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Portable Network Graphic  |  1993-02-04  |  2KB  |  512x342  |  1-bit (2 colors)

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OCR: Solution: P 00001111 P2 P3 PA +b +C + d 0 H. M = 00110011 a P3 + a + C +d 0 01010101 P2 + a + b +d = 10 1 11 1 1 1 1 1 P4 b P1 + P2 + P3 + a + P4 + b + c + d 10 C d So the conditions are py + b + c + d = 0; P3 + a + C + d = 0; P2 + a + b + d = 0; P1 + P2 + P3 + a + P4 + b + c + d = 0 (Problem continued on next card.)

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