淺談壓縮感知(二十三):壓縮感知重構(gòu)算法之壓縮采樣匹配追蹤(CoSaMP)
淺談壓縮感知(二十三):壓縮感知重構(gòu)算法之壓縮采樣匹配追蹤(CoSaMP)
主要內(nèi)容:
CoSaMP的算法流程CoSaMP的MATLAB實(shí)現(xiàn)一維信號(hào)的實(shí)驗(yàn)與結(jié)果測(cè)量數(shù)M與重構(gòu)成功概率關(guān)系的實(shí)驗(yàn)與結(jié)果一、CoSaMP的算法流程
壓縮采樣匹配追蹤(CompressiveSampling MP)是D. Needell繼ROMP之后提出的又一個(gè)具有較大影響力的重構(gòu)算法。CoSaMP也是對(duì)OMP的一種改進(jìn),每次迭代選擇多個(gè)原子,除了原子的選擇標(biāo)準(zhǔn)之外,它有一點(diǎn)不同于ROMP:ROMP每次迭代已經(jīng)選擇的原子會(huì)一直保留,而CoSaMP每次迭代選擇的原子在下次迭代中可能會(huì)被拋棄。
二、CS_CoSaMP的MATLAB實(shí)現(xiàn)(CS_CoSaMP.m)
function?[?theta?]?=?CS_CoSaMP(?y,A,K?) %???CS_CoSaOMP %???Detailed?explanation?goes?here %???y?=?Phi?*?x %???x?=?Psi?*?theta %????y?=?Phi*Psi?*?theta %???令?A?=?Phi*Psi,?則y=A*theta %???K?is?the?sparsity?level %???現(xiàn)在已知y和A,求theta %???Reference:Needell?D,Tropp?J?A.CoSaMP:Iterative?signal?recovery?from %???incomplete?and?inaccurate?samples[J].Applied?and?Computation?Harmonic? %???Analysis,2009,26:301-321. ????[m,n]?=?size(y); ????if?m<n ????????y?=?y';?%y?should?be?a?column?vector ????end ????[M,N]?=?size(A);?%傳感矩陣A為M*N矩陣 ????theta?=?zeros(N,1);?%用來(lái)存儲(chǔ)恢復(fù)的theta(列向量) ????pos_num?=?[];?%用來(lái)迭代過(guò)程中存儲(chǔ)A被選擇的列序號(hào) ????res?=?y;?%初始化殘差(residual)為y ????for?kk=1:K?%最多迭代K次 ????????%(1)?Identification ????????product?=?A'*res;?%傳感矩陣A各列與殘差的內(nèi)積 ????????[val,pos]=sort(abs(product),'descend'); ????????Js?=?pos(1:2*K);?%選出內(nèi)積值最大的2K列 ????????%(2)?Support?Merger ????????Is?=?union(pos_num,Js);?%Pos_theta與Js并集 ????????%(3)?Estimation ????????%At的行數(shù)要大于列數(shù),此為最小二乘的基礎(chǔ)(列線(xiàn)性無(wú)關(guān)) ????????if?length(Is)<=M ????????????At?=?A(:,Is);?%將A的這幾列組成矩陣At ????????else?%At的列數(shù)大于行數(shù),列必為線(xiàn)性相關(guān)的,At'*At將不可逆 ????????????if?kk?==?1 ????????????????theta_ls?=?0; ????????????end ????????????break;?%跳出for循環(huán) ????????end ????????%y=At*theta,以下求theta的最小二乘解(Least?Square) ????????theta_ls?=?(At'*At)^(-1)*At'*y;?%最小二乘解 ????????%(4)?Pruning ????????[val,pos]=sort(abs(theta_ls),'descend'); ????????%(5)?Sample?Update ????????pos_num?=?Is(pos(1:K)); ????????theta_ls?=?theta_ls(pos(1:K)); ????????%At(:,pos(1:K))*theta_ls是y在At(:,pos(1:K))列空間上的正交投影 ????????res?=?y?-?At(:,pos(1:K))*theta_ls;?%更新殘差? ????????if?norm(res)<1e-6?%Repeat?the?steps?until?r=0 ????????????break;?%跳出for循環(huán) ????????end ????end ????theta(pos_num)=theta_ls;?%恢復(fù)出的theta end
三、一維信號(hào)的實(shí)驗(yàn)與結(jié)果
%壓縮感知重構(gòu)算法測(cè)試
clear?all;close?all;clc;
M?=?64;?%觀測(cè)值個(gè)數(shù)
N?=?256;?%信號(hào)x的長(zhǎng)度
K?=?12;?%信號(hào)x的稀疏度
Index_K?=?randperm(N);
x?=?zeros(N,1);
x(Index_K(1:K))?=?5*randn(K,1);?%x為K稀疏的,且位置是隨機(jī)的
Psi?=?eye(N);?%x本身是稀疏的,定義稀疏矩陣為單位陣x=Psi*theta
Phi?=?randn(M,N);?%測(cè)量矩陣為高斯矩陣
A?=?Phi?*?Psi;?%傳感矩陣
y?=?Phi?*?x;?%得到觀測(cè)向量y
%%?恢復(fù)重構(gòu)信號(hào)x
tic
theta?=?CS_CoSaMP(?y,A,K?);
x_r?=?Psi?*?theta;?%?x=Psi?*?theta
toc
%%?繪圖
figure;
plot(x_r,'k.-');?%繪出x的恢復(fù)信號(hào)
hold?on;
plot(x,'r');?%繪出原信號(hào)x
hold?off;
legend('Recovery','Original')
fprintf('n恢復(fù)殘差:');
norm(x_r-x)?%恢復(fù)殘差四、測(cè)量數(shù)M與重構(gòu)成功概率關(guān)系的實(shí)驗(yàn)與結(jié)果
clear?all;close?all;clc;
%%?參數(shù)配置初始化
CNT?=?1000;?%對(duì)于每組(K,M,N),重復(fù)迭代次數(shù)
N?=?256;?%信號(hào)x的長(zhǎng)度
Psi?=?eye(N);?%x本身是稀疏的,定義稀疏矩陣為單位陣x=Psi*theta
K_set?=?[4,12,20,28,36];?%信號(hào)x的稀疏度集合
Percentage?=?zeros(length(K_set),N);?%存儲(chǔ)恢復(fù)成功概率
%%?主循環(huán),遍歷每組(K,M,N)
tic
for?kk?=?1:length(K_set)
????K?=?K_set(kk);?%本次稀疏度
????M_set?=?2*K:5:N;?%M沒(méi)必要全部遍歷,每隔5測(cè)試一個(gè)就可以了
????PercentageK?=?zeros(1,length(M_set));?%存儲(chǔ)此稀疏度K下不同M的恢復(fù)成功概率
????for?mm?=?1:length(M_set)
???????M?=?M_set(mm);?%本次觀測(cè)值個(gè)數(shù)
???????fprintf('K=%d,M=%dn',K,M);
???????P?=?0;
???????for?cnt?=?1:CNT?%每個(gè)觀測(cè)值個(gè)數(shù)均運(yùn)行CNT次
????????????Index_K?=?randperm(N);
????????????x?=?zeros(N,1);
????????????x(Index_K(1:K))?=?5*randn(K,1);?%x為K稀疏的,且位置是隨機(jī)的????????????????
????????????Phi?=?randn(M,N)/sqrt(M);?%測(cè)量矩陣為高斯矩陣
????????????A?=?Phi?*?Psi;?%傳感矩陣
????????????y?=?Phi?*?x;?%得到觀測(cè)向量y
????????????theta?=?CS_CoSaMP(y,A,K);?%恢復(fù)重構(gòu)信號(hào)theta
????????????x_r?=?Psi?*?theta;?%?x=Psi?*?theta
????????????if?norm(x_r-x)<1e-6?%如果殘差小于1e-6則認(rèn)為恢復(fù)成功
????????????????P?=?P?+?1;
????????????end
???????end
???????PercentageK(mm)?=?P/CNT*100;?%計(jì)算恢復(fù)概率
????end
????Percentage(kk,1:length(M_set))?=?PercentageK;
end
toc
save?CoSaMPMtoPercentage1000?%運(yùn)行一次不容易,把變量全部存儲(chǔ)下來(lái)
%%?繪圖
S?=?['-ks';'-ko';'-kd';'-kv';'-k*'];
figure;
for?kk?=?1:length(K_set)
????K?=?K_set(kk);
????M_set?=?2*K:5:N;
????L_Mset?=?length(M_set);
????plot(M_set,Percentage(kk,1:L_Mset),S(kk,:));%繪出x的恢復(fù)信號(hào)
????hold?on;
end
hold?off;
xlim([0?256]);
legend('K=4','K=12','K=20','K=28','K=36');
xlabel('Number?of?measurements(M)');
ylabel('Percentage?recovered');
title('Percentage?of?input?signals?recovered?correctly(N=256)(Gaussian)');




