I will present the code in some time but till then here is an explanation with example.
Take n=10, w = 3 and r=1 ( to say that each integer itself constitutes the bucket ) now Initially no bucket is created and the input given is 1 Hence any random representation can be taken. *1 --> 100 010 100 0* Next 2 arrives, its window lies from 0 to 4 ( both inclusive ) and everything else is outside 1 is within the window and the only bucket present So it randomly chooses a representation with 2 bits differing ( i.e, 2 1s maintain their place ) *2 --> 101 000 100 0* *------------------------------------------------------------------------------------------------------------------------------* *Present map state :* *01 --> 100 010 100 0* *02 --> 101 000 100 0* Now lets say there was a spike and 50 was given as input Its window is from 50-3+1=48 to 50+3-1=52 All the existing buckets are outside its window, and it should not have any bits in common with that *01* or *02* = 101 010 100 0 ---- the 0s are the places that this bucket should have an encoding in selecting any 3 bits in random satisfies the condition, with both the existing buckets 1 and 2 hence *50 --> 010 000 011 0* *------------------------------------------------------------------------------------------------------------------------------* *Present map state :* *01 --> 100 010 100 0* *02 --> 101 000 100 0* *50 --> 010 000 011 0* Now let the input be 5 Its window is from (5-3+1)=3 to (5+3-1)=7 All existing buckets are outside its window *01* or *02* or *50 = *111 010 111 0 ---------- the 0s are the bits this can occupy without conflict Selecting the only possible encoding *05 --> 000 101 000 1* *------------------------------------------------------------------------------------------------------------------------------* *Present map state :* *01 --> 100 010 100 0* *02 --> 101 000 100 0* *05 --> 000 101 000 1* *50 --> 010 000 011 0* Now let the input be 4 Its window is from (4-3+1)=2 to (4+3-1)=6 2 and 5 lies within the window and rest are outside *01* or *50* = 110 010 111 0 The encoding should occupy the maximum of the places containing 0s Let the possible encoding be 001 101 000 0 condition between 4 -- 1 > satisfied ( 2*3 bits differ ) condition between 4 -- 2 > satisfied ( 2*2 bits differ ) condition between 4 -- 5 > satisfied ( 2*1 bits differ ) condition between 4 -- 50 > satisfied ( 2*3 bits differ ) Hence its fine to choose this as encoding *04 --> 001 101 000 0* *------------------------------------------------------------------------------------------------------------------------------* *Present map state :* *01 --> 100 010 100 0* *02 --> 101 000 100 0* *04 --> **001 101 000 0* *05 --> 000 101 000 1* *50 --> 010 000 011 0* Let the input be 3 Its window is from (3-3+1)=1 to (3+3-1)=5 1, 2, 4, 5 lie in the window. 50 is outside 50 = 010 000 011 0 The encoding should occupy the maximum of the places containing 0s Let the randomly chosen encoding be 101 010 000 0 condition between 3 -- 1 > Not satisfied ( 2 bits differ , expected 4 ) condition between 3 -- 2 > Satisfied ( 2 bits differ ) condition between 3 -- 4 > Not satisfied ( 4 bits differ, expected 2 ) condition between 3 -- 5 > Not satisfied ( 6 bits differ, expected 4 ) condition between 3 -- 50 > Satisfied ( 6 bits differ ) Score = 2 ( #conditions satisfied ) Let the next possible encoding be 101 100 000 0 condition between 3 -- 1 > Satisfied ( 4 bits differ ) condition between 3 -- 2 > Satisfied ( 2 bits differ ) condition between 3 -- 4 > Satisfied ( 2 bits differ ) condition between 3 -- 5 > Satisfied ( 4 bits differ ) condition between 3 -- 50 > Satisfied ( 6 bits differ ) Score = 5 ( #conditions satisfied ) == number of existing buckets Hence select the encoding *3 --> 101 100 000 0* *------------------------------------------------------------------------------------------------------------------------------* *Overflow case* *Present map state :* *01 --> 100 010 100 0* *02 --> 101 000 100 0* *03 --> 101 100 000 0* *04 --> **001 101 000 0* *05 --> 000 101 000 1* *50 --> 010 000 011 0* Let the input be 6 Its window is from (6-3+1)=4 to (6+3-1)=8 4, 5 lie in the window. 1 ,2 ,3 ,50 are outside *01* or *02* or *03* or *50* = 111 110 111 0 The encoding should occupy the maximum of the places containing 0s But this is a tricky case : Encoding 100 001 000 1 gets the lowest score and is closer to bucket 1,2,3 in representation Encoding 000 001 100 1 gets median score and is closer to bucket 1,2 in representation Encoding 010 001 000 1 } Encoding 000 001 010 1 }------ gets highest score and is closer to bucket 50 in representation Encoding 000 001 001 1 } Encoding 000 011 000 1 }------ gets highest score and is closer to bucket 1 in representation. *WHICH TO CHOOSE?? * According to me the last option is better and more logical in this case but we will have to think more carefully the selection condition. This exercise did get a lot more things clear, things i was stuck at while coding. Please provide your feedback. *Regards,* *Anubhav Chaturvedi* *Birla Institute of Technology & Science, Pilani* KK Birla Goa Campus +91-9637399150 *" You can do anything if you stop doing everything "* On Wed, Feb 19, 2014 at 12:04 AM, Chetan Surpur <[email protected]> wrote: > To be clear, I don't mean necessarily in code, I mean just a simple email > explanation using a toy example. > > > On Tue, Feb 18, 2014 at 1:28 AM, Anubhav Chaturvedi < > [email protected]> wrote: > >> Sure , I can try that. I will be using the same testing technique as used >> in Fergal Byrne's example (http://fergalbyrne.github.io/rdse.html) . >> >> *Regards,* >> *Anubhav Chaturvedi* >> >> *Birla Institute of Technology & Science, Pilani* >> KK Birla Goa Campus >> +91-9637399150 >> >> *" You can do anything if you stop doing everything "* >> >> >> On Tue, Feb 18, 2014 at 11:24 AM, Chetan Surpur <[email protected]>wrote: >> >>> Hi Anubhav, >>> >>> Thanks for sharing your thoughts on this. Do you think you can show a >>> small example of your algorithm? >>> >>> Thanks, >>> Chetan >>> >>> >>> On Sat, Feb 15, 2014 at 4:35 PM, Anubhav Chaturvedi < >>> [email protected]> wrote: >>> >>>> Hi >>>> >>>> Chetan Surpur presented a great idea of RDSE. The video got me thinking >>>> if I could optimize it a bit and here is what I came up with. Please >>>> provide your feedback. >>>> >>>> URL : >>>> https://docs.google.com/document/d/1obYHE4-3q-oR1zbazUJWLTXBIzmt3GlG0wcKVT_Ohgg/edit?usp=sharing >>>> >>>> I have tried to optimize it for memory and convergence of two extremes >>>> of representation. >>>> >>>> *Regards,* >>>> *Anubhav Chaturvedi* >>>> >>>> *Birla Institute of Technology & Science, Pilani* >>>> KK Birla Goa Campus >>>> +91-9637399150 >>>> >>>> *" You can do anything if you stop doing everything "* >>>> >>>> _______________________________________________ >>>> nupic mailing list >>>> [email protected] >>>> http://lists.numenta.org/mailman/listinfo/nupic_lists.numenta.org >>>> >>>> >>> >>> _______________________________________________ >>> nupic mailing list >>> [email protected] >>> http://lists.numenta.org/mailman/listinfo/nupic_lists.numenta.org >>> >>> >> >> _______________________________________________ >> nupic mailing list >> [email protected] >> http://lists.numenta.org/mailman/listinfo/nupic_lists.numenta.org >> >> > > _______________________________________________ > nupic mailing list > [email protected] > http://lists.numenta.org/mailman/listinfo/nupic_lists.numenta.org > >
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