Your Lucas-like sequence is the answer to part (a) of that question, which is a clue that it might be useful in answering part (b).
The total length of a message with Lucas frequencies is two Lucas numbers up from the last frequency, minus one. So to get a depth d Huffman tree, the minimum message length is Ld+1 - 1. (Note that L0=2, which is broken up into the initial {1,1}. That is why I said "Lucas-like" earlier. Then L1=1, L2=3.) For a message length N, we have that N ≥ Ld+1 - 1 to be able to get a depth d tree.
The Lucas sequence (as well as the Fibonacci sequence) grows exponentially as φn, where φ is the Golden ratio. In particular Ln > φn-1.
To get depth d, we have N > φd. So log(N) > d log(φ). p(n) ≥ N, where p is a polynomial, we'll say of degree k. So log(p(n)) > d log(φ). Using the highest power of p to get the order, we have d =≤ O(k log(n) / log(φ)), or simply d =≤ O(log(n)), since k and φ are constants independent ofn. We know that d ≥ ⌈log(n)⌉, so d = O(log(n)).
Your Lucas-like sequence is the answer to part (a) of that question, which is a clue that it might be useful in answering part (b).
The total length of a message with Lucas frequencies is two Lucas numbers up from the last frequency, minus one. So to get a depth d Huffman tree, the minimum message length is Ld+1 - 1. (Note that L0=2, which is broken up into the initial {1,1}. That is why I said "Lucas-like" earlier. Then L1=1, L2=3.) For a message length N, we have that N ≥ Ld+1 - 1 to be able to get a depth d tree.
The Lucas sequence (as well as the Fibonacci sequence) grows exponentially as φn, where φ is the Golden ratio. In particular Ln > φn-1.
To get depth d, we have N > φd. So log(N) > d log(φ). p(n) ≥ N, where p is a polynomial, we'll say of degree k. So log(p(n)) > d log(φ). Using the highest power of p to get the order, we have d = O(k log(n) / log(φ)), or simply d = O(log(n)), since k and φ are constants independent of n.
Your Lucas-like sequence is the answer to part (a) of that question, which is a clue that it might be useful in answering part (b).
The total length of a message with Lucas frequencies is two Lucas numbers up from the last frequency, minus one. So to get a depth d Huffman tree, the minimum message length is Ld+1 - 1. (Note that L0=2, which is broken up into the initial {1,1}. That is why I said "Lucas-like" earlier. Then L1=1, L2=3.) For a message length N, we have that N ≥ Ld+1 - 1 to be able to get a depth d tree.
The Lucas sequence (as well as the Fibonacci sequence) grows exponentially as φn, where φ is the Golden ratio. In particular Ln > φn-1.
To get depth d, we have N > φd. So log(N) > d log(φ). p(n) ≥ N, where p is a polynomial, we'll say of degree k. So log(p(n)) > d log(φ). Using the highest power of p to get the order, we have d ≤ O(k log(n) / log(φ)), or simply d ≤ O(log(n)), since k and φ are constants independent ofn. We know that d ≥ ⌈log(n)⌉, so d = O(log(n)).
Your Lucas-like sequence is the answer to part (a) of that question, which is a clue that it might be useful in answering part (b).
The total length of a message with Lucas frequencies is two Lucas numbers up from the last frequency, minus one. So to get a depth d Huffman tree, the minimum message length is Ld+1 - 1. (Note that L0=2, which is broken up into the initial {1,1}. That is why I said "Lucas-like" earlier. Then L1=1, L2=3.) For a message length N, we have that N ≥ Ld+1 - 1 to be able to get a depth d tree.
The Lucas sequence (as well as the Fibonacci sequence) grows exponentially as φn, where φ is the Golden ratio. In particular Ln > φn-1.
To get depth d, we have N > φd+1d. So log(N) > (d+1d) log(φ). p(n) ≥ N, where p is a polynomial, we'll say of degree k. So log(p(n)) > (d+1d) log(φ). Using the highest power of p to get the order, we have d = O(k log(n) / log(φ)), or simply d = O(log(n)), since k and φ are constants independent of n.
Your Lucas-like sequence is the answer to part (a) of that question, which is a clue that it might be useful in answering part (b).
The total length of a message with Lucas frequencies is two Lucas numbers up from the last frequency, minus one. So to get a depth d Huffman tree, the minimum message length is Ld+1 - 1. (Note that L0=2, which is broken up into the initial {1,1}. That is why I said "Lucas-like" earlier. Then L1=1, L2=3.) For a message length N, we have that N ≥ Ld+1 - 1 to be able to get a depth d tree.
The Lucas sequence (as well as the Fibonacci sequence) grows exponentially as φn, where φ is the Golden ratio. In particular Ln > φn.
To get depth d, we have N > φd+1. So log(N) > (d+1) log(φ). p(n) ≥ N, where p is a polynomial, we'll say of degree k. So log(p(n)) > (d+1) log(φ). Using the highest power of p to get the order, we have d = O(k log(n) / log(φ)), or simply d = O(log(n)), since k and φ are constants independent of n.
Your Lucas-like sequence is the answer to part (a) of that question, which is a clue that it might be useful in answering part (b).
The total length of a message with Lucas frequencies is two Lucas numbers up from the last frequency, minus one. So to get a depth d Huffman tree, the minimum message length is Ld+1 - 1. (Note that L0=2, which is broken up into the initial {1,1}. That is why I said "Lucas-like" earlier. Then L1=1, L2=3.) For a message length N, we have that N ≥ Ld+1 - 1 to be able to get a depth d tree.
The Lucas sequence (as well as the Fibonacci sequence) grows exponentially as φn, where φ is the Golden ratio. In particular Ln > φn-1.
To get depth d, we have N > φd. So log(N) > d log(φ). p(n) ≥ N, where p is a polynomial, we'll say of degree k. So log(p(n)) > d log(φ). Using the highest power of p to get the order, we have d = O(k log(n) / log(φ)), or simply d = O(log(n)), since k and φ are constants independent of n.
Your Lucas-like sequence is the answer to part (a) of that question, which is a clue that it might be useful in answering part (b).
The total length of a message with Lucas frequencies is two Lucas numbers up from the last frequency, minus one. So to get a depth d Huffman tree, the minimum message length is Ld+1 - 1. (Note that L0=2, which is broken up into the initial {1,1}. That is why I said "Lucas-like" earlier. Then L1=1, L2=3.) For a message length N, we have that N ≥ Ld+1 - 1 to be able to get a depth d tree.
The Lucas sequence (as well as the Fibonacci sequence) grows exponentially as φn, where φ is the Golden ratio. In particular Ln > φn.
To get depth d, we have N > φd+1. So log(N) > (d+1) log(φ). p(n) ≥ N, where p is a polynomial, we'll say of degree k. So log(p(n)) > (d+1) log(φ). Using the highest power of p to get the order, we have d = O(k log(n) / log(φ)), or simply d = O(log(n)), since k and φ are constants independent of n.