Table of Contents

2019-09-16 :Case N1: long tail antiferromagentic

2018-07-2 :The solutions to the inverse Ising problem

Method: Boltzmann learning

Finding the external fields $h_{i}$ and the coupling parameters $J_{ij}$ can fit the same means and pairwise covariance as the experiment data, using the equations

$\left<s_{i}\right>_{Ising} = \left<s_{i}\right>_{data} $

$\left<s_{i}s_{j}\right>_{Ising} = \left<s_{i}s_{j}\right>_{data} $.

We can solve the equations by iterations $h_{i}=h_{i}+\delta h_{i}$ and $J_{ij}=J_{ij}+\delta J_{ij}$ then make changes

$\delta h_{i}=\eta \{\left<s_{i}\right>_{data} - \left<s_{i}\right>_{Ising} \} $

$\delta J_{ij}=\eta \{\left<s_{i}s_{j}\right>_{data} - \left<s_{i}s_{j}\right>_{Ising} \} $.

We use the data ACV_slice5 (6 neurons) as our sample and set $\eta = 0.1$. The cost function is

${(\left<s_{i}\right>_{data} - \left<s_{i}\right>_{Ising})}^{2} \leq 0.00000001$ && ${(\left<s_{i}s_{j}\right>_{data} - \left<s_{i}s_{j}\right>_{Ising})}^{2} \leq 0.00000001$.

The experiment result and simulation result see figure.

Mean $s_{i}$ 0.600000 0.628571 0.628571 0.600000 0.685714 0.457143
Simulation $s_{i}$ 0.590619 0.672644 0.628609 0.6095950.6476930.477471

If the number of neurons is larger than 30, it is very slow to find the suitable $h_{i}$ and $J_{ij}$ for the good fitting.

2018-07-2 :The solutions to the inverse Ising problem

7.2.18

7.2.18

2018-07-9 :The solutions to the inverse Ising problem

7.10.18

mean variance
J: 1*n(0.0,0.1) -0.789370 0.001288
J: 1*n(0.0,0.2) -0.914673 0.003162
J: 1*n(0.0,0.3) -0.901516 0.002877
J: 0.9*n(0.0,0.1)+0.1*n(0.5,0.5) -0.412155 0.000304
J: 0.9*n(0.0,0.2)+0.1*n(0.5,0.5) -0.513260 0.000366
J: 0.9*n(0.0,0.3)+0.1*n(0.5,0.5) -0.532969 0.000416
J: 0.8*n(0.0,0.1)+0.1*n(0.5,0.2)+0.1*n(1,0.2) -0.356512 0.000270
J: 0.8*n(0.0,0.2)+0.1*n(0.5,0.2)+0.1*n(1,0.2) -0.370494 0.000255
J: 0.8*n(0.0,0.3)+0.1*n(0.5,0.2)+0.1*n(1,0.2) -0.393506 0.000367
J: 0.9*n(0.0,0.1)+0.1*n(-0.5,0.5) -1.287893 0.003625
J: 0.9*n(0.0,0.2)+0.1*n(-0.5,0.5) -1.215945 0.004903
J: 0.9*n(0.0,0.3)+0.1*n(-0.5,0.5) -1.166773 0.003841

7.19.18 Inverse Problem

The coefficient of $R^{2}$ is:

$R^{2}=1-\frac{\sum_{ij} \left(J_{ij}^{\text{approx}}-J_{ij}^{\text{Exact}} \right)^{2} }{\sum_{ij} \left(J_{ij}^{\text{Exact}}-J_{ij}^{\text{$\overline{Exact}$}} \right)^{2}}$ , with

$\overline{J_{ij}^{\text{Exact}}}=\frac {\sum_{i \neq j} J_{ij}^{\text{Exact}}}{N(N-1)}$.

The rms error is: $\sqrt{\frac{1}{N(N-1)} \sum_{i \neq j} (J_{ij}^\text{approx}-J_{ij}^\text{Exact})^{2} }$.

h:n(0.0,0.1)&J:n(0.0,0.1) , n=20 at CT=0.2 rms error $R^{2}$
nMf 0.261099 -5.69658
IP 0.36969 -12.4251
SM 0.135638 -0.807203
TAP 0.225045 -3.97488

h:n(0.0,0.1)&J:n(0.0,0.1) ,n=20 at T=1 rms error $R^{2}$
nMf 0.0501364 0.753083
IP 0.0492352 0.761881
SM 0.0504632 0.749854
TAP 0.0501487 0.752963

h:n(0.0,0.2)&J:n(0.0,0.2) ,n=20 at CT=0.4 rms error $R^{2}$
nMf 0.186599 0.144925
IP 0.327701 -1.63717
SM 0.0808463 0.83949
TAP 0.146637 0.471956

h:n(0.0,0.2)&J:n(0.0,0.2) ,n=20 at T=1.0 rms error $R^{2}$
nMf 0.0989592 0.759511
IP 0.105152 0.72847
SM 0.102435 0.74232
TAP 0.0992857 0.757921

J:n(0.0,0.06) ,n=200 at T=0.4 rms error $R^{2}$ h:n(0.0,0.06) rms error $R^{2}$
nMf 0.0252805 0.820135 1.40724 -0.111763
IP 0.135993 -4.20489 14.1081 -110.741
SM 0.0429129 0.481737 3.8296 -7.23352
TAP 0.021821 0.865994 1.48845 -0.243794

J:n(0.0,0.06) ,n=200 at T=1.0 rms error $R^{2}$ h:n(0.0,0.06) rms error $R^{2}$
nMf 0.0299218 0.748029 0.0187765 0.999802
IP 0.033059 0.692421 0.0524186 0.998457
SM 0.0299849 0.746966 0.0184704 0.999808
TAP 0.0299236 0.747998 0.0354793 0.999293

7.24.18 Inverse Problem

h:n(0.0,0.07) & J:n(0.0,0.07), n=200 at T=1 $\sqrt{200}$
h:n(0.0,0.08) & J:n(0.0,0.08), n=150 at T=1 $\sqrt{150}$
h:n(0.0,0.1) & J:n(0.0,0.1) , n=100 at T=1 $\sqrt{100}$
h:n(0.0,0.14) & J:n(0.0,0.14), n=50 at T=1 $\sqrt{50} $
h:n(0.0,0.22) & J:n(0.0,0.22), n=20 at T=1 $\sqrt{20} $

We set the mean equal to zero and the variance of normal distribution of h and J from $\frac{1}{\sqrt{200}}$: $\frac{1}{\sqrt{150}}$: $\frac{1}{\sqrt{100}}$:$\frac{1}{\sqrt{50}}$:$\frac{1}{\sqrt{20}} \approx 0.07 : 0.08 : 0.1 : 0.14 : 0.22 $.

We use the normal distribution of h and J to get the coefficients of $R^{2}$ and the RMSerror at T=1.

7.24.18 Inverse Problem data

J:n(0.0,0.07) ,n=200 at CT=0.5 rms error $R^{2}$ h:n(0.0,0.07) rms error $R^{2}$
nMf 0.0123978 0.968219 0.884148 0.67757
IP 0.117457 -1.85258 10.4116 -43.7117
SM 0.0227896 0.892612 1.57309 -0.0206929
TAP 0.0107151 0.97626 1.11072 0.491141

J:n(0.0,0.07) ,n=200 at T=1.0 rms error $R^{2}$ h:n(0.0,0.07) rms error $R^{2}$
nMf 0.0348459 0.748936 0.0304221 0.999618
IP 0.0401123 0.667312 0.110992 0.994919
SM 0.0349555 0.747353 0.0295671 0.999639
TAP 0.0348501 0.748876 0.0582334 0.998601

J:n(0.0,0.08) ,n=150 at CT=0.4 rms error $R^{2}$ h:n(0.0,0.08) rms error $R^{2}$
nMf 0.0389121 0.766413 1.28183 -30.3925
IP 0.240864 -7.94997 12.1312 -2810.72
SM 0.0757059 0.115825 3.75663 -268.628
TAP 0.0338997 0.822715 1.67608 -52.6732

J:n(0.0,0.08) ,n=150 at T=1 rms error $R^{2}$ h:n(0.0,0.08) rms error $R^{2}$
nMf 0.0403046 0.749396 0.0368884 0.974002
IP 0.0464303 0.667431 0.0934541 0.833136
SM 0.0404719 0.747311 0.0355802 0.975813
TAP 0.0403135 0.749285 0.0720108 0.900926

J:n(0.0,0.1) ,n=100 at CT=0.5 rms error $R^{2}$ h:n(0.0,0.1) rms error $R^{2}$
nMf 0.0182768 0.96695 0.596538 0.18199
IP 0.200247 -2.96735 6.28192 -89.7127
SM 0.0280065 0.922396 0.603074 0.163966
TAP 0.0164129 0.973348 0.89636 -0.846922

J:n(0.0,0.1) ,n=100 at T=1 rms error $R^{2}$ h:n(0.0,0.1) rms error $R^{2}$
nMf 0.0502492 0.750181 0.0480846 0.994685
IP 0.0581101 0.665905 0.110959 0.971698
SM 0.0505891 0.74679 0.0450164 0.995342
TAP 0.0502829 0.749846 0.0932494 0.980012

J:n(0.0,0.14) ,n=50 at CT=0.2 rms error $R^{2}$ h:n(0.0,0.14) rms error $R^{2}$
nMf 0.331738 -50.2182 7.59468 -77.9424
IP 0.200528 -17.7147 8.27031 -92.6128
SM 0.276861 -34.6743 10.0695 -137.774
TAP 0.0472849 -0.0405847 0.12614 0.978223

J:n(0.0,0.14) ,n=50 at T=1 rms error $R^{2}$ h:n(0.0,0.14) rms error $R^{2}$
nMf 0.0697652 0.753093 0.0584891 0.995397
IP 0.0781925 0.689839 0.0785585 0.991696
SM 0.0705361 0.747606 0.054511 0.996002
TAP 0.0698489 0.752499 0.110346 0.983615

J:n(0.0,0.22) ,n=20 at CT=0.5 rms error $R^{2}$ h:n(0.0,0.22) rms error $R^{2}$
nMf 0.107295 0.766355 0.495427 0.715343
IP 0.264862 -0.423761 0.946502 -0.0389783
SM 0.0572563 0.933466 0.177233 0.963571
TAP 0.0846797 0.854469 0.910756 0.0380179

J:n(0.0,0.22) ,n=20 at T=1 rms error $R^{2}$ h:n(0.0,0.22) rms error $R^{2}$
nMf 0.107456 0.765655 0.0741133 0.99363
IP 0.117774 0.718488 0.103663 0.987537
SM 0.112377 0.743699 0.0658366 0.994973
TAP 0.10791 0.763667 0.132923 0.979509

7.30.18 Inverse Problem data