Search Results

Search found 1001 results on 41 pages for 'evil coder'.

Page 17/41 | < Previous Page | 13 14 15 16 17 18 19 20 21 22 23 24  | Next Page >

  • Project Euler 12: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 12.  As always, any feedback is welcome. # Euler 12 # http://projecteuler.net/index.php?section=problems&id=12 # The sequence of triangle numbers is generated by adding # the natural numbers. So the 7th triangle number would be # 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms # would be: # 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... # Let us list the factors of the first seven triangle # numbers: # 1: 1 # 3: 1,3 # 6: 1,2,3,6 # 10: 1,2,5,10 # 15: 1,3,5,15 # 21: 1,3,7,21 # 28: 1,2,4,7,14,28 # We can see that 28 is the first triangle number to have # over five divisors. What is the value of the first # triangle number to have over five hundred divisors? import time start = time.time() from math import sqrt def divisor_count(x): count = 2 # itself and 1 for i in xrange(2, int(sqrt(x)) + 1): if ((x % i) == 0): if (i != sqrt(x)): count += 2 else: count += 1 return count def triangle_generator(): i = 1 while True: yield int(0.5 * i * (i + 1)) i += 1 triangles = triangle_generator() answer = 0 while True: num = triangles.next() if (divisor_count(num) >= 501): answer = num break; print answer print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 17: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 17.  As always, any feedback is welcome. # Euler 17 # http://projecteuler.net/index.php?section=problems&id=17 # If the numbers 1 to 5 are written out in words: # one, two, three, four, five, then there are # 3 + 3 + 5 + 4 + 4 = 19 letters used in total. # If all the numbers from 1 to 1000 (one thousand) # inclusive were written out in words, how many letters # would be used? # # NOTE: Do not count spaces or hyphens. For example, 342 # (three hundred and forty-two) contains 23 letters and # 115 (one hundred and fifteen) contains 20 letters. The # use of "and" when writing out numbers is in compliance # with British usage. import time start = time.time() def to_word(n): h = { 1 : "one", 2 : "two", 3 : "three", 4 : "four", 5 : "five", 6 : "six", 7 : "seven", 8 : "eight", 9 : "nine", 10 : "ten", 11 : "eleven", 12 : "twelve", 13 : "thirteen", 14 : "fourteen", 15 : "fifteen", 16 : "sixteen", 17 : "seventeen", 18 : "eighteen", 19 : "nineteen", 20 : "twenty", 30 : "thirty", 40 : "forty", 50 : "fifty", 60 : "sixty", 70 : "seventy", 80 : "eighty", 90 : "ninety", 100 : "hundred", 1000 : "thousand" } word = "" # Reverse the numbers so position (ones, tens, # hundreds,...) can be easily determined a = [int(x) for x in str(n)[::-1]] # Thousands position if (len(a) == 4 and a[3] != 0): # This can only be one thousand based # on the problem/method constraints word = h[a[3]] + " thousand " # Hundreds position if (len(a) >= 3 and a[2] != 0): word += h[a[2]] + " hundred" # Add "and" string if the tens or ones # position is occupied with a non-zero value. # Note: routine is broken up this way for [my] clarity. if (len(a) >= 2 and a[1] != 0): # catch 10 - 99 word += " and" elif len(a) >= 1 and a[0] != 0: # catch 1 - 9 word += " and" # Tens and ones position tens_position_value = 99 if (len(a) >= 2 and a[1] != 0): # Calculate the tens position value per the # first and second element in array # e.g. (8 * 10) + 1 = 81 tens_position_value = int(a[1]) * 10 + a[0] if tens_position_value <= 20: # If the tens position value is 20 or less # there's an entry in the hash. Use it and there's # no need to consider the ones position word += " " + h[tens_position_value] else: # Determine the tens position word by # dividing by 10 first. E.g. 8 * 10 = h[80] # We will pick up the ones position word later in # the next part of the routine word += " " + h[(a[1] * 10)] if (len(a) >= 1 and a[0] != 0 and tens_position_value > 20): # Deal with ones position where tens position is # greater than 20 or we have a single digit number word += " " + h[a[0]] # Trim the empty spaces off both ends of the string return word.replace(" ","") def to_word_length(n): return len(to_word(n)) print sum([to_word_length(i) for i in xrange(1,1001)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 19: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 19.  As always, any feedback is welcome. # Euler 19 # http://projecteuler.net/index.php?section=problems&id=19 # You are given the following information, but you may # prefer to do some research for yourself. # # - 1 Jan 1900 was a Monday. # - Thirty days has September, # April, June and November. # All the rest have thirty-one, # Saving February alone, # Which has twenty-eight, rain or shine. # And on leap years, twenty-nine. # - A leap year occurs on any year evenly divisible by 4, # but not on a century unless it is divisible by 400. # # How many Sundays fell on the first of the month during # the twentieth century (1 Jan 1901 to 31 Dec 2000)? import time start = time.time() import datetime sundays = 0 for y in range(1901,2001): for m in range(1,13): # monday == 0, sunday == 6 if datetime.datetime(y,m,1).weekday() == 6: sundays += 1 print sundays print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 2: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 2.  As always, any feedback is welcome. # Euler 2 # http://projecteuler.net/index.php?section=problems&id=2 # Find the sum of all the even-valued terms in the # Fibonacci sequence which do not exceed four million. # Each new term in the Fibonacci sequence is generated # by adding the previous two terms. By starting with 1 # and 2, the first 10 terms will be: # 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... # Find the sum of all the even-valued terms in the # sequence which do not exceed four million. import time start = time.time() total = 0 previous = 0 i = 1 while i <= 4000000: if i % 2 == 0: total +=i # variable swapping removes the need for a temp variable i, previous = previous, previous + i print total print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 11: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 11.  As always, any feedback is welcome. # Euler 11 # http://projecteuler.net/index.php?section=problems&id=11 # What is the greatest product # of four adjacent numbers in any direction (up, down, left, # right, or diagonally) in the 20 x 20 grid? import time start = time.time() grid = [\ [8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8],\ [49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00],\ [81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65],\ [52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91],\ [22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],\ [24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50],\ [32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],\ [67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21],\ [24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],\ [21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95],\ [78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92],\ [16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57],\ [86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58],\ [19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40],\ [04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66],\ [88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69],\ [04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36],\ [20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16],\ [20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54],\ [01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48]] # left and right max, product = 0, 0 for x in range(0,17): for y in xrange(0,20): product = grid[y][x] * grid[y][x+1] * \ grid[y][x+2] * grid[y][x+3] if product > max : max = product # up and down for x in range(0,20): for y in xrange(0,17): product = grid[y][x] * grid[y+1][x] * \ grid[y+2][x] * grid[y+3][x] if product > max : max = product # diagonal right for x in range(0,17): for y in xrange(0,17): product = grid[y][x] * grid[y+1][x+1] * \ grid[y+2][x+2] * grid[y+3][x+3] if product > max: max = product # diagonal left for x in range(0,17): for y in xrange(0,17): product = grid[y][x+3] * grid[y+1][x+2] * \ grid[y+2][x+1] * grid[y+3][x] if product > max : max = product print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 16: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 16.  As always, any feedback is welcome. # Euler 16 # http://projecteuler.net/index.php?section=problems&id=16 # 2^15 = 32768 and the sum of its digits is # 3 + 2 + 7 + 6 + 8 = 26. # What is the sum of the digits of the number 2^1000? import time start = time.time() print sum([int(i) for i in str(2**1000)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 7: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 7.  As always, any feedback is welcome. # Euler 7 # http://projecteuler.net/index.php?section=problems&id=7 # By listing the first six prime numbers: 2, 3, 5, 7, # 11, and 13, we can see that the 6th prime is 13. What # is the 10001st prime number? import time start = time.time() def nthPrime(nth): primes = [2] number = 3 while len(primes) < nth: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes[nth - 1] print nthPrime(10001) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 4: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 4.  As always, any feedback is welcome. # Euler 4 # http://projecteuler.net/index.php?section=problems&id=4 # Find the largest palindrome made from the product of # two 3-digit numbers. A palindromic number reads the # same both ways. The largest palindrome made from the # product of two 2-digit numbers is 9009 = 91 x 99. # Find the largest palindrome made from the product of # two 3-digit numbers. import time start = time.time() def isPalindrome(s): return s == s[::-1] max = 0 for i in xrange(100, 999): for j in xrange(i, 999): n = i * j; if (isPalindrome(str(n))): if (n > max): max = n print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • 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')

    Read the article

  • Project Euler 6: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 6.  As always, any feedback is welcome. # Euler 6 # http://projecteuler.net/index.php?section=problems&id=6 # Find the difference between the sum of the squares of # the first one hundred natural numbers and the square # of the sum. import time start = time.time() square_of_sums = sum(range(1,101)) ** 2 sum_of_squares = reduce(lambda agg, i: agg+i**2, range(1,101)) print square_of_sums - sum_of_squares print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 20: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 20.  As always, any feedback is welcome. # Euler 20 # http://projecteuler.net/index.php?section=problems&id=20 # n! means n x (n - 1) x ... x 3 x 2 x 1 # Find the sum of digits in 100! import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) print sum([int(i) for i in str(factorial(100))]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • Project Euler 1: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 1.  As always, any feedback is welcome. # Euler 1 # http://projecteuler.net/index.php?section=problems&amp;id=1 # If we list all the natural numbers below 10 that are # multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of # these multiples is 23. Find the sum of all the multiples # of 3 or 5 below 1000. import time start = time.time() print sum([x for x in range(1000) if x % 3== 0 or x % 5== 0]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue') # Also cool def constraint(x): return x % 3 == 0 or x % 5 == 0 print sum(filter(constraint, range(1000)))

    Read the article

  • Project Euler 3: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 3.  As always, any feedback is welcome. # Euler 3 # http://projecteuler.net/index.php?section=problems&id=3 # The prime factors of 13195 are 5, 7, 13 and 29. # What is the largest prime factor of the number # 600851475143? import time start = time.time() def largest_prime_factor(n): max = n divisor = 2 while (n >= divisor ** 2): if n % divisor == 0: max, n = n, n / divisor else: divisor += 1 return max print largest_prime_factor(600851475143) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

    Read the article

  • RHEL Cluster FAIL after changing time on system

    - by Eugene S
    I've encountered a strange issue. I had to change the time on my Linux RHEL cluster system. I've done it using the following command from the root user: date +%T -s "10:13:13" After doing this, some message appeared relating to <emerg> #1: Quorum Dissolved however I didn't capture it completely. In order to investigate the issue I looked at /var/log/messages and I've discovered the following: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] entering GATHER state from 0. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] Creating commit token because I am the rep. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] Storing new sequence id for ring 354 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] entering COMMIT state. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] entering RECOVERY state. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] position [0] member 192.168.1.49: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] previous ring seq 848 rep 192.168.1.49 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] aru 61 high delivered 61 received flag 1 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] Did not need to originate any messages in recovery. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] Sending initial ORF token Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] CLM CONFIGURATION CHANGE Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] New Configuration: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] #011r(0) ip(192.168.1.49) Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] Members Left: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] #011r(0) ip(192.168.1.51) Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] Members Joined: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CMAN ] quorum lost, blocking activity Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] CLM CONFIGURATION CHANGE Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] New Configuration: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] #011r(0) ip(192.168.1.49) Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] Members Left: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] Members Joined: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [SYNC ] This node is within the primary component and will provide service. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] entering OPERATIONAL state. Mar 22 16:40:42 hsmsc50sfe1a kernel: dlm: closing connection to node 2 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] got nodejoin message 192.168.1.49 Mar 22 16:40:42 hsmsc50sfe1a clurgmgrd[25809]: <emerg> #1: Quorum Dissolved Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CPG ] got joinlist message from node 1 Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Cluster is not quorate. Refusing connection. Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Error while processing connect: Connection refused Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Invalid descriptor specified (-21). Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Someone may be attempting something evil. Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Error while processing disconnect: Invalid request descriptor Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] entering GATHER state from 9. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] Creating commit token because I am the rep. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] Storing new sequence id for ring 358 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] entering COMMIT state. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] entering RECOVERY state. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] position [0] member 192.168.1.49: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] previous ring seq 852 rep 192.168.1.49 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] aru f high delivered f received flag 1 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] position [1] member 192.168.1.51: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] previous ring seq 852 rep 192.168.1.51 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] aru f high delivered f received flag 1 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] Did not need to originate any messages in recovery. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] Sending initial ORF token Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] CLM CONFIGURATION CHANGE Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] New Configuration: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] #011r(0) ip(192.168.1.49) Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] Members Left: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] Members Joined: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] CLM CONFIGURATION CHANGE Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] New Configuration: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] #011r(0) ip(192.168.1.49) Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] #011r(0) ip(192.168.1.51) Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] Members Left: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] Members Joined: Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] #011r(0) ip(192.168.1.51) Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [SYNC ] This node is within the primary component and will provide service. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [TOTEM] entering OPERATIONAL state. Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [MAIN ] Node chb_sfe2a not joined to cman because it has existing state Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] got nodejoin message 192.168.1.49 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CLM ] got nodejoin message 192.168.1.51 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CPG ] got joinlist message from node 1 Mar 22 16:40:42 hsmsc50sfe1a openais[25715]: [CPG ] got joinlist message from node 2 Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Cluster is not quorate. Refusing connection. Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Error while processing connect: Connection refused Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Invalid descriptor specified (-111). Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Someone may be attempting something evil. Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Error while processing get: Invalid request descriptor Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Invalid descriptor specified (-21). Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Someone may be attempting something evil. Mar 22 16:40:42 hsmsc50sfe1a ccsd[25705]: Error while processing disconnect: Invalid request descriptor How could this be related to the time change procedure I performed?

    Read the article

  • C# Item system design approach, should I use abstract classes, interfaces or virutals?

    - by vexe
    I'm working on a Resident Evil 1/2/3/0/Remake type of game. Currently I've done a big part of the inventory system (here's a link if you wanna see my inventory, pretty outdated, added a lot of features and made a lot of enhancements) Now I'm thinking about how to approach the items system, If you've played any Resident Evil game or any of its likes you should be familiar with what I'm trying to achieve. Here's a very simple category I made for the items: So you have different items, with different operations you could perform on them, there are usable items that you could use, like for example herbs and first aid kits that 'using' them would affect your health, keys to unlock doors, and equipable items that you could 'equip' like weapons. Also, you can 'combine' two items together to get new one, like for example mixing a green and red herb would give you a new type of herb, or combining a lighter with a paper, would give you a burnt paper, or ammo with a gun, would reload the gun or something. etc. You know the usual RE drill. Not all items are 'transformable', in that, for example: lighter + paper = burnt paper (it's the paper that 'transforms' to burnt paper and not the lighter, the lighter is not transformable it will remain as it is) green herb + red herb = newHerb1 (both herbs will vanish and transform to this new type of herb) ammo + gun = reload gun (ammo state will remain as it is, it won't change but it will just decrease, nothing will happen to the gun it just gets reloaded) Also a key note to remember is that you can't just combine items randomly, each item has a 'mating' item(s). So to sum up, different items, and different operations on them. The question is, how to approach this, design-wise? I've been learning about interfaces, but it just doesn't quite get into my head, I mean, why not just use classes with the good old inheritance? I know the technical details of interfaces and that the cool thing about them is that they don't require an inheritance chain, but I just can't see how to use them properly, that is, if they were the right thing to use here. So should I go with just classes and inheritance? just like in the tree I showed you? or should I think more about how to use interfaces? (IUsable, IEquipable, ITransformable) - why not just use classes UsableItem, Equipable item, TransformableItem? I want something that won't give me headaches in the long run, something resilient/flexible to future changes. I'm OK using classes, but I smell something better here. The way I'm thinking is to possibly use both inheritance and interfaces, so that you have a branch like this: item - equipable - weapon. but then again, the weapon has methods like 'reload' 'examine' 'equip' some of them 'combine' so I'm thinking to make weapon implement ICombinable?... not all items get used the same way, using herbs will increase your health, using a key will open a door, so IUsable maybe? Should I use a big database (XML for example) for all the items, items names, mates, nRowsReq, nColsReq, etc? Thanks so much for your answers in advanced, note that demo 3 is coming after I'm done with items :D

    Read the article

  • Java in Flux: Utopia or Deuteranopia?

    - by Tori Wieldt
    What a difference a year makes, indeed. Steve Harris, Senior VP, App Server Dev, Oracle and Adam Messinger, VP, Fusion Middleware Group, Oracle presented an informative keynote at the TheServerSide Java Symposium today. With a title "Java in Flux: Utopia or Deuteranopia?" you know things are going to be interesting (see Aeon Flux if you don't get the title reference).What a YearThey started with a little background, explaining that the reactions to Oracle's acquisition of Sun (and therefore Java) one year ago varied greatly, from "Freak Out!" to "Don't Panic." From the Oracle perspective, being the steward of and key contributor to Java requires a lot of sausage making.  They admitted to Oracle's fair share of Homer Simpson-esque "D'oh" moments in the past year, which was complicated by Oracle's communication style.   "Oracle has a tradition has a saying a few things and sticking by then, in contrast to Sun who was much more open," Adam explained. "We laid out the Java roadmap and are executing on it, and we hope that speaks to our commitment."Java SEAdam talked about having a long term perspective on the Java language (20+ years), letting ideas mature in more experimental languages, then bringing them into Java. Current priorities include: JVM convergence (getting the best features of JRockit into Hotspot); support of parallel/multi-core programming, and of course, all the improvements in JDK7. The JDK7 Developer Preview is underway (please download now and report bugs!). The Oracle development team is also working on Lambda and modularity (Jigsaw) for SE 8. Less certain, but also under discussion are improvements for Java SE 9. Adam is thinking of it as a "back to basics" release. He mentioned reworking JNI, improving data integration and improved device support.Java EE To provide context about Java EE, Steve said Java EE was great at getting businesses on the internet. The success of Java EE resulted in an incredible expansion of the middleware marketplace for developers and vendors.  But with success, came more. Java EE kept piling on capabilities, but that created excess baggage.  Doing simple things was no longer so simple. That's where Java community is so valuable: "When Java EE was too complex and heavyweight, many people were happy to tell us what we were doing wrong and popularize solutions," Steve explained. Because of that feedback, the Java EE teams focused on making things simple again: POJOs and annotations, and leveraging changes in Java SE.  Steve said that "innovation doesn't happen in expert groups, it happens on the ground where developers are solving problems," and platform stewards need to pay attention and take advantage of changes that are taking place.Enter the Cloud "Developers are restless, they want cloud functionality from their own IT dept" Steve explained. With the cloud, the scope of problem has expanded to include the data center itself, with multiple tenants. To move forward, existing APIs in Java EE need to be updated to be tenant-aware, service-enabled, and EE needs to support various styles of deployment. The goal is to get all that done in Java EE 8.Adam questioned Steve about timing and schedule. "Yes, the schedule is aggressive, but it'll work" Steve said. Then Adam asked about modularization. If Java SE 8 comes out at the end of 2012, when can Java EE deliver modularization? Steve suggested that key stakeholders can come with up some pre-SE 8 agreement on how to expose the metadata about modules. He then alluded to Mark Reinhold and John Duimovich's keynote at EclipseCON next week. Stay tuned.Evil Master PlanIn conclusion, Adam finally admitted to Oracle's Evil Master Plan: 1) Invest in and improve Java SE and EE 2) Collaborate with the community 3) Broaden the marketplace for Java development. Bwaaaaaaaaahahaha! <rubs hands together>Key LinksJDK7 Developer Preview  http://jdk7.java.net/preview/Oracle Technology Network http://www.oracle.com/technetwork/java/index.htmlTheServerSide Java Symposium  http://javasymposium.techtarget.com/"Utopia or Deuteranopia?" http://en.wikipedia.org/wiki/Aeon_Flux

    Read the article

  • Countdown of Top 10 Reasons to Never Ever Use a Pie Chart

    - by Tony Wolfram
      Pie charts are evil. They represent much of what is wrong with the poor design of many websites and software applications. They're also innefective, misleading, and innacurate. Using a pie chart as your graph of choice to visually display important statistics and information demonstrates either a lack of knowledge, laziness, or poor design skills. Figure 1: A floating, tilted, 3D pie chart with shadow trying (poorly)to show usage statistics within a graphics application.   Of course, pie charts in and of themselves are not evil. This blog is really about designers making poor decisions for all the wrong reasons. In order for a pie chart to appear on a web page, somebody chose it over the other alternatives, and probably thought they were doing the right thing. They weren't. Using a pie chart is almost always a bad design decision. Figure 2: Pie Chart from an Oracle Reports User Guide   A pie chart does not do the job of effectively displaying information in an elegant visual form.  Being circular, they use up too much space while not allowing their labels to line up. Bar charts, line charts, and tables do a much better job. Expert designers, statisticians, and business analysts have documented their many failings, and strongly urge software and report designers not to use them. It's obvious to them that the pie chart has too many inherent defects to ever be used effectively. Figure 3: Demonstration of how comparing data between multiple pie charts is difficult.   Yet pie charts are still used frequently in today's software applications, financial reports, and websites, often on the opening page as a symbol of how the data inside is represented. In an attempt to get a flashy colorful graphic to break up boring text, designers will often settle for a pie chart that looks like pac man, a colored spinning wheel, or a 3D floating alien space ship.     Figure 4: Best use of a pie chart I've found yet.   Why is the pie chart so popular? Through its constant use and iconic representation as the classic chart, the idea persists that it must be a good choice, since everyone else is still using it. Like a virus or an urban legend, no amount of vaccine or debunking will slow down the use of pie charts, which seem to be resistant to logic and common sense. Even the new iPad from Apple showcases the pie chart as one of its options.     Figure 5: Screen shot of new iPad showcasing pie charts. Regardless of the futility in trying to rid the planet of this often used poor design choice, I now present to you my top 10 reasons why you should never, ever user a pie chart again.    Number 10 - Pie Charts Just Don't Work When Comparing Data Number 9 - You Have A Better Option: The Sorted Horizontal Bar Chart Number 8 - The Pie Chart is Always Round Number 7 - Some Genius Will Make It 3D Number 6 - Legends and Labels are Hard to Align and Read Number 5 - Nobody Has Ever Made a Critical Decision Using a Pie Chart Number 4 - It Doesn't Scale Well to More Than 2 Items Number 3 - A Pie Chart Causes Distortions and Errors Number 2 - Everyone Else Uses Them: Debunking the "Urban Legend" of Pie Charts Number 1 - Pie Charts Make You Look Stupid and Lazy  

    Read the article

  • Cannot find CFML template for custom tag

    - by jerrygarciuh
    Hi folks, I am not a ColdFusion coder. Doing a favor for a friend who ported his CF site from a Windows server to Unix on GoDaddy. Site is displaying error: Cannot find CFML template for custom tag jstk. ColdFusion attempted looking in the tree of installed custom tags but did not find a custom tag with this name. The site as I found it has at document root /CustomTags with the jstk.cfm file and a set of files in cf_jstk My Googling located this You must store custom tag pages in any one of the following: The same directory as the calling page; The cfusion\CustomTags directory; A subdirectory of the cfusion\CustomTags directory; A directory that you specify in the ColdFusion Administrator So I have Tried creating placing /CustomTags in /cfusion/CustomTags Tried copying /cfusion/CustomTags to above document root Tried copying jstk.cfm and subfolders into same directory as calling file(index.cfm). Update: Per GoDaddy support I have also tried adding the following to no effect: Can any one give me some tips on this or should I just tell my guy to look for a CF coder? Thanks! JG

    Read the article

  • How to protect/monitor your site from crawling by malicious user

    - by deathy
    Situation: Site with content protected by username/password (not all controlled since they can be trial/test users) a normal search engine can't get at it because of username/password restrictions a malicious user can still login and pass the session cookie to a "wget -r" or something else. The question would be what is the best solution to monitor such activity and respond to it (considering the site policy is no-crawling/scraping allowed) I can think of some options: Set up some traffic monitoring solution to limit the number of requests for a given user/IP. Related to the first point: Automatically block some user-agents (Evil :)) Set up a hidden link that when accessed logs out the user and disables his account. (Presumably this would not be accessed by a normal user since he wouldn't see it to click it, but a bot will crawl all links.) For point 1. do you know of a good already-implemented solution? Any experiences with it? One problem would be that some false positives might show up for very active but human users. For point 3: do you think this is really evil? Or do you see any possible problems with it? Also accepting other suggestions.

    Read the article

  • What harm can javascript do?

    - by The King
    I just happen to read the joel's blog here... So for example if you have a web page that says “What is your name?” with an edit box and then submitting that page takes you to another page that says, Hello, Elmer! (assuming the user’s name is Elmer), well, that’s a security vulnerability, because the user could type in all kinds of weird HTML and JavaScript instead of “Elmer” and their weird JavaScript could do narsty things, and now those narsty things appear to come from you, so for example they can read cookies that you put there and forward them on to Dr. Evil’s evil site. Since javascript runs on client end. All it can access or do is only on the client end. It can read informations stored in hidden fields and change them. It can read, write or manipulate cookies... But I feel, these informations are anyway available to him. (if he is smart enough to pass javascript in a textbox. So we are not empowering him with new information or providing him undue access to our server... Just curious to know whether I miss something. Can you list the things that a malicious user can do with this security hole. Edit : Thanks to all for enlightening . As kizzx2 pointed out in one of the comments... I was overlooking the fact that a JavaScript written by User A may get executed in the browser of User B under numerous circumstances, in which case it becomes a great risk.

    Read the article

  • UIView: how to do non-destructive drawing?

    - by Caffeine Coma
    My original question: I'm creating a simple drawing application and need to be able to draw over existing, previously drawn content in my drawRect. What is the proper way to draw on top of existing content without entirely replacing it? Based on answers received here and elsewhere, here is the deal. You should be prepared to redraw the entire rectangle whenever drawRect is called. You cannot prevent the contents from being erased by doing the following: [self setClearsContextBeforeDrawing: NO]; This is merely a hint to the graphics engine that there is no point in having it pre-clear the view for you, since you will likely need to re-draw the whole area anyway. It may prevent your view from being automatically erased, but you cannot depend on it. To draw on top of your view without erasing, do your drawing to an off-screen bitmap context (which is never cleared by the system.) Then in your drawRect, copy from this off-screen buffer to the view. Example: - (id) initWithCoder: (NSCoder*) coder { if (self = [super initWithCoder: coder]) { self.backgroundColor = [UIColor clearColor]; CGSize size = self.frame.size; drawingContext = [self createDrawingBufferContext: size]; } return self; } - (CGContextRef) createOffscreenContext: (CGSize) size { CGColorSpaceRef colorSpace = CGColorSpaceCreateDeviceRGB(); CGContextRef context = CGBitmapContextCreate(NULL, size.width, size.height, 8, size.width*4, colorSpace, kCGImageAlphaPremultipliedLast); CGColorSpaceRelease(colorSpace); CGContextTranslateCTM(context, 0, size.height); CGContextScaleCTM(context, 1.0, -1.0); return context; } - (void)drawRect:(CGRect) rect { UIGraphicsPushContext(drawingContext); CGImageRef cgImage = CGBitmapContextCreateImage(drawingContext); UIImage *uiImage = [[UIImage alloc] initWithCGImage:cgImage]; UIGraphicsPopContext(); CGImageRelease(cgImage); [uiImage drawInRect: rect]; [uiImage release]; } TODO: can anyone optimize the drawRect so that only the (usually tiny) modified rectangle region is used for the copy?

    Read the article

  • Where'd my sounds go?

    - by Dane Man
    In my Row class I have the initWithCoder method and everything restored in that method works fine, but only within the method. After the method is called I loose an array of sounds that is in my Row class. The sounds class also has the initWithCoder method, and the sound plays fine but only in the Row class initWithCoder method. After decoding the Row object, the sound array disappears completely and is unable to be called. Here's my source for the initWithCoder: - (id) initWithCoder:(NSCoder *)coder { ... soundArray = [coder decodeObjectForKey:@"soundArray"]; NSLog(@"%d",[soundArray count]); return self; } the log shows the count as 8 like it should (this is while unarchiving). Then the row object I create gets assigned. And the resulting row object no longer has a soundArray. [NSKeyedArchiver archiveRootObject:row toFile:@"DefaultRow"]; ... row = [NSKeyedUnarchiver unarchiveObjectWithFile:@"DefaultRow"]; So now whenever I call the soundArray it crashes. //ERROR IS HERE NSLog(@"%d",[[row soundArray] count]); Help please (soundArray is an NSMutableArray).

    Read the article

  • tips for fixing bad coding/dev habits ?

    - by dfafa
    i want to become a better coder....so i have decided to sign up for computing science program...maybe a formal education can assist me. i started working on smaller projects to learn but currently i have really bad coding/dev habits which is hindering my productivity as the codebase increases.... i have highlighted them and perhaps someone could make suggestions (or redirect to resources) or a more efficient method. most stuff that i made in the past were web apps. i usually develop with putty + nano...i just love the minimalist feel i use winscp and develop directly on my private web server...too lazy to do it on localhost and upload it later. i dont use subversion control...which one do i need ? sometimes ctrl +z doesn't work well. when i run out of ideas for naming variable, i use swear words instead. i swear a lot when i get stuck....how to deal with anger issue ? my codes look ugly with comments everywhere. would rather use procedural coding finds "thinking" in OO difficult and time consuming i "write first think later". refactors code only if i am getting paid for it. dislikes configuring linux distro, Apache, MySQL, scaling, designing graphics and layouts. does not like writing tests likes working alone. does not like sharing codes. has an econ degree dislikes reading other people's code would rather write it on my own it seems my only true desire is to translate my ideas to a working prototype as fast as possible....it seems like i am very uninterested in the other details...could it be that i am not cut out to be a coder after all ? is going back to study comp sci a bad idea ?

    Read the article

  • need advice on Zend framework Application architecture, or say approach dealing with modules

    - by simple
    Let me start with the things that I did and how am I using some things to get results I have set up modular structure as: application/ /configs /layouts /models /modules /users /profile /frontend /backend /controllers /views .... I write a plugin that does addes changes with FrontController-setModuleControllerDirectoryName() FrontController-addModuleDirectory() and It is all good I have a changed all the directories according weather admin page is requested in the url or not (it is some thing like /admin/some/some) Let's say I have a single layout for anything that is related to Profile viewing , in this case the "Profile" module. The Profile layout is divided into three parts In the layout I was pulling out the Profile/PhotoController 's index action with a action() $this->action('index', 'photo', 'profile'); Then I have faced few issues 1. Can get passed Params inside the Photo Controller when calling ( profile/profile/index); 2. found out that helper Action() is evil cause it starts another dispatching loop =) --- and now I am thinking that my approach on plugging in controllers modules into layout also evil =). anyhow how Should I deal with plugging in some controllers (another module controllers) into the layout ?

    Read the article

  • Remove duplicate records/objects uniquely identified by multiple attributes

    - by keruilin
    I have a model called HeroStatus with the following attributes: id user_id recordable_type hero_type (can be NULL!) recordable_id created_at There are over 100 hero_statuses, and a user can have many hero_statuses, but can't have the same hero_status more than once. A user's hero_status is uniquely identified by the combination of recordable_type + hero_type + recordable_id. What I'm trying to say essentially is that there can't be a duplicate hero_status for a specific user. Unfortunately, I didn't have a validation in place to assure this, so I got some duplicate hero_statuses for users after I made some code changes. For example: user_id = 18 recordable_type = 'Evil' hero_type = 'Halitosis' recordable_id = 1 created_at = '2010-05-03 18:30:30' user_id = 18 recordable_type = 'Evil' hero_type = 'Halitosis' recordable_id = 1 created_at = '2009-03-03 15:30:00' user_id = 18 recordable_type = 'Good' hero_type = 'Hugs' recordable_id = 1 created_at = '2009-02-03 12:30:00' user_id = 18 recordable_type = 'Good' hero_type = NULL recordable_id = 2 created_at = '2009-012-03 08:30:00' (Last two are not a dups obviously. First two are.) So what I want to do is get rid of the duplicate hero_status. Which one? The one with the most-recent date. I have three questions: How do I remove the duplicates using a SQL-only approach? How do I remove the duplicates using a pure Ruby solution? Something similar to this: http://stackoverflow.com/questions/2790004/removing-duplicate-objects. How do I put a validation in place to prevent duplicate entries in the future?

    Read the article

< Previous Page | 13 14 15 16 17 18 19 20 21 22 23 24  | Next Page >