Does anyone know how to calculate a Mod b in Casio fx-991ES Calculator. Thanks
This calculator does not have any modulo function. However there is quite simple way how to compute modulo using display mode ab/c (instead of traditional d/c).
How to switch display mode to ab/c:
Go to settings (Shift + Mode).
Press arrow down (to view more settings).
Select ab/c (number 1).
Now do your calculation (in comp mode), like 50 / 3 and you will see 16 2/3, thus, mod is 2. Or try 54 / 7 which is 7 5/7 (mod is 5).
If you don't see any fraction then the mod is 0 like 50 / 5 = 10 (mod is 0).
The remainder fraction is shown in reduced form, so 60 / 8 will result in 7 1/2. Remainder is 1/2 which is 4/8 so mod is 4.
EDIT:
As #lawal correctly pointed out, this method is a little bit tricky for negative numbers because the sign of the result would be negative.
For example -121 / 26 = -4 17/26, thus, mod is -17 which is +9 in mod 26. Alternatively you can add the modulo base to the computation for negative numbers: -121 / 26 + 26 = 21 9/26 (mod is 9).
EDIT2: As #simpatico pointed out, this method will not work for numbers that are out of calculator's precision. If you want to compute say 200^5 mod 391 then some tricks from algebra are needed. For example, using rule
(A * B) mod C = ((A mod C) * B) mod C we can write:
200^5 mod 391 = (200^3 * 200^2) mod 391 = ((200^3 mod 391) * 200^2) mod 391 = 98
As far as I know, that calculator does not offer mod functions.
You can however computer it by hand in a fairly straightforward manner.
Ex.
(1)50 mod 3
(2)50/3 = 16.66666667
(3)16.66666667 - 16 = 0.66666667
(4)0.66666667 * 3 = 2
Therefore 50 mod 3 = 2
Things to Note:
On line 3, we got the "minus 16" by looking at the result from line (2) and ignoring everything after the decimal. The 3 in line (4) is the same 3 from line (1).
Hope that Helped.
Edit
As a result of some trials you may get x.99991 which you will then round up to the number x+1.
You need 10 ÷R 3 = 1
This will display both the reminder and the quoitent
÷R
There is a switch a^b/c
If you want to calculate
491 mod 12
then enter 491 press a^b/c then enter 12. Then you will get 40, 11, 12. Here the middle one will be the answer that is 11.
Similarly if you want to calculate 41 mod 12 then find 41 a^b/c 12. You will get 3, 5, 12 and the answer is 5 (the middle one). The mod is always the middle value.
You can calculate A mod B (for positive numbers) using this:
Pol( -Rec( 1/2πr , 2πr × A/B ) , Y ) ( πr - Y ) B
Then press [CALC], and enter your values for A and B, and any value for Y.
/ indicates using the fraction key, and r means radians ( [SHIFT] [Ans] [2] )
type normal division first and then type shift + S->d
Here's how I usually do it. For example, to calculate 1717 mod 2:
Take 1717 / 2. The answer is 858.5
Now take 858 and multiply it by the mod (2) to get 1716
Finally, subtract the original number (1717) minus the number you got from the previous step (1716) -- 1717-1716=1.
So 1717 mod 2 is 1.
To sum this up all you have to do is multiply the numbers before the decimal point with the mod then subtract it from the original number.
Note: Math error means a mod m = 0
It all falls back to the definition of modulus: It is the remainder, for example, 7 mod 3 = 1.
This because 7 = 3(2) + 1, in which 1 is the remainder.
To do this process on a simple calculator do the following:
Take the dividend (7) and divide by the divisor (3), note the answer and discard all the decimals -> example 7/3 = 2.3333333, only worry about the 2. Now multiply this number by the divisor (3) and subtract the resulting number from the original dividend.
so 2*3 = 6, and 7 - 6 = 1, thus 1 is 7mod3
Calculate x/y (your actual numbers here), and press a b/c key, which is 3rd one below Shift key.
Simply just divide the numbers, it gives yuh the decimal format and even the numerical format. using S<->D
For example: 11/3 gives you 3.666667 and 3 2/3 (Swap using S<->D).
Here the '2' from 2/3 is your mod value.
Similarly 18/6 gives you 14.833333 and 14 5/6 (Swap using S<->D).
Here the '5' from 5/6 is your mod value.
Related
Kindly please help me with the problem as I need to use nlinfit function for fitting unknown parameters but it is showing some error. Although yesterday I was getting some values for parameters to be fitted but now if I am running it is having some issue for the function output to be used in fitted with NaN answer for last iteration only. X data is a concatenated matrix of three columns as independent variable and yk is dependent variable, taua is a matrix of initial guesses of number of parameters to be fitted.
function [yk]=activity_coefficientE(taua,x)
T=523;
alpha12=0.3; alpha13=0.3; alpha21=0.3; alpha23=0.3; alpha31=0.3; alpha32=0.3;
alpha18=0.2; alpha81=0.2; alpha28=0.2; alpha82=0.2; alpha38=0.2; alpha83=0.3;
alpha19=0.2; alpha91=0.2; alpha29=0.2; alpha92=0.2; alpha39=0.2; alpha93=0.2;
alpha110=0.2;alpha101=0.2;alpha210=0.2;alpha102=0.2;alpha310=0.2;alpha103=0.2;
alpha113=0.2;alpha131=0.2;alpha213=0.2;alpha132=0.2;alpha313=0.2;alpha133=0.2;
alpha114=0.2;alpha141=0.2;alpha214=0.2;alpha142=0.2;alpha314=0.2;alpha143=0.2;
alpha115=0.2;alpha151=0.2;alpha215=0.2;alpha152=0.2;alpha315=0.2;alpha153=0.2;
alpha117=0.2;alpha171=0.2;alpha217=0.2;alpha172=0.2;alpha317=0.2;alpha173=0.2;
alpha118=0.2;alpha181=0.2;alpha218=0.2;alpha182=0.2;alpha318=0.2;alpha183=0.2;
alpha810=0.2;alpha915=0.2;alpha1314=0.2;alpha108=0.2;alpha159=0.2;alpha1413=0.2;
alpha1718=0.2;alpha1817=0.2;
tau12=0; tau13=0; tau21=0; tau23=0; tau31=0; tau32=0;
%taua=randi([-5,5],1,112)
tau18=taua(1)+taua(57)/T;
tau81=taua(2)+taua(58)/T;
tau28=taua(3)+taua(59)/T;
tau82=taua(4)+taua(60)/T;
tau38=taua(5)+taua(61)/T;
tau83=taua(6)+taua(62)/T;
tau19=taua(7)+taua(63)/T;
tau91=taua(8)+taua(64)/T;
tau29=taua(9)+taua(65)/T;
tau92=taua(10)+taua(66)/T;
tau39=taua(11)+taua(67)/T;
tau93=taua(12)+taua(68)/T;
tau110=taua(13)+taua(69)/T;
tau101=taua(14)+taua(70)/T;
tau210=taua(15)+taua(71)/T;
tau102=taua(16)+taua(72)/T;
tau310=taua(17)+taua(73)/T;
tau103=taua(18)+taua(74)/T;
tau113=taua(19)+taua(75)/T;
tau131=taua(20)+taua(76)/T;
tau213=taua(21)+taua(77)/T;
tau132=taua(22)+taua(78)/T;
tau313=taua(23)+taua(79)/T;
tau133=taua(24)+taua(80)/T;
tau114=taua(25)+taua(81)/T;
tau141=taua(26)+taua(82)/T;
tau214=taua(27)+taua(83)/T;
tau142=taua(28)+taua(84)/T;
tau314=taua(29)+taua(85)/T;
tau143=taua(30)+taua(86)/T;
tau115=taua(31)+taua(87)/T;
tau151=taua(32)+taua(88)/T;
tau215=taua(33)+taua(89)/T;
tau152=taua(34)+taua(90)/T;
tau315=taua(35)+taua(91)/T;
tau153=taua(36)+taua(92)/T;
tau117=taua(37)+taua(93)/T;
tau171=taua(38)+taua(94)/T;
tau217=taua(39)+taua(95)/T;
tau172=taua(40)+taua(96)/T;
tau317=taua(41)+taua(97)/T;
tau173=taua(42)+taua(98)/T;
tau118=taua(43)+taua(99)/T;
tau181=taua(44)+taua(100)/T;
tau218=taua(45)+taua(101)/T;
tau182=taua(46)+taua(102)/T;
tau318=taua(47)+taua(103)/T;
tau183=taua(48)+taua(104)/T;
tau810=taua(49)+taua(105)/T;
tau108=taua(50)+taua(106)/T;
tau915=taua(51)+taua(107)/T;
tau159=taua(52)+taua(108)/T;
tau1314=taua(53)+taua(109)/T;
tau1413=taua(54)+taua(110)/T;
tau1718=taua(55)+taua(111)/T;
tau1817=taua(56)+taua(112)/T;
G12=exp(-(tau12*alpha12));
G21=exp(-(tau21*alpha21));
G13=exp(-(tau13*alpha13));
G31=exp(-(tau31*alpha31));
G23=exp(-(tau23*alpha23));
G32=exp(-(tau32*alpha32));
G18=exp(-(tau18*alpha18));
G81=exp(-(tau81*alpha81));
G28=exp(-(tau28*alpha28));
G82=exp(-(tau82*alpha82));
G38=exp(-(tau38*alpha83));
G83=exp(-(tau83*alpha83));
G19=exp(-(tau19*alpha19));
G91=exp(-(tau91*alpha91));
G29=exp(-(tau29*alpha29));
G92=exp(-(tau92*alpha92));
G39=exp(-(tau39*alpha39));
G93=exp(-(tau93*alpha93));
G110=exp(-(tau110*alpha110));
G101=exp(-(tau101*alpha101));
G210=exp(-(tau210*alpha210));
G102=exp(-(tau102*alpha102));
G310=exp(-(tau310*alpha310));
G103=exp(-(tau103*alpha103));
G113=exp(-(tau113*alpha113));
G131=exp(-(tau131*alpha131));
G213=exp(-(tau213*alpha213));
G132=exp(-(tau132*alpha132));
G313=exp(-(tau313*alpha313));
G133=exp(-(tau133*alpha133));
G114=exp(-(tau114*alpha114));
G141=exp(-(tau141*alpha141));
G214=exp(-(tau214*alpha214));
G142=exp(-(tau142*alpha142));
G314=exp(-(tau314*alpha314));
G143=exp(-(tau143*alpha143));
G115=exp(-(tau115*alpha115));
G151=exp(-(tau151*alpha151));
G215=exp(-(tau215*alpha215));
G152=exp(-(tau152*alpha152));
G315=exp(-(tau315*alpha315));
G153=exp(-(tau153*alpha153));
G117=exp(-(tau117*alpha117));
G171=exp(-(tau171*alpha171));
G217=exp(-(tau217*alpha217));
G172=exp(-(tau172*alpha172));
G317=exp(-(tau317*alpha317));
G173=exp(-(tau173*alpha173));
G118=exp(-(tau118*alpha118));
G181=exp(-(tau181*alpha181));
G218=exp(-(tau218*alpha218));
G182=exp(-(tau182*alpha182));
G318=exp(-(tau318*alpha318));
G183=exp(-(tau183*alpha183));
G810=exp(-(tau810*alpha810));
G108=exp(-(tau108*alpha108));
G915=exp(-(tau915*alpha915));
G159=exp(-(tau159*alpha159));
G1314=exp(-(tau1314*alpha1314));
G1413=exp(-(tau1413*alpha1413));
G1718=exp(-(tau1718*alpha1718));
G1817=exp(-(tau1817*alpha1817));
%calculating mole fractions of ionic species
x1=x(:,1);
x2=x(:,2);
x3=x(:,3);
%x1=[0.1577 0.1492 0.1462 0.1366 0.1299 0.1180 0.0863 0.0761 0.0550 ];
%x2=[0.8278 0.7945 0.7678 0.7450 0.6979 0.6309 0.4611 0.4114 0.2952 ];
%x3=[0.0145 0.0563 0.0860 0.1184 0.1722 0.2511 0.4526 0.5125 0.6498 ];
A=[0.0674243 0.0773881 0.0843400 0.0865343 0.0899223 0.0882858 0.0715087 0.0643867 0.0483658];
B=[0.0141081 0.0479814 0.0643151 0.0737477 0.0820756 0.0838701 0.0701576 0.0634457 0.0479639];
C=[0.0565665 0.0450072 0.0387724 0.0313828 0.02506094 0.0186280 0.0092734 0.0073438 0.0041595 ];
D=[0.0336447 0.0267694 0.0230611 0.0186659 0.0149058 0.0110795 0.0055157 0.0043679 0.0024739 ];
E=[0.0008148 0.0008756 0.00087131 0.0008794 0.0008711 0.0008441 0.0007384 0.0006997 0.0005980 ];
N=length(A);
x1n=zeros(N,1);x2n=zeros(N,1);x3n=zeros(N,1);
X1=zeros(N,1);X2=zeros(N,1);X3=zeros(N,1);X4=zeros(N,1);X5=zeros(N,1);X6=zeros(N,1);X7=zeros(N,1);
X12=zeros(N,1);X16=zeros(N,1);
for i=1:N
x1n(i)=(x1(i)-A(i)-D(i)-2*E(i)-C(i)+3*B(i))
x2n(i)=(x2(i)-A(i)-C(i)-D(i))
x3n(i)=(x3(i)-B(i))
X1(i)=(x1n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X2(i)=(x2n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X3(i)=(x3n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X4(i)=(A(i)+D(i)+E(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X5(i)=(C(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X6(i)=(A(i)-B(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X7(i)=(B(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X12(i)=(E(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X16(i)=(C(i)+D(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
end
yc4=X4./(X4+X5);
yc5=X5./(X4+X5);
yc6=X6./(X6+X7+X12+X16);
yc7=X7./(X6+X7+X12+X16);
yc12=X12./(X6+X7+X12+X16);
yc16=X16./(X6+X7+X12+X16);
alpha14=yc6.*alpha18+yc7.*alpha19+yc12.*alpha113+yc16.*alpha117;
%alpha41=alpha14;
alpha24=yc6.*alpha28+yc7.*alpha29+yc12.*alpha213+yc16.*alpha217;
%alpha42=alpha24;
alpha34=yc6.*alpha38+yc7.*alpha39+yc12.*alpha313+yc16.*alpha317;
%alpha43=alpha34;
alpha15=yc6.*alpha110+yc7.*alpha115+yc12.*alpha114+yc16.*alpha118;
%alpha51=alpha15;
alpha25=yc6.*alpha210+yc7.*alpha215+yc12.*alpha214+yc16.*alpha218;
%alpha52=alpha25;
alpha35=yc6.*alpha310+yc7.*alpha315+yc12.*alpha314+yc16.*alpha318;
%alpha53=alpha35;
alpha16=yc4.*alpha81+yc5.*alpha101;
%alpha61=alpha16;
alpha26=yc4.*alpha82+yc5.*alpha102;
%alpha62=alpha26;
alpha36=yc4.*alpha83+yc5.*alpha103;
%alpha63=alpha36;
alpha17=yc4.*alpha91+yc5.*alpha151;
%alpha71=alpha17;
alpha27=yc4.*alpha92+yc5.*alpha152;
%alpha72=alpha27;
alpha37=yc4.*alpha93+yc5.*alpha153;
%alpha73=alpha37;
alpha112=yc4.*alpha131+yc5.*alpha141;
%alpha121=alpha112;
alpha212=yc4.*alpha132+yc5.*alpha142;
%alpha122=alpha212;
alpha312=yc4.*alpha133+yc5.*alpha143;
%alpha123=alpha312;
alpha116=yc4.*alpha171+yc5.*alpha181;
%alpha161=alpha116;
alpha216=yc4.*alpha172+yc5.*alpha182;
%alpha162=alpha216;
alpha316=yc4.*alpha173+yc5.*alpha183;
%alpha163=alpha316;
alpha46=yc5.*alpha810;
%alpha64=alpha46;
alpha47=yc5.*alpha915;
%alpha74=alpha47;
alpha412=yc5.*alpha1314;
%alpha124=alpha412;
alpha416=yc5.*alpha1718;
%alpha164=alpha416;
alpha56=yc4.*alpha108;
%alpha65=alpha56;
alpha57=yc4.*alpha159;
%alpha75=alpha57;
alpha512=yc4.*alpha1413;
%alpha125=alpha512;
alpha516=yc4.*alpha1817;
%alpha165=alpha516;
G14=yc6.*G18+yc7.*G19+yc12.*G113+yc16.*G117;
%G41=G14;
G24=yc6.*G28+yc7.*G29+yc12.*G213+yc16.*G217;
%G42=G24;
G34=yc6.*G38+yc7.*G39+yc12.*G313+yc16.*G317;
%G43=G34;
G15=yc6.*G110+yc7.*G115+yc12.*G114+yc16.*G118;
%G51=G15;
G25=yc6.*G210+yc7.*G215+yc12.*G214+yc16.*G218;
%G52=G25;
G35=yc6.*G310+yc7.*G315+yc12.*G314+yc16.*G318;
%G53=G35;
G16=yc4.*G81+yc5.*G101;
%G61=G16;
G26=yc4.*G82+yc5.*G102;
%G62=G26;
G36=yc4.*G83+yc5.*G103;
%G63=G36;
G17=yc4.*G91+yc5.*G151;
%G71=G17;
G27=yc4.*G92+yc5.*G152;
%G72=G27;
G37=yc4.*G93+yc5.*G153;
%G73=G37;
G112=yc4.*G131+yc5.*G141;
%G121=G112;
G212=yc4.*G132+yc5.*G142;
%G122=G212;
G312=yc4.*G133+yc5.*G143;
%G123=G312;
G116=yc4.*G171+yc5.*G181;
%G161=G116;
G216=yc4.*G172+yc5.*G182;
%G162=G216;
G316=yc4.*G173+yc5.*G183;
%G163=G316;
G46=yc5.*G810;
%G64=G46;
G47=yc5.*G915;
%G74=G47;
G412=yc5.*G1314;
%G124=G412;
G416=yc5.*G1718;
%G164=G416;
G56=yc4.*G108;
%G65=G56;
G57=yc4.*G159;
%G75=G57;
G512=yc4.*G1413;
%G125=G512;
G516=yc4.*G1817;
%G165=G516;
tau14=-log(G14)./alpha14;
%tau41=tau14;
tau24=-log(G24)./alpha24;
%tau42=tau24;
tau34=-log(G34)./alpha34;
%tau43=tau34;
tau15=-log(G15)./alpha15;
%tau51=tau15;
tau25=-log(G25)./alpha25;
%tau52=tau25;
tau35=-log(G35)./alpha35;
%tau53=tau35;
tau16=-log(G16)./alpha16;
%tau61=tau16;
tau26=-log(G26)./alpha26;
%tau62=tau26;
tau36=-log(G36)./alpha36;
%tau63=tau36;
tau17=-log(G17)./alpha17;
%tau71=tau17;
tau27=-log(G27)./alpha27;
%tau72=tau27;
tau37=-log(G37)./alpha37;
%tau73=tau37;
tau112=-log(G112)./alpha112;
%tau121=tau112;
tau212=-log(G212)./alpha212;
%tau122=tau212;
tau312=-log(G312)./alpha312;
%tau123=tau312;
tau116=-log(G116)./alpha116;
%tau161=tau116;
tau216=-log(G216)./alpha216;
%tau162=tau216;
tau316=-log(G316)./alpha316;
%tau163=tau316;
tau46=-log(G46)./alpha46;
%tau64=tau46;
tau47=-log(G47)./alpha47;
%tau74=tau47;
tau412=-log(G412)./alpha412;
%tau124=tau412;
tau416=-log(G416)./alpha416;
%tau164=tau416;
tau56=-log(G56)./alpha56;
%tau65=tau56;
tau57=-log(G57)./alpha57;
%tau75=tau57;
tau512=-log(G512)./alpha512;
%tau125=tau512;
tau516=-log(G516)./alpha516;
%tau165=tau516;
ln_y1_1=G12.*X2.*tau12+ G31.*X3.*tau13+ G14.*X4.*tau14+G15.*X5.*tau15+G16.*X6.*tau16+G17.*X7.*tau17+G112.*X12.*tau112+G116.*X16.*tau116;
ln_y1_2=G12.*X2+ G13.*X3+ G14.*X4+G15.*X5+G16.*X6+G17.*X7+G112.*X12+G116.*X16;
ln_y2_1=G21.*X1.*tau12+ G32.*X3.*tau32+ G24.*X4.*tau24+G25.*X5.*tau25+G26.*X6.*tau26+G27.*X7.*tau27+G212.*X12.*tau212+G216.*X16.*tau216;
ln_y2_2=G12.*X1+ G23.*X3+G24.*X4+G25.*X5+G26.*X6+G27.*X7+G212.*X12+G216.*X16;
ln_y3_1=G13.*X1.*tau13+ G23.*X3.*tau23+ G34.*X4.*tau34+G35.*X5.*tau35+G36.*X6.*tau36+G37.*X7.*tau37+G312.*X12.*tau312+G316.*X16.*tau316;
ln_y3_2=G13.*X1+ G23.*X3+ G34.*X4+G35.*X5+G36.*X6+G37.*X7+G312.*X12+G316.*X16;
ln_y4_1=G14.*X1.*tau14+G24.*X2.*tau24+G34.*X3.*tau34+G46.*X6.*tau46+G47.*X7.*tau47+G412.*X12.*tau412+G416.*X16.*tau416;
ln_y4_2=G14.*X1+G24.*X2+G34.*X3+G46.*X6+G47.*X7+G412.*X12+G416.*X16;
ln_y5_1=G15.*X1.*tau15+G25.*X2.*tau25+G35.*X3.*tau35+G56.*X6.*tau56+G57.*X7.*tau57+G512.*X12.*tau512+G516.*X16.*tau516;
ln_y5_2=G15.*X1+G25.*X2+G35.*X3+G56.*X6+G57.*X7+G512.*X12+G516.*X16;
ln_y6_1=G16.*X1.*tau16+G26.*X2.*tau26+G36.*X3.*tau36+G46.*X4.*tau46+G56.*X5.*tau56;
ln_y6_2=G16.*X1+G26.*X2+G36.*X3+G46.*X4+G56.*X5;
ln_y7_1=G17.*X1.*tau17+G27.*X2.*tau27+G37.*X3.*tau37+G47.*X4.*tau47+G57.*X5.*tau57;
ln_y7_2=G17.*X1+G27.*X2+G37.*X3+G47.*X4+G57.*X5;
ln_y12_1=G112.*X1.*tau112+G212.*X2.*tau212+G312.*X3.*tau312+G412.*X4.*tau412+G512.*X5.*tau512;
ln_y12_2=G112.*X1+G212.*X2+G312.*X3+G412.*X4+G512.*X5;
ln_y16_1=G116.*X1.*tau116+G216.*X2.*tau216+G316.*X3.*tau316+G416.*X4.*tau416+G516.*X5.*tau516;
ln_y16_2=G116.*X1+G216.*X2+G316.*X3+G416.*X4+G516.*X5;
ln_y1_3=(((X2.*G12)./ln_y2_2).*(tau12-(ln_y2_1)./(ln_y2_2)))+(((X3.*G13)./ln_y3_2).*(tau13-(ln_y3_1)./(ln_y3_2)));
ln_y1_4=(((X6.*G16)./ln_y6_2).*(tau16- (ln_y6_1./ln_y6_2))) + (((X7.*G17)./ln_y7_2).*(tau17- (ln_y7_1./ln_y7_2)))+(((X12.*G12)./ln_y12_2).*(tau112- (ln_y12_1./ln_y12_2)))+(((X16.*G16)./ln_y16_2).*(tau116- (ln_y16_1./ln_y16_2)));
ln_y1_5=(((X4.*G14)./ln_y4_2).*(tau14- (ln_y4_1./ln_y4_2))) + (((X5.*G15)./ln_y5_2).*(tau15- (ln_y5_1./ln_y5_2)));
yk=exp((ln_y1_1./ln_y1_2) + ln_y1_3 + ln_y1_4+ ln_y1_5) % activity coefficient for H2O
end
........................................
Another function where above function to be called.....
% calling the function act_coeff to estimate the binary interaction parameters
for i=1:112
filename = 'EagelsDATA.xlsx'; %reading VLE data from excel file
Data = xlsread(filename);
x(:,1) = Data([10:15 17:19],16);
x(:,2) = Data([10:15 17:19],1);
x(:,3)= Data([10:15 17:19],2);
taua=(randi([-5,5],1,112));
yk=[0.0606 (values calculated from above function and will be used for fitting)
0.4327
0.6545
0.9417
1.2570
1.6881
1.9108
1.7777
1.3821]
% taua =[ -2 3 4 -3 -4 1 4 -2 4 -4 -1 4 5 -3 3 2 -5 3 -4
% 1 4 1 5 -1 -1 -3 2 -3 4 3 4 2 5 4 -2 4 3 -1
% 1 0 -5 -5 -5 -3 4 2 1 4 0 2 -3 -4 5 0 -3 2 5
% 1 0 5 1 -3 5 4 1 5 2 3 2 0 -5 -4 -2 1 -2 5
%-5 5 -2 -2 4 1 -1 3 -1 1 5 -1 0 -1 4 5 5 1 4
% 1 0 4 -4 4 0 -1 -2 -5 -3 -4 -5
% -5 0 -2 0 -5] (random values for which yk was calculted from the command
taua= randi([-5,5],1,112))
try % try-catch used to continue the loop without stopping on encountering an error
[taua1]= nlinfit(x,yk,#activity_coefficientE,taua)
catch exception
continue
end
end
I am not able to attach excel sheet here so data from excel sheet is as:
x =[0.1577 0.1492 0.1462 0.1366 0.1299 0.1180 0.0863 0.0761 0.0550; column 1
0.8278 0.7945 0.7678 0.7450 0.6979 0.6309 0.4611 0.4114 0.2952 ; column 2
0.0145 0.0563 0.0860 0.1184 0.1722 0.2511 0.4526 0.5125 0.6498 ]; column 3
I found 3 major problems with what you did.
Problem #1 - errors
The reason you get the error is because your function "activity_coefficientE" can sometimes return NaN or inf values. My suggestion is to look for these values and set the value of "yk" to a large value so that the optimizer in "nlinfit" will stay away from coefficients that produce infinite or NaN values. This is the code at the bottom of the function so that you avoid crashes:
if any(~isfinite(yk))
yk = 10 * ones(size(yk));
end
Problem #2 - random initial guesses
The trouble with using random numbers for your initial conditions is that every time you run it you get a different answer, so sometimes it works and sometimes it doesn't. If you set the random number generator seed, you can get the same random numbers each time you run the script. If you change you seed, you can get a different set of random numbers. I shortened your main script to this, where I try 100 different random seeds (and store the results of each attempt) to see what answers result:
for i=1:100
rng(i)
taua = randi([-5,5],1,112);
taua1(i, :) = nlinfit(x,yk,#activity_coefficientE,taua);
end
Each row of "taua1" is a set of 111 coefficients.
Problem #3 - Trying to fit 9 points with 112 coefficients
Every time nlinfit is called, you get this warning:
Warning: Rank deficient
because you have more coefficients (112) that you are asking nlinfit to find than data points you are fitting (9). It's like trying to find the 2nd order equation that best fits 2 points, there are an infinite number of solutions. When curve fitting you should have more data points than coefficients to make sure you're not fitting noise. You need more data points in "yk" and "x" and/or fewer coefficients to fit. I've done a lot of curve fitting and I've never seen an equation with 112 coefficients, so I am thinking that you are not solving the problem correctly. Perhaps the 112 coefficients aren't really independent or there are 112 data points and 9 coefficients that you want to find.
For completeness, here is my edited version of the activity_coefficientE.m function that I created to work on this solution. In general, I never see Matlab code with this many variables with similar names. Much of this code could be greatly simplified by using vector operations. Most of my changes involve formatting, adding semicolons, and the checks for non-finite values at the end.
function yk=activity_coefficientE(taua,x)
T=523;
alpha12=0.3; alpha13=0.3; alpha21=0.3; alpha23=0.3; alpha31=0.3; alpha32=0.3;
alpha18=0.2; alpha81=0.2; alpha28=0.2; alpha82=0.2; alpha38=0.2; alpha83=0.3;
alpha19=0.2; alpha91=0.2; alpha29=0.2; alpha92=0.2; alpha39=0.2; alpha93=0.2;
alpha110=0.2;alpha101=0.2;alpha210=0.2;alpha102=0.2;alpha310=0.2;alpha103=0.2;
alpha113=0.2;alpha131=0.2;alpha213=0.2;alpha132=0.2;alpha313=0.2;alpha133=0.2;
alpha114=0.2;alpha141=0.2;alpha214=0.2;alpha142=0.2;alpha314=0.2;alpha143=0.2;
alpha115=0.2;alpha151=0.2;alpha215=0.2;alpha152=0.2;alpha315=0.2;alpha153=0.2;
alpha117=0.2;alpha171=0.2;alpha217=0.2;alpha172=0.2;alpha317=0.2;alpha173=0.2;
alpha118=0.2;alpha181=0.2;alpha218=0.2;alpha182=0.2;alpha318=0.2;alpha183=0.2;
alpha810=0.2;alpha915=0.2;alpha1314=0.2;alpha108=0.2;alpha159=0.2;alpha1413=0.2;
alpha1718=0.2;alpha1817=0.2;
tau12=0; tau13=0; tau21=0; tau23=0; tau31=0; tau32=0;
tau18=taua(1)+taua(57)/T;
tau81=taua(2)+taua(58)/T;
tau28=taua(3)+taua(59)/T;
tau82=taua(4)+taua(60)/T;
tau38=taua(5)+taua(61)/T;
tau83=taua(6)+taua(62)/T;
tau19=taua(7)+taua(63)/T;
tau91=taua(8)+taua(64)/T;
tau29=taua(9)+taua(65)/T;
tau92=taua(10)+taua(66)/T;
tau39=taua(11)+taua(67)/T;
tau93=taua(12)+taua(68)/T;
tau110=taua(13)+taua(69)/T;
tau101=taua(14)+taua(70)/T;
tau210=taua(15)+taua(71)/T;
tau102=taua(16)+taua(72)/T;
tau310=taua(17)+taua(73)/T;
tau103=taua(18)+taua(74)/T;
tau113=taua(19)+taua(75)/T;
tau131=taua(20)+taua(76)/T;
tau213=taua(21)+taua(77)/T;
tau132=taua(22)+taua(78)/T;
tau313=taua(23)+taua(79)/T;
tau133=taua(24)+taua(80)/T;
tau114=taua(25)+taua(81)/T;
tau141=taua(26)+taua(82)/T;
tau214=taua(27)+taua(83)/T;
tau142=taua(28)+taua(84)/T;
tau314=taua(29)+taua(85)/T;
tau143=taua(30)+taua(86)/T;
tau115=taua(31)+taua(87)/T;
tau151=taua(32)+taua(88)/T;
tau215=taua(33)+taua(89)/T;
tau152=taua(34)+taua(90)/T;
tau315=taua(35)+taua(91)/T;
tau153=taua(36)+taua(92)/T;
tau117=taua(37)+taua(93)/T;
tau171=taua(38)+taua(94)/T;
tau217=taua(39)+taua(95)/T;
tau172=taua(40)+taua(96)/T;
tau317=taua(41)+taua(97)/T;
tau173=taua(42)+taua(98)/T;
tau118=taua(43)+taua(99)/T;
tau181=taua(44)+taua(100)/T;
tau218=taua(45)+taua(101)/T;
tau182=taua(46)+taua(102)/T;
tau318=taua(47)+taua(103)/T;
tau183=taua(48)+taua(104)/T;
tau810=taua(49)+taua(105)/T;
tau108=taua(50)+taua(106)/T;
tau915=taua(51)+taua(107)/T;
tau159=taua(52)+taua(108)/T;
tau1314=taua(53)+taua(109)/T;
tau1413=taua(54)+taua(110)/T;
tau1718=taua(55)+taua(111)/T;
tau1817=taua(56)+taua(112)/T;
G12=exp(-(tau12*alpha12));
G21=exp(-(tau21*alpha21));
G13=exp(-(tau13*alpha13));
G31=exp(-(tau31*alpha31));
G23=exp(-(tau23*alpha23));
G32=exp(-(tau32*alpha32));
G18=exp(-(tau18*alpha18));
G81=exp(-(tau81*alpha81));
G28=exp(-(tau28*alpha28));
G82=exp(-(tau82*alpha82));
G38=exp(-(tau38*alpha83));
G83=exp(-(tau83*alpha83));
G19=exp(-(tau19*alpha19));
G91=exp(-(tau91*alpha91));
G29=exp(-(tau29*alpha29));
G92=exp(-(tau92*alpha92));
G39=exp(-(tau39*alpha39));
G93=exp(-(tau93*alpha93));
G110=exp(-(tau110*alpha110));
G101=exp(-(tau101*alpha101));
G210=exp(-(tau210*alpha210));
G102=exp(-(tau102*alpha102));
G310=exp(-(tau310*alpha310));
G103=exp(-(tau103*alpha103));
G113=exp(-(tau113*alpha113));
G131=exp(-(tau131*alpha131));
G213=exp(-(tau213*alpha213));
G132=exp(-(tau132*alpha132));
G313=exp(-(tau313*alpha313));
G133=exp(-(tau133*alpha133));
G114=exp(-(tau114*alpha114));
G141=exp(-(tau141*alpha141));
G214=exp(-(tau214*alpha214));
G142=exp(-(tau142*alpha142));
G314=exp(-(tau314*alpha314));
G143=exp(-(tau143*alpha143));
G115=exp(-(tau115*alpha115));
G151=exp(-(tau151*alpha151));
G215=exp(-(tau215*alpha215));
G152=exp(-(tau152*alpha152));
G315=exp(-(tau315*alpha315));
G153=exp(-(tau153*alpha153));
G117=exp(-(tau117*alpha117));
G171=exp(-(tau171*alpha171));
G217=exp(-(tau217*alpha217));
G172=exp(-(tau172*alpha172));
G317=exp(-(tau317*alpha317));
G173=exp(-(tau173*alpha173));
G118=exp(-(tau118*alpha118));
G181=exp(-(tau181*alpha181));
G218=exp(-(tau218*alpha218));
G182=exp(-(tau182*alpha182));
G318=exp(-(tau318*alpha318));
G183=exp(-(tau183*alpha183));
G810=exp(-(tau810*alpha810));
G108=exp(-(tau108*alpha108));
G915=exp(-(tau915*alpha915));
G159=exp(-(tau159*alpha159));
G1314=exp(-(tau1314*alpha1314));
G1413=exp(-(tau1413*alpha1413));
G1718=exp(-(tau1718*alpha1718));
G1817=exp(-(tau1817*alpha1817));
%calculating mole fractions of ionic species
x1=x(:,1);
x2=x(:,2);
x3=x(:,3);
A=[0.0674243 0.0773881 0.0843400 0.0865343 0.0899223 0.0882858 0.0715087 0.0643867 0.0483658];
B=[0.0141081 0.0479814 0.0643151 0.0737477 0.0820756 0.0838701 0.0701576 0.0634457 0.0479639];
C=[0.0565665 0.0450072 0.0387724 0.0313828 0.02506094 0.0186280 0.0092734 0.0073438 0.0041595 ];
D=[0.0336447 0.0267694 0.0230611 0.0186659 0.0149058 0.0110795 0.0055157 0.0043679 0.0024739 ];
E=[0.0008148 0.0008756 0.00087131 0.0008794 0.0008711 0.0008441 0.0007384 0.0006997 0.0005980 ];
N=length(A);
x1n=zeros(N,1);x2n=zeros(N,1);x3n=zeros(N,1);
X1=zeros(N,1);X2=zeros(N,1);X3=zeros(N,1);X4=zeros(N,1);X5=zeros(N,1);X6=zeros(N,1);X7=zeros(N,1);
X12=zeros(N,1);X16=zeros(N,1);
for i=1:N
x1n(i)=(x1(i)-A(i)-D(i)-2*E(i)-C(i)+3*B(i));
x2n(i)=(x2(i)-A(i)-C(i)-D(i));
x3n(i)=(x3(i)-B(i));
X1(i)=(x1n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X2(i)=(x2n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X3(i)=(x3n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X4(i)=(A(i)+D(i)+E(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X5(i)=(C(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X6(i)=(A(i)-B(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X7(i)=(B(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X12(i)=(E(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X16(i)=(C(i)+D(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
end
yc4=X4./(X4+X5);
yc5=X5./(X4+X5);
yc6=X6./(X6+X7+X12+X16);
yc7=X7./(X6+X7+X12+X16);
yc12=X12./(X6+X7+X12+X16);
yc16=X16./(X6+X7+X12+X16);
alpha14=yc6.*alpha18+yc7.*alpha19+yc12.*alpha113+yc16.*alpha117;
alpha24=yc6.*alpha28+yc7.*alpha29+yc12.*alpha213+yc16.*alpha217;
alpha34=yc6.*alpha38+yc7.*alpha39+yc12.*alpha313+yc16.*alpha317;
alpha15=yc6.*alpha110+yc7.*alpha115+yc12.*alpha114+yc16.*alpha118;
alpha25=yc6.*alpha210+yc7.*alpha215+yc12.*alpha214+yc16.*alpha218;
alpha35=yc6.*alpha310+yc7.*alpha315+yc12.*alpha314+yc16.*alpha318;
alpha16=yc4.*alpha81+yc5.*alpha101;
alpha26=yc4.*alpha82+yc5.*alpha102;
alpha36=yc4.*alpha83+yc5.*alpha103;
alpha17=yc4.*alpha91+yc5.*alpha151;
alpha27=yc4.*alpha92+yc5.*alpha152;
alpha37=yc4.*alpha93+yc5.*alpha153;
alpha112=yc4.*alpha131+yc5.*alpha141;
alpha212=yc4.*alpha132+yc5.*alpha142;
alpha312=yc4.*alpha133+yc5.*alpha143;
alpha116=yc4.*alpha171+yc5.*alpha181;
alpha216=yc4.*alpha172+yc5.*alpha182;
alpha316=yc4.*alpha173+yc5.*alpha183;
alpha46=yc5.*alpha810;
alpha47=yc5.*alpha915;
alpha412=yc5.*alpha1314;
alpha416=yc5.*alpha1718;
alpha56=yc4.*alpha108;
alpha57=yc4.*alpha159;
alpha512=yc4.*alpha1413;
alpha516=yc4.*alpha1817;
G14=yc6.*G18+yc7.*G19+yc12.*G113+yc16.*G117;
G24=yc6.*G28+yc7.*G29+yc12.*G213+yc16.*G217;
G34=yc6.*G38+yc7.*G39+yc12.*G313+yc16.*G317;
G15=yc6.*G110+yc7.*G115+yc12.*G114+yc16.*G118;
G25=yc6.*G210+yc7.*G215+yc12.*G214+yc16.*G218;
G35=yc6.*G310+yc7.*G315+yc12.*G314+yc16.*G318;
G16=yc4.*G81+yc5.*G101;
G26=yc4.*G82+yc5.*G102;
G36=yc4.*G83+yc5.*G103;
G17=yc4.*G91+yc5.*G151;
G27=yc4.*G92+yc5.*G152;
G37=yc4.*G93+yc5.*G153;
G112=yc4.*G131+yc5.*G141;
G212=yc4.*G132+yc5.*G142;
G312=yc4.*G133+yc5.*G143;
G116=yc4.*G171+yc5.*G181;
G216=yc4.*G172+yc5.*G182;
G316=yc4.*G173+yc5.*G183;
G46=yc5.*G810;
G47=yc5.*G915;
G412=yc5.*G1314;
G416=yc5.*G1718;
G56=yc4.*G108;
G57=yc4.*G159;
G512=yc4.*G1413;
G516=yc4.*G1817;
tau14=-log(G14)./alpha14;
tau24=-log(G24)./alpha24;
tau34=-log(G34)./alpha34;
tau15=-log(G15)./alpha15;
tau25=-log(G25)./alpha25;
tau35=-log(G35)./alpha35;
tau16=-log(G16)./alpha16;
tau26=-log(G26)./alpha26;
tau36=-log(G36)./alpha36;
tau17=-log(G17)./alpha17;
tau27=-log(G27)./alpha27;
tau37=-log(G37)./alpha37;
tau112=-log(G112)./alpha112;
tau212=-log(G212)./alpha212;
tau312=-log(G312)./alpha312;
tau116=-log(G116)./alpha116;
tau216=-log(G216)./alpha216;
tau316=-log(G316)./alpha316;
tau46=-log(G46)./alpha46;
tau47=-log(G47)./alpha47;
tau412=-log(G412)./alpha412;
tau416=-log(G416)./alpha416;
tau56=-log(G56)./alpha56;
tau57=-log(G57)./alpha57;
tau512=-log(G512)./alpha512;
tau516=-log(G516)./alpha516;
ln_y1_1=G12.*X2.*tau12+ G31.*X3.*tau13+ G14.*X4.*tau14+G15.*X5.*tau15+G16.*X6.*tau16+G17.*X7.*tau17+G112.*X12.*tau112+G116.*X16.*tau116;
ln_y1_2=G12.*X2+ G13.*X3+ G14.*X4+G15.*X5+G16.*X6+G17.*X7+G112.*X12+G116.*X16;
ln_y2_1=G21.*X1.*tau12+ G32.*X3.*tau32+ G24.*X4.*tau24+G25.*X5.*tau25+G26.*X6.*tau26+G27.*X7.*tau27+G212.*X12.*tau212+G216.*X16.*tau216;
ln_y2_2=G12.*X1+ G23.*X3+G24.*X4+G25.*X5+G26.*X6+G27.*X7+G212.*X12+G216.*X16;
ln_y3_1=G13.*X1.*tau13+ G23.*X3.*tau23+ G34.*X4.*tau34+G35.*X5.*tau35+G36.*X6.*tau36+G37.*X7.*tau37+G312.*X12.*tau312+G316.*X16.*tau316;
ln_y3_2=G13.*X1+ G23.*X3+ G34.*X4+G35.*X5+G36.*X6+G37.*X7+G312.*X12+G316.*X16;
ln_y4_1=G14.*X1.*tau14+G24.*X2.*tau24+G34.*X3.*tau34+G46.*X6.*tau46+G47.*X7.*tau47+G412.*X12.*tau412+G416.*X16.*tau416;
ln_y4_2=G14.*X1+G24.*X2+G34.*X3+G46.*X6+G47.*X7+G412.*X12+G416.*X16;
ln_y5_1=G15.*X1.*tau15+G25.*X2.*tau25+G35.*X3.*tau35+G56.*X6.*tau56+G57.*X7.*tau57+G512.*X12.*tau512+G516.*X16.*tau516;
ln_y5_2=G15.*X1+G25.*X2+G35.*X3+G56.*X6+G57.*X7+G512.*X12+G516.*X16;
ln_y6_1=G16.*X1.*tau16+G26.*X2.*tau26+G36.*X3.*tau36+G46.*X4.*tau46+G56.*X5.*tau56;
ln_y6_2=G16.*X1+G26.*X2+G36.*X3+G46.*X4+G56.*X5;
ln_y7_1=G17.*X1.*tau17+G27.*X2.*tau27+G37.*X3.*tau37+G47.*X4.*tau47+G57.*X5.*tau57;
ln_y7_2=G17.*X1+G27.*X2+G37.*X3+G47.*X4+G57.*X5;
ln_y12_1=G112.*X1.*tau112+G212.*X2.*tau212+G312.*X3.*tau312+G412.*X4.*tau412+G512.*X5.*tau512;
ln_y12_2=G112.*X1+G212.*X2+G312.*X3+G412.*X4+G512.*X5;
ln_y16_1=G116.*X1.*tau116+G216.*X2.*tau216+G316.*X3.*tau316+G416.*X4.*tau416+G516.*X5.*tau516;
ln_y16_2=G116.*X1+G216.*X2+G316.*X3+G416.*X4+G516.*X5;
ln_y1_3=(((X2.*G12)./ln_y2_2).*(tau12-(ln_y2_1)./(ln_y2_2)))+(((X3.*G13)./ln_y3_2).*(tau13-(ln_y3_1)./(ln_y3_2)));
ln_y1_4=(((X6.*G16)./ln_y6_2).*(tau16- (ln_y6_1./ln_y6_2))) + (((X7.*G17)./ln_y7_2).*(tau17- (ln_y7_1./ln_y7_2)))+(((X12.*G12)./ln_y12_2).*(tau112- (ln_y12_1./ln_y12_2)))+(((X16.*G16)./ln_y16_2).*(tau116- (ln_y16_1./ln_y16_2)));
ln_y1_5=(((X4.*G14)./ln_y4_2).*(tau14- (ln_y4_1./ln_y4_2))) + (((X5.*G15)./ln_y5_2).*(tau15- (ln_y5_1./ln_y5_2)));
yk=exp((ln_y1_1./ln_y1_2) + ln_y1_3 + ln_y1_4+ ln_y1_5)'; % activity coefficient for H2O
if any(~isfinite(yk))
yk = 10 * ones(size(yk));
end
Spotify Codes are little barcodes that allow you to share songs, artists, users, playlists, etc.
They encode information in the different heights of the "bars". There are 8 discrete heights that the 23 bars can be, which means 8^23 different possible barcodes.
Spotify generates barcodes based on their URI schema. This URI spotify:playlist:37i9dQZF1DXcBWIGoYBM5M gets mapped to this barcode:
The URI has a lot more information (62^22) in it than the code. How would you map the URI to the barcode? It seems like you can't simply encode the URI directly. For more background, see my "answer" to this question: https://stackoverflow.com/a/62120952/10703868
The patent explains the general process, this is what I have found.
This is a more recent patent
When using the Spotify code generator the website makes a request to https://scannables.scdn.co/uri/plain/[format]/[background-color-in-hex]/[code-color-in-text]/[size]/[spotify-URI].
Using Burp Suite, when scanning a code through Spotify the app sends a request to Spotify's API: https://spclient.wg.spotify.com/scannable-id/id/[CODE]?format=json where [CODE] is the media reference that you were looking for. This request can be made through python but only with the [TOKEN] that was generated through the app as this is the only way to get the correct scope. The app token expires in about half an hour.
import requests
head={
"X-Client-Id": "58bd3c95768941ea9eb4350aaa033eb3",
"Accept-Encoding": "gzip, deflate",
"Connection": "close",
"App-Platform": "iOS",
"Accept": "*/*",
"User-Agent": "Spotify/8.5.68 iOS/13.4 (iPhone9,3)",
"Accept-Language": "en",
"Authorization": "Bearer [TOKEN]",
"Spotify-App-Version": "8.5.68"}
response = requests.get('https://spclient.wg.spotify.com:443/scannable-id/id/26560102031?format=json', headers=head)
print(response)
print(response.json())
Which returns:
<Response [200]>
{'target': 'spotify:playlist:37i9dQZF1DXcBWIGoYBM5M'}
So 26560102031 is the media reference for your playlist.
The patent states that the code is first detected and then possibly converted into 63 bits using a Gray table. For example 361354354471425226605 is encoded into 010 101 001 010 111 110 010 111 110 110 100 001 110 011 111 011 011 101 101 000 111.
However the code sent to the API is 6875667268, I'm unsure how the media reference is generated but this is the number used in the lookup table.
The reference contains the integers 0-9 compared to the gray table of 0-7 implying that an algorithm using normal binary has been used. The patent talks about using a convolutional code and then the Viterbi algorithm for error correction, so this may be the output from that. Something that is impossible to recreate whithout the states I believe. However I'd be interested if you can interpret the patent any better.
This media reference is 10 digits however others have 11 or 12.
Here are two more examples of the raw distances, the gray table binary and then the media reference:
1.
022673352171662032460
000 011 011 101 100 010 010 111 011 001 100 001 101 101 011 000 010 011 110 101 000
67775490487
2.
574146602473467556050
111 100 110 001 110 101 101 000 011 110 100 010 110 101 100 111 111 101 000 111 000
57639171874
edit:
Some extra info:
There are some posts online describing how you can encode any text such as spotify:playlist:HelloWorld into a code however this no longer works.
I also discovered through the proxy that you can use the domain to fetch the album art of a track above the code. This suggests a closer integration of Spotify's API and this scannables url than previously thought. As it not only stores the URIs and their codes but can also validate URIs and return updated album art.
https://scannables.scdn.co/uri/800/spotify%3Atrack%3A0J8oh5MAMyUPRIgflnjwmB
Your suspicion was correct - they're using a lookup table. For all of the fun technical details, the relevant patent is available here: https://data.epo.org/publication-server/rest/v1.0/publication-dates/20190220/patents/EP3444755NWA1/document.pdf
Very interesting discussion. Always been attracted to barcodes so I had to take a look. I did some analysis of the barcodes alone (didn't access the API for the media refs) and think I have the basic encoding process figured out. However, based on the two examples above, I'm not convinced I have the mapping from media ref to 37-bit vector correct (i.e. it works in case 2 but not case 1). At any rate, if you have a few more pairs, that last part should be simple to work out. Let me know.
For those who want to figure this out, don't read the spoilers below!
It turns out that the basic process outlined in the patent is correct, but lacking in details. I'll summarize below using the example above. I actually analyzed this in reverse which is why I think the code description is basically correct except for step (1), i.e. I generated 45 barcodes and all of them matched had this code.
1. Map the media reference as integer to 37 bit vector.
Something like write number in base 2, with lowest significant bit
on the left and zero-padding on right if necessary.
57639171874 -> 0100010011101111111100011101011010110
2. Calculate CRC-8-CCITT, i.e. generator x^8 + x^2 + x + 1
The following steps are needed to calculate the 8 CRC bits:
Pad with 3 bits on the right:
01000100 11101111 11110001 11010110 10110000
Reverse bytes:
00100010 11110111 10001111 01101011 00001101
Calculate CRC as normal (highest order degree on the left):
-> 11001100
Reverse CRC:
-> 00110011
Invert check:
-> 11001100
Finally append to step 1 result:
01000100 11101111 11110001 11010110 10110110 01100
3. Convolutionally encode the 45 bits using the common generator
polynomials (1011011, 1111001) in binary with puncture pattern
110110 (or 101, 110 on each stream). The result of step 2 is
encoded using tail-biting, meaning we begin the shift register
in the state of the last 6 bits of the 45 long input vector.
Prepend stream with last 6 bits of data:
001100 01000100 11101111 11110001 11010110 10110110 01100
Encode using first generator:
(a) 100011100111110100110011110100000010001001011
Encode using 2nd generator:
(b) 110011100010110110110100101101011100110011011
Interleave bits (abab...):
11010000111111000010111011110011010011110001...
1010111001110001000101011000010110000111001111
Puncture every third bit:
111000111100101111101110111001011100110000100100011100110011
4. Permute data by choosing indices 0, 7, 14, 21, 28, 35, 42, 49,
56, 3, 10..., i.e. incrementing 7 modulo 60. (Note: unpermute by
incrementing 43 mod 60).
The encoded sequence after permuting is
111100110001110101101000011110010110101100111111101000111000
5. The final step is to map back to bar lengths 0 to 7 using the
gray map (000,001,011,010,110,111,101,100). This gives the 20 bar
encoding. As noted before, add three bars: short one on each end
and a long one in the middle.
UPDATE: I've added a barcode (levels) decoder (assuming no errors) and an alternate encoder that follows the description above rather than the equivalent linear algebra method. Hopefully that is a bit more clear.
UPDATE 2: Got rid of most of the hard-coded arrays to illustrate how they are generated.
The linear algebra method defines the linear transformation (spotify_generator) and mask to map the 37 bit input into the 60 bit convolutionally encoded data. The mask is result of the 8-bit inverted CRC being convolutionally encoded. The spotify_generator is a 37x60 matrix that implements the product of generators for the CRC (a 37x45 matrix) and convolutional codes (a 45x60 matrix). You can create the generator matrix from an encoding function by applying the function to each row of an appropriate size generator matrix. For example, a CRC function that add 8 bits to each 37 bit data vector applied to each row of a 37x37 identity matrix.
import numpy as np
import crccheck
# Utils for conversion between int, array of binary
# and array of bytes (as ints)
def int_to_bin(num, length, endian):
if endian == 'l':
return [num >> i & 1 for i in range(0, length)]
elif endian == 'b':
return [num >> i & 1 for i in range(length-1, -1, -1)]
def bin_to_int(bin,length):
return int("".join([str(bin[i]) for i in range(length-1,-1,-1)]),2)
def bin_to_bytes(bin, length):
b = bin[0:length] + [0] * (-length % 8)
return [(b[i]<<7) + (b[i+1]<<6) + (b[i+2]<<5) + (b[i+3]<<4) +
(b[i+4]<<3) + (b[i+5]<<2) + (b[i+6]<<1) + b[i+7] for i in range(0,len(b),8)]
# Return the circular right shift of an array by 'n' positions
def shift_right(arr, n):
return arr[-n % len(arr):len(arr):] + arr[0:-n % len(arr)]
gray_code = [0,1,3,2,7,6,4,5]
gray_code_inv = [[0,0,0],[0,0,1],[0,1,1],[0,1,0],
[1,1,0],[1,1,1],[1,0,1],[1,0,0]]
# CRC using Rocksoft model:
# NOTE: this is not quite any of their predefined CRC's
# 8: number of check bits (degree of poly)
# 0x7: representation of poly without high term (x^8+x^2+x+1)
# 0x0: initial fill of register
# True: byte reverse data
# True: byte reverse check
# 0xff: Mask check (i.e. invert)
spotify_crc = crccheck.crc.Crc(8, 0x7, 0x0, True, True, 0xff)
def calc_spotify_crc(bin37):
bytes = bin_to_bytes(bin37, 37)
return int_to_bin(spotify_crc.calc(bytes), 8, 'b')
def check_spotify_crc(bin45):
data = bin_to_bytes(bin45,37)
return spotify_crc.calc(data) == bin_to_bytes(bin45[37:], 8)[0]
# Simple convolutional encoder
def encode_cc(dat):
gen1 = [1,0,1,1,0,1,1]
gen2 = [1,1,1,1,0,0,1]
punct = [1,1,0]
dat_pad = dat[-6:] + dat # 6 bits are needed to initialize
# register for tail-biting
stream1 = np.convolve(dat_pad, gen1, mode='valid') % 2
stream2 = np.convolve(dat_pad, gen2, mode='valid') % 2
enc = [val for pair in zip(stream1, stream2) for val in pair]
return [enc[i] for i in range(len(enc)) if punct[i % 3]]
# To create a generator matrix for a code, we encode each row
# of the identity matrix. Note that the CRC is not quite linear
# because of the check mask so we apply the lamda function to
# invert it. Given a 37 bit media reference we can encode by
# ref * spotify_generator + spotify_mask (mod 2)
_i37 = np.identity(37, dtype=bool)
crc_generator = [_i37[r].tolist() +
list(map(lambda x : 1-x, calc_spotify_crc(_i37[r].tolist())))
for r in range(37)]
spotify_generator = 1*np.array([encode_cc(crc_generator[r]) for r in range(37)], dtype=bool)
del _i37
spotify_mask = 1*np.array(encode_cc(37*[0] + 8*[1]), dtype=bool)
# The following matrix is used to "invert" the convolutional code.
# In particular, we choose a 45 vector basis for the columns of the
# generator matrix (by deleting those in positions equal to 2 mod 4)
# and then inverting the matrix. By selecting the corresponding 45
# elements of the convolutionally encoded vector and multiplying
# on the right by this matrix, we get back to the unencoded data,
# assuming there are no errors.
# Note: numpy does not invert binary matrices, i.e. GF(2), so we
# hard code the following 3 row vectors to generate the matrix.
conv_gen = [[0,1,0,1,1,1,1,0,1,1,0,0,0,1]+31*[0],
[1,0,1,0,1,0,1,0,0,0,1,1,1] + 32*[0],
[0,0,1,0,1,1,1,1,1,1,0,0,1] + 32*[0] ]
conv_generator_inv = 1*np.array([shift_right(conv_gen[(s-27) % 3],s) for s in range(27,72)], dtype=bool)
# Given an integer media reference, returns list of 20 barcode levels
def spotify_bar_code(ref):
bin37 = np.array([int_to_bin(ref, 37, 'l')], dtype=bool)
enc = (np.add(1*np.dot(bin37, spotify_generator), spotify_mask) % 2).flatten()
perm = [enc[7*i % 60] for i in range(60)]
return [gray_code[4*perm[i]+2*perm[i+1]+perm[i+2]] for i in range(0,len(perm),3)]
# Equivalent function but using CRC and CC encoders.
def spotify_bar_code2(ref):
bin37 = int_to_bin(ref, 37, 'l')
enc_crc = bin37 + calc_spotify_crc(bin37)
enc_cc = encode_cc(enc_crc)
perm = [enc_cc[7*i % 60] for i in range(60)]
return [gray_code[4*perm[i]+2*perm[i+1]+perm[i+2]] for i in range(0,len(perm),3)]
# Given 20 (clean) barcode levels, returns media reference
def spotify_bar_decode(levels):
level_bits = np.array([gray_code_inv[levels[i]] for i in range(20)], dtype=bool).flatten()
conv_bits = [level_bits[43*i % 60] for i in range(60)]
cols = [i for i in range(60) if i % 4 != 2] # columns to invert
conv_bits45 = np.array([conv_bits[c] for c in cols], dtype=bool)
bin45 = (1*np.dot(conv_bits45, conv_generator_inv) % 2).tolist()
if check_spotify_crc(bin45):
return bin_to_int(bin45, 37)
else:
print('Error in levels; Use real decoder!!!')
return -1
And example:
>>> levels = [5,7,4,1,4,6,6,0,2,4,3,4,6,7,5,5,6,0,5,0]
>>> spotify_bar_decode(levels)
57639171874
>>> spotify_barcode(57639171874)
[5, 7, 4, 1, 4, 6, 6, 0, 2, 4, 3, 4, 6, 7, 5, 5, 6, 0, 5, 0]
everyone,
I'm trying to do some calculations and plot the results, but it seems that these are too heavy for Maxima. When I try to calculate N1 and N2 the program crashes when parameter j is too high or when I try to plot them, the program displays the following error message: "Heap exhausted, game over." What should I do? I've seen some people saying to try to compile Maxima with ccl, but I don't know how to do it or if it will work.
I usually receive error messages like:
Message from maxima's stderr stream: Heap exhausted during garbage collection: 0 bytes available, 16 requested.
Gen Boxed Unboxed LgBox LgUnbox Pin Alloc Waste Trig WP GCs Mem-age
0 0 0 0 0 0 0 0 20971520 0 0 0,0000
1 0 0 0 0 0 0 0 20971520 0 0 0,0000
2 0 0 0 0 0 0 0 20971520 0 0 0,0000
3 16417 2 0 0 43 1075328496 707088 293986768 16419 1 0,8032
4 13432 21 0 1141 70 955593760 838624 2000000 14594 0 0,2673
5 0 0 0 0 0 0 0 2000000 0 0 0,0000
6 741 184 34 28 0 63259792 1424240 2000000 987 0 0,0000
7 0 0 0 0 0 0 0 2000000 0 0 0,0000
Total bytes allocated = 2094182048
Dynamic-space-size bytes = 2097152000
GC control variables:
*GC-INHIBIT* = true
*GC-PENDING* = true
*STOP-FOR-GC-PENDING* = false
fatal error encountered in SBCL pid 13884(tid 0000000001236360):
Heap exhausted, game over.
Here goes the code:
enter code here
a: 80$;
b: 6*a$;
h1: 80$;
t: 2$;
j: 5$;
carga: 250$;
sig: -carga/2$;
n: 2*q*%pi/b$;
m: i*%pi/a$;
i: 2*p-1$;
i1: 2*p1-1$;
/*i1: p1$;*/
Φ: a/b$;
τ: cosh(x) - (x/sinh(x))$;
σ: sinh(x) - (x/cosh(x))$;
Ψ: sinh(x)/τ$;
Χ: cosh(x)/σ$;
Λ0: 1/(((i/2)^2+Φ^2*q^2)^2)$;
Λ1: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ0, p, 1, j)$;
Λ2: sum(((q1^3*subst([x=(q1*%pi*Φ)],Χ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ1, q1, 1, j)$;
Λ3: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ2, p, 1, j)$;
Λ4: sum(((q1^3*subst([x=(q1*%pi*Φ)],Χ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ3, q1, 1, j)$;
Λ5: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ4, p, 1, j)$;
Ζ0: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ0, q, 1, j)$;
Ζ2: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ2, q, 1, j)$;
Ζ4: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ4, q, 1, j)$;
E: 200000$;
ν: 0.3$;
λ: (ν*E)/((1+ν)*(1-2*ν))$;
μ: E/(2*(1+ν))$;
a0: float(1/(b/2)*integrate(0, y, -(b/2), -h1/2)+1/b*integrate(sig, y, -h1/2, h1/2)+1/(b/2)*integrate(0, y, h1/2, (b/2)))$;
aq: float(1/(b/2)*integrate(0*cos(q*y*%pi/(b/2)), y, -(b/2), - h1/2)+1/(b/2)*integrate(sig*cos(q*y*%pi/(b/2)), y, -h1/2, h1/2)+1/(b/2)*integrate(0*cos(q*y*%pi/(b/2)), y, h1/2, (b/2)))$;
aq1: float(1/(b/2)*integrate(0*cos(q1*y*%pi/(b/2)), y, -(b/2), - h1/2)+1/(b/2)*integrate(sig*cos(q1*y*%pi/(b/2)), y, -h1/2, h1/2)+1/(b/2)*integrate(0*cos(q1*y*%pi/(b/2)), y, h1/2, (b/2)))$;
Bq: aq/((λ+μ)*subst([x=q*%pi*Φ],σ))+((16*Φ^4*q^2*(-1)^q)/((λ+μ)*%pi^2*subst([x=q*%pi*Φ],σ)))*sum(q1*aq1*(-1) ^q1*subst([x=q1*%pi*Φ],Χ)*(Λ1+(16*Φ^4/(%pi^2))*Λ3+((16*Φ^4/(%pi^2))^2)*Λ5), q1, 1, j)+(8*λ*Φ^3*q^2*(-1)^q*a0)/((λ+μ)*(λ+2*μ)*(%pi^3)*subst([x=q*%pi*Φ],σ))*sum(subst([x=i*%pi/(2*Φ)],Ψ)/(i/ 2)*(Λ0+(16*Φ^4/(%pi^2))*Λ2+((16*Φ^4/(%pi^2))^2)*Λ4), p, 1, j)$;
βp: -(2*λ*a0*(-1)^((i-1)/2))/((λ+μ)*(λ+2*μ)*(i/2)^2*%pi^2*subst([x=i*%pi/(2*Φ)],τ))-((32*λ*Φ^4*(i/2)^2*a0*(-1)^((i-1)/2))/((λ+μ)*(λ+2*μ)*%pi^2*subst([x=i*%pi/(2*Φ)],τ)))*sum(((subst([x=i1*%pi/(2*Φ)],Ψ))/(i1/2))*(Ζ0+Ζ2*((16*Φ^4)/%pi^2)+Ζ4*(((16*Φ^4)/%pi^2)^2)),p1,1,j)-((4*Φ*(i/2)^2*(-1)^((i-1)/2))/((λ+μ)*%pi*subst([x=i*%pi/(2*Φ)],τ)))*sum(q*aq*(-1)^q*subst([x=q*%pi*Φ],Χ)*(Λ0+Λ2*(16*Φ^4/%pi^2)+Λ4*(16*Φ^4/%pi^2)^2),q,1,j)$;
N1: (2*a0/a)*x+(λ+μ)*sum(Bq*((1+((n*a*sinh(n*a/2))/(2*cosh(n*a/2))))*sinh(n*x)-n*x*cosh(n*x))*cos(n*y),q,1,j)+(λ+μ)*sum(βp*((1-((m*b*cosh(m*b/2))/(2*sinh(m*b/2))))*cosh(m*y)+m*y*sinh(m*y))*sin(m*x),p,1,j)$;
N2: ((2*λ*a0)/(a*(λ+2*μ)))*x+(λ+μ)*sum(Bq*((1-((n*a*sinh(n*a/2))/(2*cosh(n*a/2))))*sinh(n*x)+n*x*cosh(n*x))*cos(n*y),q,1,j)+(λ+μ)*sum(βp*((1+((m*b*cosh(m*b/2))/(2*sinh(m*b/2))))*cosh(m*y)-m*y*sinh(m*y))*sin(m*x),p,1,j);
wxplot3d(N1, [x,-a/2,a/2], [y,-b/2,b/2])$;
wxplot3d(N2, [x,-a/2,a/2], [y,-b/2,b/2])$;
This is not a complete answer, since I don't know how this should work with wxMaxima: I would suggest that you ask the developers. However it's too long for a comment and I think might be useful to people, and it does answer the question of how you solve the heap-size limit for Maxima itself when using SBCL, at least when run on Linux or some other platform with a command-line.
As a note, I suspect that the underlying problem is not the heap size, but that the calculation is blowing up in some horrible way: the best fix is probably to understand what's blowing up and fix that. See Robert Dodier's answer, which is probably going to be a lot more helpful. However, if heap size is the problem, this is how you deal with it for Maxima.
The trick is that you can tell SBCL what the heap limit should be by passing it the --dynamic-space-size <MB> argument, and you can pas arguments through the maxima wrapper to do this.
Here is a transcript of Maxima, being run on Linux, with SBCL as a back end (this is a version built from source: the packaged version will I assume be the same):
$ maxima
Maxima 5.43.2 http://maxima.sourceforge.net
using Lisp SBCL 2.0.0
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) :lisp (sb-ext:dynamic-space-size)
1073741824
So, on this system the defaule heap limit is 1GB (this is SBCL's default limit on the platform).
Now we can pass the -X <lisp options> aka --lisp-options=<lisp options> option to the maxima wrapper to pass the appropriate option through to sbcl:
$ maxima -X '--dynamic-space-size 2000'
Lisp options: (--dynamic-space-size 2000)
Maxima 5.43.2 http://maxima.sourceforge.net
using Lisp SBCL 2.0.0
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) :lisp (sb-ext:dynamic-space-size)
2097152000
As you can see this has doubled the heap size.
If someone knows the answer for wxMaxima then please do add an edit to this answer: I can't experiment it because all my Linux VMs are headless.
Also not a complete answer here, but some more notes and pointers which I hope will help.
To make the problem easier for Maxima to digest, use only exact numbers (integers and ratios), and avoid float and numer. (Plotting functions will apply float and numer automatically.) I changed 0.3 to 3/10 and cut out the calls to float.
Also, try setting j to a smaller number (I tried j equal to 1) to try to work all the way through the problem before increasing it to 5 again.
Also, replace all sum and integrate with 'sum and 'integrate (i.e. noun expressions instead of verb expressions). Take a look at the summands and integrands to see if they look right. You can evaluate the sums and/or integrals or both via ev(expr, sum) or ev(expr, integrate) or ev(expr, nouns) to evaluate 'sum, 'integrate, or all noun expressions, respectively.
With j equal to 1, I get the following expression for N1:
(2500000*((-(13*cosh(%pi/6)
*((8503056*cosh(%pi/6)^2*sinh(3*%pi)^2)
/(9765625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(52488*cosh(%pi/6)*sinh(3*%pi))
/(15625*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+324/25))
/(120000*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))))
+(13*sinh(3*%pi)
*((2754990144*cosh(%pi/6)^3*sinh(3*%pi)^2)
/(244140625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^3
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(17006112*cosh(%pi/6)^2*sinh(3*%pi))
/(390625*%pi^2
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+(104976*cosh(%pi/6))
/(625*(sinh(%pi/6)-%pi/(6*cosh(%pi/6))))))
/(22680000*%pi^2*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+13/(35000*%pi^2*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))))
*sin((%pi*(2*p-1)*x)/80)
*((%pi*(2*p-1)*y*sinh((%pi*(2*p-1)*y)/80))/80
+(1-(3*%pi*(2*p-1)*cosh(3*%pi*(2*p-1)))
/sinh(3*%pi*(2*p-1)))
*cosh((%pi*(2*p-1)*y)/80)))
/13
+(2500000*((-(13*cosh(%pi/6)
*((344373768*cosh(%pi/6)^2*sinh(3*%pi)^3)
/(244140625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))
^3)
+(2125764*cosh(%pi/6)*sinh(3*%pi)^2)
/(390625*%pi^2
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(13122*sinh(3*%pi))
/(625*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))))
/(1620000*%pi^3*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2))
+(13*sinh(3*%pi)
*((8503056*cosh(%pi/6)^2*sinh(3*%pi)^2)
/(9765625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(52488*cosh(%pi/6)*sinh(3*%pi))
/(15625*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+324/25))
/(3780000*%pi^3*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
-13/(20000*%pi*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))))
*(((%pi*sinh(%pi/6))/(6*cosh(%pi/6))+1)
*sinh((%pi*x)/240)
-(%pi*x*cosh((%pi*x)/240))/240)*cos((%pi*y)/240))
/13-(25*x)/48$
Now in order to plot that, it should be a function of x and y only. However listofvars reports that it contains x, y, and p. Hmm. I see that βp has a summation over p1 but it contains Ζ0, which contains Λ0, which contains p. Is the summation over p1 supposed to be over p? Is the summand supposed to contain p1 instead of p?
Likewise it appears that N2, after evaluating the sums and integrals with j equal to 1, contains p.
Maybe you need to rework the formulas somewhat? I don't know what the correct form might be.
I have a text output from a program with a set format. I need to parse ~200 of them to extract an information. I tried in MATLAB with 'textscan' but did not work. Following is the input:
MOTIFS SUMMARY:
1) TTATAGCCGC (GCGGCTATAA) 1.986
2) AAACCGCCTC (GAGGCGGTTT) 1.865
DETAILED RESULTS:
1) TTATAGCCGC (GCGGCTATAA) 1.986
Matrix: MAT1 TTATAGCCGC
A 0.1249 0.177 0.7364 0.1189 0.7072 0.1149 0.09858 0.1096
C 0.0899 0.07379 0.1136 0.1298 0.08662 0.1293 0.7528 0.721
G 0.06828 0.1284 0.07195 0.1031 0.1352 0.6708 0.05556 0.0713
T 0.7169 0.6209 0.07802 0.6482 0.07096 0.08492 0.09305 0.09804
OCCURRENCES:
>GENE_1 1 TTATAGCCGC 1 561 +
>GENE_2 24 TAATAGCCGC 0.928699 762 -
>GENE_3 10 ATATAGCCGC 0.904905 185 -
>GENE_1 7 TTATAGCAGC 0.901785 726 +
**********
2) AAACCGCCTC (GAGGCGGTTT) 1.865
Matrix: MAT2 AAACCGCCTC
A 0.653 0.7401 0.7763 0.1323 0.09619 0.09134 0.07033 0.1383
C 0.1163 0.07075 0.09441 0.749 0.6347 0.1132 0.6559 0.6982
G 0.09136 0.09402 0.07385 0.04209 0.1799 0.7332 0.1241 0.07568
T 0.1393 0.09518 0.05541 0.07659 0.08921 0.06234 0.1497 0.08786
OCCURRENCES:
>GENE_1 21 AAACCGCCTC 1 963 +
>GENE_2 14 AAACGGCCTC 0.928198 212 +
>GENE_2 8 AAACCGTCTC 0.92009 170 +
>GENE_4 3 TAACCGCCTC 0.918883 370 +
**********
I am trying to count the unique() occurrence under each motif and add it to the MOTIF SUMMARY and a final average of them. My expected output is:
MOTIFS SUMMARY:
1) TTATAGCCGC (GCGGCTATAA) 1.986 3
2) AAACCGCCTC (GAGGCGGTTT) 1.865 3
AVERAGE OCCURRENCE: 3
For motif 1, unique occurrence is 3 (GENE_1, GENE_2, GENE_3). Similarly for motif 2, it is again 3 (GENE_1, GENE_2, GENE_4)
How can I use OCCURRENCES and ****** as blocks ? so that, I can regexp GENE_x to store it and count.
Kindly help.
Thanks,
AP
You better try to change the original text file so that it will be legal matlab m file code, then just use 'eval' function to run it .
Most of the job will be to find where to insert '=' and '[' ']' and '%' for ignore parts.
If all files are identical in format than it will be easy.
I have a CSV file 'XPQ12.csv' of futures tick data in the following form:
20090312 30:14.0 717.25 1 E
20090312 30:15.0 718.47 1 E
20090312 30:17.0 717.25 1 E
20090312 30:32.0 718.42 1 E
20090312 30:49.0 715.32 1 E
20090312 30:58.0 717.57 1 E
20090312 31:06.0 716.65 3 E
20090312 31:12.0 718.35 2 E
20090312 31:45.0 721.14 1 E
20090312 31:52.0 719.24 1 E
20090312 32:11.0 717.02 6 E
20090312 32:29.0 717.14 1 E
20090312 32:35.0 717.34 1 E
20090312 32:55.0 717.26 1 E
(The first column is the yearmonthdate, the second column is the minute:second:tenthofsecond, the third column is the price, the fourth column is the number of contracts traded, and the fifth indicates if the trade was electronic or in a pit). In my actual data set, I may have thousands of price quotes within any given minute.
I read the file using the following code:
fid = fopen('C:\Program Files\MATLAB\R2013a\XPQ12.csv','r');
[c] = fscanf(fid, '%d,%d:%d.%d,%f,%d,%c')
Which outputs:
20090312
30
14
0
717.25
1
69
20090312
30
15
0
718.47
3
69
.
.
.
(the 69s are the matlab representation for E I believe)
Now I want to cut this up into one minute ohlc bars, so that for each minute, I record what the first, highest, lowest, and last price was within that minute. I'd really like to know the best way to go about this.
My original idea was to store the sequence of minutes in a vector d, and while working through the data, each time the number at the end of d changed I would record the corresponding price as an open, record the previous price as a close for the last bar, and find the largest and smallest prices within each open and close.
c(2) is the first minute, so I said:
d(1)=c(2);
and then noting that I'd always be counting by 7 before getting to the next minute, I said:
Nrows = numel(textread('XPQ12.csv','%1c%*[^\n]')); % counts rows in file
for i=1:Nrows
if mod(i-2,7)== 0;
d(end+1)=c(i);
end
end
which should fill up d with all the minutes:
30
30
30
30
30
30
31
31
31
31
32
32
32
32
in the case of the example data. I'm kind of lost what to do from here, or if what I'm doing is on the right track.
From where you are:
Minutes = c(2:7:end);
MinuteValues=unique(Minutes);
Prices = c(5:7:end);
if (length(Prices)>length(Minutes))
Prices=Prices(1:length(Minutes));
elseif (length(Prices)<length(Minutes))
Minutes=Minutes(1:length(Prices));
OverflowValues=1+find(Minutes(2:end)==0 & Minutes(1:end-1)==59);
for v=length(OverflowValues):-1:1
Minutes(OverflowValues(v):end)=Minutes(OverflowValues(v):end)+60;
end
Highs=zeros(1,length(MinuteValues));
Lows=zeros(1,length(MinuteValues));
First=zeros(1,length(MinuteValues));
Last=zeros(1,length(MinuteValues));
for v=1:length(MinuteValues)
Highs(v) = max(Prices(Minutes==MinuteValues(v)));
Lows(v) = min(Prices(Minutes==MinuteValues(v)));
First(v) = Prices(find(Minutes==MinuteVales(v),1,'first'));
Last(v) = Prices(find(Minutes==MinuteVales(v),1,'last'));
end
Using textread would make this easier for you, as mentioned.
(If you are lost at this stage, I wouldn't find accumarray as mentioned in the comments is the best place to start!)
By the way, this is assuming that minutes increases above 60 and you don't have hours in there somewhere. Otherwise this won't work at all.