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• How to deal with lots of brackets in a formula?

  - by wenlibin02

  Say, I have a formula like this (in LaTeX or Maple or other text system): Result: ((6*(k2+k3))*A123*k2*k3*(A12*A13*k2^2-2*A12*A13*k2*k3+A12*A13*k3^2-A123*k2^2-2*A123*k2*k3-A123*k3^2)*(exp(-k3*(k3^2*t-x)))^2+6*A12*(-k3+k2)*k2*k3*(A12*A13*k2^2-2*A12*A13*k2*k3+A12*A13*k3^2-A123*k2^2-2*A123*k2*k3-A123*k3^2)*exp(-k3*(k3^2*t-x)))*(exp(-k2*(k2^2*t-x)))^2+(-(6*(-k3+k2))*A13*k2*k3*(A12*A13*k2^2-2*A12*A13*k2*k3+A12*A13*k3^2-A123*k2^2-2*A123*k2*k3-A123*k3^2)*(exp(-k3*(k3^2*t-x)))^2-(6*(k2+k3))*k2*k3*(A12*A13*k2^2-2*A12*A13*k2*k3+A12*A13*k3^2-A123*k2^2-2*A123*k2*k3-A123*k3^2)*exp(-k3*(k3^2*t-x)))*exp(-k2*(k2^2*t-x)) Note: the above formula is only one part of the result of a maple calculation, I just can't break them up because there are so many many terms. Apparently, It's very hard to read. What I want to do is to fold the matched brackets level by level. If all the brackets are folded, I can find out clearly how many terms there are. Then I can analyze from the top level to the details of every term. But I just don't know how to realize that. Maybe there are some existed software which can visualize this kind of complex formula. Any idea? P.S. I use Linux system. The open source alternatives are better.

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• python: sorting

  - by nabizan

  hi im doing a loop so i could get dict of data, but since its a dict it's sorting alphabetical and not as i push it trought the loop ... is it possible to somehow turn off alphabetical sorting? here is how do i do that data = {} for item in container: data[item] = {} ... for key, val in item_container.iteritems(): ... data[item][key] = val whitch give me something like this data = { A : { K1 : V1, K2 : V2, K3 : V3 }, B : { K1 : V1, K2 : V2, K3 : V3 }, C : { K1 : V1, K2 : V2, K3 : V3 } } and i want it to be as i was going throught the loop, e.g. data = { B : {K2 : V2, K3 : V3, K1 : V1}, A : {K1 : V1, K2 : V2, K3 : V3}, C : {K3 : V3, K1 : V1, K2 : V2} }

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• Ruby: rules for implicit hashes

  - by flyer

  Why second output shows me only one element of Array? Is it still Array or Hash already? def printArray(arr) arr.each { | j | k, v = j.first printf("%s %s %s \n", k, v, j) } end print "Array 1\n" printArray( [ {kk: { 'k1' => 'v1' }}, {kk: { 'k2' => 'v2' }}, {kk: { 'k3' => 'v3' }}, ]) print "Array 2\n" printArray( [ kk: { 'k1' => 'v1' }, kk: { 'k2' => 'v2' }, kk: { 'k3' => 'v3' }, ]) exit # Output: # # Array 1 # kk {"k1"=>"v1"} {:kk=>{"k1"=>"v1"}} # kk {"k2"=>"v2"} {:kk=>{"k2"=>"v2"}} # kk {"k3"=>"v3"} {:kk=>{"k3"=>"v3"}} # Array 2 # kk {"k3"=>"v3"} {:kk=>{"k3"=>"v3"}}

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• how to speed up the code??

  - by kaushik

  i have very huge code about 600 lines plus. cant post the whole thing here. but a particular code snippet is taking so much time,leading to problems. here i post that part of code please tell me what to do speed up the processing.. please suggest the part which may be the reason and measure to improve them if this small part of code is understandable. using_data={} def join_cost(a , b): global using_data #print a #print b save_a=[] save_b=[] print 1 #for i in range(len(m)): #if str(m[i][0])==str(a): save_a=database_index[a] #for i in range(len(m)): # if str(m[i][0])==str(b): #print 'save_a',save_a #print 'save_b',save_b print 2 save_b=database_index[b] using_data[save_a[0]]=save_a s=str(save_a[1]).replace('phone','text') s=str(s)+'.pm' p=os.path.join("c:/begpython/wavnk/",s) x=open(p , 'r') print 3 for i in range(6): x.readline() k2='a' j=0 o=[] while k2 is not '': k2=x.readline() k2=k2.rstrip('\n') oj=k2.split(' ') o=o+[oj] #print o[j] j=j+1 #print j #print o[2][0] temp=long(1232332) end_time=save_a[4] #print end_time k=(j-1) for i in range(k): diff=float(o[i][0])-float(end_time) if diff<0: diff=diff*(-1) if temp>diff: temp=diff pm_row=i #print pm_row #print temp #print o[pm_row] #pm_row=3 q=[] print 4 l=str(p).replace('.pm','.mcep') z=open(l ,'r') for i in range(pm_row): z.readline() k3=z.readline() k3=k3.rstrip('\n') q=k3.split(' ') #print q print 5 s=str(save_b[1]).replace('phone','text') s=str(s)+'.pm' p=os.path.join("c:/begpython/wavnk/",s) x=open(p , 'r') for i in range(6): x.readline() k2='a' j=0 o=[] while k2 is not '': k2=x.readline() k2=k2.rstrip('\n') oj=k2.split(' ') o=o+[oj] #print o[j] j=j+1 #print j #print o[2][0] temp=long(1232332) strt_time=save_b[3] #print strt_time k=(j-1) for i in range(k): diff=float(o[i][0])-float(strt_time) if diff<0: diff=diff*(-1) if temp>diff: temp=diff pm_row=i #print pm_row #print temp #print o[pm_row] #pm_row=3 w=[] l=str(p).replace('.pm','.mcep') z=open(l ,'r') for i in range(pm_row): z.readline() k3=z.readline() k3=k3.rstrip('\n') w=k3.split(' ') #print w cost=0 for i in range(12): #print q[i] #print w[i] h=float(q[i])-float(w[i]) cost=cost+math.pow(h,2) j_cost=math.sqrt(cost) #print cost return j_cost def target_cost(a , b): a=(b+1)*3 b=(a+1)*2 t_cost=(a+b)*5/2 return t_cost r1='shht:ra_77' r2='grx_18' g=[] nodes=[] nodes=nodes+[[r1]] for i in range(len(y_in_db_format)): g=y_in_db_format[i] #print g #print g[0] g.remove(str(g[0])) nodes=nodes+[g] nodes=nodes+[[r2]] print nodes print "lenght of nodes",len(nodes) lists=[] #lists=lists+[r1] for i in range(len(nodes)): for j in range(len(nodes[i])): lists=lists+[nodes[i][j]] #lists=lists+[r2] print lists distance={} for i in range(len(lists)): if i==0: distance[str(lists[i])]=0 else: distance[str(lists[i])]=long(123231223) #print distance group_dist=[] infinity=long(123232323) for i in range(len(nodes)): distances=[] for j in range(len(nodes[i])): #distances=[] if i==0: distances=distances+[[nodes[i][j], 0]] else: distances=distances+[[nodes[i][j],infinity]] group_dist=group_dist+[distances] #print distances print "group_distances",group_dist #print "check",group_dist[0][0][1] #costs={} #for i in range(len(lists)): #if i==0: # costs[str(lists[i])]=1 #else: # costs[str(lists[i])]=get_selfcost(lists[i]) path=[] for i in range(len(nodes)): mini=[] if i!=(len(nodes)-1): #temp=long(123234324) #Now calculate the cost between the current node and each of its neighbour for k in range(len(nodes[(i+1)])): for j in range(len(nodes[i])): current=nodes[i][j] #print "current_node",current j_distance=join_cost( current , nodes[i+1][k]) #t_distance=target_cost( current , nodes[i+1][k]) t_distance=34 #print distance #print "distance between current and neighbours",distance total_distance=(.5*(float(group_dist[i][j][1])+float(j_distance))+.5*(float(t_distance))) #print "total distance between the intial_nodes and current neighbour",total_distance if int(group_dist[i+1][k][1]) > int(total_distance): group_dist[i+1][k][1]=total_distance #print "updated distance",group_dist[i+1][k][1] a=current #print "the neighbour",nodes[i+1][k],"updated the value",a mini=mini+[[str(nodes[i+1][k]),a]] print mini

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• Runge-Kutta Method with adaptive step

  - by infoholic_anonymous

  I am implementing Runge-Kutta method with adaptive step in matlab. I get different results as compared to matlab's own ode45 and my own implementation of Runge-Kutta method with fixed step. What am I doing wrong in my code? Is it possible? function [ result ] = rk4_modh( f, int, init, h, h_min ) % % f - function handle % int - interval - pair (x_min, x_max) % init - initial conditions - pair (y1(0),y2(0)) % h_min - lower limit for h (step length) % h - initial step length % x - independent variable ( for example time ) % y - dependent variable - vertical vector - in our case ( y1, y2 ) function [ k1, k2, k3, k4, ka, y ] = iteration( f, h, x, y ) % core functionality performed within loop k1 = h * f(x,y); k2 = h * f(x+h/2, y+k1/2); k3 = h * f(x+h/2, y+k2/2); k4 = h * f(x+h, y+k3); ka = (k1 + 2*k2 + 2*k3 + k4)/6; y = y + ka; end % constants % relative error eW = 1e-10; % absolute error eB = 1e-10; s = 0.9; b = 5; % initialization i = 1; x = int(1); y = init; while true hy = y; hx = x; %algorithm [ k1, k2, k3, k4, ka, y ] = iteration( f, h, x, y ); % error estimation for j=1:2 [ hk1, hk2, hk3, hk4, hka, hy ] = iteration( f, h/2, hx, hy ); hx = hx + h/2; end err(:,i) = abs(hy - y); % step adjustment e = abs( hy ) * eW + eB; a = min( e ./ err(:,i) )^(0.2); mul = a * s; if mul >= 1 % step length admitted keepH(i) = h; k(:,:,i) = [ k1, k2, k3, k4, ka ]; previous(i,:) = [ x+h, y' ]; %' i = i + 1; if floor( x + h + eB ) == int(2) break; else h = min( [mul*h, b*h, int(2)-x] ); x = x + keepH(i-1); end else % step length requires further adjustments h = mul * h; if ( h < h_min ) error('Computation with given precision impossible'); end end end result = struct( 'val', previous, 'k', k, 'err', err, 'h', keepH ); end The function in question is: function [ res ] = fun( x, y ) % res(1) = y(2) + y(1) * ( 0.9 - y(1)^2 - y(2)^2 ); res(2) = -y(1) + y(2) * ( 0.9 - y(1)^2 - y(2)^2 ); res = res'; %' end The call is: res = rk4( @fun, [0,20], [0.001; 0.001], 0.008 ); The resulting plot for x1 : The result of ode45( @fun, [0, 20], [0.001, 0.001] ) is:

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• Unix sort keys cause performance problems

  - by KenFar

  My data: It's a 71 MB file with 1.5 million rows. It has 6 fields All six fields combine to form a unique key - so that's what I need to sort on. Sort statement: sort -t ',' -k1,1 -k2,2 -k3,3 -k4,4 -k5,5 -k6,6 -o output.csv input.csv The problem: If I sort without keys, it takes 30 seconds. If I sort with keys, it takes 660 seconds. I need to sort with keys to keep this generic and useful for other files that have non-key fields as well. The 30 second timing is fine, but the 660 is a killer. More details using unix time: sort input.csv -o output.csv = 28 seconds sort -t ',' -k1 input.csv -o output.csv = 28 seconds sort -t ',' -k1,1 input.csv -o output.csv = 64 seconds sort -t ',' -k1,1 -k2,2 input.csv -o output.csv = 194 seconds sort -t ',' -k1,1 -k2,2 -k3,3 input.csv -o output.csv = 328 seconds sort -t ',' -k1,1 -k2,2 -k3,3 -k4,4 input.csv -o output.csv = 483 seconds sort -t ',' -k1,1 -k2,2 -k3,3 -k4,4 -k5,5 input.csv -o output.csv = 561 seconds sort -t ',' -k1,1 -k2,2 -k3,3 -k4,4 -k5,5 -k6,6 input.csv -o output.csv = 660 seconds I could theoretically move the temp directory to SSD, and/or split the file into 4 parts, sort them separately (in parallel) then merge the results, etc. But I'm hoping for something simpler since looks like sort is just picking a bad algorithm. Any suggestions? Testing Improvements using buffer-size: With 2 keys I got a 5% improvement with 8, 20, 24 MB and best performance of 8% improvement with 16MB, but 6% worse with 128MB With 6 keys I got a 5% improvement with 8, 20, 24 MB and best performance of 9% improvement with 16MB. Testing improvements using dictionary order (just 1 run each): sort -d --buffer-size=8M -t ',' -k1,1 -k2,2 input.csv -o output.csv = 235 seconds (21% worse) sort -d --buffer-size=8M -t ',' -k1,1 -k2,2 input.csv -o ouput.csv = 232 seconds (21% worse) conclusion: it makes sense that this would slow the process down, not useful Testing with different file system on SSD - I can't do this on this server now. Testing with code to consolidate adjacent keys: def consolidate_keys(key_fields, key_types): """ Inputs: - key_fields - a list of numbers in quotes: ['1','2','3'] - key_types - a list of types of the key_fields: ['integer','string','integer'] Outputs: - key_fields - a consolidated list: ['1,2','3'] - key_types - a list of types of the consolidated list: ['string','integer'] """ assert(len(key_fields) == len(key_types)) def get_min(val): vals = val.split(',') assert(len(vals) <= 2) return vals[0] def get_max(val): vals = val.split(',') assert(len(vals) <= 2) return vals[len(vals)-1] i = 0 while True: try: if ( (int(get_max(key_fields[i])) + 1) == int(key_fields[i+1]) and key_types[i] == key_types[i+1]): key_fields[i] = '%s,%s' % (get_min(key_fields[i]), key_fields[i+1]) key_types[i] = key_types[i] key_fields.pop(i+1) key_types.pop(i+1) continue i = i+1 except IndexError: break # last entry return key_fields, key_types While this code is just a work-around that'll only apply to cases in which I've got a contiguous set of keys - it speeds up the code by 95% in my worst case scenario.

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• SQL Server 2008 ContainsTable, CTE, and Paging

  - by David Murdoch

  I'd like to perform efficient paging using containstable. The following query selects the top 10 ranked results from my database using containstable when searching for a name (first or last) that begins with "Joh". DECLARE @Limit int; SET @Limit = 10; SELECT TOP @Limit c.ChildID, c.PersonID, c.DOB, c.Gender FROM [Person].[vFullName] AS v INNER JOIN CONTAINSTABLE( [Person].[vFullName], (FullName), IS ABOUT ( "Joh*" WEIGHT (.4), "Joh" WEIGHT (.6)) ) AS k3 ON v.PersonID = k3.[KEY] JOIN [Child].[Details] c ON c.PersonID = v.PersonID JOIN [Person].[Details] p ON p.PersonID = c.PersonID ORDER BY k3.RANK DESC, FullName ASC, p.Active DESC, c.ChildID ASC I'd like to combine it with the following CTE which returns the 10th-20th results ordered by ChildID (the primary key): DECLARE @Start int; DECLARE @Limit int; SET @Start = 10; SET @Limit = 10; WITH ChildEntities AS ( SELECT ROW_NUMBER() OVER (ORDER BY ChildID) AS Row, ChildID FROM Child.Details ) SELECT c.ChildID, c.PersonID, c.DOB, c.Gender FROM ChildEntities cte INNER JOIN Child.Details c ON cte.ChildID = c.ChildID WHERE cte.Row BETWEEN @Start+1 AND @Start+@Limit ORDER BY cte.Row ASC

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• Append class="external" to external links.

  - by K3

  What is the php code to append class="external" to links that are posted and are not the domain. For example my site is www.mysite.com and you post a link to www.mysite.com/news and a link to www.yoursite.com How do I set it so only non "mysite.com" links have the class specified?

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• Why are there 3 conflicting OpenCV camera calibration formulas?

  - by John

  I'm having a problem with OpenCV's various parameterization of coordinates used for camera calibration purposes. The problem is that three different sources of information on image distortion formulae apparently give three non-equivalent description of the parameters and equations involved: (1) In their book "Learning OpenCV…" Bradski and Kaehler write regarding lens distortion (page 376): xcorrected = x * ( 1 + k1 * r^2 + k2 * r^4 + k3 * r^6 ) + [ 2 * p1 * x * y + p2 * ( r^2 + 2 * x^2 ) ], ycorrected = y * ( 1 + k1 * r^2 + k2 * r^4 + k3 * r^6 ) + [ p1 * ( r^2 + 2 * y^2 ) + 2 * p2 * x * y ], where r = sqrt( x^2 + y^2 ). Assumably, (x, y) are the coordinates of pixels in the uncorrected captured image corresponding to world-point objects with coordinates (X, Y, Z), camera-frame referenced, for which xcorrected = fx * ( X / Z ) + cx and ycorrected = fy * ( Y / Z ) + cy, where fx, fy, cx, and cy, are the camera's intrinsic parameters. So, having (x, y) from a captured image, we can obtain the desired coordinates ( xcorrected, ycorrected ) to produced an undistorted image of the captured world scene by applying the above first two correction expressions. However... (2) The complication arises as we look at OpenCV 2.0 C Reference entry under the Camera Calibration and 3D Reconstruction section. For ease of comparison we start with all world-point (X, Y, Z) coordinates being expressed with respect to the camera's reference frame, just as in #1. Consequently, the transformation matrix [ R | t ] is of no concern. In the C reference, it is expressed that: x' = X / Z, y' = Y / Z, x'' = x' * ( 1 + k1 * r'^2 + k2 * r'^4 + k3 * r'^6 ) + [ 2 * p1 * x' * y' + p2 * ( r'^2 + 2 * x'^2 ) ], y'' = y' * ( 1 + k1 * r'^2 + k2 * r'^4 + k3 * r'^6 ) + [ p1 * ( r'^2 + 2 * y'^2 ) + 2 * p2 * x' * y' ], where r' = sqrt( x'^2 + y'^2 ), and finally that u = fx * x'' + cx, v = fy * y'' + cy. As one can see these expressions are not equivalent to those presented in #1, with the result that the two sets of corrected coordinates ( xcorrected, ycorrected ) and ( u, v ) are not the same. Why the contradiction? It seems to me the first set makes more sense as I can attach physical meaning to each and every x and y in there, while I find no physical meaning in x' = X / Z and y' = Y / Z when the camera focal length is not exactly 1. Furthermore, one cannot compute x' and y' for we don't know (X, Y, Z). (3) Unfortunately, things get even murkier when we refer to the writings in Intel's Open Source Computer Vision Library Reference Manual's section Lens Distortion (page 6-4), which states in part: "Let ( u, v ) be true pixel image coordinates, that is, coordinates with ideal projection, and ( u ~, v ~ ) be corresponding real observed (distorted) image coordinates. Similarly, ( x, y ) are ideal (distortion-free) and ( x ~, y ~ ) are real (distorted) image physical coordinates. Taking into account two expansion terms gives the following: x ~ = x * ( 1 + k1 * r^2 + k2 * r^4 ) + [ 2 p1 * x * y + p2 * ( r^2 + 2 * x^2 ) ] y ~ = y * ( 1 + k1 * r^2 + k2 * r^4 ] + [ 2 p2 * x * y + p2 * ( r^2 + 2 * y^2 ) ], where r = sqrt( x^2 + y^2 ). ... "Because u ~ = cx + fx * u and v ~ = cy + fy * v , … the resultant system can be rewritten as follows: u ~ = u + ( u – cx ) * [ k1 * r^2 + k2 * r^4 + 2 * p1 * y + p2 * ( r^2 / x + 2 * x ) ] v ~ = v + ( v – cy ) * [ k1 * r^2 + k2 * r^4 + 2 * p2 * x + p1 * ( r^2 / y + 2 * y ) ] The latter relations are used to undistort images from the camera." Well, it would appear that the expressions involving x ~ and y ~ coincided with the two expressions given at the top of this writing involving xcorrected and ycorrected. However, x ~ and y ~ do not refer to corrected coordinates, according to the given description. I don't understand the distinction between the meaning of the coordinates ( x ~, y ~ ) and ( u ~, v ~ ), or for that matter, between the pairs ( x, y ) and ( u, v ). From their descriptions it appears their only distinction is that ( x ~, y ~ ) and ( x, y ) refer to 'physical' coordinates while ( u ~, v ~ ) and ( u, v ) do not. What is this distinction all about? Aren't they all physical coordinates? I'm lost! Thanks for any input!

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• how to write the code for this program specially in mathematica? [closed]

  - by asd

  I implemented a solution to the problem below in Mathematica, but it takes a very long time (hours) to compute f of kis or the set B for large numbers. Somebody suggested that implementing this in C++ resulted in a solution in less than 10 minutes. Would C++ be a good language to learn to solve these problems, or can my Mathematica code be improved to fix the performance issues? I don't know anything about C or C++ and it should be difficult to start to learn this languages. I prefer to improve or write new code in mathematica. Problem Description Let $f$ be an arithmetic function and A={k1,k2,...,kn} are integers in increasing order. Now I want to start with k1 and compare f(ki) with f(k1). If f(ki)f(k1), put ki as k1. Now start with ki, and compare f(kj) with f(ki), for ji. If f(kj)f(ki), put kj as ki, and repeat this procedure. At the end we will have a sub sequence B={L1,...,Lm} of A by this property: f(L(i+1))f(L(i)), for any 1<=i<=m-1 For example, let f is the divisor function of integers. Here I put some part of my code and this is just a sample and the question in my program could be more larger than these: «««««««««««««««««««««««««««««««««««« f[n_] := DivisorSigma[0, n]; g[n_] := Product[Prime[i], {i, 1, PrimePi[n]}]; k1 = g[67757] g[353] g[59] g[19] g[11] g[7] g[5]^2 6^3 2^7; k2 = g[67757] g[353] g[59] g[19] g[11] g[7] g[5] 6^5 2^7; k3 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^7; k4 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5] 6^5 2^6; k5 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^8; k6 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^7; k7 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^5 2^6; k8 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^9; k9 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^7; k10 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^5 2^7; k11 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^6; k12 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^8; k13 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^6; k14 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^9; k15 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^7; k16 = g[67757] g[359] g[53] g[23] g[11] g[7] g[5] 6^4 2^8; k17 = g[67757] g[359] g[59] g[19] g[11] g[7] g[5] 6^4 2^7; k18 = g[67757] g[359] g[53] g[23] g[11] g[7] g[5] 6^4 2^9; k19 = g[67759] g[353] g[53] g[19] g[11] g[7] g[5] 6^4 2^6; k20 = g[67763] g[347] g[53] g[19] g[11] g[7] g[5] 6^4 2^7; k = Table[k1, k2, k3, k4, k5, k6, k7, k8, k9, k10, k11, k12, k13, k14, k15, k16, k17, k18, k19, k20]; i = 1; count = 0; For[j = i, j <= 20, j++, If[f[k[[j]]] - f[k[[i]]] > 0, i = j; Print["k",i]; count = count + 1]]; Print["count= ", count] ««««««««««««««««««««««««««««««««««««

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• html to text with domdocument class

  - by turbod

  How to get a html page source code without htl tags? For example: <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <meta http-equiv="content-language" content="hu"/> <title>this is the page title</title> <meta name="description" content="this is the description" /> <meta name="keywords" content="k1, k2, k3, k4" /> start the body content <!-- <div>this is comment</div> --> <a href="open.php" title="this is title attribute">open</a> End now one noframes tag. <noframes><span>text</span></noframes> <select name="select" id="select"><option>ttttt</option></select> <div class="robots-nocontent"><span>something</span></div> <img src="url.png" alt="this is alt attribute" /> I need this result: this is the page title this is the description k1, k2, k3, k4 start the body content this is title attribute open End now one noframes tag. text ttttt something this is alt attribute I need too the title and the alt attributes. Idea?

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• Rewrite arrays using collections

  - by owca

  I have a task, which I was able to do with the use of simplest methods - arrays. Now I'd like to go further and redo it using some more complicated java features like collections, but I've never used anything more complicated than 2d matrix. What should I look at and how to start with it. Should Tower become a Collection ? And here's the task : We have two classes - Tower and Block. Towers are built from Blocks. Ande here's sample code for testing: Block k1=new Block("yellow",1,5,4); Block k2=new Block("blue",2,2,6); Block k3=new Block("green",3,4,2); Block k4=new Block("yellow",1,5,4); Tower tower=new Tower(); tower.add(k1,k2,k3); "Added 3 blocks." System.out.println(tower); "block: green, base: 4cm x 3cm, thicknes: 2 cm block: blue, base: 6cm x 2cm, thicknes: 2 cm block: yellow, base: 5cm x 4cm, thicknes: 1 cm" tower.add(k2); "Tower already contains this block." tower.add(k4); "Added 1 block." System.out.println(tower); "block: green, base: 4cm x 3cm, thicknes: 2 cm block: blue, base: 6cm x 2cm, thicknes: 2 cm block: yellow, base: 5cm x 4cm, thicknes: 1 cm block: yellow, base: 5cm x 4cm, thicknes: 1 cm" tower.delete(k1); "Deleted 1 block" tower.delete(k1); "Block not in tower" System.out.println(tower); "block: blue, base: 6cm x 2cm, thicknes: 2 cm block: yellow, base: 5cm x 4cm, thicknes: 1 cm block: yellow, base: 5cm x 4cm, thicknes: 1 cm" Let's say I will treat Tower as a collection of blocks. How to perform search for specific block among whole collection ? Or should I use other interface ?

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• Symfony2 bindrequest error

  - by user1321109

  namespace Topxia\LiftBundle\Form; use Symfony\Component\Validator\Constraint as Assert; class AddUser { /** * @Assert\NotBlank(message="???????") */ public $username; /** * @Assert\NotBlank(message="????????") */ public $name; /** * @Assert\NotBlank(message="??????") * @Assert\MinLength(limit=4,message="?????") * @Assert\MaxLength(limit=24,message="?????") */ public $password; /** * @Assert\NotBlank(message="??????") */ public $rpassword; /** * @Assert\NotBlank(message="???????") * @Assert\Email(message="???????") */ public $email; /** * @Assert\NotBlank(message="k3????????") */ public $num; /** * @Assert\NotBlank() * @Assert\Choice({"0", "1", "2", "3", "4"}) */ public $roles; public $changePassword; } namespace Topxia\LiftBundle\Form; use Symfony\Component\Form\AbstractType; use Symfony\Component\Form\FormBuilder; class AddUserType extends AbstractType{ public function buildForm(FormBuilder $builder, array $options){ $builder->add('username', 'text', array('label' => '????')); $builder->add('name', 'text', array('label' => '???')); $builder->add('password', 'password', array('label' => '??')); $builder->add('rpassword', 'password', array('label' => '????')); $builder->add('email', 'email', array('label' => '??')); $builder->add('num', 'text', array('label' => 'K3????')); $builder->add('roles', 'choice', array('label' => '???', 'multiple' => false, 'expanded' => true, 'choices' => array( '0' => '?????', '1' => '?????', '2' => '?????', '3' => '???', '4' => '???' ), )); $builder->add('changePassword', 'checkbox', array('label' => '??????', 'value' => '1')); } public function getName(){ return 'add_user'; } } when i use $form-bindRequest($request); in conctollor there is an error : [Semantical Error] The annotation "@Symfony\Component\Validator\Constraint\NotBlank" in property Topxia\LiftBundle\Form\AddUser::$username does not exist, or could not be auto-loaded. 500 Internal Server Error - AnnotationException I have no idea about this. Thanks

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• How would I change this Query builder from Kohana 2.4 to Kohana 3?

  - by Thorpe Obazee

  I have had this website built for a few months and I am just getting on Kohana 3. I'd just like to convert this K2.4 query builder to K3 query builder. return DB::select(array('posts.id', 'posts.created', 'posts.uri', 'posts.price', 'posts.description', 'posts.title', 'image_count' => db::expr('COUNT(images.id)'))) ->from('posts') ->join('images')->on('images.post_id', '=', 'posts.id') ->group_by(array('posts.id')) ->order_by('posts.id', 'DESC') ->limit($limit) ->offset($offset) ->execute();

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• how to add enum value to list

  - by netmajor

  I have enum: public enum SymbolWejsciowy { K1 , K2 , K3 , K4 , K5 , K6 , K7 , K8 } and I want to create list of this enum public List<SymbolWejsciowy> symbol; but in my way to add enum to List: 1. SymbolWejsciowy symbol ; symbol.Add(symbol = SymbolWejsciowy.K1); 2. symbol.Add(SymbolWejsciowy.K1); I always get Exception Object reference not set to an instance of an object. What i do wrong :/ Please help :)

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• Oracle Systems and Solutions at OpenWorld Tokyo 2012

  - by ferhat

  Oracle OpenWorld Tokyo and JavaOne Tokyo will start next week April 4th. We will cover Oracle systems and Oracle Optimized Solutions in several keynote talks and general sessions. Full schedule can be found here. Come by the DemoGrounds to learn more about mission critical integration and optimization of complete Oracle stack. Our Oracle Optimized Solutions experts will be at hand to discuss 1-1 several of Oracle's systems solutions and technologies. Oracle Optimized Solutions are proven blueprints that eliminate integration guesswork by combing best in class hardware and software components to deliver complete system architectures that are fully tested, and include documented best practices that reduce integration risks and deliver better application performance. And because they are highly flexible by design, Oracle Optimized Solutions can be implemented as an end-to-end solution or easily adapted into existing environments. Oracle Optimized Solutions, Servers,  Storage, and Oracle Solaris  Sessions, Keynotes, and General Session Talks DAY TIME TITLE Notes Session Wednesday  April 4 9:00 - 11:15 Keynote: ENGINEERED FOR INNOVATION - Engineered Systems Mark Hurd,  President, Oracle Takao Endo, President & CEO, Oracle Corporation Japan John Fowler, EVP of Systems, Oracle Ed Screven, Chief Corporate Architect, Oracle English Session K1-01 11:50 - 12:35 Simplifying IT: Transforming the Data Center with Oracle's Engineered Systems Robert Shimp, Group VP, Product Marketing, Oracle English Session S1-01 15:20 - 16:05 Introducing Tiered Storage Solution for low cost Big Data Archiving S1-33 16:30 - 17:15 Simplifying IT - IT System Consolidation that also Accelerates Business Agility S1-42 Thursday  April 5 9:30 - 11:15 Keynote: Extreme Innovation Larry Ellison, Chief Executive Officer, Oracle English Session K2-01 11:50 - 13:20 General Session: Server and Storage Systems Strategy John Fowler, EVP of Systems, Oracle English Session G2-01 16:30 - 17:15 Top 5 Reasons why ZFS Storage appliance is "The cloud storage" by SAKURA Internet Inc L2-04 16:30 - 17:15 The UNIX based Exa* Performance IT Integration Platform - SPARC SuperCluster S2-42 17:40 - 18:25 Full stack solutions of hardware and software with SPARC SuperCluster and Oracle E-Business Suite  to minimize the business cost while maximizing the agility, performance, and availability S2-53 Friday April 6 9:30 - 11:15 Keynote: Oracle Fusion Applications & Cloud Robert Shimp, Group VP, Product Marketing Anthony Lye, Senior VP English Session K3-01 11:50 - 12:35 IT at Oracle: The Art of IT Transformation to Enable Business Growth English Session S3-02 13:00-13:45 ZFS Storagge Appliance: Architecture of high efficient and high performance S3-13 14:10 - 14:55 Why "Niko Niko doga" chose ZFS Storage Appliance to support their growing requirements and storage infrastructure By DWANGO Co, Ltd. S3-21 15:20 - 16:05 Osaka University: Lower TCO and higher flexibility for student study by Virtual Desktop By Osaka University S3-33 Oracle Developer Sessions with Oracle Systems and Oracle Solaris DAY TIME TITLE Notes LOCATION Friday April 6 13:00 - 13:45 Oracle Solaris 11 Developers D3-03 13:00 - 14:30 Oracle Solaris Tuning Contest Hands-On Lab D3-04 14:00 - 14:35 How to build high performance and high security Oracle Database environment with Oracle SPARC/Solaris English Session D3-13 15:00 - 15:45 IT Assets preservation and constructive migration with Oracle Solaris virtualization D3-24 16:00 - 17:30 The best packaging system for cloud environment - Creating an IPS package D3-34 Follow Oracle Infrared at Twitter, Facebook, Google+, and LinkedIn  to catch the latest news, developments, announcements, and inside views from  Oracle Optimized Solutions.

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• How do I get the F1-F12 keys to switch screens in gnu screen in cygwin when connecting via SSH?

  - by Mikey

  I'm connecting to a desktop running cygwin via SSH from the terminal app in Mac OS X. I have already started screen on the cygwin side and can connect to it over the SSH session. Furthermore, I have the following in the .screenrc file: bindkey -k k1 select 1 # F1 = screen 1 bindkey -k k2 select 2 # F2 = screen 2 bindkey -k k3 select 3 # F3 = screen 3 bindkey -k k4 select 4 # F4 = screen 4 bindkey -k k5 select 5 # F5 = screen 5 bindkey -k k6 select 6 # F6 = screen 6 bindkey -k k7 select 7 # F7 = screen 7 bindkey -k k8 select 8 # F8 = screen 8 bindkey -k k9 select 9 # F9 = screen 9 bindkey -k F1 prev # F11 = prev bindkey -k F2 next # F12 = next However, when I start multiple windows in screen and attempt to switch between them via the function keys, all I get is a beep. I have tried various settings for $TERM (e.g. ansi, cygwin, xterm-color, vt100) and they don't really seem to affect anything. I have verified that the terminal app is in fact sending the escape sequence for the function key that I'm expecting and that my bash shell (running inside screen) is receiving it. For example, for F1, it sends the following (hexdump is a perl script I wrote that takes STDIN in binmode and outputs it as a hexadecimal/ascii dump): % hexdump [press F1 and then hit ^D to terminate input] 00000000: 1b4f50 .OP If things were working correctly, I don't think bash should receive the escape sequence because screen should have caught it and turned it into a command. How do I get the function keys to work?

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• Unix sort 10x slower with keys specified

  - by KenFar

  My data: It's a 71 MB file with 1.5 million rows. It has 6 fields, four of which are strings of avg. 15 characters, two are integers. Three of the fields are sometimes empty. All six fields combine to form a unique key - and that's what I need to sort on. Sort statement: sort -t ',' -k1,1 -k2,2 -k3,3 -k4,4 -k5,5 -k6,6 -o a_out.csv a_in.csv The problem: If I sort without keys, it takes 30 seconds. If I sort with keys, it takes 660 seconds. I need to sort with keys to keep this generic and useful for other files that have non-key fields as well. The 30 second timing is fine, but the 660 is a killer. I could theoretically move the temp directory to SSD, and/or split the file into 4 parts, sort them separately (in parallel) then merge the results, etc. But I'm hoping for something simpler since these results are so bad as-is. Any suggestions?

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• how to store data in ram in verilog

  - by anum

  i am having a bit stream of 128 bits @ each posedge of clk,i.e.total 10 bit streams each of length 128 bits. i want to divide the 128 bit stream into 8, 8 bits n hve to store them in a ram / memory of width 8 bits. i did it by assigning 8, 8 bits to wires of size 8 bit.in this way there are 16 wires. and i am using dual port ram...wen i cal module of memory in stimulus.i don know how to give input....as i am hving 16 different wires naming from k1 to k16. **codeeee** // this is stimulus file module final_stim; reg [7:0] in,in_data; reg clk,rst_n,rd,wr,rd_data,wr_data; wire [7:0] out,out_wr, ouut; wire[7:0] d; integer i; //wire[7:0] xor_out; reg kld,f; reg [127:0]key; wire [127:0] key_expand; wire [7:0]out_data; reg [7:0] k; //wire [7:0] k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12,k13,k14,k15,k16; wire [7:0] out_data1; **//key_expand is da output which is giving 10 streams of size 128 bits.** assign k1=key_expand[127:120]; assign k2=key_expand[119:112]; assign k3=key_expand[111:104]; assign k4=key_expand[103:96]; assign k5=key_expand[95:88]; assign k6=key_expand[87:80]; assign k7=key_expand[79:72]; assign k8=key_expand[71:64]; assign k9=key_expand[63:56]; assign k10=key_expand[55:48]; assign k11=key_expand[47:40]; assign k12=key_expand[39:32]; assign k13=key_expand[31:24]; assign k14=key_expand[23:16]; assign k15=key_expand[15:8]; assign k16=key_expand[7:0]; **// then the module of memory is instanciated. //here k1 is sent as input.but i don know how to save the other values of k. //i tried to use for loop but it dint help** memory m1(clk,rst_n,rd, wr,k1,out_data1); aes_sbox b(out,d); initial begin clk=1'b1; rst_n=1'b0; #20 rst_n = 1; //rd=1'b1; wr_data=1'b1; in=8'hd4; #20 //rst_n=1'b1; in=8'h27; rd_data=1'b0; wr_data=1'b1; #20 in=8'h11; rd_data=1'b0; wr_data=1'b1; #20 in=8'hae; rd_data=1'b0; wr_data=1'b1; #20 in=8'he0; rd_data=1'b0; wr_data=1'b1; #20 in=8'hbf; rd_data=1'b0; wr_data=1'b1; #20 in=8'h98; rd_data=1'b0; wr_data=1'b1; #20 in=8'hf1; rd_data=1'b0; wr_data=1'b1; #20 in=8'hb8; rd_data=1'b0; wr_data=1'b1; #20 in=8'hb4; rd_data=1'b0; wr_data=1'b1; #20 in=8'h5d; rd_data=1'b0; wr_data=1'b1; #20 in=8'he5; rd_data=1'b0; wr_data=1'b1; #20 in=8'h1e; rd_data=1'b0; wr_data=1'b1; #20 in=8'h41; rd_data=1'b0; wr_data=1'b1; #20 in=8'h52; rd_data=1'b0; wr_data=1'b1; #20 in=8'h30; rd_data=1'b0; wr_data=1'b1; #20 wr_data=1'b0; #380 rd_data=1'b1; #320 rd_data = 1'b0; /////////////// #10 kld = 1'b1; key=128'h 2b7e151628aed2a6abf7158809cf4f3c; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b0; #10 wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 kld = 1'b0; key = 128'h 2b7e151628aed2a6abf7158809cf4f3c; wr = 1'b1; rd = 1'b1; #20 wr = 1'b0; #20 rd = 1'b1; #4880 f=1'b1; ///////////////////////////////////////////////// // out_data[i] end /*always@(*) begin while(i) mem[i]^mem1[i] ; i<=16; break; end*/ always #10 clk=~clk; always@(posedge clk) begin //$monitor($time," out_wr=%h,out_rd=%h\n ",out_wr,out); #10000 $stop; end endmodule

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• How do I get the F1-F12 keys to switch screens in gnu screen in cygwin when connecting via SSH?

  - by Mikey

  I'm connecting to a desktop running cygwin via SSH from the terminal app in Mac OS X. I have already started screen on the cygwin side and can connect to it over the SSH session. Furthermore, I have the following in the .screenrc file: bindkey -k k1 select 1 # F1 = screen 1 bindkey -k k2 select 2 # F2 = screen 2 bindkey -k k3 select 3 # F3 = screen 3 bindkey -k k4 select 4 # F4 = screen 4 bindkey -k k5 select 5 # F5 = screen 5 bindkey -k k6 select 6 # F6 = screen 6 bindkey -k k7 select 7 # F7 = screen 7 bindkey -k k8 select 8 # F8 = screen 8 bindkey -k k9 select 9 # F9 = screen 9 bindkey -k F1 prev # F11 = prev bindkey -k F2 next # F12 = next However, when I start multiple windows in screen and attempt to switch between them via the function keys, all I get is a beep. I have tried various settings for $TERM (e.g. ansi, cygwin, xterm-color, vt100) and they don't really seem to affect anything. I have verified that the terminal app is in fact sending the escape sequence for the function key that I'm expecting and that my bash shell (running inside screen) is receiving it. For example, for F1, it sends the following (hexdump is a perl script I wrote that takes STDIN in binmode and outputs it as a hexadecimal/ascii dump): % hexdump [press F1 and then hit ^D to terminate input] 00000000: 1b4f50 .OP If things were working correctly, I don't think bash should receive the escape sequence because screen should have caught it and turned it into a command. How do I get the function keys to work?

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• L2TP connection fails!

  - by a.toraby

  I've installed l2tp-ipsec-vpn but when I try to connect to the vpn server I get error 500. Here are the logs: Jun 17 12:54:37.449 ipsec_setup: Stopping Openswan IPsec... Jun 17 12:54:38.858 Stopping xl2tpd: xl2tpd. Jun 17 12:54:38.859 xl2tpd[1511]: death_handler: Fatal signal 15 received Jun 17 12:54:38.872 ipsec_setup: Starting Openswan IPsec U2.6.37/K3.2.0-23-generic... Jun 17 12:54:39.027 ipsec__plutorun: Starting Pluto subsystem... Jun 17 12:54:39.033 ipsec__plutorun: adjusting ipsec.d to /etc/ipsec.d Jun 17 12:54:39.037 recvref[30]: Protocol not available Jun 17 12:54:39.038 xl2tpd[2442]: This binary does not support kernel L2TP. Jun 17 12:54:39.038 xl2tpd[2444]: xl2tpd version xl2tpd-1.3.1 started on atp-ThinkPad-SL410 PID:2444 Jun 17 12:54:39.038 xl2tpd[2444]: Written by Mark Spencer, Copyright (C) 1998, Adtran, Inc. Jun 17 12:54:39.038 xl2tpd[2444]: Forked by Scott Balmos and David Stipp, (C) 2001 Jun 17 12:54:39.038 xl2tpd[2444]: Inherited by Jeff McAdams, (C) 2002 Jun 17 12:54:39.039 xl2tpd[2444]: Forked again by Xelerance (www.xelerance.com) (C) 2006 Jun 17 12:54:39.039 xl2tpd[2444]: Listening on IP address 0.0.0.0, port 1701 Jun 17 12:54:39.040 Starting xl2tpd: xl2tpd. Jun 17 12:54:39.062 ipsec__plutorun: 002 added connection description "L2TP" Jun 17 12:55:30.753 104 "L2TP" #1: STATE_MAIN_I1: initiate Jun 17 12:55:30.754 010 "L2TP" #1: STATE_MAIN_I1: retransmission; will wait 20s for response Jun 17 12:55:30.754 010 "L2TP" #1: STATE_MAIN_I1: retransmission; will wait 40s for response Jun 17 12:55:30.754 003 "L2TP" #1: ignoring Vendor ID payload [MS NT5 ISAKMPOAKLEY 00000008] Jun 17 12:55:30.754 003 "L2TP" #1: received Vendor ID payload [RFC 3947] method set to=109 Jun 17 12:55:30.754 003 "L2TP" #1: received Vendor ID payload [draft-ietf-ipsec-nat-t-ike-02_n] meth=106, but already using method 109 Jun 17 12:55:30.755 003 "L2TP" #1: ignoring Vendor ID payload [FRAGMENTATION] Jun 17 12:55:30.755 003 "L2TP" #1: ignoring Vendor ID payload [MS-Negotiation Discovery Capable] Jun 17 12:55:30.755 003 "L2TP" #1: ignoring Vendor ID payload [IKE CGA version 1] Jun 17 12:55:30.755 106 "L2TP" #1: STATE_MAIN_I2: sent MI2, expecting MR2 Jun 17 12:55:30.755 010 "L2TP" #1: STATE_MAIN_I2: retransmission; will wait 20s for response Jun 17 12:55:30.755 003 "L2TP" #1: NAT-Traversal: Result using RFC 3947 (NAT-Traversal): i am NATed Jun 17 12:55:30.755 108 "L2TP" #1: STATE_MAIN_I3: sent MI3, expecting MR3 Jun 17 12:55:30.756 004 "L2TP" #1: STATE_MAIN_I4: ISAKMP SA established {auth=OAKLEY_PRESHARED_KEY cipher=oakley_3des_cbc_192 prf=oakley_sha group=modp1024} Jun 17 12:55:30.756 117 "L2TP" #2: STATE_QUICK_I1: initiate Jun 17 12:55:30.756 010 "L2TP" #2: STATE_QUICK_I1: retransmission; will wait 20s for response Jun 17 12:55:30.756 003 "L2TP" #2: ignoring informational payload, type IPSEC_RESPONDER_LIFETIME msgid=6b03ff69 Jun 17 12:55:30.756 003 "L2TP" #2: NAT-Traversal: received 2 NAT-OA. ignored because peer is not NATed Jun 17 12:55:30.756 003 "L2TP" #2: our client subnet returned doesn't match my proposal - us:192.168.1.3/32 vs them:109.162.174.235/32 Jun 17 12:55:30.757 003 "L2TP" #2: Allowing questionable proposal anyway [ALLOW_MICROSOFT_BAD_PROPOSAL] Jun 17 12:55:30.757 004 "L2TP" #2: STATE_QUICK_I2: sent QI2, IPsec SA established transport mode {ESP=>0x23af21f8 <0xdb4a87b6 xfrm=AES_128-HMAC_SHA1 NATOA=none NATD=none DPD=none} Jun 17 12:55:31.759 xl2tpd[2444]: Connecting to host x.x.x.x, port 1701 Jun 17 12:55:32.021 xl2tpd[2444]: Connection established to x.x.x.x, 1701. Local: 4720, Remote: 200 (ref=0/0). Jun 17 12:55:32.023 xl2tpd[2444]: Calling on tunnel 4720 Jun 17 12:55:32.454 xl2tpd[2444]: Call established with x.x.x.x, Local: 9667, Remote: 3, Serial: 1 (ref=0/0) Jun 17 12:55:32.456 xl2tpd[2444]: start_pppd: I'm running: Jun 17 12:55:32.456 xl2tpd[2444]: "/usr/sbin/pppd" Jun 17 12:55:32.457 xl2tpd[2444]: "passive" Jun 17 12:55:32.458 xl2tpd[2444]: "nodetach" Jun 17 12:55:32.458 xl2tpd[2444]: ":" Jun 17 12:55:32.459 xl2tpd[2444]: "file" Jun 17 12:55:32.459 xl2tpd[2444]: "/etc/ppp/L2TP.options.xl2tpd" Jun 17 12:55:32.460 xl2tpd[2444]: "ipparam" Jun 17 12:55:32.461 xl2tpd[2444]: "x.x.x.x" Jun 17 12:55:32.462 xl2tpd[2444]: "/dev/pts/1" Jun 17 12:55:32.583 pppd[2711]: Plugin passprompt.so loaded. Jun 17 12:55:32.583 pppd[2711]: pppd 2.4.5 started by root, uid 0 Jun 17 12:55:32.619 pppd[2711]: Using interface ppp0 Jun 17 12:55:32.620 pppd[2711]: Connect: ppp0 <--> /dev/pts/1 Jun 17 12:55:33.693 pppd[2711]: /usr/bin/L2tpIPsecVpn exited with code 0 Jun 17 12:55:34.454 [ERROR 404] Authentication failed: closing connection to 'L2TP' Jun 17 12:55:34.456 pppd[2711]: MS-CHAP authentication failed: E=691 Authentication failure Jun 17 12:55:34.457 pppd[2711]: CHAP authentication failed Jun 17 12:55:34.461 Stopping xl2tpd: xl2tpd. Jun 17 12:55:34.462 xl2tpd[2444]: death_handler: Fatal signal 15 received Jun 17 12:55:34.463 pppd[2711]: Modem hangup Jun 17 12:55:34.463 pppd[2711]: Connection terminated. Jun 17 12:55:34.474 ipsec_setup: Stopping Openswan IPsec... Jun 17 12:55:34.482 pppd[2711]: Exit. Jun 17 12:55:35.587 ipsec_setup: ERROR: Module xfrm4_mode_transport is in use Jun 17 12:55:35.665 ipsec_setup: ERROR: Module esp4 is in use I had this problem by ubuntu 11.10 though I can easily connect to the server from windows. I use ubuntu 12.0 64bit

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• Algorithm to Render a Horizontal Binary-ish Tree in Text/ASCII form

  - by Justin L.

  It's a pretty normal binary tree, except for the fact that one of the nodes may be empty. I'd like to find a way to output it in a horizontal way (that is, the root node is on the left and expands to the right). I've had some experience expanding trees vertically (root node at the top, expanding downwards), but I'm not sure where to start, in this case. Preferably, it would follow these couple of rules: If a node has only one child, it can be skipped as redundant (an "end node", with no children, is always displayed) All nodes of the same depth must be aligned vertically; all nodes must be to the right of all less-deep nodes and to the left of all deeper nodes. Nodes have a string representation which includes their depth. Each "end node" has its own unique line; that is, the number of lines is the number of end nodes in the tree, and when an end node is on a line, there may be nothing else on that line after that end node. As a consequence of the last rule, the root node should be in either the top left or the bottom left corner; top left is preferred. For example, this is a valid tree, with six end nodes (node is represented by a name, and its depth): [a0]------------[b3]------[c5]------[d8] \ \ \----------[e9] \ \----[f5] \--[g1]--------[h4]------[i6] \ \--------------------[j10] \-[k3] Which represents the horizontal, explicit binary tree: 0 a / \ 1 g * / \ \ 2 * * * / \ \ 3 k * b / / \ 4 h * * / \ \ \ 5 * * f c / \ / \ 6 * i * * / / \ 7 * * * / / \ 8 * * d / / 9 * e / 10 j (branches folded for compactness; * representing redundant, one-child nodes; note that *'s are actual nodes, storing one child each, just with names omitted here for presentation sake) (also, to clarify, I'd like to generate the first, horizontal tree; not this vertical tree) I say language-agnostic because I'm just looking for an algorithm; I say ruby because I'm eventually going to have to implement it in ruby anyway. Assume that each Node data structure stores only its id, a left node, and a right node. A master Tree class keeps tracks of all nodes and has adequate algorithms to find: A node's nth ancestor A node's nth descendant The generation of a node The lowest common ancestor of two given nodes Anyone have any ideas of where I could start? Should I go for the recursive approach? Iterative?

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• OpenSwan IPSec phase #2 complications

  - by XXL

  Phase #1 (IKE) succeeds without any problems (verified at the target host). Phase #2 (IPSec), however, is erroneous at some point (apparently due to misconfiguration on localhost). This should be an IPSec-only connection. I am using OpenSwan on Debian. The error log reads the following (the actual IP-addr. of the remote endpoint has been modified): pluto[30868]: "x" #2: initiating Quick Mode PSK+ENCRYPT+PFS+UP+IKEv2ALLOW+SAREFTRACK {using isakmp#1 msgid:5ece82ee proposal=AES(12)_256-SHA1(2)_160 pfsgroup=OAKLEY_GROUP_DH22} pluto[30868]: "x" #1: ignoring informational payload, type NO_PROPOSAL_CHOSEN msgid=00000000 pluto[30868]: "x" #1: received and ignored informational message pluto[30868]: "x" #1: the peer proposed: 0.0.0.0/0:0/0 - 0.0.0.0/0:0/0 pluto[30868]: "x" #3: responding to Quick Mode proposal {msgid:a4f5a81c} pluto[30868]: "x" #3: us: 192.168.1.76<192.168.1.76[+S=C] pluto[30868]: "x" #3: them: 222.222.222.222<222.222.222.222[+S=C]===10.196.0.0/17 pluto[30868]: "x" #3: transition from state STATE_QUICK_R0 to state STATE_QUICK_R1 pluto[30868]: "x" #3: STATE_QUICK_R1: sent QR1, inbound IPsec SA installed, expecting QI2 pluto[30868]: "x" #1: ignoring informational payload, type NO_PROPOSAL_CHOSEN msgid=00000000 pluto[30868]: "x" #1: received and ignored informational message pluto[30868]: "x" #3: next payload type of ISAKMP Hash Payload has an unknown value: 97 X pluto[30868]: "x" #3: malformed payload in packet pluto[30868]: | payload malformed after IV I am behind NAT and this is all coming from wlan2. Here are the details: default via 192.168.1.254 dev wlan2 proto static 169.254.0.0/16 dev wlan2 scope link metric 1000 192.168.1.0/24 dev wlan2 proto kernel scope link src 192.168.1.76 metric 2 Output of ipsec verify: Checking your system to see if IPsec got installed and started correctly: Version check and ipsec on-path [OK] Linux Openswan U2.6.37/K3.2.0-24-generic (netkey) Checking for IPsec support in kernel [OK] SAref kernel support [N/A] NETKEY: Testing XFRM related proc values [OK] [OK] [OK] Checking that pluto is running [OK] Pluto listening for IKE on udp 500 [OK] Pluto listening for NAT-T on udp 4500 [OK] Two or more interfaces found, checking IP forwarding [OK] Checking NAT and MASQUERADEing [OK] Checking for 'ip' command [OK] Checking /bin/sh is not /bin/dash [WARNING] Checking for 'iptables' command [OK] Opportunistic Encryption Support [DISABLED] This is what happens when I run ipsec auto --up x: 104 "x" #1: STATE_MAIN_I1: initiate 003 "x" #1: received Vendor ID payload [RFC 3947] method set to=109 106 "x" #1: STATE_MAIN_I2: sent MI2, expecting MR2 003 "x" #1: received Vendor ID payload [Cisco-Unity] 003 "x" #1: received Vendor ID payload [Dead Peer Detection] 003 "x" #1: ignoring unknown Vendor ID payload [502099ff84bd4373039074cf56649aad] 003 "x" #1: received Vendor ID payload [XAUTH] 003 "x" #1: NAT-Traversal: Result using RFC 3947 (NAT-Traversal): i am NATed 108 "x" #1: STATE_MAIN_I3: sent MI3, expecting MR3 004 "x" #1: STATE_MAIN_I4: ISAKMP SA established {auth=OAKLEY_PRESHARED_KEY cipher=aes_128 prf=oakley_sha group=modp1024} 117 "x" #2: STATE_QUICK_I1: initiate 010 "x" #2: STATE_QUICK_I1: retransmission; will wait 20s for response 010 "x" #2: STATE_QUICK_I1: retransmission; will wait 40s for response 031 "x" #2: max number of retransmissions (2) reached STATE_QUICK_I1. No acceptable response to our first Quick Mode message: perhaps peer likes no proposal 000 "x" #2: starting keying attempt 2 of at most 3, but releasing whack I have enabled NAT traversal in ipsec.conf accordingly. Here are the settings relative to the connection in question: version 2.0 config setup plutoopts="--perpeerlog" plutoopts="--interface=wlan2" dumpdir=/var/run/pluto/ nat_traversal=yes virtual_private=%v4:10.0.0.0/8,%v4:192.168.0.0/16,%v4:172.16.0.0/12 oe=off protostack=netkey conn x authby=secret pfs=yes auto=add phase2alg=aes256-sha1;dh22 keyingtries=3 ikelifetime=8h type=transport left=192.168.1.76 leftsubnet=192.168.1.0/24 leftprotoport=0/0 right=222.222.222.222 rightsubnet=10.196.0.0/17 rightprotoport=0/0 Here are the specs provided by the other end that must be met for Phase #2: encryption algorithm: AES (128 or 256 bit) hash algorithm: SHA local ident1 (addr/mask/prot/port): (10.196.0.0/255.255.128.0/0/0) local ident2 (addr/mask/prot/port): (10.241.0.0/255.255.0.0/0/0) remote ident (addr/mask/prot/port): (x.x.x.x/x.x.x.x/0/0) (internal network or localhost) Security association lifetime: 4608000 kilobytes/3600 seconds PFS: DH group2 So, finally, what might be the cause of the issue that I am experiencing? Thank you.

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• Why does Google mark one e-mail as spam while does not the other?

  - by nKn

  I've a Postfix installation which works fine, I don't get any trouble with mails sent through a mail client (in my case, Thunderbird or RoundCube) when the To: address is a GMail account. However, I recently needed to use the PHPMailer tool to send some e-mails to some GMail accounts, so I configured an account to be used via SASL authentication + TLS. I don't mean mass mailing, just 2-3 mails. If I send the e-mail from the Thunderbird or RoundCube clients, the mail is not marked as spam. However, if I use PHPMailer, it always gets catalogued as spam. So I compared both headers and I just can't find the reason why the second is marked as spam while the first one is just ok. The first header sent from a mail client which is not marked as spam: Delivered-To: [email protected] Received: by 10.76.153.102 with SMTP id vf6csp230573oab; Tue, 19 Aug 2014 11:08:19 -0700 (PDT) X-Received: by 10.60.23.39 with SMTP id j7mr45544050oef.20.1408471699715; Tue, 19 Aug 2014 11:08:19 -0700 (PDT) Return-Path: <[email protected]> Received: from mail.mydomain.com (X.ip-92-222-X.eu. [92.222.X.X]) by mx.google.com with ESMTPS id t5si27115082oej.10.2014.08.19.11.08.18 for <[email protected]> (version=TLSv1.2 cipher=ECDHE-RSA-AES128-GCM-SHA256 bits=128/128); Tue, 19 Aug 2014 11:08:19 -0700 (PDT) Received-SPF: pass (google.com: domain of [email protected] designates 92.222.X.X as permitted sender) client-ip=92.222.X.X; Authentication-Results: mx.google.com; spf=pass (google.com: domain of [email protected] designates 92.222.X.X as permitted sender) [email protected]; dkim=pass (test mode) [email protected] Received: by mail.mydomain.com (Postfix, from userid 111) id D8F69120293D; Tue, 19 Aug 2014 19:08:17 +0100 (BST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mydomain.com; s=mail; t=1408471697; bh=wKMX9gkQ7tCLv8ezrG5t4bICm/SSLQsNfTdZMToksWw=; h=Date:From:To:Subject:From; b=qRNcYVdmk+n3D1uuv0FInTx7/LzH2ojck9DgCmabFPvfke233lkojUOjezCUGx7iV DL8EayZ28mzzzHpB7ETeMzop/5OS3BmvFtGKVD9gzc78cDIFXTDoRFAnkRWDR2IOxI SOn5tiyODTFpkbDgJOndzQ6qL5K0S9ASNGCZrNL4= X-Spam-Checker-Version: SpamAssassin 3.4.0 (2014-02-07) on vpsX.ovh.net X-Spam-Level: X-Spam-Status: No, score=-1.0 required=3.0 tests=ALL_TRUSTED,T_DKIM_INVALID autolearn=ham autolearn_force=no version=3.4.0 Received: from [192.168.1.111] (unknown [77.231.X.X]) (using TLSv1 with cipher ECDHE-RSA-AES128-SHA (128/128 bits)) (No client certificate requested) (Authenticated sender: [email protected]) by mail.mydomain.com (Postfix) with ESMTPSA id 910341202624 for <[email protected]>; Tue, 19 Aug 2014 19:08:17 +0100 (BST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mydomain.com; s=mail; t=1408471697; bh=wKMX9gkQ7tCLv8ezrG5t4bICm/SSLQsNfTdZMToksWw=; h=Date:From:To:Subject:From; b=qRNcYVdmk+n3D1uuv0FInTx7/LzH2ojck9DgCmabFPvfke233lkojUOjezCUGx7iV DL8EayZ28mzzzHpB7ETeMzop/5OS3BmvFtGKVD9gzc78cDIFXTDoRFAnkRWDR2IOxI SOn5tiyODTFpkbDgJOndzQ6qL5K0S9ASNGCZrNL4= Message-ID: <[email protected]> Date: Tue, 19 Aug 2014 19:08:24 +0100 From: My Name <[email protected]> User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64; rv:24.0) Gecko/20100101 Thunderbird/24.6.0 MIME-Version: 1.0 To: My other account <[email protected]> Subject: . Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit . The second header sent from PHPMailer which is always marked as spam: Delivered-To: [email protected] Received: by 10.76.153.102 with SMTP id vf6csp230832oab; Tue, 19 Aug 2014 11:12:10 -0700 (PDT) X-Received: by 10.60.121.67 with SMTP id li3mr44086252oeb.17.1408471930520; Tue, 19 Aug 2014 11:12:10 -0700 (PDT) Return-Path: <[email protected]> Received: from mail.mydomain.com (X.ip-92-222-X.eu. [92.222.X.X]) by mx.google.com with ESMTPS id w8si27103806obn.30.2014.08.19.11.12.10 for <[email protected]> (version=TLSv1.2 cipher=ECDHE-RSA-AES128-GCM-SHA256 bits=128/128); Tue, 19 Aug 2014 11:12:10 -0700 (PDT) Received-SPF: pass (google.com: domain of [email protected] designates 92.222.X.X as permitted sender) client-ip=92.222.X.X; Authentication-Results: mx.google.com; spf=pass (google.com: domain of [email protected] designates 92.222.X.X as permitted sender) [email protected]; dkim=pass (test mode) [email protected] Received: by mail.mydomain.com (Postfix, from userid 111) id 1999D120293D; Tue, 19 Aug 2014 19:12:09 +0100 (BST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mydomain.com; s=mail; t=1408471929; bh=N1JuHq1S+8GrjHcEK3xn8P1JS+ygEBv5LKe0BiXuVJo=; h=Date:To:From:Reply-to:Subject:From; b=K7tcPyArzSTY91VEw6mAAFtDurSGwgTLGkfUZdC5mqsg0g/1LzmZkgwdjj4NdJa6M E2kDz3dwYN8FcZmbampJYFXxj4NQVtSnzjiWV40rpfOFqD2rXDGNIyB2QOjBZZ4WK3 7s4lyoJ/BrdQH4en8ctLVsDHed/KpHD4iGFEl67E= X-Spam-Checker-Version: SpamAssassin 3.4.0 (2014-02-07) on vpsX.ovh.net X-Spam-Level: X-Spam-Status: No, score=-1.0 required=3.0 tests=ALL_TRUSTED,T_DKIM_INVALID autolearn=ham autolearn_force=no version=3.4.0 Received: from rpi.mydomain.com (unknown [77.231.X.X]) (using TLSv1 with cipher ECDHE-RSA-AES256-SHA (256/256 bits)) (No client certificate requested) (Authenticated sender: [email protected]) by mail.mydomain.com (Postfix) with ESMTPSA id B42AF1202624 for <[email protected]>; Tue, 19 Aug 2014 19:12:08 +0100 (BST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mydomain.com; s=mail; t=1408471928; bh=N1JuHq1S+8GrjHcEK3xn8P1JS+ygEBv5LKe0BiXuVJo=; h=Date:To:From:Reply-to:Subject:From; b=iXPM0tS36swudPTT4FOHHtPi5Ll6LbR60kNqCinZ8utcWoFE31SFTpoMEq5aCM5ux wQMdFiN8c6vkjRGabmvqFTTIbwJsrToHo/4+Lt5HEBoQQE2Y3T+xGmnmGAHCS6stKB yb7SVmtrIAsVtSMKA8VYIbmu2oYqV3afYt7g0OMQ= Date: Tue, 19 Aug 2014 20:12:07 +0200 To: [email protected] From: Trying another account <[email protected]> Reply-to: Trying another account <[email protected]> Subject: . Message-ID: <[email protected]> X-Priority: 3 X-Mailer: PHPMailer 5.1 (phpmailer.sourceforge.net) MIME-Version: 1.0 Content-Transfer-Encoding: 8bit Content-Type: text/plain; charset="UTF-8" . I also tried: Adding a User-Agent header to match the first one. Removing the X-Mailer header. No one of them made a difference. Is there some significant difference which is making the second e-mail to be marked as spam by Google?

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• Can't get the L2TP IPSEC up and running

  - by Maciej Swic

  i have an Ubuntu 11.10 (oneiric) server running on a ReadyNAS. Im planning to use this to accept ipsec+l2tp connections through a router. However, the connection is failing somewhere half through. Using Openswan IPsec U2.6.28/K3.0.0-12-generic and trying to connect with an iOS 5 iPhone 4S. This is how far i can get: auth.log: Jan 19 13:54:11 ubuntu pluto[1990]: added connection description "PSK" Jan 19 13:54:11 ubuntu pluto[1990]: added connection description "L2TP-PSK-NAT" Jan 19 13:54:11 ubuntu pluto[1990]: added connection description "L2TP-PSK-noNAT" Jan 19 13:54:11 ubuntu pluto[1990]: added connection description "passthrough-for-non-l2tp" Jan 19 13:54:11 ubuntu pluto[1990]: listening for IKE messages Jan 19 13:54:11 ubuntu pluto[1990]: NAT-Traversal: Trying new style NAT-T Jan 19 13:54:11 ubuntu pluto[1990]: NAT-Traversal: ESPINUDP(1) setup failed for new style NAT-T family IPv4 (errno=19) Jan 19 13:54:11 ubuntu pluto[1990]: NAT-Traversal: Trying old style NAT-T Jan 19 13:54:11 ubuntu pluto[1990]: adding interface eth0/eth0 192.168.19.99:500 Jan 19 13:54:11 ubuntu pluto[1990]: adding interface eth0/eth0 192.168.19.99:4500 Jan 19 13:54:11 ubuntu pluto[1990]: adding interface lo/lo 127.0.0.1:500 Jan 19 13:54:11 ubuntu pluto[1990]: adding interface lo/lo 127.0.0.1:4500 Jan 19 13:54:11 ubuntu pluto[1990]: adding interface lo/lo ::1:500 Jan 19 13:54:11 ubuntu pluto[1990]: adding interface eth0/eth0 2001:470:28:81:a00:27ff:* Jan 19 13:54:11 ubuntu pluto[1990]: loading secrets from "/etc/ipsec.secrets" Jan 19 13:54:11 ubuntu pluto[1990]: loading secrets from "/var/lib/openswan/ipsec.secrets.inc" Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: received Vendor ID payload [RFC 3947] method set to=109 Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: received Vendor ID payload [draft-ietf-ipsec-nat-t-ike] method set to=110 Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: ignoring unknown Vendor ID payload [8f8d83826d246b6fc7a8a6a428c11de8] Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: ignoring unknown Vendor ID payload [439b59f8ba676c4c7737ae22eab8f582] Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: ignoring unknown Vendor ID payload [4d1e0e136deafa34c4f3ea9f02ec7285] Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: ignoring unknown Vendor ID payload [80d0bb3def54565ee84645d4c85ce3ee] Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: ignoring unknown Vendor ID payload [9909b64eed937c6573de52ace952fa6b] Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: received Vendor ID payload [draft-ietf-ipsec-nat-t-ike-03] meth=108, but already using method 110 Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: received Vendor ID payload [draft-ietf-ipsec-nat-t-ike-02] meth=107, but already using method 110 Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: received Vendor ID payload [draft-ietf-ipsec-nat-t-ike-02_n] meth=106, but already using method 110 Jan 19 14:04:31 ubuntu pluto[1990]: packet from 95.*.*.233:500: received Vendor ID payload [Dead Peer Detection] Jan 19 14:04:31 ubuntu pluto[1990]: "PSK"[1] 95.*.*.233 #1: responding to Main Mode from unknown peer 95.*.*.233 Jan 19 14:04:31 ubuntu pluto[1990]: "PSK"[1] 95.*.*.233 #1: transition from state STATE_MAIN_R0 to state STATE_MAIN_R1 Jan 19 14:04:31 ubuntu pluto[1990]: "PSK"[1] 95.*.*.233 #1: STATE_MAIN_R1: sent MR1, expecting MI2 Jan 19 14:04:33 ubuntu pluto[1990]: "PSK"[1] 95.*.*.233 #1: NAT-Traversal: Result using draft-ietf-ipsec-nat-t-ike (MacOS X): both are NATed Jan 19 14:04:33 ubuntu pluto[1990]: "PSK"[1] 95.*.*.233 #1: transition from state STATE_MAIN_R1 to state STATE_MAIN_R2 Jan 19 14:04:33 ubuntu pluto[1990]: "PSK"[1] 95.*.*.233 #1: STATE_MAIN_R2: sent MR2, expecting MI3 Jan 19 14:05:03 ubuntu pluto[1990]: ERROR: asynchronous network error report on eth0 (sport=500) for message to 95.*.*.233 port 500, complainant 95.*.*.233: Connection refused [errno 111, origin ICMP type 3 code 3 (not authenticated)] Router config UDP 500, 1701 and 4500 forwarded to 192.168.19.99 (Ubuntu server for ipsec). Ipsec passthrough enabled. /etc/ipsec.conf # /etc/ipsec.conf - Openswan IPsec configuration file # This file: /usr/share/doc/openswan/ipsec.conf-sample # # Manual: ipsec.conf.5 version 2.0 # conforms to second version of ipsec.conf specification config setup nat_traversal=yes #charonstart=yes #plutostart=yes protostack=netkey conn PSK authby=secret forceencaps=yes pfs=no auto=add keyingtries=3 dpdtimeout=60 dpdaction=clear rekey=no left=192.168.19.99 leftnexthop=192.168.19.1 leftprotoport=17/1701 right=%any rightprotoport=17/%any rightsubnet=vhost:%priv,%no dpddelay=10 #dpdtimeout=10 #dpdaction=clear include /etc/ipsec.d/l2tp-psk.conf /etc/ipsec.d/l2tp-psk.conf conn L2TP-PSK-NAT rightsubnet=vhost:%priv also=L2TP-PSK-noNAT conn L2TP-PSK-noNAT # # PreSharedSecret needs to be specified in /etc/ipsec.secrets as # YourIPAddress %any: "sharedsecret" authby=secret pfs=no auto=add keyingtries=3 # we cannot rekey for %any, let client rekey rekey=no # Set ikelifetime and keylife to same defaults windows has ikelifetime=8h keylife=1h # l2tp-over-ipsec is transport mode type=transport # left=192.168.19.99 # # For updated Windows 2000/XP clients, # to support old clients as well, use leftprotoport=17/%any leftprotoport=17/1701 # # The remote user. # right=%any # Using the magic port of "0" means "any one single port". This is # a work around required for Apple OSX clients that use a randomly # high port, but propose "0" instead of their port. rightprotoport=17/%any dpddelay=10 dpdtimeout=10 dpdaction=clear conn passthrough-for-non-l2tp type=passthrough left=192.168.19.99 leftnexthop=192.168.19.1 right=0.0.0.0 rightsubnet=0.0.0.0/0 auto=route /etc/ipsec.secrets include /var/lib/openswan/ipsec.secrets.inc %any %any: PSK "my-key" 192.168.19.99 %any: PSK "my-key" /etc/xl2tpd/xl2tpd.conf [global] debug network = yes debug tunnel = yes ipsec saref = no listen-addr = 192.168.19.99 [lns default] ip range = 192.168.19.201-192.168.19.220 local ip = 192.168.19.99 require chap = yes refuse chap = no refuse pap = no require authentication = no ppp debug = yes pppoptfile = /etc/ppp/options.xl2tpd length bit = yes /etc/ppp/options.xl2tpd pcp-accept-local ipcp-accept-remote noccp auth crtscts idle 1800 mtu 1410 mru 1410 defaultroute debug lock proxyarp connect-delay 5000 ipcp-accept-local /etc/ppp/chap-secrets # Secrets for authentication using CHAP # client server secret IP addresses maciekish * my-secret * * maciekish my-secret * I can't seem to find the problem. Other ipsec connections to other hosts work from the network im currently at.

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