Table of Contents

7.19.18 Inverse Problem

The mean field theory (nMF):

$\mathbf{J}^\text{nMF}=-\mathbf{C}^{-1}, (i \neq j) $,

$h_{i}^\text{nMF}=\tanh^{-1}\left< s_{i} \right>-\sum_{j=1}^{N} J{ij}^\text{nMF}\left< s_{j} \right> $ .

The independent-pair theory (IP):

$J_{ij}^\text{pair}=0.25*\ln\left[ \frac{(1+m_{i}+m_{j}+C_{ij}^{*}) (1-m_{i}-m_{j}+C_{ij}^{*})} {(1-m_{i}+m_{j}-C_{ij}^{*})(1+m_{i}-m_{j}-C_{ij}^{*})} \right]$, with $C_{ij}^{*}=C_{ij}+m_{i}m_{j} , (i \neq j)$

$h_{i}^\text{pair}=0.5*\ln\left( \frac{1+m_{i}} {1-m_{i}} \right)-\sum_{j=1}^{N} J_{ij}^\text{pair} m_{j}$.

The Thouless-Anderson-Palmer theory (TAP):

$\mathbf{C}^{-1}+J_{ij}^\text{TAP}+2(J_{ij}^\text{TAP})^{2}m_{i}m_{j}=0 , (i \neq j)$

$h_{i}^\text{TAP}=h_{i}^\text{nMF}-m_{i}\sum_{i=1}^{N} (J_{ij}^\text{TAP})^{2} (1-m_{j})^{2}$.

The Sessak and Monasson theory (SM):

$J_{ij}^\text{SM}=J_{ij}^\text{nMF}+J_{ij}^\text{pair}-\frac{C_{ij}}{(1-m_{i}^{2})(1-m_{j}^{2})-C_{ij}^{2}} , (i \neq j)$

$h_{i}^\text{SM}=0.5*\ln\left( \frac{1+m_{i}} {1-m_{i}} \right)-\sum_{j=1}^{N} J_{ij}^\text{SM} m_{j}$.

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

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

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(0.0,0.10) , 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 DATA of Inverse Problem

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 0.01 0.96 0.88 0.67
IP 0.11 -1.85 10.41 -43.71
SM 0.02 0.89 1.57 -0.02
TAP 0.01 0.97 1.11 0.49

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 0.03 0.74 0.03 0.99
IP 0.04 0.66 0.11 0.99
SM 0.03 0.74 0.02 0.99
TAP 0.03 0.74 0.05 0.99

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 0.03 0.76 1.28 -30.39
IP 0.24 -7.94 12.13 -2810.72
SM 0.07 0.11 3.75 -268.62
TAP 0.03 0.82 1.67 -52.67

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 0.04 0.74 0.03 0.97
IP 0.04 0.66 0.09 0.83
SM 0.04 0.74 0.03 0.97
TAP 0.04 0.74 0.07 0.90

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 0.01 0.96 0.59 0.18
IP 0.20 -2.96 6.28 -89.71
SM 0.02 0.92 0.60 0.16
TAP 0.01 0.97 0.89 -0.84

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 0.05 0.75 0.04 0.99
IP 0.05 0.66 0.11 0.97
SM 0.05 0.74 0.04 0.99
TAP 0.05 0.74 0.09 0.98

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 0.33 -50.21 7.59 -77.94
IP 0.20 -17.71 8.27 -92.61
SM 0.27 -34.67 10.06 -137.77
TAP 0.04 -0.04 0.12 0.97

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 0.06 0.75 0.05 0.99
IP 0.07 0.68 0.07 0.99
SM 0.07 0.74 0.05 0.99
TAP 0.06 0.75 0.11 0.98

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 0.10 0.76 0.49 0.71
IP 0.26 -0.42 0.94 -0.03
SM 0.05 0.93 0.17 0.96
TAP 0.08 0.85 0.91 0.03

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 0.10 0.76 0.07 0.99
IP 0.11 0.71 0.10 0.98
SM 0.11 0.74 0.06 0.99
TAP 0.10 0.76 0.13 0.97

7.30.18 DATA of Inverse Problem

11-24-18 Gaussian in spins(-1,1) distributions

We consider the cases in the beta=1.0.

The external fields are h~n(0.1,0.1)

The coupling distributions are case 1: J~ 0.8*n(-0.2,0.1) +0.1*n(0,0.2)+0.1*n(0.5,0.5)

case2: J~ 0.5*n(-0.1,0.1) +0.3*n(0,0.3)+0.2*n(0.5,0.5)

case3: J~ 1*n(0,0.1)

case4: J~ 0.5*n(-0.2,0.1) +0.5*n(0.2,0.1)

case5: J~ 1*uniform(-0.15,0.1)