Política-Parlamentaria y Dret-Polític, Química

Política-Parlamentaria:

Oposició:

Cal que el señor president dagi explicacions,

perque pot baishar del núvol,

y explicar la veritat.

Govern:

No cal que el señor president dagi explicacions,

perque pot pujjar al núvol,

y no explicar la veritat.

Oposició:

Vagi-se'n señor president,

que Catalunya ya necesita un canvi.

Govern:

No se'n vagi señor president,

que Catalunya encara no necesita un canvi.

Oposició:

Aquets son els pitxors presupostos que es podien proposar,

perque no son els presupostos que Catalunya necesita.

Govern:

Aquets son els millors presupostos que es podien proposar,

perque son els presupostos que Catalunya necesita.

Oposició:

Totes-algunes decisions d'aquet govern,

no porten a ningún lloc,

per molt que el govern ens vulgui fer creure el contrari.

Govern:

Totes decisions d'aquet govern,

porten a algún lloc,

per molt que l'oposició ens vulgui fer creure el contrari.

Oposició:

Vusté es un mal president señor president,

perque creu que al soci se'l pot engañar.

Govern:

Vusté és un bon president señor president,

perque creu que al soci no se'l pot engañar.

Dret-Polític:

Si no paren aquet-çes hostilitats els fatxes españols,

hi haurá un txoc de trens.

Pero:

Si paren aquet-çes hostilitats els fatxes españols,

no hi haurá un txoc de trens.

Si la presidenta del congrés no deisha repetir la votació,

pot caure en un delicte de prevaricació.

Pero:

Si la presidenta del congrés deisha repetir la votació,

no pot caure en un delicte de prevaricació.

El que tingui de decidir el bloc confederal,

ho decidirá el bloc confederal.

Y:

El que tingui de decidir el bloc federal,

ho decidirá el bloc federal.

Si el señor diputat torna els diners desviats a l'estat,

tindrá la causa artxivada,

y ya no la tindrá pendent de judici.

Pero:

Si el señor diputat no torna els diners desviats a l'estat,

no tindrá la causa artxivada,

y encara la tindrá pendent de judici.

Química:

4·H_{2}+O_{4} <==> 4·H_{2}O

[4·H_{2}]·[O_{4}] <==> [4e^{(-1)}]·[4·H_{2}O]

PH[e^{(-1)}] = log_{2}(4) = 2

3·H_{2}+O_{6} <==> 3·H_{2}O_{2}

[3·H_{2}]·[O_{6}] <==> [2e^{(-1)}]·[3·H_{2}O_{2}]

PH[e^{(-1)}] = log_{2}(2) = 1

CH_{4}+O_{4} <==> CO_{4}H_{4}

[CH_{4}]·[O_{4}] <==> [4e^{(-1)}]·[CO_{4}H_{4}]

PH[e^{(-1)}] = log_{2}(4) = 2

C_{4}H_{8}+O_{4} <==> C_{4}O_{4}H_{8}

[C_{4}H_{8}]·[O_{4}] <==> [4e^{(-1)}]·[C_{4}O_{4}H_{8}]

PH[e^{(-1)}] = log_{2}(4) = 2

C_{4}H_{8}+2·O_{4} <==> C_{4}O_{8}H_{8}

[C_{4}H_{8}]·[2·O_{4}] <==> [8e^{(-1)}]·[C_{4}O_{8}H_{8}]

PH[e^{(-1)}] = log_{2}(8) = 3

C_{3}H_{8}+O_{4} <==> C_{3}O_{4}H_{8}

[C_{3}H_{8}]·[O_{4}] <==> [4e^{(-1)}]·[C_{3}O_{4}H_{8}]

PH[e^{(-1)}] = log_{2}(4) = 2

C_{3}H_{8}+2·O_{4} <==> C_{3}O_{8}H_{8}

[C_{3}H_{8}]·[2·O_{4}] <==> [8e^{(-1)}]·[C_{3}O_{8}H_{8}]

PH[e^{(-1)}] = log_{2}(8) = 3

C_{3}H_{8}+O_{6} <==> C_{3}O_{6}H_{8}

[C_{3}H_{8}]·[O_{6}] <==> [6e^{(-1)}]·[C_{3}O_{6}H_{8}]

PH[e^{(-1)}] = log_{2}(6) = 1+log_{2}(3) = 1+log_{2}(2+1) = 1+[1]

log_{2}(2) = 1

log_{2}(4) = 2

log_{2}(6) = 1+[1]

log_{2}(8) = 3

log_{2}(10) = 1+log_{2}(5) = 1+log_{2}(4+1) = 1+[2]

log_{2}(12) = 2+log_{2}(3) = 2+log_{2}(2+1) = 2+[1]

log_{2}(14) = 1+log_{2}(7) = 1+log_{2}(8+(-1)) = 1+]3[

Operadors Integrals:

A[y(x)] = int[y(x)]d[x]+y(x)

A[y(x)] = f(x) <==> y(x) = d_{x}[ e^{(-x)}·int[f(x)·e^{x}]d[x] ]

B[y(x)] = int[ g(x)·d_{x}[y(x)] ]d[x]+(-1)·g(x)·y(x)

B[y(x)] = f(x) <==> y(x) = (-1)·( 1/d_{x}[g(x)] )·d_{x}[f(x)]

Teorema:

C[y(x)] = int[ x·d_{x}[y(x)] ]d[x]+int[y(x)]d[x]

C[y(x)] = f(x) <==> y(x) = ( f(x)/x )

Demostració:

C[y(x)] = x·y(x) = f(x)

La definición de poder oscuro es:

Odiar a los fieles del Gestalt,

y usar infieles para odiar a fieles,

sin reacción negativa,

pero es lo que más duele limpiando los centros andando.

La definición de poder claro es:

Amar a los fieles del Gestalt,

y usar infieles para amar a fieles,

sin reacción positiva,

pero es lo que menos duele limpiando los centros andando

Si en la universitat hi ha algún fiel,

dan el títol universitari als fiels

encara que potser resen amb poder foscur que no.

Si en la universitat no hi ha ningún fiel,

no dan el títol universitari als fiels

perque resen amb poder foscur que no.

x^{(1/n)}+x = c <==> x = c^{( 1/(1+[(1/n)+(-1)]) )}

c^{( (1/n)/(1+[(1/n)+(-1)]) )}+c^{( 1/(1+[(1/n)+(-1)]) )} = ...

... c^{( 1/(1+[(1/n)+(-1)]) )}·( c^{( ((1/n)+(-1))/(1+[(1/n)+(-1)]) )}+1 ) = ...

... c^{( 1/(1+[(1/n)+(-1)]) )}·c^{( [(1/n)+(-1)]/(1+[(1/n)+(-1)]) )} = c

x^{n}+1 = c

x^{[n]} = x^{[n]/1} = c

x^{[n]/[1]} = c

x^{n}+1 = x^{[n]} = c^{[1]} = c+1

x^{n} = c

electro-magnetisme de ecuacions de fluxe-zero

Principi:

E_{e}(x,y,z) = qk_{e}·(1/r^{2})·( < x,y,z >/r )

B_{e}(x,y,z) = (-1)·qk_{e}·(1/r^{2})·( < d_{t}[x],d_{t}[y],d_{t}[z] >/r )

Lley: [ de anti-gravetat ]

(-r)·(2pi/T) = pq·( k_{e}/m )·(1/r^{2})·( T/(2pi) )

r·(2pi/T) = (-1)·pq·( k_{e}/m )·(1/r^{2})·( T/(2pi) )

T = [ segon ]·[ Radiá ]

Deducció:

d_{tt}^{2}[ r·cos(vt) ] = (-r)·cos(vt)·v^{2}

vT = 2·pi

d_{tt}[r·cos(ut)·sin(vt)] = 0

d_{tt}[r·sin(ut)·sin(vt)] = 0

u = 0

Ecuacions de Maxwell de Fluxe-Zero:

Lley: [ de Maxwell-Coulomb en forma integral ]

anti-potencial[ rot[ E_{e}(r·f(t),r·g(t),r·h(t)) ] ] = ...

... qk_{e}+...

... (1/3)·anti-potencial[ ( 1/(f(t)·g(t)·h(t)) )·int[ B_{e}(r·f(t),r·g(t),r·h(t)) ]d[t] ]

Lley: [ de Maxwell-Ampere en forma integral ]

anti-potencial[ rot[ B_{e}(r·f(t),r·g(t),r·h(t)) ] ] = ...

... d_{t}[q(t)]·k_{e}+...

... (-1)·(1/3)·anti-potencial[ ( 1/(f(t)·g(t)·h(t)) )·d_{t}[ E_{e}(r·f(t),r·g(t),r·h(t),q(t)) ] ]+...

... (-1)·(1/3)·anti-potencial[ ( 1/(f(t)·g(t)·h(t)) )·B_{e}(r·f(t),r·g(t),r·h(t),q(t)) ]

Lley: [ de Maxwell-Coulomb en forma diferencial ]

rot[ E_{e}(r·f(t),r·g(t),r·h(t)) ] = ...

... H_{e}(r·f(t),r·g(t),r·h(t))+(1/3)·( 1/(f(t)·g(t)·h(t)) )·int[ B_{e}(r·f(t),r·g(t),r·h(t)) ]d[t]

Lley: [ de Maxwell-Ampere en forma diferencial ]

rot[ B_{e}(r·f(t),r·g(t),r·h(t)) ] = ...

... J_{e}(r·f(t),r·g(t),r·h(t),q(t))+...

... (-1)·(1/3)·( 1/(f(t)·g(t)·h(t)) )·d_{t}[ E_{e}(r·f(t),r·g(t),r·h(t),q(t)) ]+...

... (-1)·(1/3)·( 1/(f(t)·g(t)·h(t)) )·B_{e}(r·f(t),r·g(t),r·h(t),q(t))

Lley: [ de Maxwell-Coulomb de l'inducció eléctrica ]

H_{e}(r·f(t),r·g(t),r·h(t)) = ...

... rot[ E_{e}(r·f(t),r·g(t),r·h(t)) ]+...

... (1/3)·qk_{e}(1/r^{2})·( 1/(f(t)·g(t)·h(t)) )·< f(t),g(t),h(t) >

Lley: [ de Maxwell-Ampere de l'inducció magnética ]

J_{e}(r·f(t),r·g(t),r·h(t),q(t)) = ...

... rot[ B_{e}(r·f(t),r·g(t),r·h(t)) ]+...

... (1/3)·d_{t}[q(t)]·k_{e}(1/r^{2})·( 1/(f(t)·g(t)·h(t)) )·< f(t),g(t),h(t) >

Lley: [ de Coulomb de l'inducció eléctrica ]

anti-potencial[ H_{e}(r·f(t),r·g(t),r·h(t)) ] = qk_{e}

Lley: [ de Ampere de l'inducció magnética ]

anti-potencial[ J_{e}(r·f(t),r·g(t),r·h(t),q(t)) ] = d_{t}[q(t)]·k_{e}

Lley: [ de Gauss en forma integral ]

anti-potencial[ E_{e}(r·f(t),r·g(t),r·h(t)) ] = ...

... 3qk_{e}·f(t)·g(t)·h(t)

anti-potencial[ B_{e}(r·f(t),r·g(t),r·h(t)) ] = ...

... (-1)·qk_{e}·( d_{t}[f(t)]·g(t)·h(t)+f(t)·d_{t}[g(t)]·h(t)+f(t)·g(t)·d_{t}[h(t)] )

Lley: [ de Gauss en forma diferencial ]

div[ E_{e}(r·f(t),r·g(t),r·h(t)) ] = ...

... 3qk_{e}·(1/r^{3})

div[ B_{e}(r·f(t),r·g(t),r·h(t)) ] = ...

... (-1)·qk_{e}·(1/r^{3})·( ...

... ( d_{tt}^{2}[f(t)]/d_{t}[f(t)] )+...

... ( d_{tt}^{2}[g(t)]/d_{t}[g(t)] )+...

... ( d_{tt}^{2}[h(t)]/d_{t}[h(t)] ) ...

... )

Deducció:

int-int-int[ ( d_{tt}^{2}[x]/d_{t}[x] ) ]( d_{t}[x]·d[t] )d[y]d[z] = d_{t}[x]·yz

Ecucions de Maxwell-Gauss originals:

Lley: [ de Gauss integral ]

anti-potencial[ E_{e}(r,r,r) ] = 3q(r^{s})·k_{e}

anti-potencial[ B_{e}(r,r,r) ] = 0

Lley: [ de Gauss diferencial ]

div[ E_{e}(r,r,r) ] = 3·d_{rrr}^{3}[q(r^{s})·(a/b)]·k_{e}

div[ B_{e}(r,r,r) ] = 0

Deducció:

div[ E_{e}(r,r,r) ] = 3·q(r^{s})·k_{e}(1/r^{3})

div[ E_{e}(r,r,r) ] = 3·q(r^{s})·k_{e}·d_{111}^{3}[(a/b)]·(1/r^{3})

div[ E_{e}(r,r,r) ] = 3·d_{rrr}^{3}[q(r^{s})·(a/b)]·k_{e}

Lley: [ de Maxwell en forma integral ]

anti-potencial[ rot[ E_{e}(r,r,r) ] ] = ...

... qk_{e}+(1/3)·anti-potencial[ int[ B_{e}(r,r,r) ]d[t] ]

anti-potencial[ rot[ B_{e}(r,r,r) ] ] = ...

... d_{t}[q(t)]·k_{e}+(-1)·(1/3)·anti-potencial[ d_{t}[ E_{e}(r,r,r,q(t)) ] ]

Lley: [ de Maxwell en forma diferencial ]

rot[ E_{e}(r,r,r) ] = H_{e}(r,r,r)+(1/3)·int[ B_{e}(r,r,r) ]d[t]

rot[ B_{e}(r,r,r) ] = J_{e}(r,r,r)+(-1)·(1/3)·d_{t}[ E_{e}(r,r,r) ]

Densitats de carga:

Lley:

anti-potencial[ E_{e}(r,r,r) ] = 3q·(a_{1}·r)·k_{e}

div[ E_{e}(r,r,r) ] = 3·d_{rrr}^{3}[q·(a_{1}·r)]·k_{e}

Lley:

anti-potencial[ E_{e}(r,r,r) ] = 3q·(a_{2}·r^{2})·k_{e}

div[ E_{e}(r,r,r) ] = 3·d_{rrr}^{3}[q·(a_{2}·r^{2})·(1/2)]·k_{e}

Lley:

n >] 3

anti-potencial[ E_{e}(r,r,r) ] = 3q·(a_{n}·r^{n})·k_{e}

div[ E_{e}(r,r,r) ] = 3·d_{rrr}^{3}[q·(a_{n}·r^{n})·(1/(n·(n+(-1))·(n+(-2))))]·k_{e}

Lley:

anti-potencial[ E_{e}(r,r,r) ] = 3q·(a_{n}·r^{(-n)})·k_{e}

div[ E_{e}(r,r,r) ] = 3·d_{rrr}^{3}[q·(a_{n}·r^{(-n)})·(1/((-n)·(n+1)·(n+2)))]·k_{e}

Ecuacions de fluxe-zero para un rectangle cúbic:

Lley:

anti-potencial[ rot[ E_{e}(ra,rb,rc) ] ] = ...

... qk_{e}+...

... (1/3)·anti-potencial[ ( 1/(abc) )·int[ B_{e}(ra,rb,rc) ]d[t] ]

anti-potencial[ rot[ B_{e}(ra,rb,rc) ] ] = ...

... d_{t}[q(t)]·k_{e}+...

... (-1)·(1/3)·anti-potencial[ ( 1/(abc) )·d_{t}[ E_{e}(ra,rb,rc,q(t)) ] ]+...

... (-1)·(1/3)·anti-potencial[ ( 1/(abc) )·B_{e}(ra,rb,rc,q(t)) ]

Lley:

rot[ E_{e}(ra,rb,rc) ] = ...

... H_{e}(ra,rb,rc)+(1/3)·( 1/(abc) )·int[ B_{e}(ra,rb,rc) ]d[t]

rot[ B_{e}(ra,rb,rc) ] = ...

... J_{e}(ra,rb,rc,q(t))+...

... (-1)·(1/3)·( 1/(abc) )·d_{t}[ E_{e}(ra,rb,rc,q(t)) ]+...

... (-1)·(1/3)·( 1/(abc) )·B_{e}(ra,rb,rc,q(t))

Lley:

H_{e}(ra,rb,rc) = ...

... rot[ E_{e}(ra,rb,rc) ]+(1/3)·qk_{e}·(1/r^{2})·( 1/(abc) )·< a,b,c >

J_{e}(ra,rb,rc,q(t)) = ...

... rot[ B_{e}(ra,rb,rc) ]+(1/3)·d_{t}[q(t)]·k_{e}·(1/r^{2})·( 1/(abc) )·< a,b,c >

Lley:

Vectors de inducció en un cub:

H_{e}(rd,rd,rd) = ...

... rot[ E_{e}(rd,rd,rd) ]+(1/3)·qk_{e}·(1/r^{2})·( 1/d^{2} )·< 1,1,1 >

J_{e}(rd,rd,rd,q(t)) = ...

... rot[ B_{e}(rd,rd,rd) ]+(1/3)·d_{t}[q(t)]·k_{e}·(1/r^{2})·( 1/d^{2} )·< 1,1,1 >

Ecuacions de fluxe-zero para un cilindre:

Lley:

anti-potencial[ rot[ E_{e}(r·cos(s),r·sin(s),r·h) ] ] = ...

... qk_{e}+...

... (1/3)·anti-potencial[ ( 1/(cos(s)·sin(s)·h) )·int[ B_{e}(r·cos(s),r·sin(s),r·h) ]d[t] ]

anti-potencial[ rot[ B_{e}(r·cos(s),r·sin(s),r·h) ] ] = ...

... d_{t}[q(t)]·k_{e}+...

... (-1)·(1/3)·anti-potencial[ (1/(cos(s)·sin(s)·h))·d_{t}[ E_{e}(r·cos(s),r·sin(s),r·h,q(t)) ] ]+...

... (-1)·(1/3)·anti-potencial[ (1/(cos(s)·sin(s)·h))·B_{e}(r·cos(s),r·sin(s),r·h,q(t)) ]

Lley:

rot[ E_{e}(r·cos(s),r·sin(s),r·h) ] = ...

... H_{e}(r·cos(s),r·sin(s),r·h)+...

... (1/3)·( 1/(cos(s)·sin(s)·h) )·int[ B_{e}(r·cos(s),r·sin(s),r·h) ]d[t]

rot[ B_{e}(r·cos(s),r·sin(s),r·h) ] = ...

... J_{e}(r·cos(s),r·sin(s),r·h,q(t))+...

... (-1)·(1/3)·( 1/(cos(s)·sin(s)·h) )·d_{t}[ E_{e}(r·cos(s),r·sin(s),r·h,q(t)) ]+...

... (-1)·(1/3)·( 1/(cos(s)·sin(s)·h) )·B_{e}(r·cos(s),r·sin(s),r·h,q(t))

Lley:

Vectors de inducció en els eishos de coordenades de un cilindre que no existeishen:

s = 0+2pi·k <==> ( [1] & [2] )

[1] H_{e}(r,0,r·h) = ...

... rot[ E_{e}(r,0,r·h) ]+(1/3)·qk_{e}·(1/r^{2})·(1/h)·< oo,1,oo·h >

[2] J_{e}(r,0,r·h,q(t)) = ...

... rot[ B_{e}(r,0,r·h) ]+(1/3)·d_{t}[q(t)]·k_{e}·(1/r^{2})·(1/h)·< oo,1,oo·h >

s = pi+2pi·k <==> ( [1] & [2] )

[1] H_{e}((-r),0,r·h) = ...

... rot[ E_{e}((-r),0,r·h) ]+(-1)·(1/3)·qk_{e}·(1/r^{2})·(1/h)·< (-oo),1,oo·h >

[2] J_{e}((-r),0,r·h,q(t)) = ...

... rot[ B_{e}((-r),0,r·h) ]+(-1)·(1/3)·d_{t}[q(t)]·k_{e}·(1/r^{2})·(1/h)·< (-oo),1,oo·h >

s = (pi/2)+2pi·k <==> ( [1] & [2] )

[1] H_{e}(0,r,r·h) = ...

... rot[ E_{e}(0,r,r·h) ]+(1/3)·qk_{e}·(1/r^{2})·(1/h)·< 1,oo,oo·h >

[2] J_{e}(0,r,r·h,q(t)) = ...

... rot[ B_{e}(0,r,r·h) ]+(1/3)·d_{t}[q(t)]·k_{e}·(1/r^{2})·(1/h)·< 1,oo,oo·h >

s = (-1)·(pi/2)+2pi·k <==> ( [1] & [2] )

[1] H_{e}(0,(-r),r·h) = ...

... rot[ E_{e}(0,(-r),r·h) ]+(-1)·(1/3)·qk_{e}·(1/r^{2})·(1/h)·< 1,(-oo),oo·h >

[2] J_{e}(0,(-r),r·h,q(t)) = ...

... rot[ B_{e}(0,(-r),r·h) ]+(-1)·(1/3)·d_{t}[q(t)]·k_{e}·(1/r^{2})·(1/h)·< 1,(-oo),oo·h >

evangelio stronikián y física

Tenemos 3 panes y 2 peces,

que no son pares pan-pez,

y vamos a tener más de 3 pares pan-pez.

No tenemos 3 panes o 2 peces,

que son pares pan-pez,

y no vamos a tener más de 3 pares pan-pez.

tenemos 3 panes y 3 peces.

tenemos 2 panes y 2 peces.

tenemos 1 pan y 1 pez.

Lley:

Sea A(x,y) un hombre que está de pié y está inclinado,

y sea F la fuerza de resistencia del suelo del pié al centro de masas:

m·d_{tt}^{2}[y] = (-1)·qg+F·cos(s)

m·d_{tt}^{2}[x] = F·sin(s)+(-1)·k·F·cos(s)

A(x,y) está en equilibri <==> ( tan(s) [< k & F = ( (qg)/cos(s) ) )

Deducció:

Sea cos(s) > 0

F·cos(s) = qg

F = ( (qg)/cos(s) )

F·sin(s) [< k·F·cos(s)

sin(s) [< k·cos(s)

tan(s) [< k

Lley:

Sea A(x,y) una caja en un plano inclinado,

y sea F la fuerza de resistencia del suelo:

m·d_{tt}^{2}[y] = (-1)·qg·cos(s)+F

m·d_{tt}^{2}[x] = qg·sin(s)+(-1)·k·F

A(x,y) está en equilibri <==> ( tan(s) [< k & F = qg·cos(s) )

Deducció:

Sea cos(s) > 0

F = qg·cos(s)

qg·sin(s) [< k·F

qg·sin(s) [< k·qg·cos(s)

sin(s) [< k·cos(s)

tan(s) [< k

Lley:

Sea A(x,y) una caja en lo plano y una caja colgada de una polea,

y sea F la fuerza de resistencia del suelo con la caja del plano:

m_{1}·d_{tt}^{2}[y] = (-1)·qg+F

(m_{1}+m_{2})·d_{tt}^{2}[x] = pg+(-1)·k·F

A(x,y) está en equilibri <==> ( (p/q) [< k & F = qg )

T = ( (p·m_{1}+kq·m_{2})/(m_{1}+m_{2}) )·g

protón:

m = (0.9)·uma

neutrón:

m = (0.9)·uma

electrón:

m = (0.2)·uma

Átomo de hidrógeno:

m = 2 uma

1 mol de hidrógeno tiene masa 2 Kg.

1 Kg = 10^{23} uma

1 g = 10^{20} uma

quark:

m = (0,3) uma

Potencial de fusión nuclear:

E(t)·e^{(-1)}+E(t)·e^{(-1)} = 2·E(t)·e^{(-1)} = 2·E(t)·e^{(-2)}+G_{2}(t)

2·E(t)·e^{(-2)}+2·E(t)·e^{(-2)} = 4·E(t)·e^{(-2)} = 4·E(t)·e^{(-4)}+G_{4}(t)

G_{2}(t) = 2·E(t)·e^{(-2)}( e+(-1) )

G_{4}(t) = 4·E(t)·e^{(-4)}( e^{2}+(-1) )

Ecuacions de Maxwell electro-magnétiques en un cub:

Principi:

E_{e}(x,y,z) = qk_{e}·(1/r^{2})·(< x,y,z >/r)

B_{e}(x,y,z) = (-1)·qk_{e}·(1/r^{2})·(< d_{t}[x],d_{t}[y],d_{t}[z] >/r)

Lley:

anti-potencial[ E_{e}(r,r,r) ] = 3qk_{e}

anti-potencial[ B_{e}(r,r,r) ] = (-0)

Lley:

anti-potencial[ rot[ E_{e}(x,y,z) ] ] = ...

... qk_{e}+int[ anti-potencial[ B_{e}(r,r,r) ] ]d[t]

anti-potencial[ rot[ B_{e}(x,y,z) ] ] = ...

... d_{t}[q(t)]·k_{e}+(-1)·(1/3)·d_{t}[ anti-potencial[ E_{e}(r,r,r,q(t)) ] ]

Lley:

rot[ E_{e}(x,y,z) ] = H_{e}(x,y,z)+int[ B_{e}(r,r,r) ]d[t]

rot[ B_{e}(x,y,z) ] = J_{e}(x,y,z,q(t))+(-1)·(1/3)·d_{t}[ E_{e}(r,r,r,q(t)) ]

Lley:

H_{e}(x,y,z) = rot[ E_{e}(x,y,z) ]+qk_{e}·(1/r^{2})·< 1,1,1 >

J_{e}(x,y,z,q(t)) = rot[ B_{e}(x,y,z) ]+(1/3)·d_{t}[q(t)]·k_{e}·(1/r^{2})·< 1,1,1 >

Lley:

Si ( E_{e}(x,y,z) = 0 & hi ha inducció magnética ) ==> ...

... hi ha cárrega. [ T = R·q ]

Si ( B_{e}(x,y,z) = 0 & hi ha corrent de desplaçament de diferencia de cárrega ) ==> ...

... hi ha intensitat del corrent. [ A = R·d_{t}[q(t)] ]

Lley:

m·d_{tt}^{2}[x] = p·( E_{e}(x)+int[ B_{e}(x) ]d[t] )

x(t) = vt

Ecuacions de Maxwell gravito-magnétiques en un cub:

Principi:

E_{g}(x,y,z) = (-1)·qk_{g}·(1/r^{2})·(< x,y,z >/r)

B_{g}(x,y,z) = qk_{g}·(1/r^{2})·(< d_{t}[x],d_{t}[y],d_{t}[z] >/r)

Lley:

anti-potencial[ E_{g}(r,r,r) ] = (-3)·qk_{g}

anti-potencial[ B_{g}(r,r,r) ] = 0

Lley:

anti-potencial[ rot[ E_{g}(x,y,z) ] ] = ...

... qk_{g}+(-1)·int[ anti-potencial[ B_{g}(r,r,r) ] ]d[t]

anti-potencial[ rot[ B_{g}(x,y,z) ] ] = ...

.. d_{t}[q(t)]·k_{g}+(1/3)·d_{t}[ anti-potencial[ E_{g}(r,r,r,q(t)) ] ]

Lley:

rot[ E_{g}(x,y,z) ] = H_{g}(x,y,z)+(-1)·int[ B_{g}(r,r,r) ]d[t]

rot[ B_{g}(x,y,z) ] = J_{g}(x,y,z,q(t))+(1/3)·d_{t}[ E_{g}(r,r,r,q(t)) ]

Lley:

H_{g}(x,y,z) = rot[ E_{g}(x,y,z) ]+qk_{g}·(1/r^{2})·< 1,1,1 >

J_{g}(x,y,z,q(t)) = rot[ B_{g}(x,y,z) ]+(1/3)·d_{t}[q(t)]·k_{g}·(1/r^{2})·< 1,1,1 >

Lley:

m·d_{tt}^{2}[x] = p·( E_{g}(x)+int[ B_{g}(x) ]d[t] )

x(t) = vt

mecánica clásica

l = longitud del hombre de pié.

somier:

m·d_{tt}^{2}[z(t)] = P·l·( x+y )

z(t) = A·e^{2^{(1/2)}·( (P·l)/m )^{(1/2)}·t}

colchón:

m·d_{tt}^{2}[z(t)] = (-1)·P·l·( x+y )

z(t) = A·e^{2^{(1/2)}·( (P·l)/m )^{(1/2)}·it}

váter:

m·d_{tt}^{2}[z(t)] = P·( x^{2}+y^{2} )

z(t) = ( (1/3)^{(1/2)}·(P/m)^{(1/2)}·t )^{(-2)}

ducha:

m·d_{tt}^{2}[z(t)] = (-P)·( x^{2}+y^{2} )

z(t) = ( (1/3)^{(1/2)}·(P/m)^{(1/2)}·it )^{(-2)}

sofá-derecho

m·d_{tt}^{2}[z(t)] = (P/l)·( x^{3}+y^{3} )

z(t) = ( ( P/(m·l) )^{(1/2)}·t )^{(-1)}

sofá-izquierdo

m·d_{tt}^{2}[z(t)] = (-1)·(P/l)·( x^{3}+y^{3} )

z(t) = ( ( P/(m·l) )^{(1/2)}·it )^{(-1)}

No desearás nada que le pertenezca al prójimo:

Si no sois del Gestalt,

no hagáis modus ponens de este blog,

que se convierte en destrocter ponens contra vosotros.

Desearás algo que le pertenezca al próximo:

Si sois del Gestalt,

haced modus ponens de este blog,

que no se convierte en destrocter ponens contra vosotros.

d_{x}[u(x,y)]+d_{y}[u(x,y)] = f(x)+g(y)

u(x,y) = int[ f(x) ]d[x]+int[ g(y) ]d[y]

d_{x}[u(x,y)]+d_{y}[u(x,y)] = x+y

u(x,y) = ( (1/2)·x^{2}+(1/2)·y^{2} )

d_{x}[u(x,y)] = x

d_{y}[u(x,y)] = y

d_{x}[u(x,y)]+d_{y}[u(x,y)] = x·y

u(x,y) = (1/4)·( yx^{2}+(-1)·(1/3)·x^{3}+xy^{2}+(-1)·(1/3)·y^{3} )

d_{x}[u(x,y)] = 2yx+(-1)·x^{2}+y^{2}

d_{y}[u(x,y)] = 2xy+(-1)·y^{2}+x^{2}

curvas elípticas:

elipses de coordenada: < cos[n](t),sin[n](t) >

cos[n](0) = n

sin[n](0) = 0

cos[n](pi/2) = ( n^{n+1}+(-1) )^{(1/(n+1))}

sin[n](pi/2) = 1

cos[n](pi) = (-n)

sin[n](pi) = 0

cos[n]( (-1)·(pi/2) ) = ( (-1)·(-n)^{n+1}+1 )^{(1/(n+1))}

sin[n]( (-1)·(pi/2) ) = (-1)

sin[n](x) = sum[ (-1)^{k_{1}...k_{n}}·( 1/(2·(k_{1}...k_{n})+1)! )·...

... (x/n)^{2·(k_{1}...k_{n})+1} ]

cos[n](x) = sum[ (-1)^{k_{1}...k_{n}}·( 1/(2·(k_{1}...k_{n}))! )·...

... (x/n)^{2·(k_{1}...k_{n})} ]

lim[ x --> 0 ][ ( sin[n](x)/x ) ] = 1

física de infieles y condenación

Juego amar ganando lo máximo.

porque es positivo lo juego 2n.

amar a un fiel:

< n,n >

F(n) = n^{2}+2n

amar a un infiel:

< (-n),n >

F(n) = (-1)·n^{2}

Juego odiar ganando lo mínimo,

porque es negativo lo juego (-2)·n.

odiar a un fiel:

< (-n),(-n) >

F(n) = n^{2}+(-2)·n

odiar a un infiel:

< n,(-n) >

F(n) = (-1)·n^{2}

Condenación de Vox:

Juego de las autonomías ganando lo máximo,

porque es positivo lo juego (n+2):

n autonomías

n países soberanos

< n,n >

F(n) = n^{2}+2n

1 autonomía

(-n)+1 países no soberanos

< 1,(-n)+1 >

F(n) = (-2)·n+3

Fuerza membrana-eléctrica:

H_{h,e}(x,y,z) = qk_{h,e}·( 1/r^{4} )·( < x,y,z >/r )

Fuerza membrana-gravitatoria:

H_{h,g}(x,y,z) = (-1)·qk_{h,g}·( 1/r^{4} )·( < x,y,z >/r )

Anti-gravedad real.

Anti-gravedad imaginaria.

Fuerza eléctrica:

E_{e}(x,y,z) = qk_{e}·( 1/r^{2} )·( < x,y,z >/r )

Fuerza gravitatoria:

E_{g}(x,y,z) = (-1)·qk_{g}·( 1/r^{2} )·( < x,y,z >/r )

Anti-gravedad real.

Anti-gravedad imaginaria de Otelo. [ espejo de interrogatorio ]

budismo stronikiano

Caminad con la luz,

mientras tengáis luz,

para que no vos sorprendan las tinieblas,

y lleguéis a la iluminación.

Hasta que l'hombre fiel sienta que no es,

y haga imposible lo sufrimiento de un hombre esclavo infiel,

y de todo lo masculino que no es.

Caminad con lo sonido,

mientras tengáis sonido,

para que no vos sorprenda lo silencio,

y lleguéis a la sonorización.

Hasta que la mujer fiel sienta que no es,

y haga imposible lo sufrimiento de una mujer esclava infiel,

y de todo lo femenino que no es.

Mis amigos:

Áragorn de corona de Aragón:

Països Catalans.

Italia.

Grecia.

Rohan-Occitania.

Bilbao-Bolsón-de-Euskal-Herria.

Mis enemigos:

Mordor-Fachas-Españoles.

Isengard-Fachas-Franceses.

Si ye vule ye-de-muá surtire-dom,

elet-pú a-vot-má,

avec ye-de-muá,

que suy-pas tun mesier.

Si tú vule tú-de-tuá surtire-dom,

elet-pé a-vot-má,

avec tú-de-tuá,

que nets-pas mun madam.

Lley:

Sigui A(x) una escala de longitud = d , recolzada en una pared,

amb fregament P amb el terra y amb fregament Q amb la pared.

Si m·d_{tt}^{2}[x] = (-1)·qg+P+Q ==> ...

... ( A(x) está en equilibri <==> P+Q = qg )

Deducció:

d·P = paralel a (-1)·qg.

(d/2)·( (-1)·qg )+d·P+( (-1)·(d/2) )·qg+( (-1)·d )·( (-1)·Q ) = 0

Lley:

A una taula A(y) de longitud = d , se li apliquen dues forces en els extrems.

Si m·d_{tt}^{2}[y] = 2F+(-T) ==> ...

... ( A(y) está en equilibri <==> T = 2F )

Deducció:

d·( (-1)·F )+( (-1)·d )·F+(d/2)·( (-1)·T )+( (-1)·(d/2) )·T = 0

Sigensmés hi ha seleccions autonómiques

pero sin-embarg no hi ha selecció nacional,

en el model confederal.

( P(x),, & ¬Q(y) )

P(x),, <==> ¬¬( ¬¬P(x) )

Nogensmenys hi ha seleccions autonómiques

y aleshores áduc hi ha selecció nacional,

en el model federal.

( P(x);; ==> Q(y) )

P(x);; <==> ¬¬¬( ¬¬¬P(x) )

ye tingue ye-de-muá que treballare-dom,

y elet-pú tambén.

tú tingue tú-de-tuá que treballare-dom,

y elet-pé tambén.

condicionel:

ye treballe ye-de-muá,

y elet-pú a-vot-má de-le-tóm tambén. [ tindríes que treballar també ]

tú treballe tú-de-tuá,

y elet-pé a-vot-má de-le-tóm tambén. [ tindría que treballar també ]

ye parle ye-de-muá le Françé de le Patuá,

y elet-pú a-vot-má,

que ets-de-puá de Le Franç.

tú parle tú-de-tuá le Françé de le Patuá,

y elet-pé a-vot-má,

que suí-de-puá de Le Franç.

ye sé-pont de-le-com és-de-puá la verité.

ye ne sé-pont de-le-com és-de-puá la verité.

nus soms-de-puá les-des campiuns.

vus soz-de-puá les-des campiuns.

sere-dom [o] estare-dom:

suí-de-puá [o] estuí-de-puá

ets-de-puá [o] estás-de-puá

és-de-puá [o] está-de-puá

soms-de-puá [o] estoms-de-puá

soz-de-puá [o] estoz-de-puá

son-de-puá [o] están-de-puá

contacto conmigo y ventas de video-juegos

Datos para contactar:

Jûan Garriga Peralta-Peraltotzak

Correo electrónico:

drgarriga303@gmail.com

Teléfono:

625 19 87 72

Solo respondo a mensajes.

Vendo:

Zelda de Súper-Nintendo,

caja, juego, instrucciones y mapa,

más 140 años de espíritu santo de videojuego por:

Precio = 1,024,000€ + (-1)·102,400€ = ...

... 200·(llama-violeta)·256·(byte/llama-violeta)·1·(día/byte)·20·(€/día)

Vendo:

Zelda de Nintendo-64,

caja, juego y instrucciones,

más 140 años de espíritu santo de videojuego por:

Precio = 1,024,000€ + (-1)·102,400€ = ...

... 200·(llama-violeta)·256·(byte/llama-violeta)·1·(día/byte)·20·(€/día)

Mentiras que te hacen malo:

se tiene poder ilimitado.

no hay condenación.

no hay esclavos infieles que no son.

se puede conquistar un planeta extraterrestre.

se puede proyectar la mente en un extraterrestre.

[ cuerpo de hombre || de mujer & alma extraterrestre <==> ...

... los infieles hombre || mujer le sacan la luz al extraterrestre ]

se puede recordar las palabras y las experiencias de un fiel que es.

se puede saltar-se algún mandamiento con un fiel que es.

No puede haber ningún fiel en un planeta extraterrestre que no sea un dios,

porque los infieles le sacan toda la luz y muere.

Tetris-Colums-Kúkons:

[x] = 0

[y] = 1

[status] = 1

[m] = 2·[L]+1;

for( [j] = not(1) ; [j] >] not([n]) ; [j]-- )

{

put-grafic( grafic[i][j] , ( centre-x+[i]+2·[x] )·[m] , ( [j]+2·[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+[x] )·[m] , ( [j]+[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i] )·[m] , [j]·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+not([x]) )·[m] , ( [j]+not([y]) )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+2·not([x]) )·[m] , ( [j]+2·not([y]) )·[m] );

if( kbhit-positiu() == 1 & [status] == 1 & ...

... ( [i]+3 [< [p] || [i]+not(3) >] [q] ) )

{

put-grafic( not-grafic[i][j] , ( centre-x+[i]+2·[x] )·[m] , ( [j]+2·[y] )·[m] );

put-grafic( not-tgrafic[i][j] , ( centre-x+[i]+[x] )·[m] , ( [j]+[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i] )·[m] , [j]·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i]+not([x]) )·[m] , ( [j]+not([y]) )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i]+2·not([x]) )·[m] , ( [j]+2·not([y]) )·[m] );

[x]++;

[y]--;

put-grafic( grafic[i][j] , ( centre-x+[i]+2·[x] )·[m] , ( [j]+2·[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+[x] )·[m] , ( [j]+[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i] )·[m] , [j]·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+not([x]) )·[m] , ( [j]+not([y]) )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+2·not([x]) )·[m] , ( [j]+2·not([y]) )·[m] );

[status] = not(1);

}

if( kbhit-negatiu() == not(1) & [status] == not(1) & ...

... ( [i]+1 [< [p] || [i]+not(1) >] [q] ) )

{

put-grafic( not-grafic[i][j] , ( centre-x+[i]+2·[x] )·[m] , ( [j]+2·[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i]+[x] )·[m] , ( [j]+[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i] )·[m] , [j]·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i]+not([x]) )·[m] , ( [j]+not([y]) )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i]+2·not([x]) )·[m] , ( [j]+2·not([y]) )·[m] );

[x]--;

[y]++;

put-grafic( grafic[i][j] , ( centre-x+[i]+2·[x] )·[m] , ( [j]+2·[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+[x] )·[m] , ( [j]+[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i] )·[m] , [j]·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+not([x]) )·[m] , ( [j]+not([y]) )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+2·not([x]) )·[m] , ( [j]+2·not([y]) )·[m] );

[status] = 1;

}

if( [status] == 1 )

{

if( contacte[i][j+not(3)] == 1 )

{

contacte[i][j+2] = 1;

contacte[i][j+1] = 1;

contacte[i][j] = 1;

contacte[i][j+not(1)] = 1;

contacte[i][j+not(2)] = 1;

break;

}

}

if( [status] == not(1) )

{

if( contacte[i+not(2)][j+not(1)] == 1 || ...

... contacte[i+not(1)][j+not(1)] == 1 || ...

... contacte[i][j+not(1)] == 1 || ...

... contacte[i+1][j+not(1)] == 1 || ...

... contacte[i+2][j+not(1)] == 1 || ...

... )

{

contacte[i+2][j] = 1;

contacte[i+1][j] = 1;

contacte[i][j] = 1;

contacte[i+not(1)][j] = 1;

contacte[i+not(2)][j] = 1;

break;

}

}

put-grafic( not-grafic[i][j] , ( centre-x+[i]+2·[x] )·[m] , ( [j]+2·[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i]+[x] )·[m] , ( [j]+[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i] )·[m] , [j]·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i]+not([x]) )·[m] , ( [j]+not([y]) )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+[i]+2·not([x]) )·[m] , ( [j]+2·not([y]) )·[m] );

}

put-grafic( grafic[i][j] , ( centre-x+[i]+2·[x] )·[m] , ( [j]+2·[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+[x] )·[m] , ( [j]+[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i] )·[m] , [j]·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+not([x]) )·[m] , ( [j]+not([y]) )·[m] );

put-grafic( grafic[i][j] , ( centre-x+[i]+2·not([x]) )·[m] , ( [j]+2·not([y]) )·[m] );

for( [j] = 1 ; [j] [< [n] ; [j]++ )

{

put-grafic( grafic[i][j] , ( centre-x+not([i])+2·[x] )·[m] , ( not([j])+2·[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+[x] )·[m] , ( not([j])+[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i]) )·[m] , not([j])·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+not([x]) )·[m] , ( not([j])+not([y]) )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+2·not([x]) )·[m] , ( not([j])+2·not([y]) )·[m] );

if( kbhit-positiu() == 1 & [status] == 1 & ...

... ( not([i])+not(3) >] not([p]) || not([i])+3 [< not([q]) ) )

{

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+2·[x] )·[m] , ( not([j])+2·[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+[x] )·[m] , ( not([j])+[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i]) )·[m] , not([j])·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+not([x]) )·[m] , ( not([j])+not([y]) )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+2·not([x]) )·[m] , ( not([j])+2·not([y]) )·[m] );

[x]++;

[y]--;

put-grafic( grafic[i][j] , ( centre-x+not([i])+2·[x] )·[m] , ( not([j])+2·[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+[x] )·[m] , ( not([j])+[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i]) )·[m] , not([j])·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+not([x]) )·[m] , ( not([j])+not([y]) )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+2·not([x]) )·[m] , ( not([j])+2·not([y]) )·[m] );

[status] = not(1);

}

if( kbhit-negatiu() == not(1) & [status] == not(1) & ...

... ( not([i])+not(1) >] not([p]) || not([i])+1 [< not([q]) )  )

{

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+2·[x] )·[m] , ( not([j])+2·[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+[x] )·[m] , ( not([j])+[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i]) )·[m] , not([j])·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+not([x]) )·[m] , ( not([j])+not([y]) )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+2·not([x]) )·[m] , ( not([j])+2·not([y]) )·[m] );

[x]--;

[y]++;

put-grafic( grafic[i][j] , ( centre-x+not([i])+2·[x] )·[m] , ( not([j])+2·[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+[x] )·[m] , ( not([j])+[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i]) )·[m] , not([j])·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+not([x]) )·[m] , ( not([j])+not([y]) )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+2·not([x]) )·[m] , ( not([j])+2·not([y]) )·[m] );

[status] = 1;

}

if( [status] == 1 )

{

if( contacte[not(i)][not(j)+not(3)] == 1 )

{

contacte[not(i)][not(j)+2] = 1;

contacte[not(i)][not(j)+1] = 1;

contacte[not(i)][not(j)] = 1;

contacte[not(i)][not(j)+not(1)] = 1;

contacte[not(i)][not(j)+not(2)] = 1;

break;

}

}

if( [status] == not(1) )

{

if( contacte[not(i)+not(2)][not(j)+not(1)] == 1 || ...

... contacte[not(i)+not(1)][not(j)+not(1)] == 1 || ...

... contacte[not(i)][not(j)+not(1)] == 1 || ...

... contacte[not(i)+1][not(j)+not(1)] == 1 || ...

... contacte[not(i)+2][not(j)+not(1)] == 1 || ...

... )

{

contacte[not(i)+2][not(j)] = 1;

contacte[not(i)+1][not(j)] = 1;

contacte[not(i)][not(j)] = 1;

contacte[not(i)+not(1)][not(j)] = 1;

contacte[not(i)+not(2)][not(j)] = 1;

break;

}

}

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+2·[x] )·[m] , ( not([j])+2·[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+[x] )·[m] , ( not([j])+[y] )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i]) )·[m] , not([j])·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+not([x]) )·[m] , ( not([j])+not([y]) )·[m] );

put-grafic( not-grafic[i][j] , ( centre-x+not([i])+2·not([x]) )·[m] , ( not([j])+2·not([y]) )·[m] );

}

put-grafic( grafic[i][j] , ( centre-x+not([i])+2·[x] )·[m] , ( not([j])+2·[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+[x] )·[m] , ( not([j])+[y] )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i]) )·[m] , not([j])·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+not([x]) )·[m] , ( not([j])+not([y]) )·[m] );

put-grafic( grafic[i][j] , ( centre-x+not([i])+2·not([x]) )·[m] , ( not([j])+2·not([y]) )·[m] );

moviment-horitzontal-dreta( int i , int status , int p )

{

if( kbhit-positu() == "==>" )

{

if( [status] == 1 & [i]+1 [< [p] )

{

[i]++;

}

if( [status] == not(1) & [i]+3 [< [p] )

{

[i]++;

}

}

}

moviment-horitzontal-esquerra( int i , int status , int q )

{

if( kbhit-negatiu() == "<==" )

{

if( [status] == 1 & [i]+not(1) >] [q] )

{

[i]--;

}

if( [status] == not(1) & [i]+not(3) >] [q] )

{

[i]--;

}

}

}

for( [j] = not(1) ; [j] >] not([n]) ; [j]-- )

{

for( [i] = [p] ; [i] [< [q] ; [i]++ )

{

for( [v] = 0 ; [v] [< [L] ; [v]++ )

{

for( [u] = 0 ; [u] [< [L] ; [u]++ )

{

contacte-pixel[(i·L)+v][(j·L)+u] = contacte[i][j];

}

for( [u] = not(0) ; [u] >] not([L]) ; [u]-- )

{

contacte-pixel[(i·L)+v][(j·L)+u] = contacte[i][j];

}

}

for( [v] = not(0) ; [v] >] not([L]) ; [v]-- )

{

for( [u] = 0 ; [u] [< [L] ; [u]++ )

{

contacte-pixel[(i·L)+v][(j·L)+u] = contacte[i][j];

}

for( [u] = not(0) ; [u] >] not([L]) ; [u]-- )

{

contacte-pixel[(i·L)+v][(j·L)+u] = contacte[i][j];

}

}

}

}

for( [j] = 1 ; [j] [< [n] ; [j]++ )

{

for( [i] = not([p]) ; [i] >] not([q]) ; [i]-- )

{

for( [v] = not(0) ; [v] >] not([L]) ; [v]-- )

{

for( [u] = not(0) ; [u] >] not([L]) ; [u]-- )

{

contacte-pixel[(not(i)·L)+v][(not(j)·L)+u] = contacte[not(i)][not(j)];

}

for( [u] = 0 ; [u] [< [L] ; [u]++ )

{

contacte-pixel[(not(i)·L)+v][(not(j)·L)+u] = contacte[not(i)][not(j)];

}

}

for( [v] = 0 ; [v] [< [L] ; [v]++ )

{

for( [u] = not(0) ; [u] >] not([L]) ; [u]-- )

{

contacte-pixel[(not(i)·L)+v][(not(j)·L)+u] = contacte[not(i)][not(j)];

}

for( [u] = 0 ; [u] [< [L] ; [u]++ )

{

contacte-pixel[(not(i)·L)+v][(not(j)·L)+u] = contacte[not(i)][not(j)];

}

}

}

}

}

files-negatiu-grafic( int contacte-pixel[i][j] )

{

for( [j] = not(1) ; [j] >] not([n])·[m] ; [j]-- )

{

[s] = ([m]·[p])+not(L);

for( [i] = ([m]·[p])+not(L) ; [i] [< ([m]·[q])+L ; [i]++ )

{

if( contacte-pixel[i][j] == 0 ) ==> break;

[s]++;

}

if( [s] == ([m]·[q])+L )

{

for( [u] = not(1) ; [u] >] not([j]) ; [u]-- )

{

for( [v] = ([m]·[p])+not(L) ; [v] [< ([m]·[q])+L ; [v]++ )

{

contacte-pixel-nou[v][u+not(1)] = contacte-pixel[v][u];

}

}

for( [k] = ([m]·[p])+not(L) ; [k] [< ([m]·[q])+L ; [k]++ )

{

contacte-pixel-nou[k][not(1)] = 0;

}

for( [u] = not(1) ; [u] >] not([j]) ; [u]-- )

{

for( [v] = ([m]·[p])+not(L) ; [v] [< ([m]·[q])+L ; [v]++ )

{

contacte-pixel[v][u] = contacte-pixel-nou[v][u];

}

}

}

}

}

files-positiu-grafic( int contacte-pixel[i][j] )

{

for( [j] = 1 ; [j] [< [n]·[m] ; [j]++ )

{

[s] = (not([p])·[m])+L;

for( [i] = (not([p])·[m])+L ; [i] >] (not([q])·[m])+not(L) ; [i]-- )

{

if( contacte-pixel[i][not(j)] == 0 ) ==> break;

[s]--;

}

if( [s] == (not([q])·[m])+not(L) )

{

for( [u] = 1 ; [u] [< [j] ; [u]++ )

{

for( [v] = (not([p])·[m])+L ; [v] >] (not([q])·[m])+not(L) ; [v]-- )

{

contacte-pixel-nou[v][not(u)+not(1)] = contacte-pixel[v][not(u)];

}

}

for( [k] = (not([p])·[m])+L ; [k] >] (not([q])·[m])+not(L) ; [k]-- )

{

contacte-pixel-nou[k][not(1)] = 0;

}

for( [u] = 1 ; [u] [< [j] ; [u]++ )

{

for( [v] = (not([p])·[m])+L ; [v] >] (not([q])·[m])+not(L) ; [v]-- )

{

contacte-pixel[v][not(u)] = contacte-pixel-nou[v][not(u)];

}

}

}

}

}

ye parle ye-de-muá,

pero ne elet-vut a-vot-má,

de-le-du ye-de-muá.

tú parle tú-de-tuá,

pero ne elet-nut a-vot-má,

de-le-du tú-de-tuá.