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  • Can NP-Intermediate exist if P = NP?

    - by Jason Baker
    My understanding is that Ladner's theorem is basically this: P != NP implies that there exists a set NPI where NPI is not in P and NPI is not NP-complete What happens to this theorem if we assume that P = NP rather than P != NP? We know that if NP Intermediate doesn't exist, then P = NP. But can NP Intermediate exist if P = NP?

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  • Why is numpy's einsum faster than numpy's built in functions?

    - by Ophion
    Lets start with three arrays of dtype=np.double. Timings are performed on a intel CPU using numpy 1.7.1 compiled with icc and linked to intel's mkl. A AMD cpu with numpy 1.6.1 compiled with gcc without mkl was also used to verify the timings. Please note the timings scale nearly linearly with system size and are not due to the small overhead incurred in the numpy functions if statements these difference will show up in microseconds not milliseconds: arr_1D=np.arange(500,dtype=np.double) large_arr_1D=np.arange(100000,dtype=np.double) arr_2D=np.arange(500**2,dtype=np.double).reshape(500,500) arr_3D=np.arange(500**3,dtype=np.double).reshape(500,500,500) First lets look at the np.sum function: np.all(np.sum(arr_3D)==np.einsum('ijk->',arr_3D)) True %timeit np.sum(arr_3D) 10 loops, best of 3: 142 ms per loop %timeit np.einsum('ijk->', arr_3D) 10 loops, best of 3: 70.2 ms per loop Powers: np.allclose(arr_3D*arr_3D*arr_3D,np.einsum('ijk,ijk,ijk->ijk',arr_3D,arr_3D,arr_3D)) True %timeit arr_3D*arr_3D*arr_3D 1 loops, best of 3: 1.32 s per loop %timeit np.einsum('ijk,ijk,ijk->ijk', arr_3D, arr_3D, arr_3D) 1 loops, best of 3: 694 ms per loop Outer product: np.all(np.outer(arr_1D,arr_1D)==np.einsum('i,k->ik',arr_1D,arr_1D)) True %timeit np.outer(arr_1D, arr_1D) 1000 loops, best of 3: 411 us per loop %timeit np.einsum('i,k->ik', arr_1D, arr_1D) 1000 loops, best of 3: 245 us per loop All of the above are twice as fast with np.einsum. These should be apples to apples comparisons as everything is specifically of dtype=np.double. I would expect the speed up in an operation like this: np.allclose(np.sum(arr_2D*arr_3D),np.einsum('ij,oij->',arr_2D,arr_3D)) True %timeit np.sum(arr_2D*arr_3D) 1 loops, best of 3: 813 ms per loop %timeit np.einsum('ij,oij->', arr_2D, arr_3D) 10 loops, best of 3: 85.1 ms per loop Einsum seems to be at least twice as fast for np.inner, np.outer, np.kron, and np.sum regardless of axes selection. The primary exception being np.dot as it calls DGEMM from a BLAS library. So why is np.einsum faster that other numpy functions that are equivalent? The DGEMM case for completeness: np.allclose(np.dot(arr_2D,arr_2D),np.einsum('ij,jk',arr_2D,arr_2D)) True %timeit np.einsum('ij,jk',arr_2D,arr_2D) 10 loops, best of 3: 56.1 ms per loop %timeit np.dot(arr_2D,arr_2D) 100 loops, best of 3: 5.17 ms per loop The leading theory is from @sebergs comment that np.einsum can make use of SSE2, but numpy's ufuncs will not until numpy 1.8 (see the change log). I believe this is the correct answer, but have not been able to confirm it. Some limited proof can be found by changing the dtype of input array and observing speed difference and the fact that not everyone observes the same trends in timings.

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  • P-NP-Problem: What are the most promising methods?

    - by phimuemue
    Hello everybody, I know that P=NP has not been solved up to now, but can anybody tell me something about the following: What are currently the most promising mathematical / computer scientific methods that could be helpful to tackle this problem? Or are there even none such methods known to be potentially helpful up to now? Is there any (free) compendium on this topic where I can find all / most of the research done in this area?

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  • P=NP?-Problem: What are the most promising methods?

    - by phimuemue
    Hello everybody, I know that P=NP has not been solved up to now, but can anybody tell me something about the following: What are currently the most promising mathematical / computer scientific methods that could be helpful to tackle this problem? Or are there even none such methods known to be potentially helpful up to now? Is there any (free) compendium on this topic where I can find all / most of the research done in this area?

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  • Simple reduction (NP completeness)

    - by Allen
    hey guys I'm looking for a means to prove that the bicriteria shortest path problem is np complete. That is, given a graph with lengths and weights, I need to know if a there exists a path in the graph from s to t with total length <= L and weight <= W. I know that i must take an NP complete problem and reduce it to this one. We have at our disposal the following problems to choose from: 3-SAT, independent set, vertex cover, hamiltonian cycle, and 3-dimensional matching. Any ideas on which may be viable? thanks

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  • The subsets-sum problem and the solvability of NP-complete problems

    - by G.E.M.
    I was reading about the subset-sums problem when I came up with what appears to be a general-purpose algorithm for solving it: (defun subset-contains-sum (set sum) (let ((subsets) (new-subset) (new-sum)) (dolist (element set) (dolist (subset-sum subsets) (setf new-subset (cons element (car subset-sum))) (setf new-sum (+ element (cdr subset-sum))) (if (= new-sum sum) (return-from subset-contains-sum new-subset)) (setf subsets (cons (cons new-subset new-sum) subsets))) (setf subsets (cons (cons element element) subsets))))) "set" is a list not containing duplicates and "sum" is the sum to search subsets for. "subsets" is a list of cons cells where the "car" is a subset list and the "cdr" is the sum of that subset. New subsets are created from old ones in O(1) time by just cons'ing the element to the front. I am not sure what the runtime complexity of it is, but appears that with each element "sum" grows by, the size of "subsets" doubles, plus one, so it appears to me to at least be quadratic. I am posting this because my impression before was that NP-complete problems tend to be intractable and that the best one can usually hope for is a heuristic, but this appears to be a general-purpose solution that will, assuming you have the CPU cycles, always give you the correct answer. How many other NP-complete problems can be solved like this one?

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  • Is the board game "Go" NP complete?

    - by sharkin
    There are plenty of Chess AI's around, and evidently some are good enough to beat some of the world's greatest players. I've heard that many attempts have been made to write successful AI's for the board game Go, but so far nothing has been conceived beyond average amateur level. Could it be that the task of mathematically calculating the optimal move at any given time in Go is an NP-complete problem?

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  • Proving that P <= NP

    - by Gail
    As most people know, P = NP is unproven and seems unlikely to be true. The proof would prove that P <= NP and NP <= P. Only one of those is hard, though. P <= NP is almost by definition true. In fact, that's the only way that I know how to state that P <= NP. It's just intuitive. How would you prove that P <= NP?

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  • Solving the NP-complete problem in XKCD

    - by Adam Tuttle
    The problem/comic in question: http://xkcd.com/287/ I'm not sure this is the best way to do it, but here's what I've come up with so far. I'm using CFML, but it should be readable by anyone. <cffunction name="testCombo" returntype="boolean"> <cfargument name="currentCombo" type="string" required="true" /> <cfargument name="currentTotal" type="numeric" required="true" /> <cfargument name="apps" type="array" required="true" /> <cfset var a = 0 /> <cfset var found = false /> <cfloop from="1" to="#arrayLen(arguments.apps)#" index="a"> <cfset arguments.currentCombo = listAppend(arguments.currentCombo, arguments.apps[a].name) /> <cfset arguments.currentTotal = arguments.currentTotal + arguments.apps[a].cost /> <cfif arguments.currentTotal eq 15.05> <!--- print current combo ---> <cfoutput><strong>#arguments.currentCombo# = 15.05</strong></cfoutput><br /> <cfreturn true /> <cfelseif arguments.currentTotal gt 15.05> <cfoutput>#arguments.currentCombo# > 15.05 (aborting)</cfoutput><br /> <cfreturn false /> <cfelse> <!--- less than 15.05 ---> <cfoutput>#arguments.currentCombo# < 15.05 (traversing)</cfoutput><br /> <cfset found = testCombo(arguments.currentCombo, arguments.currentTotal, arguments.apps) /> </cfif> </cfloop> </cffunction> <cfset mf = {name="Mixed Fruit", cost=2.15} /> <cfset ff = {name="French Fries", cost=2.75} /> <cfset ss = {name="side salad", cost=3.35} /> <cfset hw = {name="hot wings", cost=3.55} /> <cfset ms = {name="moz sticks", cost=4.20} /> <cfset sp = {name="sampler plate", cost=5.80} /> <cfset apps = [ mf, ff, ss, hw, ms, sp ] /> <cfloop from="1" to="6" index="b"> <cfoutput>#testCombo(apps[b].name, apps[b].cost, apps)#</cfoutput> </cfloop> The above code tells me that the only combination that adds up to $15.05 is 7 orders of Mixed Fruit, and it takes 232 executions of my testCombo function to complete. Is there a better algorithm to come to the correct solution? Did I come to the correct solution?

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  • NP-complete problem in Prolog

    - by Ashley
    I saw this ECLiPSe solution to the problem mentioned in this XKCD comic. I tried to convert this to pure Prolog. go:- Total = 1505, Prices = [215, 275, 335, 355, 420, 580], length(Prices, N), length(Amounts, N), totalCost(Prices, Amounts, 0, Total), writeln(Total). totalCost([], [], TotalSoFar, TotalSoFar). totalCost([P|Prices], [A|Amounts], TotalSoFar, EndTotal):- between(0, 10, A), Cost is P*A, TotalSoFar1 is TotalSoFar + Cost, totalCost(Prices, Amounts, TotalSoFar1, EndTotal). I don't think that this is the best / most declarative solution that one can come up with. Does anyone have any suggestions for improvement? Thanks in advance!

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  • Rewriting a for loop in pure NumPy to decrease execution time

    - by Statto
    I recently asked about trying to optimise a Python loop for a scientific application, and received an excellent, smart way of recoding it within NumPy which reduced execution time by a factor of around 100 for me! However, calculation of the B value is actually nested within a few other loops, because it is evaluated at a regular grid of positions. Is there a similarly smart NumPy rewrite to shave time off this procedure? I suspect the performance gain for this part would be less marked, and the disadvantages would presumably be that it would not be possible to report back to the user on the progress of the calculation, that the results could not be written to the output file until the end of the calculation, and possibly that doing this in one enormous step would have memory implications? Is it possible to circumvent any of these? import numpy as np import time def reshape_vector(v): b = np.empty((3,1)) for i in range(3): b[i][0] = v[i] return b def unit_vectors(r): return r / np.sqrt((r*r).sum(0)) def calculate_dipole(mu, r_i, mom_i): relative = mu - r_i r_unit = unit_vectors(relative) A = 1e-7 num = A*(3*np.sum(mom_i*r_unit, 0)*r_unit - mom_i) den = np.sqrt(np.sum(relative*relative, 0))**3 B = np.sum(num/den, 1) return B N = 20000 # number of dipoles r_i = np.random.random((3,N)) # positions of dipoles mom_i = np.random.random((3,N)) # moments of dipoles a = np.random.random((3,3)) # three basis vectors for this crystal n = [10,10,10] # points at which to evaluate sum gamma_mu = 135.5 # a constant t_start = time.clock() for i in range(n[0]): r_frac_x = np.float(i)/np.float(n[0]) r_test_x = r_frac_x * a[0] for j in range(n[1]): r_frac_y = np.float(j)/np.float(n[1]) r_test_y = r_frac_y * a[1] for k in range(n[2]): r_frac_z = np.float(k)/np.float(n[2]) r_test = r_test_x +r_test_y + r_frac_z * a[2] r_test_fast = reshape_vector(r_test) B = calculate_dipole(r_test_fast, r_i, mom_i) omega = gamma_mu*np.sqrt(np.dot(B,B)) # write r_test, B and omega to a file frac_done = np.float(i+1)/(n[0]+1) t_elapsed = (time.clock()-t_start) t_remain = (1-frac_done)*t_elapsed/frac_done print frac_done*100,'% done in',t_elapsed/60.,'minutes...approximately',t_remain/60.,'minutes remaining'

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  • Relating NP-Complete problems to real world problems

    - by terru
    I have a decent grasp of NP Complete problems; that's not the issue. What I don't have is a good sense of where they turn up in "real" programming. Some (like knapsack and traveling salesman) are obvious, but others don't seem obviously connected to "real" problems. I've had the experience several times of struggling with a difficult problem only to realize it is a well known NP Complete problem that has been researched extensively. If I had recognized the connection more quickly I could have saved quite a bit of time researching existing solutions to my specific problem. Are there any resources (online or print) that specifically connect NP Complete to real world instances?

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  • wget recursively with -np option still ascends to parent directory

    - by vectra
    tl;dr: will `wget --no-parrent -r ' download from a directory above the given url's directory? when using wget to download, say images, recursively from example.com/a/b with the -r and -np options, will a picture that is under example.com/a/c/ be downloaded when example.com/a/b/ delivers a html-file containing a link to the picture? if so, how do i get all pictures, that are in a folder and it's subfolders and only those? the description of the option --no-parent says "Do not ever ascend to the parent directory when retrieving recursively". anyway directory browsing delivers a link to the parent directory, which wget will follow, despite mentioned option. now what did i miss? edit: using GNU Wget 1.12

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  • How can I solve NP complete problems in erlang?

    - by Yadira Suazo
    Hi, I already have my operators for, by example, eat banana problem [#op{ action = [climb, on, {object}], preconds = [[at, {place}, {object}], [at, {place}, me], [on, floor, me], [on, floor, {object}], [large, {object}]], add_list = [[on, {object}, me]], del_list = [[on, floor, me]] }, But how can I use it in the function solve(Problem, depth_first, []). And depth_first (Problem, Start) - search_tree(Problem, container.stack, Start).

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  • Incremental PCA

    - by smichak
    Hi, Lately, I've been looking into an implementation of an incremental PCA algorithm in python - I couldn't find something that would meet my needs so I did some reading and implemented an algorithm I found in some paper. Here is the module's code - the relevant paper on which it is based is mentioned in the module's documentation. I would appreciate any feedback from people who are interested in this. Micha #!/usr/bin/env python """ Incremental PCA calculation module. Based on P.Hall, D. Marshall and R. Martin "Incremental Eigenalysis for Classification" which appeared in British Machine Vision Conference, volume 1, pages 286-295, September 1998. Principal components are updated sequentially as new observations are introduced. Each new observation (x) is projected on the eigenspace spanned by the current principal components (U) and the residual vector (r = x - U(U.T*x)) is used as a new principal component (U' = [U r]). The new principal components are then rotated by a rotation matrix (R) whose columns are the eigenvectors of the transformed covariance matrix (D=U'.T*C*U) to yield p + 1 principal components. From those, only the first p are selected. """ __author__ = "Micha Kalfon" import numpy as np _ZERO_THRESHOLD = 1e-9 # Everything below this is zero class IPCA(object): """Incremental PCA calculation object. General Parameters: m - Number of variables per observation n - Number of observations p - Dimension to which the data should be reduced """ def __init__(self, m, p): """Creates an incremental PCA object for m-dimensional observations in order to reduce them to a p-dimensional subspace. @param m: Number of variables per observation. @param p: Number of principle components. @return: An IPCA object. """ self._m = float(m) self._n = 0.0 self._p = float(p) self._mean = np.matrix(np.zeros((m , 1), dtype=np.float64)) self._covariance = np.matrix(np.zeros((m, m), dtype=np.float64)) self._eigenvectors = np.matrix(np.zeros((m, p), dtype=np.float64)) self._eigenvalues = np.matrix(np.zeros((1, p), dtype=np.float64)) def update(self, x): """Updates with a new observation vector x. @param x: Next observation as a column vector (m x 1). """ m = self._m n = self._n p = self._p mean = self._mean C = self._covariance U = self._eigenvectors E = self._eigenvalues if type(x) is not np.matrix or x.shape != (m, 1): raise TypeError('Input is not a matrix (%d, 1)' % int(m)) # Update covariance matrix and mean vector and centralize input around # new mean oldmean = mean mean = (n*mean + x) / (n + 1.0) C = (n*C + x*x.T + n*oldmean*oldmean.T - (n+1)*mean*mean.T) / (n + 1.0) x -= mean # Project new input on current p-dimensional subspace and calculate # the normalized residual vector g = U.T*x r = x - (U*g) r = (r / np.linalg.norm(r)) if not _is_zero(r) else np.zeros_like(r) # Extend the transformation matrix with the residual vector and find # the rotation matrix by solving the eigenproblem DR=RE U = np.concatenate((U, r), 1) D = U.T*C*U (E, R) = np.linalg.eigh(D) # Sort eigenvalues and eigenvectors from largest to smallest to get the # rotation matrix R sorter = list(reversed(E.argsort(0))) E = E[sorter] R = R[:,sorter] # Apply the rotation matrix U = U*R # Select only p largest eigenvectors and values and update state self._n += 1.0 self._mean = mean self._covariance = C self._eigenvectors = U[:, 0:p] self._eigenvalues = E[0:p] @property def components(self): """Returns a matrix with the current principal components as columns. """ return self._eigenvectors @property def variances(self): """Returns a list with the appropriate variance along each principal component. """ return self._eigenvalues def _is_zero(x): """Return a boolean indicating whether the given vector is a zero vector up to a threshold. """ return np.fabs(x).min() < _ZERO_THRESHOLD if __name__ == '__main__': import sys def pca_svd(X): X = X - X.mean(0).repeat(X.shape[0], 0) [_, _, V] = np.linalg.svd(X) return V N = 1000 obs = np.matrix([np.random.normal(size=10) for _ in xrange(N)]) V = pca_svd(obs) print V[0:2] pca = IPCA(obs.shape[1], 2) for i in xrange(obs.shape[0]): x = obs[i,:].transpose() pca.update(x) U = pca.components print U

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  • fit a ellipse in Python given a set of points xi=(xi,yi)

    - by Gianni
    I am computing a series of index from a 2D points (x,y). One index is the ratio between minor and major axis. To fit the ellipse i am using the following post when i run these function the final results looks strange because the center and the axis length are not in scale with the 2D points center = [ 560415.53298363+0.j 6368878.84576771+0.j] angle of rotation = (-0.0528033467597-5.55111512313e-17j) axes = [0.00000000-557.21553487j 6817.76933256 +0.j] thanks in advance for help import numpy as np from numpy.linalg import eig, inv def fitEllipse(x,y): x = x[:,np.newaxis] y = y[:,np.newaxis] D = np.hstack((x*x, x*y, y*y, x, y, np.ones_like(x))) S = np.dot(D.T,D) C = np.zeros([6,6]) C[0,2] = C[2,0] = 2; C[1,1] = -1 E, V = eig(np.dot(inv(S), C)) n = np.argmax(np.abs(E)) a = V[:,n] return a def ellipse_center(a): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] num = b*b-a*c x0=(c*d-b*f)/num y0=(a*f-b*d)/num return np.array([x0,y0]) def ellipse_angle_of_rotation( a ): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] return 0.5*np.arctan(2*b/(a-c)) def ellipse_axis_length( a ): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] up = 2*(a*f*f+c*d*d+g*b*b-2*b*d*f-a*c*g) down1=(b*b-a*c)*( (c-a)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a)) down2=(b*b-a*c)*( (a-c)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a)) res1=np.sqrt(up/down1) res2=np.sqrt(up/down2) return np.array([res1, res2]) if __name__ == '__main__': points = [(560036.4495758876, 6362071.890493258), (560036.4495758876, 6362070.890493258), (560036.9495758876, 6362070.890493258), (560036.9495758876, 6362070.390493258), (560037.4495758876, 6362070.390493258), (560037.4495758876, 6362064.890493258), (560036.4495758876, 6362064.890493258), (560036.4495758876, 6362063.390493258), (560035.4495758876, 6362063.390493258), (560035.4495758876, 6362062.390493258), (560034.9495758876, 6362062.390493258), (560034.9495758876, 6362061.390493258), (560032.9495758876, 6362061.390493258), (560032.9495758876, 6362061.890493258), (560030.4495758876, 6362061.890493258), (560030.4495758876, 6362061.390493258), (560029.9495758876, 6362061.390493258), (560029.9495758876, 6362060.390493258), (560029.4495758876, 6362060.390493258), (560029.4495758876, 6362059.890493258), (560028.9495758876, 6362059.890493258), (560028.9495758876, 6362059.390493258), (560028.4495758876, 6362059.390493258), (560028.4495758876, 6362058.890493258), (560027.4495758876, 6362058.890493258), (560027.4495758876, 6362058.390493258), (560026.9495758876, 6362058.390493258), (560026.9495758876, 6362057.890493258), (560025.4495758876, 6362057.890493258), (560025.4495758876, 6362057.390493258), (560023.4495758876, 6362057.390493258), (560023.4495758876, 6362060.390493258), (560023.9495758876, 6362060.390493258), (560023.9495758876, 6362061.890493258), (560024.4495758876, 6362061.890493258), (560024.4495758876, 6362063.390493258), (560024.9495758876, 6362063.390493258), (560024.9495758876, 6362064.390493258), (560025.4495758876, 6362064.390493258), (560025.4495758876, 6362065.390493258), (560025.9495758876, 6362065.390493258), (560025.9495758876, 6362065.890493258), (560026.4495758876, 6362065.890493258), (560026.4495758876, 6362066.890493258), (560026.9495758876, 6362066.890493258), (560026.9495758876, 6362068.390493258), (560027.4495758876, 6362068.390493258), (560027.4495758876, 6362068.890493258), (560027.9495758876, 6362068.890493258), (560027.9495758876, 6362069.390493258), (560028.4495758876, 6362069.390493258), (560028.4495758876, 6362069.890493258), (560033.4495758876, 6362069.890493258), (560033.4495758876, 6362070.390493258), (560033.9495758876, 6362070.390493258), (560033.9495758876, 6362070.890493258), (560034.4495758876, 6362070.890493258), (560034.4495758876, 6362071.390493258), (560034.9495758876, 6362071.390493258), (560034.9495758876, 6362071.890493258), (560036.4495758876, 6362071.890493258)] a_points = np.array(points) x = a_points[:, 0] y = a_points[:, 1] from pylab import * plot(x,y) show() a = fitEllipse(x,y) center = ellipse_center(a) phi = ellipse_angle_of_rotation(a) axes = ellipse_axis_length(a) print "center = ", center print "angle of rotation = ", phi print "axes = ", axes from pylab import * plot(x,y) plot(center[0:1],center[1:], color = 'red') show() each vertex is a xi,y,i point plot of 2D point and center of fit ellipse

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  • mpirun -np N, what if N is larger than my core number?

    - by Daniel
    Say I have a 4-core workstation, what would linux (Ubuntu) do if I execute mpirun -np 9 XXX Q1. Will 9 run immediately together, or they will run 4 after 4? Q2. I suppose that using 9 is not good, because the remainder 1, it will make the computer confused, (I don't know is it going to be confused at all, or the "head" of the computer will decide which core among the 4 cores will be used?) Or it will be randomly picked. Who decide which one core to call? Q3. If I feel my cpu is not bad and my ram is okay and large enough, and my case is not very big. Is it a good idea in order to fully use my cpu and ram, that I do mpirun -np 8 XXX, or even mpirun -np 12 XXX. Q4. Who decides all of these effciency optimization, Ubuntu, or linux, or motherboard or cpu? Your enlightenment would be really appreciated.

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  • Simplified knapsack in PHP

    - by Mikhail
    I have two instances where I'd like to display information in a "justified" alignment - but I don't care if the values are switched in order. One example being displaying the usernames of people online: Anton Brother68 Commissar Dougheater Elflord Foobar Goop Hoo Iee Joo Rearranging them we could get exactly 22 characters long on each line: Anton Brother68 Foobar Commissar Elflord Goop Dougheater Hoo Iee Joo This is kind of a knapsack, except seems like there ought to be a P solution since I don't care about perfection, and I have multiple lines. Second instance is identical, except instead of names and character count I would be displaying random images and use their width.

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  • Pass data in np.dnarray to Highcharts

    - by F.N.B
    I'm working with python 2.7, jinja2, flask and Highcharts. I create two numpy array (x1 and x2, type = numpy.dnarray) and I pass to Highcharts. My problems is, Highcharts don't recognize the commas in the vector. This is my jinja2 code: <script> $(function () { $('#container').highcharts({ series: [{ name: 'Tokyo', data: {{ x1 }} }, { name: 'London', data: {{ x2 }} }] }); }); And this is the error that I look with network chrome dev tools: series: [{ name: 'Tokyo', data: [1 4 5 2 3] }, { name: 'London', data: [3 6 7 4 1] }] I need change the numpy array to python list to pass to Highcharts or there is a better way to do?? Thanks

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  • async handler deleted by the wrong thread in django

    - by user3480706
    I'm run this algorithm in my django application.when i run several time from my GUI django local server will stopped and i got this error Exception RuntimeError: RuntimeError('main thread is not in main loop',) in ignored Tcl_AsyncDelete: async handler deleted by the wrong thread Aborted (core dumped) code print "Learning the sin function" network =MLP.MLP(2,10,1) samples = np.zeros(2000, dtype=[('x', float, 1), ('y', float, 1)]) samples['x'] = np.linspace(-5,5,2000) samples['y'] = np.sin(samples['x']) #samples['y'] = np.linspace(-4,4,2500) for i in range(100000): n = np.random.randint(samples.size) network.propagate_forward(samples['x'][n]) network.propagate_backward(samples['y'][n]) plt.figure(figsize=(10,5)) # Draw real function x = samples['x'] y = samples['y'] #x=np.linspace(-6.0,7.0,50) plt.plot(x,y,color='b',lw=1) samples1 = np.zeros(2000, dtype=[('x1', float, 1), ('y1', float, 1)]) samples1['x1'] = np.linspace(-4,4,2000) samples1['y1'] = np.sin(samples1['x1']) # Draw network approximated function for i in range(samples1.size): samples1['y1'][i] = network.propagate_forward(samples1['x1'][i]) plt.plot(samples1['x1'],samples1['y1'],color='r',lw=3) plt.axis([-2,2,-2,2]) plt.show() plt.close() return HttpResponseRedirect('/charts/charts') how can i fix this error ?need a quick help

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  • mpirun -np N, what if N is larger than my core number?

    - by Daniel
    Say I have a 4-core workstation, what would linux (Ubuntu) do if I execute mpirun -np 9 XXX Q1. Will 9 run immediately together, or they will run 4 after 4? Q2. I suppose that using 9 is not good, because the remainder 1, it will make the computer confused, (I don't know is it going to be confused at all, or the "head" of the computer will decide which core among the 4 cores will be used?) Or it will be randomly picked. Who decide which one core to call? Q3. If I feel my cpu is not bad and my ram is okay and large enough, and my case is not very big. Is it a good idea in order to fully use my cpu and ram, that I do mpirun -np 8 XXX, or even mpirun -np 12 XXX. Q4. Who decides all of these effciency optimization, Ubuntu, or linux, or motherboard or cpu? Your enlightenment would be really appreciated.

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