JP3746804B2 - Image compression device - Google Patents

Description

により求められる。スキャンＡＣ_i におけるラン長が（ｋ＋１）であるブロック数Ｚ〔ｋ＋１〕（すなわちラン長がｋでカテゴリが０であるブロック数）は、

により求められる。
【００９０】
修正係数ＺＷは、０から１０００までの値をとりうる。例えばＺＷ＝０のとき、
Ｚ〔ｋ＋１〕＝ＣＴ_j × Ｚ〔ｋ〕／Ｚ_ALL
となり、これは第２の予測方法と同じ結果になる。ＺＷ＝１０００とすると、
Ｚ〔ｋ＋１〕＝Ｚ〔ｋ〕
となり、これは、スキャンＡＣ_i-1 のラン長ｋのブロックが、スキャンＡＣ_i において全てカテゴリ「０」となることを示している。以下に説明する例では、ＺＷ＝５０としている。
【００９１】
ここで、スキャンＡＣ₃ において、Ｚ〔０〕＝１０３４、Ｚ〔１〕＝１４４０、Ｚ〔２〕＝１０４１、Ｚ〔３〕＝１８８５であり、またスキャンＡＣ₄ におけるカテゴリ分布として、図３０のような表が得られているとする。すなわち、スキャンＡＣ₃ がカテゴリ「０」である４３６６ブロックにおいて、スキャンＡＣ₄ では、カテゴリ「０」のブロック数ＣＴ₀ は３３２１、カテゴリ「１」のブロック数ＣＴ₁ は６５５、カテゴリ「２」のブロック数ＣＴ₂ は２７６、カテゴリ「３」のブロック数ＣＴ₃ は１０１、カテゴリ「４」のブロック数ＣＴ₄ は１３、カテゴリ「５」以上のブロック数は０であるとする。
【００９２】
次に述べるように、ラン長が１であるデータからブロック数が求められ、図３１に示すようなラン長・カテゴリの分布表が求められる。
【００９３】
スキャンＡＣ₄ において、ラン長が１であり、かつカテゴリ「１」であるブロック数は、（１）式より、

となる。同様に、ラン長が１であり、かつカテゴリ「２」であるブロック数は、（１）式より、

となる。
【００９４】
ラン長が１であり、かつカテゴリ「３」であるブロック数については、（１）式より、

となるが、図３０に示されるようにＣＴ₃ ＝Ｃ〔３〕＝１０１である。したがって、１０４ではなく、強制的に１０１に定められる。
【００９５】
ラン長が１であり、かつカテゴリ「４」であるブロック数についても同様に、Ｂ〔１，４〕は強制的に１３に定められる。
【００９６】
スキャンＡＣ₄ においてラン長が２であるブロック数Ｚ〔２〕（すなわちラン長が１であり、かつカテゴリが０であるブロック数）は、
Ｚ〔２〕＝Ｚ〔１〕−（277 + 158 + 101 + 13) ＝ 891
となる。
【００９７】
ラン長が２であり、かつカテゴリ「１」であるブロック数については、

となる。ラン長が２であり、かつカテゴリ「２」であるブロック数については、Ｂ〔２，２〕＝ 66 + 49 ＝ 115
となる。
【００９８】
ラン長が２であり、かつカテゴリ「３」であるブロック数Ｂ〔２，３〕は、Ｂ〔１，３〕の結果に従って０であり、またラン長が２であり、かつカテゴリ「４」であるブロック数Ｂ〔２，４〕は、Ｂ〔１，４〕の結果に従って０である。
【００９９】
スキャンＡＣ₄ においてラン長が３であるブロック数Ｚ〔３〕（すなわちラン長が２であり、かつカテゴリが０であるブロック数）は、
Ｚ〔３〕＝Ｚ〔２〕−（200 + 115 + 0 + 0)＝ 726
となる。
【０１００】
ラン長が３であり、かつカテゴリ「１」であるブロック数については、

となる。しかし、カテゴリ「１」であるブロック数ＣＴ₁ の総数が６５５であるため、
Ｂ〔３，１〕＝ 655 - 277 - 200＝ 178
となる。
【０１０１】
ラン長が３であり、かつカテゴリ「２」であるブロック数については、

となる。しかし、カテゴリ「２」であるブロック数ＣＴ₂ の総数が２７６であるため、
Ｂ〔３，２〕＝ 276 - 158 - 115＝ 3
となる。
【０１０２】
ラン長が３であり、かつカテゴリ「３」であるブロック数は、Ｂ〔１，３〕、Ｂ〔２，３〕の結果に従って０であり、またラン長が３であり、かつカテゴリ「４」であるブロック数は、Ｂ〔１，４〕、Ｂ〔２，４〕の結果に従って０である。
【０１０３】
スキャンＡＣ₄ においてラン長が４であるブロック数Ｚ〔４〕（すなわちラン長が３であり、かつカテゴリが０であるブロック数）は、
Ｚ〔４〕＝Ｚ〔３〕−（178 + 3 + 0 + 0)＝ 1704
となる。
【０１０４】
ラン長が０であるブロック数において、カテゴリ「１」〜「４」のブロック数は、図３０より、１５５、６５、２４、３となる。また、ラン長が１であるブロック数Ｚ〔１〕（すなわちラン長が０であり、かつカテゴリが０であるブロック数）は、図３０より、
Ｚ〔１〕＝787
となる。
【０１０５】
図３２は、このようにして求められたラン長・カテゴリの分布の表であり、第１の予測方法における図２３に対応している。また図３１に示されるように、ラン長が４であるブロック数（ラン長が３であり、かつカテゴリが０であるブロック数）は１７０４であり、これは第１の予測方法において得られたＺ〔４〕＝１４３４よりも大きくなっている。すなわち第３の予測方法によれば、ラン長が伸びることが理解される。
【０１０６】
次に、図３０に示されるようなカテゴリ分布表において、スキャンＡＣ_i-1 のカテゴリが「０」で、かつスキャンＡＣ_i のカテゴリが「１」以上のブロック数を図２９の折れ線ａｂｆｇｈを用いて求める方法を説明する。
【０１０７】
この場合、図２９において横軸はスキャンＡＣ_i-1 のカテゴリ「０」のブロック数を示し、縦軸はスキャンＡＣ_i におけるカテゴリ「０」とカテゴリ「１」のブロック数の和を示す。まず、縦軸のＯ_max をＣ〔０〕とＣ〔１〕の和（＝４１０８＋８１０＝４９１８）に定め、これに伴いＯ_high、Ｏ_mid 、Ｏ_low を求める。次に、横座標の４３６６に対応した折れ線上の点（例えばＰ３）を読み、この点に対応した縦座標を読むと、この値はスキャンＡＣ₄ におけるカテゴリ「０」とカテゴリ「１」のブロック数の和である。したがって、この和から、既に求められているカテゴリ「０」の値を引けば、スキャンＡＣ_i-1 のカテゴリが「０」でかつスキャンＡＣ_i のカテゴリが「１」のブロック数が求められる。以下、同様にして、スキャンＡＣ_i-1 のカテゴリが「０」でかつスキャンＡＣ_i のカテゴリが「２」以上のブロック数が求められる。
【０１０８】
次に、図３３のフローチャートを参照して、図２５のステップ１０６において用いられた設定符号量Ｓ_i の分布（図１３参照）の生成について説明する。
【０１０９】
ステップ２０１では、１つの画像の全ブロックに関して、ＤＣ成分とスキャンＡＣ₆₃の符号化データの最小限の合計符号量（ビット数）が求められる。ＤＣ成分の合計符号量ＣＤ₀ は、ハフマン符号量とこれに対応する付加ビット数との合計値の最小値ＨＮ（０）＝２に、全ブロック数を乗じることにより得られる。なお、通常は、カテゴリ「０」においてハフマン符号量と付加ビット数との合計値が最小値をとるので、ここでは最小値ＨＮ（０）という符号を用いている。一方、スキャンＡＣ₆₃の合計符号量ＣＤ₆₃は、ハフマン符号化における終端データ（ＥＯＢ）の符号量ＨＮ（“ＥＯＢ”）＝４に全ブロック数を乗じることにより得られる。
【０１１０】
ステップ２０２では、デフォルトの量子化テーブルを用いた場合における、スキャンＡＣ₆₃を除いた全てのブロックに関する合計符号量ＣＤＡが
ＣＤＡ＝Σ（ＣＯ_i ／ＦＴ_i ) （３）
により求められる。ただし、ＣＯ_i は、デフォルトの量子化テーブルを用いた時のスキャンＡＣ_i の符号量（ビット数）（図１２参照）であり、ＦＴ_i は、スキャンＡＣ_i のフィルタリングテーブルＦＬ（図３４参照）のテーブル値、すなわちフィルタリング係数を示す。なお、Σはパラメータｉについて０から６２まで加算することを示す。
【０１１１】
フィルタリングテーブルＦＬは、例えば図３４に示すように各空間周波数に対応したフィルタリング係数ＦＴ_i から成る。フィルタリング係数ＦＴ₀ はＤＣ成分に対応し、フィルタリング係数ＦＴ₁ 、ＦＴ₂ ・・・ＦＴ₆₃はスキャンＡＣ₁ 、ＡＣ₂ ・・・ＡＣ₆₃に対応する。またフィルタリング係数は、データ圧縮の度合いが大きい空間周波数ほど大きい値を有し、したがって図３４の例は高周波成分をカットするフィルタを示している。すなわち、高周波成分のフィルタリング係数は２００００あるいは３００００であり、高周波成分は、（３）式によって大きく圧縮される。
【０１１２】
ステップ２０３では、修正係数ＣＤＲが
ＣＤＲ＝（ＳＥＴＩＭ−ＣＤ₀ −ＣＤ₆₃）／ＣＤＡ （４）
により求められる。ただしＳＥＴＩＭは設定合計符号量であり、例えば５２４２８８ビット（６４Ｋbyte）である。修正係数ＣＤＲの値は、（３）式によって得られた合計符号量ＣＤＡが小さいほど大きくなり、例えば約１００である。
【０１１３】
ステップ２０４においてパラメータｉが０にクリアされた後、ステップ２０５において、
ＣＯＤＥ_i ＝（ＣＯ_i ／ＦＴ_i ）×ＣＤＲ （５）
により、各空間周波数（スキャンＡＣ_i ）における圧縮データの量（ビット数）が求められる。（５）式に示されるように圧縮データ量ＣＯＤＥ_i は、デフォルトの量子化テーブルを用いた時の符号量ＣＯ_i をフィルタリング係数ＦＴ_i で割った結果に、修正係数ＣＤＲを乗じることによって得られる。
【０１１４】
ステップ２０６では、圧縮データ量ＣＯＤＥ_i がデフォルトの量子化テーブルを用いた時の符号量ＣＯ_i よりも大きいか否かが判定される。圧縮データ量ＣＯＤＥ_i の方が大きい場合、ステップ２０７において符号量ＣＯ_i が圧縮データ量ＣＯＤＥ_i として設定される。換言すると、この場合、量子化係数ｑがいくらであっても圧縮符号量は符号量ＣＯ_i を越える可能性が低いため、その符号量ＣＯ_i が圧縮データ量ＣＯＤＥ_i として定められる。
【０１１５】
これに対し、圧縮データ量ＣＯＤＥ_i の方が小さい場合、ステップ２０８においてパラメータｉが１だけインクリメントされた後、ステップ２０９においてパラメータが６２より大きいか否かが判定される。パラメータｉが６２を越えていない場合、ステップ２０５に戻り、上述した処理が再び実行される。このようにしてスキャンＡＣ₆₂までの圧縮データ量ＣＯＤＥ_i が設定されると、ステップ２１０においてスキャンＡＣ₆₃の圧縮データ量ＣＯＤＥ₆₃がＣＤ₆₃に設定され、このプログラムは終了する。
【０１１６】
以上のようにして得られた圧縮データ量ＣＯＤＥ_i が、図２５のステップ１０６において用いられる設定符号量Ｓ_i である。
【０１１７】
図３５は、図３４に示すフィルタリングテーブルＦＬにおける各スキャンＡＣ_i の割当比率を示している。この割当比率は、各ＡＣ成分のフィルタリング係数ＦＴ_i とＤＣ成分のフィルタリング係数ＦＴ₀ との比の逆数に１００を乗じたものであり、例えばスキャンＡＣ₂ の場合、フィルタリング係数ＦＴ₂ は９８であるので、割当比率は符号ＲＡにより示すように１００よりも大きい。この図から理解されるように、図３４のフィルタリングテーブルＦＬは、高周波成分に対する割当比率が小さく、ローパスフィルタを示している。逆に、ハイパスフィルタのテーブルＦＬを生成するには、図３５の割当比率において、低周波成分に対する割当比率を小さくするようにしてフィルタリング係数を決定すればよい。
【０１１８】
図３６は、第１のローパスフィルタに対応したフィルタリングテーブルを示している。符号（ａ）は輝度用のフィルタリングテーブルを示し、これは図３４のフィルタリングテーブルと同じである。符号（ｂ）は色差用のフィルタリングテーブルを示している。図３７は、平均化フィルタに対応したフィルタリングテーブルであり、符号（ａ）は輝度用のフィルタリングテーブルを示し、符号（ｂ）は色差用のフィルタリングテーブルを示している。図３８は、第２のローパスフィルタに対応したフィルタリングテーブルを示している。この例では、輝度用と色差用のフィルタリングテーブルは共通である。図３９は、ハイパスフィルタに対応したフィルタリングテーブルを示し、このテーブルも、輝度用と色差用において共通である。
【０１１９】
【発明の効果】
以上のように本発明によれば、個々の画像の画質に応じた画像圧縮を達成することができるという効果が得られる。
【図面の簡単な説明】
【図１】本発明の一実施例の画像圧縮装置を示すブロック図である。
【図２】画像データＰ(Y)xy 、ＤＣＴ変換係数Ｓ(Y)uv 、量子化ＤＣＴ係数Ｒ(Y)uv 、量子化テーブルＱ(Y)uv の例を示す図である。
【図３】ＤＣ係数の差分値のカテゴリを示す図である。
【図４】ＤＣ係数の符号化テーブルを示す図である。
【図５】量子化ＡＣ係数の符号化を行う処理ルーチンを示すフローチャートである。
【図６】ＡＣ係数をハフマン符号化する際に行われるジグザグスキャンを示す図である。
【図７】ＡＣ係数のカテゴリを示す図である。
【図８】ＪＰＥＧにより推奨されたハフマンテーブルを示す図である。
【図９】ハフマン符号化による符号化データの一例を示す図である。
【図１０】符号化データの構成要素を示す図である。
【図１１】全ての量子化係数が「１」である量子化テーブルを用いた時の所定の空間周波数でのカテゴリ分布の例を示す図である。
【図１２】全ての量子化係数が「１」である量子化テーブルを用いた時の、各スキャンにおける符号化データのビット数の分布の例を示す図である。
【図１３】空間周波数データ量設定部において設定される各空間周波数毎（すなわちスキャン毎）の符号量の分布の例を示す図である。
【図１４】図１１に示すカテゴリ分布を各スキャンについて同時に示す図である。
【図１５】全ての量子化係数が「１６」である量子化テーブルを用いた時のカテゴリ分布の表を示す図である。
【図１６】スキャンＡＣ₁ のカテゴリ「０」のブロック数と、カテゴリ「１」以上のブロック数を示す図である。
【図１７】スキャンＡＣ₂ までのラン長が２、１、０であるブロック数を示す図である。
【図１８】スキャンＡＣ₃ までのラン長が３、２、１、０であるブロック数を示す図である。
【図１９】スキャンＡＣ₄ までのラン長が４、３、２、１、０であるブロック数を示す図である。
【図２０】スキャンＡＣ₁ のラン長・カテゴリの表を示す図である。
【図２１】スキャンＡＣ₂ のラン長・カテゴリの表を示す図である。
【図２２】スキャンＡＣ₃ のラン長・カテゴリの表を示す図である。
【図２３】スキャンＡＣ₄ のラン長・カテゴリの表を示す図である。
【図２４】ＪＰＥＧにより推奨されたハフマンテーブルの各符号語の符号長を示す図である。
【図２５】量子化テーブルを生成するプログラムのフローチャートである。
【図２６】「００」データの分布を示す図である。
【図２７】スキャンＡＣ₁ とスキャンＡＣ₂ の「００」データの正規化を示す図である。
【図２８】正規化された「００」データの分布の表を示す図である。
【図２９】スキャンＡＣ_i-1 のカテゴリ「０」のブロック数とスキャンＡＣ_i-1 とスキャンＡＣ_i の「００」データとの関係を示す図である。
【図３０】スキャンＡＣ₄ におけるカテゴリ分布の例を示す図である。
【図３１】第３の予測方法によって得られたラン長・カテゴリの分布の例を示す図である。
【図３２】図３１のラン長・カテゴリの分布の表を図２３に対応した形式で表した図である。
【図３３】設定符号量Ｓ_i の分布を生成するプログラムのフローチャートである。
【図３４】フィルタリングテーブルの一例を示す図である。
【図３５】図３４のフィルタリングテーブルにおける各スキャンの割当比率を示す図である。
【図３６】第１のローパスフィルタのフィルタリングテーブルを示す図である。
【図３７】平均化フィルタのフィルタリングテーブルを示す図である。
【図３８】第２のローパスフィルタのフィルタリングテーブルを示す図である。
【図３９】ハイパスフィルタのフィルタリングテーブルを示す図である。
【符号の説明】
Ｍ ＩＣメモリカード[0001]
[Industrial application fields]
The present invention relates to an image compression apparatus that compresses information of a color still image according to a JPEG algorithm.
[0002]
[Prior art]
A standardized algorithm that encodes a high-resolution image and exchanges information via a communication transmission path is recommended by JPEG (Joint Photographic Expert Group). In the JPEG recommended algorithm, that is, the baseline process of the JPEG algorithm, in order to perform significant information compression, the original image data is first decomposed into components on the spatial frequency axis by two-dimensional DCT transformation, and Each data represented on the spatial frequency axis is quantized using a quantization table, and each quantized data is encoded. JPEG recommends a predetermined Huffman table for this encoding.
[0003]
In a conventional image compression apparatus, a default quantization table is usually used, and a modified quantization table is created as necessary by multiplying each quantization coefficient of the quantization table by a single coefficient. Yes.
[0004]
[Problems to be solved by the invention]
In this way, the modified quantization table is obtained by uniformly multiplying each spatial frequency by a coefficient. Therefore, characteristics of individual images (for example, there are many high frequency components compared to low frequency components, etc.) Therefore, it is difficult to say that the image data is compressed without degrading the image quality.
[0005]
In view of the above problems, an object of the present invention is to provide an image compression apparatus that can achieve image compression according to the image quality of each image.
[0006]
[Means for Solving the Problems]
An image compression apparatus according to the present invention performs orthogonal transformation on original image data to obtain an orthogonal transformation coefficient for each spatial frequency, and quantizes the orthogonal transformation coefficient by a quantization table composed of predetermined quantization coefficients. Quantization means for obtaining a quantized orthogonal transform coefficient, and after rearranging the quantized orthogonal transform coefficient into predetermined one-dimensional array data with respect to the spatial frequency, encoding is performed based on the quantized orthogonal transform coefficient. Based on the encoding means to be obtained, the quantized orthogonal transform coefficient obtained by quantization with the default quantization table in which all the quantization coefficients are 1, and a predetermined filtering table, the encoded data for each spatial frequency A means for setting a target value for the data amount and a quantizer for determining a quantization coefficient corresponding to the spatial frequency so that the data amount for each spatial frequency is less than the target value. It is characterized in that an arithmetic unit.
[0007]
【Example】
Hereinafter, the present invention will be described based on illustrated embodiments.
FIG. 1 is a block diagram of an image compression apparatus according to an embodiment of the present invention.
[0008]
Light coming from the subject S is collected by the condenser lens 11, and a subject image is formed on a light receiving surface of a CCD (solid-state imaging device) 12. A large number of photoelectric conversion elements are arranged on the light receiving surface of the CCD 12, and a color filter made up of, for example, R, G, and B color filter elements is provided on the upper surface of the photoelectric conversion elements. Each photoelectric conversion element corresponds to one pixel. The subject image is converted into an electrical signal corresponding to a predetermined color by each photoelectric conversion element and input to the A / D converter 13. In the configuration of FIG. 1, only one CCD 12 is provided, but a configuration in which two or more CCDs are provided may be used.
[0009]
The signal A / D converted by the A / D converter 13 is converted into a luminance signal Y and color difference signals Cb and Cr by a signal processing circuit (not shown) and input to the image memory 14. The image memory 14 is divided into mutually independent memory areas for storing the luminance signal Y and the color difference signals Cb and Cr, and each memory area has a storage capacity for one image.
[0010]
The luminance signal Y and the color difference signals Cb and Cr read from the image memory 14 are input to the DCT processing circuit 21 for data compression processing. In the DCT processing circuit 21, the original image data such as the luminance signal Y is subjected to discrete cosine transform (hereinafter referred to as DCT). That is, in this embodiment, DCT transformation is used as orthogonal transformation of original image data. In FIG. 1, the DCT processing circuit 21 is shown as one processing circuit, but actually, an independent DCT processing circuit is provided for each of the luminance signal Y and the color difference signals Cb and Cr.
[0011]
The image compression apparatus includes a DCT processing circuit 21, a quantization processing circuit 22, a Huffman coding processing circuit 23, a spatial frequency data amount setting unit 24, a quantization table generation unit 25, and the like. In the DCT processing circuit 21, the quantization processing circuit 22, and the Huffman coding processing circuit 23, image data such as the luminance signal Y is divided into a plurality of blocks for one screen and processed in units of blocks. Each block is composed of 8 × 8 pixel data.
[0012]
The luminance signal Y and the DCT coefficients of the color difference signals Cb and Cr obtained in the DCT processing circuit 21 are respectively input to the quantization processing circuit 22. Similarly to the DCT processing circuit 21, the quantization processing circuit 22 is also provided for each signal. The DCT coefficients of the luminance signal Y and the color difference signals Cb and Cr input to the quantization processing circuit 22 are respectively quantized by a quantization table Q1 configured by 8 × 8 quantization coefficients. This quantization is linear quantization, i.e. each DCT coefficient is divided by the corresponding quantization coefficient.
[0013]
In this embodiment, the quantization table Q1 for quantizing the DCT coefficients of the luminance signal Y and the quantization table Q1 for quantizing the DCT coefficients of the color difference signals Cb and Cr are different from each other in accordance with the JPEG algorithm. However, the same quantization table Q1 may be used for each signal. As will be described later, these quantization tables Q1 are generated by the spatial frequency data amount setting unit 24 and the quantization table generation unit 25 according to the properties such as the spatial frequency distribution of the original image data.
[0014]
The quantized DCT coefficients of the luminance signal Y and the color difference signals Cb and Cr output from the quantization processing circuit 22 are input to the Huffman encoding processing circuit 23 and are Huffman encoded by a predetermined algorithm using the Huffman table Q2.
[0015]
An image signal (compressed image data) obtained by Huffman coding is recorded on a recording medium such as an IC memory card.
[0016]
FIG. 2 shows, as an example, image data P (Y) xy of a block of 8 × 8 pixels, DCT coefficient S (Y) uv, quantized DCT coefficient R (Y) uv, and quantization table Q (Y). uv.
[0017]
The image data P (Y) xy in FIG. 2A is converted into 8 × 8 = 64 DCT coefficients S (Y) uv shown in FIG. 2B by two-dimensional DCT conversion. Among these DCT coefficients, the DCT coefficient S (Y) at the position (0, 0)₀₀Is a DC component, and the remaining 63 DCT coefficients S (Y) uv are AC components. The AC component is the coefficient S (Y)₀₁Or coefficient S (Y)_TenTo coefficient S (Y)₇₇FIG. 9 shows how much higher spatial frequency components exist in the image data of the 8 × 8 pixel block. The DC component represents an average value (DC component) of pixel values of the entire 8 × 8 pixel block. That is, each DCT coefficient S (Y) uv corresponds to a predetermined spatial frequency.
[0018]
FIG. 2D shows an example of the quantization table Q (Y) uv used in the quantization processing circuit 21. Such a quantization table Q (Y) uv may be different for the luminance signal Y and the color difference signals Cb and Cr as described above. The quantization table Q (Y) uv is stored in the position corresponding to each signal when the image data in the JPEG format is recorded on the recording medium, in the quantization table Q (Y) uv used for quantization of the signal. The contents are recorded.
[0019]
An expression for quantizing the DCT coefficient S (Y) uv using the quantization table Q (Y) uv is defined as follows.
R (Y) uv = round (S (Y) uv / Q (Y) uv) {0 ≦ u, v ≦ 7}
Round in this equation means an approximation to the nearest integer. That is, the quantized DCT coefficient R (Y) uv as shown in FIG. 2C is obtained by dividing and rounding the elements of the DCT coefficient S (Y) uv and the quantization table Q (Y) uv. It is done.
[0020]
The quantized DCT coefficients R (Y) uv, R (Cb) uv, and R (Cr) uv obtained by the quantization processing circuit 22 in this way are input to the Huffman coding processing circuit 23.
[0021]
Next, Huffman coding in the Huffman coding processing circuit 23 will be described with reference to FIGS. In the following description, a quantized DC coefficient refers to a quantized DC component, and a quantized AC coefficient refers to a quantized AC component.
[0022]
Quantized DC coefficient R (Y)₀₀And quantized AC coefficient (quantized DC coefficient R (Y)₀₀Quantization DCT coefficients (R (Y) uv) other than are different in encoding method. Quantized DC coefficient R (Y)₀₀Is encoded as follows.
[0023]
First, the quantized DC coefficient R (Y) of the block to be encoded at present₀₀And the quantized DC coefficient R (Y) of the previously coded block₀₀The difference is obtained. It is determined to which of the categories shown in FIG. 3 this difference value belongs, and a code word representing the category is obtained from the code table (DC component coding table) shown in FIG. For example, the quantized DC coefficient R (Y) of the block to be encoded at present₀₀Is “16”, and the quantized DC coefficient R (Y) of the block previously encoded is₀₀3 is “−9”, the category to which the difference value = −9 belongs is determined to be “4” from the category table of FIG. From the code table of 4, it is determined as “101”.
[0024]
Next, in the category table of FIG. 3, the difference value is what number in the category is represented by an additional bit. For example, since the difference value = −9 is the seventh from the smallest in the category = 4 group, the additional bit is “0110”. That is, the quantized DC coefficient R (Y) of the currently encoded block₀₀The Huffman codeword is “1010110”.
[0025]
On the other hand, encoding of the quantized AC coefficient is performed by a processing routine shown in FIG. First, in step 120, 63 quantized AC coefficients are zigzag scanned in the order shown in FIG. 6 and rearranged into one-dimensional array data. The quantized AC coefficients are scanned according to the zigzag scan order.₁, AC₂... AC₆₃I will call it. Next, in step 122, it is determined whether or not each quantized AC coefficient arranged in one dimension is “0”. When the quantized AC coefficient is “0”, the number of consecutive quantized AC coefficients that are “0” is counted in step 124. Thus, a length in which “0” continues, that is, a run length is obtained.
[0026]
On the other hand, when it is determined in step 122 that the quantized AC coefficient is not “0”, in step 126, grouping similar to the quantized DC coefficient is performed and additional bits are obtained. This grouping of quantized AC coefficients is performed on the quantized AC coefficients themselves, unlike the grouping of quantized DC coefficients. That is, when the quantized AC coefficient is “4”, for example, the category “3” is obtained with reference to the table shown in FIG. Further, since the quantized AC coefficient “4” is the fifth smallest from the category = 3 group, the additional bit is “100”.
[0027]
In step 130, referring to the AC code table of the Huffman table (FIG. 8), for example, when the run length of the data immediately before the quantized AC coefficient “4” is “0”, the run length and category = 3 are set. Based on this, the code word “100” is obtained. Then, the two-dimensional Huffman code word “100100” is obtained by combining this code word “100” and the additional bit “100” obtained in step 126.
[0028]
The result of Huffman encoding the quantized DCT coefficient in FIG. 2C is shown as encoded data HF in FIG.
[0029]
FIG. 10 shows the encoded data HF shown in FIG. 9 again. Such encoded data HF is obtained for each block. When one screen is composed of 5400 blocks, only 5400 encoded data HF is obtained. As described above, the encoded data HF includes encoded data relating to one quantized DC coefficient and encoded data relating to 63 quantized AC coefficients.
[0030]
The encoded data related to the quantized DC coefficient includes a category code word FA0 and an additional bit FB0. The encoded data related to the quantized AC coefficient includes a run length / category code word and additional bits. Next, the encoded data regarding the quantized AC coefficient will be described in more detail.
[0031]
In the example of FIG. 9, the scan AC₁Is encoded data based on the fact that the run length is 0 and the run length / category code word FA1 and the additional bit FB1 are generated, and the scan AC₂Is encoded data based on the fact that the run length is 0 and the run length / category code word FA2 and the additional bit FB2 are generated. Scan AC_ThreeIs 0, and therefore, after the additional bit FB2, the scan AC_FourAs the encoded data based on the fact that the run length is 1 and the run length is 1, the run length / category code word FA4 and the additional bit FB4 are generated. Similarly, a run length / category code word FA5 and additional bits FB5, a run length / category code word FA8 and additional bits FB8, and a run length / category code word FA9 and additional bits FB9 are generated. End data (EOB) is scan AC_TenThereafter, all indicate that 0 continues.
[0032]
Next, generation of the quantization table Q1 will be described.
As shown in FIG. 1, the quantization table Q1 is generated based on the set total code amount, the DCT data statistic, and the filtering table FL.
[0033]
The set total code amount is the total number of bits of the encoded data HF for one screen recorded on the recording medium, and is, for example, 524288 bits (64 Kbytes). The DCT data statistic is obtained based on the output data of the DCT processing circuit 21. The output data of the DCT processing circuit 21 is obtained by DCT transforming the original image data, and a quantization table Q1 (hereinafter referred to as a default quantization table) in which 8 × 8 quantization coefficients are all “1” is used. Is equivalent to the quantized one. The output data is zigzag scanned (see FIG. 6), the run length and the category are obtained (see FIG. 5), and then the bit length for each spatial frequency is obtained with reference to the table shown in FIG. Data statistics are obtained. That is, the DCT data statistics are a category distribution as shown in FIG. 11 and a code amount distribution for each scan as shown in FIG. The filtering table FL determines the degree of data compression at each spatial frequency. As will be described later, the filtering table FL has a filtering coefficient corresponding to each spatial frequency, and each filtering coefficient has a larger value as the spatial frequency has a higher degree of data compression.
[0034]
FIG. 11 shows a predetermined spatial frequency (for example, scan AC) when the default quantization table is used.₁) Category distribution, that is, an example of the number of blocks classified into each category. As for the category distribution as shown in this figure, in one image, those relating to DC components and those relating to 63 AC components are generated, that is, only 64 category distributions are generated in total. In this embodiment, the original image data is divided into 5400 blocks. In the example of FIG. 11, the number of blocks whose category is 0 is 602, and the number of blocks whose category is 1 is 1088.
[0035]
FIG. 12 is an example of the distribution of the number of bits of the encoded data HF in each scan when the default quantization table is used. This is the total number of bits for each spatial frequency for all blocks of one image. Show. For example, in the case of a DC component, the total number of bits of the category code word FA0 and the additional bits FB0 (see FIG. 10) is 42238 (code DB0). Scan AC₁In this case, the total number of bits of the run length / category code word FA1 and the additional bit FB1 is 34,010 (code DB1), and scan AC₂In this case, the total number of bits of the run length / category code word FA2 and the additional bits FB2 is 33833 (code DB2). In the example of FIG. 10, scan AC_ThreeThere are no run-length / category codewords and additional bits for scan AC in other blocks_ThreeTherefore, the total number of bits is 25920 (code DB3). In this way, the scan AC₆₃The total number of bits up to 20909 (code DB 63) is generated.
[0036]
The set total code amount, DCT data statistic, and filtering table FL are input to the spatial frequency data amount setting unit 24. The spatial frequency data amount setting unit 24 sets the distribution of code amounts for each spatial frequency (ie, for each scan) as shown in FIG. 13, for example, based on the set total code amount, the DCT data statistic, and the filtering table FL. Is done. That is, the code amount of the DC component is 44240 bits (code SB0), scan AC₁Code amount is 33008 bits (code SB1), scan AC₂The code amount is 35629 bits (code SB2), and the scan AC₆₃Is set up to the code amount (reference SB63). The total value of these code amounts, that is, the set total code amount is determined to be a predetermined value (524288 bits in the example of FIG. 13).
[0037]
This code amount (data amount) is a target value, and in this image compression apparatus, the quantization table Q1 is generated so that the code amount for each spatial frequency becomes this target value, as will be described later. That is, this code amount distribution is the target value of the code amount for each spatial frequency in the Huffman encoded data when the finally obtained quantization table Q1 is used. For example, when it is desired to cut a high frequency component, the code amount related to the high frequency component is determined to be relatively small, and the distribution of the code amount is set based on the filtering table FL.
[0038]
Further, the spatial frequency data amount setting unit 24 may determine the upper limit value of the code amount for each spatial frequency by using the code amount distribution for each scan (see FIG. 12) in the DCT data statistics. For example, in FIG._FourIs defined as 28844 (code SB4), but may be limited to 28461 bits based on the code amount distribution of the DCT data statistics (see code DB4 in FIG. 12).
[0039]
In the quantization table generation unit 25, each quantization coefficient constituting the quantization table Q1 is generated based on the DCT data statistic and the input data from the spatial frequency data amount setting unit 24.
[0040]
A method for obtaining the quantization coefficient for the DC component will be described.
First, DC components of all blocks are quantized using a default quantization table (a quantization table in which all quantization coefficients are 1). Then, the difference value between the quantized DC coefficient of the block for which the current quantized coefficient is to be obtained and the quantized DC coefficient of the previous block is obtained.
[0041]
It is determined to which of the categories shown in FIG. 3 this difference value belongs, and a code word representing the category is obtained from the code table (DC component coding table) shown in FIG. Further, the number of additional bits corresponding to the difference value is obtained from the category table of FIG. For example, when the category of the difference value is “2”, the code length is 3 bits and the number of additional bits is 2 bits. Therefore, the code amount of the quantized DC coefficient whose difference value category is “2” is 5 bits.
[0042]
In this way, the code amount of the quantized DC coefficient is obtained for each block, and the total code amount (number of bits) is obtained. If this total code amount is equal to or less than the DC component code amount (code SB0) shown in FIG. 13, the quantization coefficient at that time is determined as the final one. If the total value is larger than the code amount (code SB0) in FIG. 13, then the quantization coefficient is changed to 2, and the total number of bits as described above is examined using this quantization coefficient. Is called.
[0043]
In order to obtain the total number of bits when this new quantization coefficient is used, it is necessary to predict the number of blocks corresponding to each category. That is, it is necessary to obtain a category distribution table as shown in FIG. An example of how to obtain this category distribution table will now be described with reference to FIG.
[0044]
For example, the difference values belonging to the category “2” are −3, −2, 2, and 3. When the quantization coefficient is 2, since the difference value “2” is 2/2 = 1, the process moves to the category “1”. The same applies to the difference value “−2”. On the other hand, since the difference value “3” is 3/2 = 1.5≈2, the category “2” remains unchanged. The same applies to the difference value “−3”. Accordingly, in this example, of the difference values belonging to the category “2” when the quantization coefficient is 1, half changes to the category “1” and half remains the category “2”. In this way, the category distribution when the quantization coefficient changes is predicted, and the total code amount is obtained by using this category distribution table.
[0045]
Next, how to obtain the quantization coefficient for the AC component will be described.
As for the AC component, data relating to run length is included in the Huffman encoded data (eg, codes FA1, FA2, FA4, etc. in FIG. 10), and the run length changes when the quantization coefficient is changed. Therefore, the data amount of each scan when the quantization coefficient is changed cannot be predicted based only on the data amount of the scan before changing the quantization coefficient. Therefore, in this embodiment, as described below, the data amount of each scan when the quantization coefficient is changed is predicted using a category distribution table, and the quantization coefficient is determined based on the predicted value. ing.
[0046]
FIG. 14 is an example of a table that simultaneously shows the category distribution shown in FIG. 11 for each scan. That is, this category distribution table indicates the number of blocks classified into each category at each spatial frequency when the default quantization table is used. In FIG. 14, the DC component and the scan AC₁To scan AC₁₁Scan AC shown₁₂To scan AC₆₃Up to is omitted. In FIG. 14, the uppermost number indicates a category. For example, scan AC₁, The number of blocks in category “0” is 602, and the number of blocks in category “1” is 1088.
[0047]
Unlike FIG. 14, FIG. 15 shows a table of category distribution when using a quantization table in which all quantization coefficients are “16”. As can be understood from FIG. 7, the AC component values in the categories “0” to “3” when the quantization coefficient “1” is used are 0 when the quantization coefficient “16” is used. The category of the AC component value is “0”. That is, the scan AC in FIG.₁The 3479 category “0” blocks correspond to the blocks 602, 1088, 1184, and 605 of categories “0” to “3” in FIG. 14.
[0048]
Similarly, when the quantization coefficient “1” is used, the AC component value of the category “4” is −1 or 1 when the quantization coefficient “16” is used. Therefore, the category of the AC component value is “1”. It becomes. On the other hand, the AC component value of the category “5” when the quantization coefficient “1” is used may be 1 or 2 when the quantization coefficient “16” is used. Therefore, when the quantization coefficient “16” is used, the category of AC component values of some blocks is “1”, and the category of AC component values of other blocks is “2”. That is, the scan AC in FIG.₁The 866 category “1” blocks correspond to the 529 block of the category “4” and the partial block of the category “5” in FIG. As described above, when the quantization coefficient is changed, a block belonging to a certain category does not always move to another category but may move to a different category.
[0049]
Next, scan AC_i-1, AC_iThe prediction of the number of blocks in which both categories are “0” will be described with reference to FIGS.
[0050]
16 to 19, C [0] is the number of blocks of category “0” in the scan, and C [1] is the number of blocks of category “1” or more in the scan. Z [k] is a predicted value of the number of blocks whose run length is k. C ′ [0] is the number of blocks of category “0” in the previous scan, and Z ′ [0] is the number of blocks whose run length is 0 in the previous scan.
[0051]
FIG. 16 shows scan AC₁The number of blocks in category “0” and the number of blocks in category “1” or more are shown. Scan AC₁As shown in FIG. 15, the number of blocks C [0] in the category “0” is 3479, and the number of blocks C [1] in the category “1” or more is 1921 (= 5400-3479). Therefore, scan AC₁The number of blocks Z [1] whose run length is 1 is 3479, and the number of blocks Z [0] whose run length is 0 is 1921.
[0052]
FIG. 17 shows a scan AC₂The number of blocks whose run lengths are 2, 1, 0 are shown. Scan AC₂As shown in FIG. 15, the number of blocks C [0] in the category “0” is 3619, and the number of blocks C [1] in the category “1” or more is 1781. Therefore, it is clear that the number of blocks Z [0] whose run length is 0 is 1781, but the number of blocks Z [2] whose run length is 2 and the number of blocks Z [1] whose run length is 1 Is not clear. Therefore, in this embodiment, the ratio of the number of blocks Z [2] to the number of blocks Z [1]₁Assuming that the number of blocks Z [1] whose run length is 1 and the number of blocks Z ′ [0] whose run length is 0 is equal to the ratio of the number of blocks Z [2] and the number of blocks Z [1] ] Is expected. That is, in this embodiment, the scan AC₁, AC₂The number of blocks whose category is “0” in both₁Is related to the number of blocks of category “0” in FIG.
[0053]
The number of blocks Z [2] obtained in this way is the scan AC₁, AC₂Scan AC when both categories are “0”₂It corresponds to. The number of blocks Z [1] is the scan AC₁Scan AC with category “0” when category is other than “0”₂It corresponds to.
[0054]
FIG. 18 shows a scan AC_ThreeThe number of blocks with run lengths up to 3, 2, 1, 0 is shown. Scan AC_ThreeAs shown in FIG. 15, the number of blocks C [0] in the category “0” is 4366, and the number of blocks C [1] in the category “1” or more is 1034. Therefore, the number of blocks Z [0] is 1034. The number of blocks Z [1] is the scan AC₂Assuming that it depends on the ratio of the number of blocks Z ′ [0] at
Z [1] = 4366 × 1781/5400 = 1440
It becomes. On the other hand, the ratio of the number of blocks Z [2], [3]₂Is equal to the ratio of the number of blocks Z [1], [2] at
Z [2] = 4366 × 1287/5400 = 1041
Z [3] = 4366 × 2332/5400 = 1888
It becomes.
[0055]
The number of blocks Z [3] is the scan AC₁, AC₂, AC_ThreeScan AC when both categories are “0”_ThreeIt corresponds to. The number of blocks Z [2] is the scan AC₁Category other than "0" and scan AC₂Scan AC with category “0” when category is “0”_ThreeIt corresponds to. The number of blocks Z [1] is the scan AC₂Scan AC with category “0” when category is other than “0”_ThreeIt corresponds to.
[0056]
Figure 19 shows Scan AC_FourBlocks with run lengths up to 4, 3, 2, 1, 0 are shown. Scan AC_FourThereafter, the above-described processing is performed, and the number of blocks Z [k] whose run length is k is obtained.
[0057]
In this way, scan AC₆₃When the run lengths up to are calculated, run length / category tables as shown in FIGS. 20 to 23 are created for each category.
[0058]
FIG. 20 shows a scan AC₁The run length / category table of FIG. 5 shows a category distribution other than the category “0” in which the run length is 0. Scan AC₁The total number of blocks other than the category “0” is 1921 (= Z [0]) as shown in FIG. As shown in FIG. 15, the category distribution is 866, 519, 300, 212, 24 in the order of categories “1”, “2”... “5”, and the number of blocks of category “6” or more is 0. It is. By transcribing this numerical value as it is, the table of FIG. 20 is created.
[0059]
FIG. 21 shows a scan AC₂The run length / category table of FIG. 2 shows a category distribution with run lengths of 0 and 1 and a category other than category “0”. Scan AC₂The total number of blocks other than the category “0” is 1781 (= Z [0]) as shown in FIG. As shown in FIG. 15, the category distribution is 850, 487, 315, 119, 10 in the order of categories “1”, “2”... “5”, and the number of blocks of category “6” or more is 0. It is. The 850 block of category “1” has a ratio of Z [1] and Z [2] shown in FIG.₁Assuming that the number of blocks Z ′ [0] with a run length of 0 is equal to the ratio of the number of blocks Z [1] with a run length of 1, the number of blocks with a run length of 0 is 302, The number of blocks with 1 is set to 548. Similarly, for the 487 blocks in category “2”, the number of blocks with a run length of 0 is 173, and the number of blocks with a run length of 1 is 314. In this way, the table of FIG. 21 is created.
[0060]
FIG. 22 shows a scan AC_ThreeThe run length / category table of FIG. 1 shows a category distribution with run lengths of 0, 1, and 2 and a category other than “0”. Scan AC_ThreeThe number of blocks other than the category “0” is 1034 (= Z [0]) in total as shown in FIG. As shown in FIG. 15, the category distribution is 675, 220, 121, 18 in the order of categories “1”, “2”... “4”, and the number of blocks of category “5” or more is 0. . The number of blocks in each category is distributed to 0, 1 and 2 blocks according to the ratio of Z [1], Z [2] and Z [3] shown in FIG.
[0061]
FIG. 23 shows a scan AC_FourIt is a table of run length and category. This table is also created with reference to FIGS. 15 and 19 by the processing described above. In this way, scan AC₆₃A run length / category table is created.
[0062]
Next, the data amount (number of bits) in each scan is calculated with reference to the run length / category tables as shown in FIGS. For this calculation, the code length table shown in FIG. 24 is used. Each numerical value in this table indicates the code length of each codeword of the Huffman table recommended by JPEG. For example, the code length of a codeword having a run length of 1 and a category of 2 is 5.
[0063]
Scan AC₁For FIG. 20 and FIG. 24, the run length / category codeword bits in all blocks are
866 x 2 + 519 x 2 + 300 x 3 + 212 x 4 + 24 x 5 = 4638
It becomes. On the other hand, for additional bits, the category number directly corresponds to the number of bits, so the number of additional bits is
866 x 1 + 519 x 2+ 300 x 3 + 212 x 4 + 24 x 5 = 3772
It becomes. Therefore scan AC₁Expected data volume is
  4638 + 3772 = 8410 (bits)
It is.
[0064]
Scan AC₂~ Scan AC₆₃Also about scan AC₁The amount of data can be obtained by the same calculation as.
[0065]
The amount of data in each scan obtained in this way must be less than or equal to the set code amount shown in FIG. For example, scan AC₁Must be limited to 33008 bits or less. In the above example, since the quantization coefficient is set to “16” for the purpose of explaining the embodiment, the data amount is 8410 bits, which is considerably smaller than the set code amount. Therefore, in practice, a numerical value smaller than “16” is appropriate for the quantization coefficient. That is, if the expected data amount when the quantized coefficient is “1” is larger than 33008 bits, the quantized coefficient is increased by 1 and the expected data amount is recalculated. A quantization coefficient that is less than the code amount is obtained.
[0066]
The generation of the quantization table described above will be described with reference to the flowchart of FIG. This flowchart shows the scan AC₂~ AC₆₃Shows the generation of quantization coefficients for the DC component and the scan AC₁It is assumed that the quantization coefficient of is already obtained.
[0067]
In step 101, the parameter i is set to 2. In step 102, the quantization coefficient q is set to 1 as an initial value.
[0068]
In step 103, the scan AC as shown in FIG._iA category distribution is generated. For example, when the parameter i is 3 and the quantization coefficient q is 16, this category distribution is as shown in FIG.
  4366, 675, 220, 121, 18, 0, 0, 0, 0, 0, 0
It is.
[0069]
Before step 103 is executed, as shown by reference numeral P11, scan AC_i-1The number of blocks of category “0” is calculated. For example, parameter i is 3 and scan AC₂If the quantization coefficient q of the scan AC is 16, the scan AC₂The number of blocks in category “0” is 3619 in the example of FIG. Further, as indicated by reference numeral P12, the scan AC_i-1The distribution of the number of blocks of category “0” is calculated. For example, parameter i is 3 and scan AC₂If the quantization coefficient q of the scan AC is 16, the scan AC₂In the example of FIG. 17, the number of blocks Z [2] is 2332, the number of blocks Z [1] is 1287, and the number of blocks Z [0] is 1781.
[0070]
In step 104, the scan AC obtained in step 103 is obtained._iCategory distribution and scan AC_i-1Number of blocks of category “0” (symbol P11) and scan AC_i-1Based on the distribution of the number of blocks of the category “0” (symbol P12)_iThe distribution of the number of blocks of category “0” in is predicted. For example, when the parameter i is 3 and the quantization coefficient q is 16, the scan AC_ThreeIn the example of FIG. 18, the number of blocks Z [3] is 1885, the number of blocks Z [2] is 1041, and the number of blocks Z [1] is 1440 based on the data of FIGS.
[0071]
In step 105, the scan AC_iThe predicted data amount of the compressed data is calculated. For example, when the parameter i is 3 and the quantization coefficient q is 16, the scan AC_ThreePredicted data amount D_iFrom the run length / category table of FIG. 22 and the code length table of FIG.
    223 x 2 + 73 x 2 + 40 x 3 + 6 x 4
+ 161 x 4 + 52 x 5 + 29 x 7 + 4 x 9
+ 291 x 5 + 95 x 8 + 52 x 10 + 8 x 12
+ (223 + 161 + 291) + (73 + 52 + 95) x 2
+ (40 + 29 + 52) x 3 + (6 + 4 + 8) x 4
= 6260 (bit)
It becomes.
[0072]
In step 106, the predicted data amount D calculated in step 105 is displayed._iIs the set code amount S as shown in FIG._iIt is determined whether or not. In the example of the description of step 105 described above, since the quantization coefficient q is 16, the predicted data amount D_iBecomes 6260 bits, and the set code amount S in FIG._Three= 24042 (reference SB3). However, in practice, since the quantization coefficient q is started from 1, the predicted data amount D is initially set._iIs the set code amount S_iBigger than. Therefore, after it is confirmed in step 107 that the quantization coefficient q has not yet reached 255, the quantization coefficient q is incremented by 1 in step 108 and the process returns to step 103 again.
[0073]
Then, Steps 103 to 105 are executed again. In Step 106, the predicted data amount D_iIs the set code amount S_iIf it is determined that it is the following, the quantization coefficient q is determined thereby. That is, the process proceeds from step 106 to step 110, and it is determined whether or not the parameter i has reached 63. When the parameter i does not reach 63, in step 111, the set code amount S for the next scan_{i + 1}Is the set code amount S_iAnd predicted data amount D_iOnly the difference is added. That is, the set code amount S of the next scan_{i + 1}Is the previous scan set code amount S_iThe amount not used in is allocated.
[0074]
Next, in step 112, the parameter i is incremented by 1, and the process returns to step 102, and steps 102 to 108 are executed for the next scan to obtain the quantization coefficient.
[0075]
If it is determined in step 110 that the parameter i has reached 63, the process proceeds to step 113, where a quantized coefficient is generated based on the quantized coefficient q, and this program ends.
[0076]
Each scan AC described with reference to FIGS._iThe first method of predicting the amount of data is a relatively simple example, and is not necessarily a highly accurate prediction. Therefore, the second prediction method with higher accuracy will be described next with reference to FIGS.
[0077]
FIG. 26 shows the distribution of “00” data, that is, scan AC._i-1And scan AC_iBoth indicate the number of blocks of category “0”, and corresponds to the category distribution table of FIG. For example, scan AC₁, AC₂The number of blocks in category “0” is 120 (reference code J₁₂), Scan AC₂, AC_ThreeAre 159 (symbol J_{twenty three}), Scan AC_Three, AC_FourAre 186 (symbol J₃₄), Scan AC₆₂, AC₆₃The number of blocks in category “0” is 327 (reference code J₆₂₆₃).
[0078]
Scan AC_i-1, AC_iWhen the number of blocks each of which is in category “0” changes, scan AC_i-1, AC_iThe number of blocks whose category is category “0” is scan AC_i-1Rate of change in the number of blocks and scan AC_iIs proportional to the product of the rate of change of For example, scan AC_i-1, AC_iAssuming that the number of blocks has doubled, scan AC_i-1, AC_iThe number of blocks in which both are in category “0” is quadrupled. Under such assumption, scan AC_i-1And AC_iBased on the number of blocks of category “0”, the scan AC after quantization_i-1, Scan AC after quantization_iBoth predict the number of blocks of category “0”.
[0079]
The distribution of “00” data is normalized by setting the reference value of the number of blocks to 3500, and is stored in the memory. For example, scan AC₁, AC₂The number of blocks in which both are in category “0” is converted into 3823 as shown in FIG. 27 by normalization. FIG. 28 shows a table of normalized “00” data distribution, eg, scan AC₁, AC₂“00” data of 3823 (symbol K₁₂), Scan AC₂, AC_Three"00" data of 3577 (reference code K_{twenty three}), Scan AC_Three, AC_Four“00” data of 3280 (code K₃₄), Scan AC₆₂, AC₆₃"00" data is 2262 (code KJ₆₂₆₃).
[0080]
FIG. 29 shows a scan AC of category “0”._i-1The number of blocks on the horizontal axis and scan AC_i-1, AC_iAre graphs with the vertical axis representing the number of blocks of category “0” (ie, “00” data). In this graph, the maximum value of the abscissa is the total number of blocks (5400), and S_lowIs the above-mentioned reference value 3500, S_midFor example 4800, S_highFor example, 5150. Maximum value of ordinate O_maxIs variable, maximum value O_maxWhen the reference value is 3500, O_highFor example 3403, O_midFor example 3144, O_lowFor example, 2463. The break points f, g, and h are obtained by trial and error, for example, so that the quality of the reproduced image is close to that of the original image, but the break point b is the scan AC as described below._i-1, AC_iVaries according to the number of blocks of category “0”.
[0081]
The method for obtaining the break point b will be described by taking the case of FIG. 15 as an example.
For example, scan AC_ThreeThe number of blocks in category “0” is 4366, and scan AC_FourThe number of blocks in category “0” is 4108.
[0082]
First, referring to the table of FIG._Three, AC_FourAs “00” data of K,₃₄= 3280 is obtained. Next, in FIG. 29, the maximum value O_maxWith the reference value 3500 set to S_lowReading the ordinate corresponding to (3500) gives point c. A straight line L1 connecting the origin a and the point c, and O_lowWhen the intersection point b with the straight line L2 extending in the horizontal direction from is obtained, a broken line abfgh is obtained.
[0083]
Next, the maximum ordinate O_maxScan AC_FourTo match the number of blocks 4108, O_highIs 3994, O_midIs 3690, O_lowBecomes 2891. In this state, scan AC in the abscissa_ThreeWhen reading the ordinate of the point P3 on the polygonal line bf corresponding to the number of blocks 4366, 3530 is a scan AC_Three, AC_FourObtained as “00” data. This numerical value corresponds to the sum of Z [2], Z [3], and Z [4] (= 3321) in FIG. 19, and is larger than the example of FIG. That is, according to the second prediction method, scan AC_Three, AC_FourThere are more “00” data than in the first prediction method.
[0084]
The “00” data obtained in this way is distributed to Z [2], Z [3] and Z [4] by the method described below.
[0085]
For this distribution, a scan AC as shown in FIG.₂, AC_ThreeZ [1], Z [2], Z [3] values related to the "00" data are required, and in order to obtain these Z [1], Z [2], Z [3] Scan AC as shown in FIG.₁, AC₂Z [1] and Z [2] values related to “00” data are required.
[0086]
Scan AC₁, AC₂Z [1] and Z [2] are obtained by creating a broken line abfgh in the same manner as in FIG. 29 and reading the points on the broken line. Z [2] is obtained directly from the broken line, and Z [1] is obtained as the difference between C [0] and Z [2]. Scan AC₂, AC_ThreeFor Z, the sum of Z [2] and Z [3] is obtained by creating a broken line abfgh and reading the points on this broken line, and Z [1] is the Z [2] and Z [3] It is obtained by subtracting the sum from C [0]. Each value of Z [2] and Z [3]₁, AC₂Are distributed according to the ratio of Z [1] and Z [2]. That is, scan AC₂, AC_ThreeEach value of Z [1], Z [2], Z [3] is obtained.
[0087]
Similarly, scan AC_Three, AC_FourZ [2], Z [3], and Z [4] values of the scan AC₂, AC_ThreeAre distributed according to the ratio of Z [1], Z [2] and Z [3]. As a result, a run length / category table as shown in FIG. 23 is obtained._FourThe predicted data amount of the compressed data at is calculated. Similar calculations are performed for other scans to obtain the predicted data amount.
[0088]
Next, scan AC_iA third method for predicting the data amount will be described.
Scan AC_i-1The number of blocks whose run length is k in Z [k] and scan AC_i-1The total number of blocks with category “0” in Z_ALL, Scan AC_i-1Category “0” and scan AC_iThe number of blocks of category “j” in CT_jAnd The correction coefficient is ZW.
[0089]
Scan AC_iThe number of blocks B [k, j] whose run length is k and category is j is

It is calculated by. Scan AC_iThe number of blocks Z [k + 1] whose run length is (k + 1) in (i.e., the number of blocks whose run length is k and whose category is 0) is

It is calculated by.
[0090]
The correction coefficient ZW can take a value from 0 to 1000. For example, when ZW = 0,
Z [k + 1] = CT_j× Z [k] / Z_ALL
This is the same result as the second prediction method. If ZW = 1000,
Z [k + 1] = Z [k]
This is a scan AC_i-1The block of run length k is Scan AC_iIndicates that all of them are in category “0”. In the example described below, ZW = 50.
[0091]
Where scan AC_Three, Z [0] = 1034, Z [1] = 1440, Z [2] = 1041, Z [3] = 1888, and scan AC_FourAssume that a table as shown in FIG. 30 is obtained as the category distribution in FIG. That is, scan AC_ThreeIn block 4366 where is category “0”, scan AC_FourThen, the number of blocks of category “0” CT₀Is the number of blocks of 3321, category “1” CT₁Is the number of blocks CT of 655, category “2”₂Is 276, the number of blocks in category “3” CT_ThreeIs the number of blocks in category 101 and category “4” CT_Four13 and the number of blocks of category “5” or more is 0.
[0092]
As will be described below, the number of blocks is obtained from data having a run length of 1, and a run length / category distribution table as shown in FIG. 31 is obtained.
[0093]
Scan AC_Four, The number of blocks whose run length is 1 and category “1” is

It becomes. Similarly, the number of blocks whose run length is 1 and whose category is “2” is as follows from the equation (1):

It becomes.
[0094]
For the number of blocks whose run length is 1 and category “3”, from equation (1):

However, as shown in FIG._Three= C [3] = 101. Therefore, it is forcibly set to 101, not 104.
[0095]
Similarly, B [1,4] is forcibly set to 13 for the number of blocks having a run length of 1 and a category of “4”.
[0096]
Scan AC_FourThe number of blocks Z [2] in which the run length is 2 (that is, the number of blocks in which the run length is 1 and the category is 0) is
Z [2] = Z [1]-(277 + 158 + 101 + 13) = 891
It becomes.
[0097]
For the number of blocks with run length 2 and category “1”,

It becomes. For the number of blocks with run length 2 and category “2”, B [2,2] = 66 + 49 = 115
It becomes.
[0098]
The number of blocks B [2,3] having a run length of 2 and category “3” is 0 according to the result of B [1,3], the run length is 2, and category “4”. The number of blocks B [2,4] is 0 according to the result of B [1,4].
[0099]
Scan AC_FourThe number of blocks Z [3] whose run length is 3 (that is, the number of blocks whose run length is 2 and whose category is 0) is
Z [3] = Z [2]-(200 + 115 + 0 + 0) = 726
It becomes.
[0100]
For the number of blocks with run length 3 and category “1”,

It becomes. However, the number of blocks in category “1” CT₁Because the total number of
B [3,1] = 655-277-200 = 178
It becomes.
[0101]
For the number of blocks with a run length of 3 and category “2”,

It becomes. However, the number of blocks in category “2” CT₂Because the total number of is 276,
B [3,2] = 276-158-115 = 3
It becomes.
[0102]
The number of blocks whose run length is 3 and category “3” is 0 according to the results of B [1,3] and B [2,3], run length is 3, and category “4”. The number of blocks is “0” according to the results of B [1,4] and B [2,4].
[0103]
Scan AC_FourThe number of blocks Z [4] whose run length is 4 (that is, the number of blocks whose run length is 3 and whose category is 0) is
Z [4] = Z [3]-(178 + 3 + 0 + 0) = 1704
It becomes.
[0104]
In the number of blocks whose run length is 0, the numbers of blocks of categories “1” to “4” are 155, 65, 24, and 3 from FIG. Further, the number of blocks Z [1] whose run length is 1 (that is, the number of blocks whose run length is 0 and whose category is 0) is as shown in FIG.
Z [1] = 787
It becomes.
[0105]
FIG. 32 is a table of run length / category distributions obtained in this manner, and corresponds to FIG. 23 in the first prediction method. As shown in FIG. 31, the number of blocks having a run length of 4 (the number of blocks having a run length of 3 and a category of 0) is 1704, which was obtained in the first prediction method. It is larger than Z [4] = 1434. That is, according to the third prediction method, it is understood that the run length increases.
[0106]
Next, in the category distribution table as shown in FIG._i-1Category is “0” and scan AC_iA method for obtaining the number of blocks whose category is “1” or more using the broken line abfgh in FIG. 29 will be described.
[0107]
In this case, the horizontal axis in FIG._i-1Indicates the number of blocks of category “0”, and the vertical axis indicates scan AC_iThe sum of the number of blocks of category “0” and category “1” in FIG. First, the vertical axis O_maxIs defined as the sum of C [0] and C [1] (= 4108 + 810 = 4918)._high, O_mid, O_lowAsk for. Next, when a point on the polygonal line corresponding to the abscissa 4366 (for example, P3) is read and the ordinate corresponding to this point is read, this value is calculated as scan AC._FourIs the sum of the number of blocks of category “0” and category “1”. Therefore, if the value of the category “0” already obtained is subtracted from this sum, the scan AC_i-1Category is “0” and scan AC_iThe number of blocks whose category is “1” is obtained. Similarly, scan AC_i-1Category is “0” and scan AC_iThe number of blocks whose category is “2” or more is obtained.
[0108]
Next, referring to the flowchart of FIG. 33, the set code amount S used in step 106 of FIG._iThe generation of the distribution (see FIG. 13) will be described.
[0109]
In step 201, the DC component and the scan AC for all blocks of one image.₆₃The minimum total code amount (number of bits) of the encoded data is obtained. Total code amount CD of DC component₀Is obtained by multiplying the minimum value HN (0) = 2 of the total value of the Huffman code amount and the number of additional bits corresponding thereto by the total number of blocks. Normally, the total value of the Huffman code amount and the number of additional bits takes the minimum value in category “0”, and therefore, the code of the minimum value HN (0) is used here. On the other hand, scan AC₆₃Total code amount CD₆₃Is obtained by multiplying the code amount HN (“EOB”) = 4 of the end data (EOB) in Huffman coding by the total number of blocks.
[0110]
In step 202, the scan AC when using the default quantization table is used.₆₃The total code amount CDA for all blocks except for
CDA = Σ (CO_i/ FT_i(3)
It is calculated by. However, CO_iIs the scan AC when using the default quantization table_iCode amount (number of bits) (see FIG. 12), and FT_iScan AC_iThe table value of the filtering table FL (see FIG. 34), that is, the filtering coefficient. Note that Σ indicates that the parameter i is added from 0 to 62.
[0111]
The filtering table FL includes, for example, a filtering coefficient FT corresponding to each spatial frequency as shown in FIG._iConsists of. Filtering coefficient FT₀Corresponds to the DC component and the filtering coefficient FT₁, FT₂... FT₆₃Is scan AC₁, AC₂... AC₆₃Corresponding to Further, the filtering coefficient has a larger value as the spatial frequency with a higher degree of data compression. Therefore, the example of FIG. 34 shows a filter that cuts high frequency components. That is, the filtering coefficient of the high frequency component is 20000 or 30000, and the high frequency component is greatly compressed by the equation (3).
[0112]
In step 203, the correction coefficient CDR is
CDR = (SETIM-CD₀-CD₆₃/ CDA (4)
It is calculated by. However, SETIM is a set total code amount, for example, 524288 bits (64 Kbytes). The value of the correction coefficient CDR becomes larger as the total code amount CDA obtained by the equation (3) is smaller, and is about 100, for example.
[0113]
After parameter i is cleared to 0 in step 204, in step 205,
CODE_i= (CO_i/ FT_i) X CDR (5)
By means of each spatial frequency (scan AC_i) For the amount of compressed data (number of bits). As shown in the equation (5), the compressed data amount CODE_iIs the code amount CO when the default quantization table is used._iFiltering coefficient FT_iThe result of dividing by is multiplied by the correction factor CDR.
[0114]
In step 206, the compressed data amount CODE_iCode amount CO when using the default quantization table_iIt is determined whether or not the value is greater than. Compressed data amount CODE_iIs larger, in step 207, the code amount CO_iCompressed data amount CODE_iSet as In other words, in this case, the compression code amount is equal to the code amount CO regardless of the quantization coefficient q._iThe code amount CO is low._iCompressed data amount CODE_iIt is determined as
[0115]
On the other hand, the compressed data amount CODE_iIf it is smaller, after the parameter i is incremented by 1 in step 208, it is determined in step 209 whether or not the parameter is greater than 62. If the parameter i does not exceed 62, the process returns to step 205 and the above-described processing is executed again. In this way, scan AC₆₂Compressed data volume up to CODE_iIs set, in step 210, the scan AC₆₃Compressed data amount CODE₆₃Is CD₆₃This program ends.
[0116]
Compressed data amount CODE obtained as described above_iIs the set code amount S used in step 106 of FIG._iIt is.
[0117]
FIG. 35 shows each scan AC in the filtering table FL shown in FIG._iShows the allocation ratio. This allocation ratio is the filtering coefficient FT of each AC component._iAnd DC component filtering coefficient FT₀Is obtained by multiplying the reciprocal of the ratio by 100 with, for example, scan AC₂In the case of the filtering coefficient FT₂Is 98, the allocation ratio is greater than 100 as indicated by the symbol RA. As can be understood from this figure, the filtering table FL in FIG. 34 shows a low-pass filter with a small allocation ratio for high-frequency components. On the other hand, in order to generate the table FL of the high-pass filter, it is only necessary to determine the filtering coefficient so as to reduce the allocation ratio for the low frequency component in the allocation ratio of FIG.
[0118]
FIG. 36 shows a filtering table corresponding to the first low-pass filter. Reference numeral (a) denotes a luminance filtering table, which is the same as the filtering table of FIG. Reference numeral (b) represents a filtering table for color difference. FIG. 37 shows a filtering table corresponding to the averaging filter, where symbol (a) indicates a luminance filtering table and symbol (b) indicates a color difference filtering table. FIG. 38 shows a filtering table corresponding to the second low-pass filter. In this example, the luminance and color difference filtering tables are common. FIG. 39 shows a filtering table corresponding to the high-pass filter, and this table is also common for luminance and color difference.
[0119]
【The invention's effect】
As described above, according to the present invention, it is possible to achieve the effect that image compression corresponding to the image quality of each image can be achieved.
[Brief description of the drawings]
FIG. 1 is a block diagram illustrating an image compression apparatus according to an embodiment of the present invention.
FIG. 2 is a diagram showing an example of image data P (Y) xy, DCT transform coefficient S (Y) uv, quantized DCT coefficient R (Y) uv, and quantization table Q (Y) uv.
FIG. 3 is a diagram showing categories of DC coefficient difference values;
FIG. 4 is a diagram illustrating a coding table of DC coefficients.
FIG. 5 is a flowchart showing a processing routine for encoding quantized AC coefficients.
FIG. 6 is a diagram illustrating a zigzag scan performed when Huffman coding is performed on an AC coefficient.
FIG. 7 is a diagram showing categories of AC coefficients.
FIG. 8 is a diagram showing a Huffman table recommended by JPEG.
FIG. 9 is a diagram illustrating an example of encoded data by Huffman encoding.
FIG. 10 is a diagram illustrating components of encoded data.
FIG. 11 is a diagram illustrating an example of category distribution at a predetermined spatial frequency when a quantization table having all quantization coefficients of “1” is used.
FIG. 12 is a diagram illustrating an example of a distribution of the number of bits of encoded data in each scan when a quantization table in which all quantization coefficients are “1” is used.
FIG. 13 is a diagram illustrating an example of code amount distribution for each spatial frequency (that is, for each scan) set in a spatial frequency data amount setting unit;
14 is a diagram showing the category distribution shown in FIG. 11 for each scan at the same time. FIG.
FIG. 15 is a diagram showing a table of category distribution when using a quantization table in which all quantization coefficients are “16”.
FIG. 16 Scan AC₁It is a figure which shows the number of blocks of category "0" and the number of blocks of category "1" or more.
FIG. 17: Scan AC₂It is a figure which shows the number of blocks whose run length is 2, 1, and 0.
FIG. 18 Scan AC_ThreeIt is a figure which shows the number of blocks whose run length is 3, 2, 1, 0.
FIG. 19: Scan AC_FourIt is a figure which shows the number of blocks whose run length is 4, 3, 2, 1, 0.
FIG. 20 Scan AC₁It is a figure which shows the table | surface of run length and category.
FIG. 21: Scan AC₂It is a figure which shows the table | surface of run length and category.
FIG. 22 Scan AC_ThreeIt is a figure which shows the table | surface of run length and category.
FIG. 23: Scan AC_FourIt is a figure which shows the table | surface of run length and category.
FIG. 24 is a diagram illustrating the code length of each codeword in the Huffman table recommended by JPEG.
FIG. 25 is a flowchart of a program for generating a quantization table.
FIG. 26 is a diagram showing a distribution of “00” data.
FIG. 27: Scan AC₁And scan AC₂It is a figure which shows normalization of "00" data.
FIG. 28 is a diagram showing a distribution table of normalized “00” data.
FIG. 29: Scan AC_i-1Of category “0” and scan AC_i-1And scan AC_iIt is a figure which shows the relationship with "00" data.
FIG. 30: Scan AC_FourIt is a figure which shows the example of the category distribution in.
FIG. 31 is a diagram showing an example of run length / category distribution obtained by the third prediction method;
32 is a diagram showing the run length / category distribution table of FIG. 31 in a format corresponding to FIG.
FIG. 33: Setting code amount S_iIt is a flowchart of the program which produces | generates distribution of.
FIG. 34 is a diagram illustrating an example of a filtering table.
35 is a diagram showing an allocation ratio of each scan in the filtering table of FIG. 34. FIG.
FIG. 36 is a diagram illustrating a filtering table of a first low-pass filter.
FIG. 37 is a diagram showing a filtering table of an averaging filter.
FIG. 38 is a diagram illustrating a filtering table of a second low-pass filter.
FIG. 39 is a diagram illustrating a filtering table of a high-pass filter.
[Explanation of symbols]
M IC memory card

Claims (3)

Orthogonal transformation means for dividing the two-dimensional original image data into a plurality of blocks and performing orthogonal transformation to obtain orthogonal transformation coefficients for each spatial frequency;
Quantization means for quantizing the orthogonal transform coefficient with a two-dimensional quantization table including predetermined quantization coefficients to obtain a quantized orthogonal transform coefficient;
By scanning the quantized orthogonal transform coefficients in an order corresponding to the spatial frequency, the quantized orthogonal transform coefficients are rearranged into one-dimensional array data, and then encoded based on the quantized orthogonal transform coefficients according to the scan order. Encoding means for obtaining encoded data;
The coded data obtained by the coding means based on the quantized orthogonal transform coefficient quantized by a default quantization table in which all the quantization coefficients are 1, and the spatial frequency with a high degree of data compression From the filtering table having the filtering coefficient for each spatial frequency having a large value, the total code amount is obtained by summing the values obtained by dividing the code amount of the encoded data by the filtering coefficient for each spatial frequency. A correction coefficient that is a ratio between the total code amount and a set total code amount that is a target value of the total number of bits for one screen of the encoded data, and a quantization quantized by the default quantization table orthogonal transform coefficients, based on said filtering coefficients, by dividing the code amount of the coded data in the filtering coefficients, and Wherein by multiplying the correction coefficient, and means for setting a target value of the code amount of the coded data for each of the spatial frequencies,
Quantization coefficient calculation means for determining a quantization coefficient corresponding to the spatial frequency so that the code amount of each spatial frequency is equal to or less than the target value ;
Classifying the quantized orthogonal transform coefficients quantized by the default quantization table into categories having a predetermined amount of data, obtaining a category distribution table indicating the number of blocks classified into each category in each spatial frequency, and For the spatial frequency located at the head of the AC component in the one-dimensional array data, the code amount of the encoded data is predicted by changing the quantization coefficient based on the category distribution table. For spatial frequencies located at positions other than the head of the AC component, the quantization coefficient is changed based on the category distribution table and at least the statistic of the encoded data of the first spatial frequency, and 0 at the first spatial frequency is obtained. Using the number of blocks of category “0” corresponding to the quantized orthogonal transform coefficient of the second spatial frequency, And a prediction means for predicting the amount,
The quantization coefficient calculation means is a quantization coefficient corresponding to the second spatial frequency so that a code amount of the encoded data of the second spatial frequency predicted by the prediction means is equal to or less than the target value. image compression apparatus characterized by determining the. For the spatial frequency located at a position other than the head of the AC component in the one-dimensional array data, the prediction means encodes the category distribution table, the first spatial frequency, and a spatial frequency lower than the first spatial frequency. The image compression apparatus according to claim 1 , wherein the code amount of the encoded data of the second spatial frequency is predicted based on a data statistic . The quantized orthogonal transform coefficients corresponding to a second spatial frequency, claim, characterized in that in the one-dimensional array data, it is adjacent to the quantized orthogonal transformation coefficients corresponding to the first spatial frequency 1 The image compression apparatus described in 1.

Images