SURF Logo detection with opencv python

I am trying to detect some logos of known software related things in pictures. I use Opencv 2.4.5 with python 2.7. I would like to use SURF detector implemented in opencv but the problem is that I don't obtain good results. There are lots of false negative and false positive. My code is :

import cv2 import numpy as np def detectLogo(template, img): templateg = cv2.cvtColor(template, cv2.COLOR_BGR2GRAY) imgg = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY) # SURF extraction hessian_threshold = 30 surf = cv2.SURF(hessian_threshold) kp, desc = surf.detect(imgg, None, useProvidedKeypoints = False) # KNN samples = np.array(desc) responses = np.arange(len(kp), dtype = np.float32) knn = cv2.KNearest() knn.train(samples, responses) # Loading template and searching for similar kp kp2, desc2 = surf.detect(templateg, None, useProvidedKeypoints = False) matched = 0 total = 0 for h,des in enumerate(desc2): des = np.array(des,np.float32).reshape((1,128)) retval, results, neigh_resp, dists = knn.find_nearest(des,1) res,dist = int(results[0][0]),dists[0][0] total += 1 if dist<0.1: # draw matched keypoints in red color color = (0,0,255) matched += 1 else: color = (255,0,0) #Draw matched key points on original image x,y = kp[res].pt center = (int(x),int(y)) cv2.circle(img,center,2,color,-1) #Draw matched key points on template image x,y = kp2[h].pt center = (int(x),int(y)) cv2.circle(template,center,2,color,-1) cv2.imwrite("../resources/template.jpg", template) cv2.imwrite("../resources/image.jpg", img) return matched / float(total) template = cv2.imread("../resources/pictures/appleLogo.jpg") img = cv2.imread("../resources/pictures/pic2.jpg") print detectLogo(template, img) 

Below are results.

_{(source: hostingpics.net)}

Template url.

_{(source: hostingpics.net)}

Image url.

Matched points don't correspond at all to the loho and I got same results for two completely different images.

I think it's the only solution to perform this task but where is the problem with this dectection ? Thank you in advance.

Alexandre