2 * Licensed under the Apache License, Version 2.0 (the "License");
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3 * you may not use this file except in compliance with the License.
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4 * You may obtain a copy of the License at
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6 * http://www.apache.org/licenses/LICENSE-2.0
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8 * Unless required by applicable law or agreed to in writing, software
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9 * distributed under the License is distributed on an "AS IS" BASIS,
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10 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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11 * See the License for the specific language governing permissions and
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12 * limitations under the License.
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16 namespace com.google.zxing.common
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19 /// <summary> <p>Encapsulates logic that estimates the optimal "black point", the luminance value
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20 /// which is the best line between "white" and "black" in a grayscale image.</p>
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22 /// <p>For an interesting discussion of this issue, see
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23 /// <a href="http://webdiis.unizar.es/~neira/12082/thresholding.pdf">http://webdiis.unizar.es/~neira/12082/thresholding.pdf</a>.
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27 /// <author> srowen@google.com (Sean Owen)
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29 /// <author> dswitkin@google.com (Daniel Switkin)
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31 public sealed class BlackPointEstimator
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33 private BlackPointEstimator()
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38 * <p>Given an array of <em>counts</em> of luminance values (i.e. a histogram), this method
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39 * decides which bucket of values corresponds to the black point -- which bucket contains the
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40 * count of the brightest luminance values that should be considered "black".</p>
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42 * @param histogram an array of <em>counts</em> of luminance values
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43 * @return index within argument of bucket corresponding to brightest values which should be
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44 * considered "black"
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45 * @throws ReaderException if "black" and "white" appear to be very close in luminance in the image
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47 public static int estimate(int[] histogram)
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51 int numBuckets = histogram.Length;
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52 int maxBucketCount = 0;
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53 // Find tallest peak in histogram
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55 int firstPeakSize = 0;
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56 for (int i = 0; i < numBuckets; i++) {
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57 if (histogram[i] > firstPeakSize) {
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59 firstPeakSize = histogram[i];
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61 if (histogram[i] > maxBucketCount) {
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62 maxBucketCount = histogram[i];
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66 // Find second-tallest peak -- well, another peak that is tall and not
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67 // so close to the first one
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69 int secondPeakScore = 0;
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70 for (int i = 0; i < numBuckets; i++) {
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71 int distanceToBiggest = i - firstPeak;
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72 // Encourage more distant second peaks by multiplying by square of distance
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73 int score = histogram[i] * distanceToBiggest * distanceToBiggest;
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74 if (score > secondPeakScore) {
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76 secondPeakScore = score;
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80 // Put firstPeak first
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81 if (firstPeak > secondPeak) {
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82 int temp = firstPeak;
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83 firstPeak = secondPeak;
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87 // Kind of aribtrary; if the two peaks are very close, then we figure there is so little
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88 // dynamic range in the image, that discriminating black and white is too error-prone.
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89 // Decoding the image/line is either pointless, or may in some cases lead to a false positive
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90 // for 1D formats, which are relatively lenient.
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91 // We arbitrarily say "close" is "<= 1/16 of the total histogram buckets apart"
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92 if (secondPeak - firstPeak <= numBuckets >> 4) {
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93 throw new ReaderException("");
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96 // Find a valley between them that is low and closer to the white peak
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97 int bestValley = secondPeak - 1;
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98 int bestValleyScore = -1;
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99 for (int i = secondPeak - 1; i > firstPeak; i--) {
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100 int fromFirst = i - firstPeak;
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101 // Favor a "valley" that is not too close to either peak -- especially not the black peak --
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102 // and that has a low value of course
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103 int score = fromFirst * fromFirst * (secondPeak - i) * (maxBucketCount - histogram[i]);
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104 if (score > bestValleyScore) {
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106 bestValleyScore = score;
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112 catch (Exception e)
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114 throw (ReaderException) e;
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