Algorithms and recursive functions by A.I. Mal'cev

By A.I. Mal'cev

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This process is repeated until we generate. size(); i++) std::cout << gray[i] << std::endl; } 25. Represent unsigned integers with variable length encoding using the continuation bit Solution The key idea is to take a 64bit unsigned integer and represent it with a list of bytes. For each byte seven bits are used for storing the integers with variable length encoding. In addition the most significant bit is the continuation bit and it is used to signal whether or not we need an additional byte for encoding n.

The solution is therefore 16. Compute parity for a 32 bit number Solution Given a number, we can drop the last bit with the expression. Then if the number of bits set to 1 is even, we return 1, otherwise we return 0. In other words, we need to maintain a bit which changes its status from 0 to 1 as many times as the number of bits set to 1 in. This is achieved in our code with the variable. In many practical implementations we can avoid to process every single bit in isolation by storing a number of pre-computed parity tables for a number of bits.

Code unsigned int swapBits(unsigned int x) { unsigned int evenBits = x & 0xAAAAAAAA; unsigned int oddBits = x & 0x55555555; evenBits >>= 1; oddBits <<= 1; return (evenBits | oddBits); } 2. Print the binary representation of an unsigned int Solution An easy solution is to AND the bit with the number Code void bin(unsigned n) { for (unsigned int i = 1 << 31; i > 0; i = i >> 1) if (n & i) std::cout << 1; else std::cout << 0; } 3. Compute whether or not an unsigned number is a power of two Solution Suppose that the number is nonzero.

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