function QT_parCorrVsRegression
% QT_parCorrVsRegression.m
% When comparing a reference RDM to a candidate RDM, we might want to
% partial out the contribution of another RDM on the correlation between
% the candidate and the reference RDMs.
% this could be done via partial correlations or linear regression.
% this scripts demonstrates that the two approaches are identical.
% clear;clc;close all
%% control parameters
nCond = 72;
nVox = 100;

%% simulating the RDMs
rdm_ref = pdist(randn(nCond,nVox))';
rdm_cand1 = pdist(randn(nCond,nVox))';
rdm_cand2 = pdist(randn(nCond,nVox))';
%% partial correlation
r1 = partialcorr(rdm_ref,rdm_cand1,rdm_cand2);
fprintf('partial correlation coefficient = %.3f \n',r1)

% for understanding it's better to code it up yourself:
% z score so linear fit is equivalent to pearson r
rdm_ref_z = zscore(rdm_ref);
rdm_cand1_z = zscore(rdm_cand1);
rdm_cand2_z = zscore(rdm_cand2);
% remove the fitted contribution of cand2 from data and cand1
projectout = @(y,c) y - (c * (c \ y));
rdm_cand1_zp = projectout(rdm_cand1_z,rdm_cand2_z);
rdm_ref_zp = projectout(rdm_ref_z,rdm_cand2_z);
% now partial r is
r1_alt = rdm_cand1_zp \ rdm_ref_zp;
fprintf('partial correlation coefficient (alt) = %.3f \n',r1)

% from the above it becomes pretty obvious that the reason why the below
% might not produce similar results is that you haven't Z-scored the
% predictors and data.

%% linear regression
% regress-out rdm2 from the reference rdm
b1 = regress(rdm_ref,rdm_cand2);
% you use b before defined here. Not sure if you were planning to use b or
% b1. Assume b1.
res_ref = rdm_ref-b1*rdm_cand2;
% regress-out rdm2 from the candidate rdm
b = regress(rdm_cand1,rdm_cand2);
res_cand = rdm_cand1-b*rdm_cand2;
r2 = corr(res_cand,res_ref);
fprintf('regression-based results = %.3f \n',r2)