Ответы на вопрос:
Используем формулу i = log(2)1/p i1 = log(2)[1/(128/512)] = log(2)[4] = 2 бита (2^2 = 4) i2 = log(2)[1/(256/512)] = log(2)[2] = 1 бит (2^1 = 2) i3 = log(2)[1/(64/512)] = log(2)[8] = 3 бита (2^3 = 8) i4 = log(2)[1/(64/512)] = log(2)[8] = 3 бита (2^3 = 8)
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