This supplementary material contains
I.
Comparison with Monte Carlo simulations
This section compares the mean square displacement (MSD) and probability density
function (PDF) obtained from the analytical model with that of Monte Carlo simulations.
Table S1. shows MSD values from analytical and Monte Carlo method for volume occupancy
of 0.4 and taking biasing parameter a=0.7.
N
10
Table S1.
Analytical (MSD)
2.9796
Monte Carlo simulation (MSD)
2.9795
50
100
150
200
260
13.2653
26.1224
38.9795
51.8367
67.2653
13.2662
26.1221
38.9710
51.8150
67.2280
Figure S1. Shows probability distribution plotted for random walk of length N=100, obtained
from analytical and Monte Carlo simulations at volume occupancy of 0.4 and biasing parameter
a=0.7.
Figure S2. Shows probability distribution plotted for random walk of length N=260, obtained
from analytical and Monte Carlo simulations at volume occupancy of 0.4 and biasing parameter
a=0.7.
II: Figures related to variation of probability with volume fraction.
Figure S3: Figure shows variation of probabilities with volume fraction at different positions for
random walk of step length (N) 10 for a fixed biasing parameter a=0.7.
Figure S4: Figure shows variation of mean square displacement (MSD) with volume fraction, at
biasing parameter a=0.7, taking contributions from positions starting from ±1.
Figure S5: Figure shows variation of probabilities with biasing parameter at different positions
for random walk of step N=10 at volume occupancy of 0.4.
Figure S6: Figure shows variation of mean square displacement (MSD) with biasing
parameters, at volume occupancy of 0.4, taking contributions from positions starting from ±1.
III.
Table of Probability Distribution Function
Table S2. Shows errors in parameters while fitting Gaussian distribution for walks of length N=10,
50, 100.
 = 0.4; P(R|L)=0.7*(1-  ); N=10, 50, 100
Table S2.
N
Error in c
Error in sd
10
0.9920 (0.4095%)
1.6762 (0.4732%)
50
0.9975(0.0462%)
3.6106(0.0534%)
100
0.9986 (0.0165%)
5.0880(0.0190%)
From the table S2. it can be seen that percentage error decreases with increase in N. The
distribution reaches to Gaussian with increase in N.
Table S3. Shows errors in parameters while fitting Gaussian distribution for walk of length N=50 at
different values of biasing parameter.
N=50;  =0.4; P(R|L)=a*(1-  ) where a=0.5, 0.7, 0.9
Table S3.
N=50
Error in c
Error in sd
P(R|L)= 0.5*(1-  )
1.0009 (0.0169%)
5.4956(0.0196%)
P(R|L)= 0.7*(1-  )
0.9975 (0.0462%)
3.6106(0.0534%)
P(R|L)= 0.9*(1-  )
0.9920 (0.1625%)
1.8975(0.1878%)
From the table S3. it can be seen that percentage error increases with increase in biasing parameter.
The distribution reaches close to Gaussian with decrease in biasing parameter.
Table S4. Shows errors in parameters while fitting Gaussian distribution for a walk of length N=50
at different values of volume fraction (  )
N=50; P(R|L)=0.7*(1-  );  =0.2, 0.4, 0.6
Table S4.
N=50
Error in c
Error in sd
 =0.2
0.9985(0.0255%)
4.1692(0.0294%)
 =0.4
0.9975(0.0461%)
3.6106(0.0534%)
 =0.6
0.9956 (0.0867%)
2.9488(0.1002%)
From the table S4. it can be seen that percentage error increases with increase in volume fraction.
The distribution reaches close to normal with decrease in volume fraction.
IV.
Plot of Kurtosis of the probability distribution function
Figure S7. Figure shows kurtosis values varying with walk length (N) for a particular value of
biasing parameter and volume fraction.
Figure also shows the effect of biasing probability and volume fraction on the kurtosis. Both the
factors increase the kurtosis with increase in their values. However, the quantitative effect of
both of them is not same.
V.
Velocity autocorrelation function
The velocity autocorrelation function has been calculated from the Monte Carlo trajectories at
different values of volume fraction and biasing parameters taken in the analytical model. For
normalizing the autocorrelation function we have not considered the correlation between the
moves when either or both entries are zero.
Figure S8. Figure shows velocity autocorrelation function plotted against N at different values
of volume fraction and biasing parameter.
From the figure we can see the negative correlation between the velocities at time difference of
1 step with increase in walk length (N). The negative correlation is because of the biasing which
gives more probability to motion in opposite direction of the previous. The correlation in each
case decays to zero showing the onset of normal diffusion. From the figure it is also seen that
when volume fraction is kept at 0.4 and biasing parameter has been increased from 0.7 to 0.9,
the correlation becomes more and more negative. The more the correlation is negative more
will be subdiffusive behavior. However for a particular value of biasing parameter increasing
volume fraction increases the time of convergence to pure normal diffusion.
VI.
Derivation of Magnitude biased walk for different cases (meaning the symbols are
same as in the main text).
Let flip1 and flip2 be the maximum number of flips in right direction and left direction
respectively. Let combination1 and combination2 be the maximum number of sequences
corresponding to number of flips in right and left direction respectively.
case1: If flip1>0 and flip2>0
flip1flip2
P=  
RF+RH +LF+LH
i =1 j=1
C RF+RH comb1i comb2  j0.251  φ
2 a 1  φ i+ j b1  φ RF+RH +LF+LHi j2 N CSφ
case2: If flip1=0 and flip2>0 and (RF>0 || RH>0)
flip2
P 
RF  RH  LF LH
j1
C RF RH comb2(j) * (0.25 * (1  φ)) 2 (a(1  φ)) j (b(1  φ)) RF RH  LF LH j2 N C S φ S
case3: If flip1=0 and flip2>0 and (RF=0 and RH=0)
flip2
P 
RF  RH  LF LH
j1
C RF RH comb2(j)(0 .25(1  φ)) (a(1  φ)) j (b(1  φ)) RF RH  LF LH j1 N C S φ S
case4: If flip1>0 and flip2=0 and (LF>0 || LH>0)
flip1
P 
RF  RH  LF LH
i 1
C RF RH comb1(i)(0 .25(1  φ)) 2 (a(1  φ)) i (b(1  φ)) RF RH  LF LHi 2 N C S φ S
case5: If flip1>0 and flip2=0 and (LF=0 and LH=0)
flip1
P 
i 1
RF  RH  LF LH
C RF RH comb1(i)(0 .25 * (1  φ)) (a(1  φ)) i (b(1  φ)) RF RH  LF LHi 1 N C S φ S
case6: If flip1=0 and flip2=0 and (RF >0 || RH>0) and (LF>0 || LH>0)
S
P= RF+RH +LF+LH C RF+RH 0.251  φ
2 b1  φ RF+RH+LF+LH2 N CS φ
S
case7: If flip1=0 and flip2=0 and (RF>0 || RH>0) and (LF=0 and LH=0)
P= RF+RH +LF+LH C RF+RH 0.251  φ
b1  φ RF+RH+LF+LH1 N CSφS
case8: If flip1=0 and flip2=0 and (LF>0 || LH>0) and (RF=0 and RH=0)
P= RF+RH +LF+LH C RF+RH 0.251  φ
case9: If S=N
P  φS
b1  φ RF+RH+LF+LH1 N CS φ
S