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devel / comp.theory / ((space 5m) + (time 5n over m)) identity recognition

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* ((space 5m) + (time 5n over m)) identity recognitionDaniel Pehoushek
`* ((space 5m) + (time 5n over m)) identity recognitionDaniel Pehoushek
 `* ((space 5m) + (time 5n over m)) identity recognitionDaniel Pehoushek
  `* ((space 5m) + (time 5n over m)) identity recognitionDaniel Pehoushek
   `- ((space 5m) + (time 5n over m)) identity recognitionDaniel Pehoushek

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((space 5m) + (time 5n over m)) identity recognition

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Subject: ((space 5m) + (time 5n over m)) identity recognition
From: pehoush...@gmail.com (Daniel Pehoushek)
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 by: Daniel Pehoushek - Wed, 1 Jun 2022 11:15 UTC

pattern m stream n
big oh of n over m
specifically 5n/m
time recognition
in space size 5m
daniel2380++

Re: ((space 5m) + (time 5n over m)) identity recognition

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Subject: Re: ((space 5m) + (time 5n over m)) identity recognition
From: pehoush...@gmail.com (Daniel Pehoushek)
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 by: Daniel Pehoushek - Wed, 1 Jun 2022 12:06 UTC

5m + 5n / m
mental recognition
daniel2380++

/// ((space 5m) + (time 5n over m)) identity recognition
num iziv = five; // ( empty bubble ) languages
num agenum(nums& b, num z) { // build ans num for small pattern
num a=zero;for(num h=zero;h<iziv&&z+h<b.y;h++)a=(a<<iziv)+b.v[z+h];return(a);}// thirtytwo bits
num age = zero;// just count all finds one by one (page faults)
num recoggen(nums& bzmo, nums& gzmo, numnums* mem) { num m = bz.size(); // m + n / m
for (num g = (*mem).size(); g < (one << iziv); g++)(*mem).add(new nums);//zero
for(num g=zero;g+ iziv <m;g++)(*(*mem)[agenum(bzmo,g)]).add(g);// bit shifts
num n = gzmo.size(); num mmm = Minus(m + one, iziv); for (num j = mmm; j + iziv < n; j = j + mmm) {// 5n over (m less four)
nums* p=(*mem)[agenum(gzmo,j)];for(num o=zero;o<(*p).size();o++){num be=Minus(j,(*p)[o]);num h=one;for(num g=zero;g<m;g++)
{when(gzmo[be+g]==bzmo[g])continue;h=zero;break;}when(h)age++;}}
for(num g=zero;g+iziv<m;g++)(*(*mem)[agenum(bzmo,g)]).clear();return age;}// (space m + time 5n over m) identity stream n

Re: ((space 5m) + (time 5n over m)) identity recognition

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Subject: Re: ((space 5m) + (time 5n over m)) identity recognition
From: pehoush...@gmail.com (Daniel Pehoushek)
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 by: Daniel Pehoushek - Wed, 1 Jun 2022 12:54 UTC

On Wednesday, June 1, 2022 at 8:06:50 AM UTC-4, Daniel Pehoushek wrote:
> 5m + 5n / m
> mental recognition
> daniel2380++
>

recog is better now at 5n over m versus 5m + 5n with dekmp
i have removed negation from good thoughts so
good thoughts now rule very many (two to the en) sentences
wisdom is deeply monotone over two to the en forms
commonsense is monotone over many forms
through gritted teeth i say say so simply
abortion is bad but justifiable
weapons are bad but justifiable
daniel2380++

num minusbit(num bits, num g) { when (g < one + fifthtau) { when (bits & ones[g]) bits &= diagovreason[g]; otherwise { bits += ones[g]; bits = minusbit(bits, g + one); } } return bits;}
num minus(num more, num less) { for(num g = fifthtau; g > zero; g = Lessone(g)) when (less & ones[g]) more = minusbit(more, g); when (less & one) more = minusbit(more, zero); return more; }
typedef void joy;
template < class tt > class set { // fibolacci 0 1 2 4 7 11 17 26 /// space growth of ordered sets
public: joy allocay(num get) {if(have<get){tt* tmp = new tt[(num)get]; if (tmp == (tt*)zero) {} else { for (num h = zero; h < have; h++)
{ tmp[h] = v[h]; v[h] = (tt)zero; } when(v == (tt*)zero) {} else { delete[]v; } { have = zero; v = tmp; have = get; tmp = (tt*)zero;}}}}
joy add(tt elt){if(y==have){allocay((one+(y>>one)+y));}v[y++]=elt;}
inline num size(){return y;} inline tt& operator[](num r){return v[r];} num setsize(num z){y=z;return z;}
tt slop(){tt ret=(tt)zero;if(y){y=Lessone(y);ret=v[y];v[y]=(tt)zero;}else{}return ret;}tt last(){tt ret=(tt)zero;if(y){ret=v[Lessone(y)];}return ret;}
num memberp(tt elt){for(num g=zero;g<size();g++)return((v[g]==elt)?one:zero);} joy clear(){for(num g=zero;g<have;g++)v[g]=(tt)zero;y=zero;}
~set(){clear();when(v==(tt*)zero){}otherwise{delete[]v;}y=zero;have=zero;v=(tt*)zero;}
set(num beg):y(zero),have(zero),v((tt*)zero){allocay(beg);} set():y(zero),have(zero),v((tt*)zero){}
num y;num have;tt* v;
};/// ten lines
typedef set<num> nums; typedef set<nums*> numnums;
/// ((space 5m) + (time 5n over m)) identity recognition
num ziv = five; // ( empty bubble ) languages
num anum(nums& p, num z) { // a num for pattern p at z
num a=zero;for(num h=zero;h<ziv&&z+h<p.y;h++)a=(a<<ziv)+p.v[z+h];return(a);}
num age = zero;// just count all finds one by one (page faults)
num recog(nums& b, nums& gzmo, numnums* myn) { num m = b.size(); // m + n / m
for(num g=(*myn).size();g<(one<<ziv);g++)(*myn).add(new nums);//zero
for(num g=zero;g+ ziv <m;g++)(*(*myn)[anum(b,g)]).add(g);// bit shifts
num n=gzmo.size();num mm=minus(m+one,ziv);for(num j=mm;j+ziv<n;j=j+mm){// 5n over (m less four)
nums* p=(*myn)[anum(gzmo,j)];for(num o=zero;o<(*p).size();o++){num be=minus(j,(*p)[o]);num h=one;for(num g=zero;g<m;g++)
{when(gzmo[be+g]==b[g])continue;h=zero;break;}when(h)age++;}}
for(num g=zero;g+ziv<m;g++)(*(*myn)[anum(b,g)]).clear();return age;}

Re: ((space 5m) + (time 5n over m)) identity recognition

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Subject: Re: ((space 5m) + (time 5n over m)) identity recognition
From: pehoush...@gmail.com (Daniel Pehoushek)
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 by: Daniel Pehoushek - Wed, 1 Jun 2022 13:00 UTC

// recognition iz (n/(m less four))
typedef unsigned long num; // //jdp2380++
const num one = (num)true; // fundamental unity of one unioned with truth
const num zero = one >> one; // one down one bit is zero
#define when if /// the concept when has more than if
const num two = one + one + (one >> one); // two is one plus one plus zero // zero counts most cleanly //
const num three = two + one; //let three be light numbers 0123
const num four = two + two; //at last after all the theoretical work on small numbera we have every child knows
const num five = three + two; //bobs favorite number is five // childrens recursive theory of equality begins
const num siv = three + two + one; //awvul awvul letter gone
const num sev = three + three + one; //favorite number in heaven
const num eigh = sev + one; //monotone digestion of acceptable input
const num nine = eigh + one; //triple nines favorite number
const num ten = five + five; //my favorite number is ten
num Lessone(num g) {num p=zero; for(p=zero;(p+one)<g;p=((p+g)>>one)){} return p;}// clear as a silver church bell ringing for mass
num Times(num a, num b){ if (a) return b + (Times(Lessone(a), b)); else return zero;}// times doan change and people do
const num Truth = 323;// three tothepower ofthree
const num onehundred = Times(ten, ten); // squarely written but interpretted as limitted line with whole number of segments onehundred
const num sivteen = siv + ten; /// i seen tv
const num thirtyone = sivteen+ten+five; /// i seen tv at ten and at five years old tv black and white /// the shorter suit of t h a t aaai where he told me his favorite impressive line /// it may all be a grand misconception /// and so i believed him /// ineedtwogoldfrontteeth really ///
num ones[] = { one, two, four, eigh, sivteen, thirtyone + one, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288,
1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648 };/// handmade reasoning table//the true by definition bits //
num tautologies [] ={ zero, one, three, sev, sev+sev+one, thirtyone, thirtyone+one+thirtyone, Times(ten,ten)+Times(Times(three,three),three), onehundred+onehundred+Times(five,ten)+five,
Times(five,onehundred)+ten+one, Times(ten,onehundred)+ten+ten+three, 2047, 4095, 8191, 16383, 32767, 65535 };
const num fifthtau = tautologies [five];
num diagovreason[] = { // the beauty ov beauty iz tau nearly
(num)4294967294, (num)4294967293, (num)4294967291, (num)4294967287, // four horses
(num)4294967279, (num)4294967263, (num)4294967231, (num)4294967167, // four cars
(num)4294967039, (num)4294966783, (num)4294966271, (num)4294965247, // four coin
(num)4294963199, (num)4294959103, (num)4294950911, (num)4294934527, // four food
(num)4294901759, (num)4294836223, (num)4294705151, (num)4294443007, // four love
(num)4293918719, (num)4292870143, (num)4290772991, (num)4286578687, // four pure
(num)4278190079, (num)4261412863, (num)4227858431, (num)4160749567, // four ....
(num)4026531839, (num)3758096383, (num)3221225471, (num)2147483647 }; // ...nit //
#define otherwise else /// otherwise is length nine but here now i define it to be length ten by green writing while else is just four
num minusbit(num bits, num g) { when (g < one + fifthtau) { when (bits & ones[g]) bits &= diagovreason[g]; otherwise { bits += ones[g]; bits = minusbit(bits, g + one); } } return bits;}
num minus(num more, num less) { for(num g = fifthtau; g > zero; g = Lessone(g)) when (less & ones[g]) more = minusbit(more, g); when (less & one) more = minusbit(more, zero); return more; }
typedef void joy;
template < class tt > class set { // fibolacci 0 1 2 4 7 11 17 26 /// space growth of ordered sets
public: joy allocay(num get) {if(have<get){tt* tmp = new tt[(num)get]; if (tmp == (tt*)zero) {} else { for (num h = zero; h < have; h++)
{ tmp[h] = v[h]; v[h] = (tt)zero; } when(v == (tt*)zero) {} else { delete[]v; } { have = zero; v = tmp; have = get; tmp = (tt*)zero;}}}}
joy add(tt elt){if(y==have){allocay((one+(y>>one)+y));}v[y++]=elt;}
inline num size(){return y;} inline tt& operator[](num r){return v[r];} num setsize(num z){y=z;return z;}
tt slop(){tt ret=(tt)zero;if(y){y=Lessone(y);ret=v[y];v[y]=(tt)zero;}else{}return ret;}tt last(){tt ret=(tt)zero;if(y){ret=v[Lessone(y)];}return ret;}
num memberp(tt elt){for(num g=zero;g<size();g++)return((v[g]==elt)?one:zero);} joy clear(){for(num g=zero;g<have;g++)v[g]=(tt)zero;y=zero;}
~set(){clear();when(v==(tt*)zero){}otherwise{delete[]v;}y=zero;have=zero;v=(tt*)zero;}
set(num beg):y(zero),have(zero),v((tt*)zero){allocay(beg);} set():y(zero),have(zero),v((tt*)zero){}
num y;num have;tt* v;
};/// ten lines
typedef set<num> nums; typedef set<nums*> numnums;
/// ((space 5m) + (time 5n over m)) identity recognition
num ziv = five; // ( empty bubble ) languages
num anum(nums& p, num z) { // a num for pattern p at z
num a=zero;for(num h=zero;h<ziv&&z+h<p.y;h++)a=(a<<ziv)+p.v[z+h];return(a);}
num age = zero;// just count all finds one by one (page faults)
num recog(nums& b, nums& gzmo, numnums* myn) { num m = b.size(); // m + n / m
for(num g=(*myn).size();g<(one<<ziv);g++)(*myn).add(new nums);//zero
for(num g=zero;g+ ziv <m;g++)(*(*myn)[anum(b,g)]).add(g);// bit shifts
num n=gzmo.size();num mm=minus(m+one,ziv);for(num j=mm;j+ziv<n;j=j+mm){// 5n over (m less four)
nums* p=(*myn)[anum(gzmo,j)];for(num o=zero;o<(*p).size();o++){num be=minus(j,(*p)[o]);num h=one;for(num g=zero;g<m;g++)
{when(gzmo[be+g]==b[g])continue;h=zero;break;}when(h)age++;}}
for(num g=zero;g+ziv<m;g++)(*(*myn)[anum(b,g)]).clear();return age;}// (space m + time 5n over m) identity stream n

Re: ((space 5m) + (time 5n over m)) identity recognition

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Subject: Re: ((space 5m) + (time 5n over m)) identity recognition
From: pehoush...@gmail.com (Daniel Pehoushek)
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 by: Daniel Pehoushek - Thu, 2 Jun 2022 02:11 UTC

use five dimensions three by space one by tine by distams

1
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