Does this work? Could be extended to negative values of y.
NB. Dyad, giving i.!.x~ y for large x. y must be nonnegative
tidr =. tolerantidotreflex =. ((% -.)~ I. ]) { /:@]
0.2 (tidr ,: ]) 0 3 0.3 3.2 0.6 2 2.3 2.6 3
0 8 2 8 2 7 7 1 8
0 3 0.3 3.2 0.6 2 2.3 2.6 3
0.5 (tidr ,: ]) 0 3 0.3 3.2 0.6 2 2.3 2.6 3
0 7 2 7 2 7 7 7 7
0 3 0.3 3.2 0.6 2 2.3 2.6 3
Henry Rich
On 1/16/2012 2:37 PM, Roger Hui wrote:
> You can use ": as part of the hashing function and yes you do have to hash
> x*1-t and x%1-t .
>
>
>
> On Mon, Jan 16, 2012 at 11:19 AM, Raul Miller<rauldmil...@gmail.com> wrote:
>
>> If hashing would work, then keying on ": would work. I expect though
>> that I would need to hash at least twice (adding epsilon before the
>> second hash) and I expect that I would need to do something similar if
>> I used ":
>>
>> --
>> Raul
>>
>> On Mon, Jan 16, 2012 at 2:14 PM, Roger Hui<rogerhui.can...@gmail.com>
>> wrote:
>>> Hashing has expected O(n) time.
>>>
>>>
>>>
>>> On Mon, Jan 16, 2012 at 11:11 AM, Raul Miller<rauldmil...@gmail.com>
>> wrote:
>>>
>>>> I think that the "monster case" would be a case where all values are
>>>> similar enough that they all map to the same index but different
>>>> enough that they cannot be recognized as literal equivalents.
>>>>
>>>> In the case I am currently interested in, the original values would be
>>>> 32 bit floating point numbers and only a relatively few bits of the
>>>> available precision would allow values to be treated as "tolerantly
>>>> equal". This would suggest that the size of the "monster case" is
>>>> limited based on the number of bits being ignored.
>>>>
>>>> So, in principle at least, this should limit the size of the
>>>> "quadratic part" of the problem, for the cases I am trying to address.
>>>>
>>>> --
>>>> Raul
>>>>
>>>> On Mon, Jan 16, 2012 at 12:04 PM, Roger Hui<rogerhui.can...@gmail.com>
>>>> wrote:
>>>>> The paper I cited, *Hashing for Tolerant
>>>>> Index-Of<http://www.jsoftware.com/papers/Hashing.htm>
>>>>> * , presents a "monster" that defeats a sorting algorithm. (Defeat
>> in
>>>> the
>>>>> sense of causing it take quadratic time.)
>>>>>
>>>>>
>>>>>
>>>>> On Mon, Jan 16, 2012 at 8:07 AM, Henry Rich<henryhr...@nc.rr.com>
>>>> wrote:
>>>>>
>>>>>> You can sort the lists and then compare adjacent values; find
>>>>>> superfluous ones; then i.!.0 to find them in the original list.
>>>>>>
>>>>>> A tricky part is that proximity is not a transitive property. If the
>>>>>> tolerance is 2, and the data is
>>>>>>
>>>>>> 1 2 3 4 5 6 7
>>>>>>
>>>>>> what should the result of the i.~ be?
>>>>>>
>>>>>> Henry Rich
>>>>>>
>>>>>> On 1/16/2012 10:06 AM, Raul Miller wrote:
>>>>>>> First: I like Roger Hui's response. And, in essence, it's doing
>>>>>>> exactly what you suggest. However, this requires comparing every
>>>>>>> number in the left list with every number in the right list. I am
>>>>>>> currently pondering algorithms which rely on I. so that when the
>> lists
>>>>>>> are long computation times are still reasonable (perhaps with
>> 100000
>>>>>>> members in each list).
>>>>>>>
>>>>>>> Second: I would want the three PI values in my original message
>> to be
>>>>>>> treated as equal. I want to be able to specify a magnitude of
>>>>>>> acceptable difference which is greater than any of the differences
>> in
>>>>>>> that data sample.
>>>>>>>
>>>>>>> FYI,
>>
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