/* * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. */ using System; namespace com.google.zxing.common { ///

Encapsulates logic that estimates the optimal "black point", the luminance value /// which is the best line between "white" and "black" in a grayscale image.

/// ///

For an interesting discussion of this issue, see /// http://webdiis.unizar.es/~neira/12082/thresholding.pdf. ///

/// ///
/// srowen@google.com (Sean Owen) /// /// dswitkin@google.com (Daniel Switkin) /// public sealed class BlackPointEstimator { private BlackPointEstimator() { } /** *

Given an array of counts of luminance values (i.e. a histogram), this method * decides which bucket of values corresponds to the black point -- which bucket contains the * count of the brightest luminance values that should be considered "black".

* * @param histogram an array of counts of luminance values * @return index within argument of bucket corresponding to brightest values which should be * considered "black" * @throws ReaderException if "black" and "white" appear to be very close in luminance in the image */ public static int estimate(int[] histogram) { try{ int numBuckets = histogram.Length; int maxBucketCount = 0; // Find tallest peak in histogram int firstPeak = 0; int firstPeakSize = 0; for (int i = 0; i < numBuckets; i++) { if (histogram[i] > firstPeakSize) { firstPeak = i; firstPeakSize = histogram[i]; } if (histogram[i] > maxBucketCount) { maxBucketCount = histogram[i]; } } // Find second-tallest peak -- well, another peak that is tall and not // so close to the first one int secondPeak = 0; int secondPeakScore = 0; for (int i = 0; i < numBuckets; i++) { int distanceToBiggest = i - firstPeak; // Encourage more distant second peaks by multiplying by square of distance int score = histogram[i] * distanceToBiggest * distanceToBiggest; if (score > secondPeakScore) { secondPeak = i; secondPeakScore = score; } } // Put firstPeak first if (firstPeak > secondPeak) { int temp = firstPeak; firstPeak = secondPeak; secondPeak = temp; } // Kind of aribtrary; if the two peaks are very close, then we figure there is so little // dynamic range in the image, that discriminating black and white is too error-prone. // Decoding the image/line is either pointless, or may in some cases lead to a false positive // for 1D formats, which are relatively lenient. // We arbitrarily say "close" is "<= 1/16 of the total histogram buckets apart" if (secondPeak - firstPeak <= numBuckets >> 4) { throw new ReaderException(""); } // Find a valley between them that is low and closer to the white peak int bestValley = secondPeak - 1; int bestValleyScore = -1; for (int i = secondPeak - 1; i > firstPeak; i--) { int fromFirst = i - firstPeak; // Favor a "valley" that is not too close to either peak -- especially not the black peak -- // and that has a low value of course int score = fromFirst * fromFirst * (secondPeak - i) * (maxBucketCount - histogram[i]); if (score > bestValleyScore) { bestValley = i; bestValleyScore = score; } } return bestValley; } catch (Exception e) { throw (ReaderException) e; } } } }