Sparse fast DCT for vectors with one-block support

Abstract

In this paper, we present a new fast and deterministic algorithm for the inverse discrete cosine transform of type II that reconstructs the vector $\mathbf {x}\in \mathbb {R}^{N}$, N = 2^J− 1, with short support of length m from its discrete cosine transform $\mathbf {x}^{\widehat {\text {II}}}=\mathbf {C}_N^{\text {II}}\mathbf {x}$, while no a priori knowledge of m is required. The resulting algorithm has a runtime of $\mathcal {O}\left (m\log m\log \frac {2N}{m}\right )$ and uses $\mathcal {O}\left (m\log \frac {2N}{m}\right )$ samples of $\mathbf {x}^{\widehat {\text {II}}}$. When applied to an arbitrary $\mathbf {x}^{\widehat {\text {II}}}\in \mathbb {R}^{N}$, the algorithm automatically recognizes and exploits a possible short support. Hence, its runtime and sampling complexity approach those of a regular IDCT-II for m → N. In order to derive this algorithm, we also develop a new fast and deterministic inverse FFT algorithm that reconstructs the vector $\mathbf {y}\in \mathbb {R}^{2N}$ with reflected block support of block length m from $\widehat {\mathbf {y}}$ with the same runtime and sampling complexities as our IDCT-II algorithm, while also computing m on the fly. The runtime and the numerical stability of both algorithms are also illustrated by numerical experiments.

Appendix: Proof of Lemma 1

• 1. Note that y is symmetric by construction,

  $$\mathbf{J}_{2N}\mathbf{y}=\mathbf{J}_{2N}\left( \begin{array}{ccccccc} \mathbf{x} \\ \mathbf{J}_{N}\mathbf{x} \end{array}\right) =\left( \begin{array}{cccc} \mathbf{x} \\ \mathbf{J}_{N}\mathbf{x} \end{array}\right)=\mathbf{y}. $$

• 2. Let k ∈{0,…,N − 1}. We find that

  $$\begin{array}{@{}rcl@{}} {{x}}^{\widehat{\text{II}}}_{k}&=&\sqrt{\frac{2}{N}}\varepsilon_{N}(k)\sum\limits_{l = 0}^{N-1}\cos\left( \frac{2\cdot k(2l + 1)\pi}{2\cdot2N}\right)x_{l} \\ &=&\frac{\varepsilon_{N}(k)}{\sqrt{2N}}\sum\limits_{l = 0}^{N-1}\left( \omega_{4N}^{k(2l + 1)}+\omega_{4N}^{-k(2l + 2-1)}\right)x_{l} \\ &=&\frac{\varepsilon_{N}(k)}{\sqrt{2N}}\omega_{4N}^{k}\sum\limits_{l = 0}^{N-1}\left( \omega_{2N}^{kl}+\omega_{2N}^{k(2N-1-l)}\right)x_{l} \\ &=&\frac{\varepsilon_{N}(k)}{\sqrt{2N}}\omega_{4N}^{k}\left( \sum\limits_{l = 0}^{N-1}\omega_{2N}^{kl}x_{l}+\sum\limits_{l^{\prime}=N}^{2N-1}\omega_{2N}^{kl^{\prime}}x_{2N-1-l^{\prime}}\right) =\frac{\varepsilon_{N}(k)}{\sqrt{2N}}\omega_{4N}^{k}\cdot\widehat{y}_{k}. \end{array} $$

• 3. For k ∈{0,…,N − 1}, the claim in (??) follows directly from (??). For k ∈{N,…, 2N − 1} the symmetry property, (??), implies that

  $$ \widehat{y}_{k}=\left( \widehat{\mathbf{J}_{2N}\mathbf{y}}\right)_{k}=\sum\limits_{l^{\prime}= 0}^{2N-1}\omega_{2N}^{kl^{\prime}}y_{2N-1-l^{\prime}} =\sum\limits_{l = 0}^{2N-1}\omega_{2N}^{k(2N-1-l)}y_{l} =\omega_{2N}^{-k}\sum\limits_{l = 0}^{2N-1}\omega_{2N}^{-kl}y_{l}. $$

  (1)

  If k = N, we obtain from (1) that

  $$\widehat{{y}}_{N}=\omega_{2N}^{-N}\sum\limits_{l = 0}^{2N-1}\omega_{2N}^{-Nl}y_{l} =-\sum\limits_{l = 0}^{2N-1}\omega_{2N}^{Nl}y_{l}=-\widehat{{y}}_{N}, $$

  so $\widehat {{y}}_{N}= 0$. If k ∈{N + 1,…, 2N − 1}, then 2N − k ∈{1,…,N − 1}, and (1) yields

  $$\begin{array}{@{}rcl@{}} \widehat{y}_{k}&=&\omega_{2N}^{-k}\sum\limits_{l = 0}^{2N-1}\omega_{2N}^{-kl} y_{l} =\omega_{2N}^{-k}\sum\limits_{l = 0}^{2N-1}\omega_{2N}^{(2N-k)l}y_{l} \\ &=&\omega_{2N}^{-k}\cdot\widehat{{y}}_{2N-k} =-\frac{\sqrt{2N}}{\varepsilon_{N}(2N-k)}\omega_{4N}^{-k}\cdot{{x}}^{\widehat{\text{II}}}_{2N-k}. \end{array} $$