平成24年度 秋期 ITパスポート試験 問94

※ Our check while writing this explanation could not fully confirm the published answer; scoring follows IPA's official key.

• ×Incorrectア

  12 bits corresponds to representing the three runs “white 10, black 34, white 20” with 4 bits each without dividing them or forming pairs, that is 3 x 4 = 12. It applies neither the division within a maximum of 15 in step (1) nor the “always an 8-bit pair of white and black” of step (2). Besides, 34 and 20 cannot be expressed in 4 bits (maximum 15) in the first place, so this way of counting itself breaks the stated rules.

• ×Incorrectイ

  24 bits is 3 x 8 = 24, obtained by counting the three runs “white 10, black 34, white 20” at 8 bits per run without dividing them. That is the answer to Question 95, which encodes the same Figure 2 with a different method, not the method of this question. Here step (1) divides values within a maximum of 15, so black 34 splits into 15, 15 and 4 and white 20 splits into 15 and 5; there are simply more than 3 runs.

• ○ウ

  正解 Read the bit string in Figure 2 as one continuous sequence, from the top row down, left to right: 00000000 / 00111111 / 11111111 / 11111111 / 11111111 / 11110000 / 00000000 / 00000000 Counting 0 as white and 1 as black: - there are ten 0s (the 8 bits of row 1 + the first 2 bits of row 2) - there are thirty-four 1s (the remaining 6 bits of row 2 + the 24 bits of rows 3 to 5 + the first 4 bits of row 6) - there are twenty 0s (the remaining 4 bits of row 6 + the 16 bits of rows 7 and 8) and 10 + 34 + 20 = 64 pixels = 8 x 8, which matches. So the run lengths are “white 10 -> black 34 -> white 20”. Step (1): Divide the values within a maximum range of 15. White 10 stays as it is, black 34 splits into 15 + 15 + 4, and white 20 splits into 15 + 5, giving white 10 / black 15 / black 15 / black 4 / white 15 / white 5 which is 6 run lengths. Step (2): Turn these into 8-bit pairs that are always “white and black”. The sequence starts with white, so no white 0 is needed at the front. Pair 1: white 10, black 15 ... this is a white-then-black order, so it forms one pair as it is Pair 2: white 0, black 15 ... black continues, so a white run length of 0 is inserted between them Pair 3: white 0, black 4 ... black continues again, so a white 0 is inserted Pair 4: white 15, black 0 ... white continues, so a black run length of 0 is inserted between them Pair 5: white 5, black 0 ... it ends with white, so a black run length of 0 is added at the end That makes 5 pairs. One pair is 8 bits (4 bits for white + 4 bits for black), so 5 pairs x 8 bits = 40 bits Therefore the correct answer is ウ, 40. The actual bit string is 10101111 00001111 00000100 11110000 01010000 (white 10 = 1010, black 15 = 1111, white 0 = 0000, black 4 = 0100, white 15 = 1111, black 0 = 0000), which is indeed 40 bits. The original 64 bits have shrunk to 40 bits, and that is exactly what makes this method a compression.

• ×Incorrectエ

  48 bits comes from treating each of the 6 run lengths obtained after the division in step (1) (white 10 / black 15 / black 15 / black 4 / white 15 / white 5) as an independent pair and mechanically counting 6 x 8 = 48. But step (2) inserts a run length of 0 as padding only in three cases: when the same color continues (a 0 of the opposite color goes between them), when the sequence begins with black (a white 0 goes at the front), and when it ends with white (a black 0 goes at the end). Where the color switches naturally from white to black, no padding is needed. In this question the leading white 10 and black 15 are exactly such a white-then-black order, so those two combine into one pair. Because of that one pair, the number of pairs is 5 rather than 6, and the answer is 40 rather than 48.