Re^2: Hamming Sequences and Lazy Lists

Roy Johnson,
I don't think this is right. It is my understanding that the results should be the multiples of the factors merged minus duplicates. If you change $iterations_to_print = 23 and @arr = (2,3,5), where is 15 in the results for instance? See Re: Hamming Sequences and Lazy Lists for my understanding and implementation of the problem.

Cheers - L~R

Comment on Re^2: Hamming Sequences and Lazy Lists

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Re^3: Hamming Sequences and Lazy Lists by Roy Johnson (Monsignor) on Mar 17, 2005 at 18:01 UTC
I did not implement Hamming here; only merging of two list iterators. But I have just posted a lazy Hamming solution. Caution: Contents may have been coded under pressure.	[reply]
Re^4: Hamming Sequences and Lazy Lists by Limbic~Region (Chancellor) on Mar 17, 2005 at 18:04 UTC
Roy Johnson, Ok - my mistake. In fact, I completely missed the boat on the entire problem. I was so focused on lazy evaluation that I misunderstood "how do you generate the series of numbers composed of a given list of prime factors, where each can be used an unlimited number of times?" To me, that any positive multiple of any factor was valid. Cheers - L~R	[reply]