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@Astroneko404
Created October 13, 2019 21:51
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Astroneko404 commented Oct 13, 2019

eq1
eq5
eq4
eq3
eq2
CodeCogsEqn
20191013171934
CodeCogsEqn
CodeCogsEqn(1)

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Astroneko404 commented Oct 14, 2019

CodeCogsEqn(2)
CodeCogsEqn
CodeCogsEqn(1)
CodeCogsEqn(2)
CodeCogsEqn(3)
CodeCogsEqn
CodeCogsEqn(1)

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Astroneko404 commented Oct 15, 2019

CodeCogsEqn(6)
CodeCogsEqn(7)
CodeCogsEqn(8)
CodeCogsEqn(9)
CodeCogsEqn(10)
CodeCogsEqn(11)
CodeCogsEqn(12)
CodeCogsEqn(13)
CodeCogsEqn(14)

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Astroneko404 commented Nov 8, 2019

CodeCogsEqn(1)
CodeCogsEqn

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Astroneko404 commented Nov 20, 2019

network
MICH_10
SRI_4

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Astroneko404 commented Nov 26, 2019

20191126015033
20191126160508
20191126161942
20191126162319

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Astroneko404 commented Dec 4, 2019

20191203233807
20191203234115
20191204134138

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20191204173227

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bigbridge_720p

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20200724100852

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Astroneko404 commented Dec 23, 2020

pmi
20210112093243

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1
2
3
4
5
6
7
8
9

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Astroneko404 commented Apr 12, 2021

QQ截图20210414175630
QQ截图20210414175637

DataClustering_ElbowCriterion
3
1
2
4
QQ截图20210413110815
QQ截图20210413111156
QQ截图20210414175213

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20220608171128

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20220923-0

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0_LeK_gmCf3DfO3gj_

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20221002-2
20221002-1
20221002-0

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IMG_2838(20221122-232903)
IMG_2839(20221122-232917)
IMG_2840(20221122-232927)

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20221209123304

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Astroneko404 commented Feb 8, 2023

gamegrid

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Astroneko404 commented Aug 10, 2023

S(k_1, k_2) = \sum_{n_1=0}^{N_1-1}\sum_{n_2=0}^{N_2-1} s(n_1, n_2)g(n_1, n_2, k_1, k_2)
equation
image
image

s(n_1, n_2) = \sum_{k_1=0}^{N_1-1}\sum_{k_2=0}^{N_2-1} S(k_1, k_2)h(n_1, n_2, k_1, k_2)
image

g(n_1, n_2, k_1, k_2) &= exp(-j\frac{2\pi(n_1k_1+n_2k_2)}{N^2}) \
&=exp(-j\frac{2\pi n_1k_1}{N^2})exp(-j\frac{2\pi n_2k_2}{N^2})
image

S = \sum_{k_1=0}^{N-1}\sum_{k_2=0}^{N-1}S(k_1, k_2)H_{k_1, k_2}
equation
image

H_{k_1, k_2} =
\begin{bmatrix}
h(0, 0, k_1, k_2) & h(0, 1, k_1, k_2) & \dots & h(0, N-1, k_1, k_2) \
h(1, 0, k_1, k_2) & & & \vdots \
\vdots & & & \vdots \
h(N-1, 0, k_1, k_2) & \dots & \dots & h(N-1, N-1, k_1, k_2)
\end{bmatrix}
image
equation (1)

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Astroneko404 commented Aug 11, 2023

S(k_1, k_2) &= \sqrt{\frac{4}{N^2}} C(k_1)C(k_2) \sum_{n_1=0}^{N-1} \sum_{n_2=0}^{N-1}s(n_1, n_2) \
& \times{cos(\frac{\pi (2n_1+1)k_1}{2N})} \times{cos(\frac{\pi (2n_2+1)k_2}{2N})}
image

C(k)=
\begin{cases}
\frac{1}{\sqrt{2}},& \text{for } k=0\
1, & \text{otherwise}
\end{cases}
image

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Astroneko404 commented Aug 11, 2023

image
image

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image

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20230925213611

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IMG_3447(20230927-031136)

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Sound Debug Pokemon drawio

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istockphoto-1184661675-612x612

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my-image

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F9FA96CF42491275347F05E66B5297EC

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