497806860984\ 48403098129077791799088218795327364475675590848030\ 87086987551392711854517078544161852424320693150332\ 59959406895756536782107074926966537676326235447210\ 69793950679652694742597709739166693763042633987085\ 41052684708299085211399427365734116182760315001271\ 65378607361501080857009149939512557028198746004375\ 35829035317434717326932123578154982629742552737307\ 94953759765105305946966067683156574377167401875275\ 88902802571733229619176668713819931811048770190271\ 25267680276078003013678680992525463401061632866526\ 36270218540497705585629946580636237993140746255962\ 24074486908231174977792365466257246923322810917141\ 91430288197103288597806669760892938638285025333403\ 34413065578016127815921815005561868836468420090470\ 23053081172816430487623791969842487255036638784583\ 11487696932154902810424020138335124462181441773470\ 63783299490636259666498587618221225225512486764533\ 67720186971698544312419572409913959008952310058822\ 95548255300263520781532296796249481641953868218774\ 76085327132285723110424803456124867697064507995236\ 37774242535411291684276865538926205024910326572967\ 23701913275725675285653248258265463092207058596522\ 29798860272258331913126375147341994889534765745501\ 18495701454879288984856827726077713721403798879715\ 38298203783031473527721580348144513491373226651381\ 34829543829199918180278916522431027392251122869539\ 40957953066405232632538044100059654939159879593635\ 29746152185502371307642255121183693803580388584903\ 41698116222072977186158236678424689157993532961922\ 62467957194401269043877107275048102390895523597457\ 23189706772547915061505504953922979530901129967519\ 86188088225875314529584099251203829009407770775672\ 11306739708304724483816533873502340845647058077308\ 82959174767140363198008187129011875491310547126581\ 97623331044818386269515456334926366572897563400500\ 42846280183517070527831839425882145521227251250327\ 55121603546981200581762165212827652751691296897789\ 32238195734329339946437501907836945765883352399886\ 75506164965184775180738168837861091527357929701337\ 62177842752192623401942399639168044983993173312731\ 32924185707147349566916674687634660915035914677504\ 99518671430235219628894890102423325116913619626622\ 73267460800591547471830798392868535206946944540724\ 76841822524674417161514036427982273348055556214818\ 97142617910342598647204516893989422179826088076852\ 87783646182799346313767754307809363333018982642090\ 10848802521674670883215120185883543223812876952786\ 71329612474782464538636993009049310363619763878039\ 62184073572399794223406235393808339651327408011116\ 66627891981488087797941876876144230030984490851411\ 60661826293682836764744779239180335110989069790714\ 85786944089552990653640447425576083659976645795096\ 66024396409905389607120198219976047599490197230297\ 64913982680032973156037120041377903785566085089252\ 16730939319872750275468906903707539413042652315011\ 94809377245048795150954100921645863754710598436791\ 78639167021187492431995700641917969777599028300699\ 15368713711936614952811305876380278410754449733078\ 40789923115535562561142322423255033685442488917353\ 44889911501440648020369068063960672322193204149535\ 41503128880339536053299340368006977710650566631954\ 81234880673210146739058568557934581403627822703280\ 82616570773948327592232845941706525094512325230608\ 22918802058777319719839450180888072429661980811197\ 77158542502016545090413245809786882778948721859617\ 72107838435069186155435662884062257473692284509516\ 20849603980134001723930671666823555245252804609722\ 53503534226472524250874054075591789781264330331690' total = 0 for i in xrange(0, 100 * 50 - 1, 50): total += int(number_string[i:i+49]) print str(total)[:10] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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