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★OceanKTV 應用程式，隨點隨唱，打造專屬 KTV 環境?
★Linux Station 支援 Ubuntu? 正式授權作業系統，可搭配 HDMI 輸出桌面?
★支援網路虛擬儲存擴充櫃 (VJBOD)
?
QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器
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標題:
演算法題目求詳解
發問:
V. Pan has discovered a way of multiplying 68X 68 matrices using 132,464 multiplications, a way of multiplying 70 X70 matrices using 143,640 multiplications, and a way of multiplying 72 X72 matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer... 顯示更多 V. Pan has discovered a way of multiplying 68X 68 matrices using 132,464 multiplications, a way of multiplying 70 X70 matrices using 143,640 multiplications, and a way of multiplying 72 X72 matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? Compare it with the running time for Strassen's algorithm.
最佳解答:
SOLUTION: The strassen's divide-and-conquer algorithm for matrix multiplication has the recurrence for its time complexity (considering only the counts of scalar multiplications): T(n) = 7 T(n/2), with the solution T(n) = O(n^log(2,7)) = O(n^2.807355...) If, just like Strassen's algorithm, we divide an nxn matrix into smaller matrices (recursively), but each with size (n/68)x(n/68), then when two 68x68 matrices can be multiplied by 132464 scalar multiplications, we'll have the recurrence T(n) = 132464 T(n/68). Similarly, the other two ways will have their recurrence: T(n) = 143640 T(n/70) and T(n) = 155424 T(n/72), respectively. The solutions are (respectively) (with big O sign omitted) n^log(68, 132464) = n^2.795128..., n^log(70, 143640) = n^2.795122..., and n^log(72, 155424) = n^2.795147.... So, the 2nd method is the fastest (in theory), and all of the three Victor Y. Pan's algorithms are better than Strassen's. Notes: (1) I used log(b, v) to mean "log v with base b". (2) The solution to the recurrence T(n) = a T(n/c) is n^(log(c, a)) in our case, which is covered by the "Master Theorem". See Theorem 3.1 on http://chern.ie.nthu.edu.tw/alg2003/Suppl_3_rec-sol.PDF or http://en.wikipedia.org/wiki/Master_theorem. 圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_35.gif
其他解答:
34F8E429322BE148
平時我自己上網搜尋資料就還蠻喜歡看 QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器
的
因為可以一網打盡真的是太方便!!!!!
就算沒買過肯定逛過聽過看過 QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器
吧!!!
QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器
功能:
QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器
描述:
★搭載 Intel? Celeron? N3060 雙核心 1.6GHz 處理器?
★硬體加密引擎?，AES-NI?
★4K 影片硬體解碼，影片線上即時轉檔與離線背景轉檔?
★內建 USB QuickAccess 接口隨插即用，不用網路也可存取資料?
★OceanKTV 應用程式，隨點隨唱，打造專屬 KTV 環境?
★Linux Station 支援 Ubuntu? 正式授權作業系統，可搭配 HDMI 輸出桌面?
★支援網路虛擬儲存擴充櫃 (VJBOD)
?
QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器
相關 QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器
商品推薦
標題:
演算法題目求詳解
發問:
V. Pan has discovered a way of multiplying 68X 68 matrices using 132,464 multiplications, a way of multiplying 70 X70 matrices using 143,640 multiplications, and a way of multiplying 72 X72 matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer... 顯示更多 V. Pan has discovered a way of multiplying 68X 68 matrices using 132,464 multiplications, a way of multiplying 70 X70 matrices using 143,640 multiplications, and a way of multiplying 72 X72 matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? Compare it with the running time for Strassen's algorithm.
最佳解答:
SOLUTION: The strassen's divide-and-conquer algorithm for matrix multiplication has the recurrence for its time complexity (considering only the counts of scalar multiplications): T(n) = 7 T(n/2), with the solution T(n) = O(n^log(2,7)) = O(n^2.807355...) If, just like Strassen's algorithm, we divide an nxn matrix into smaller matrices (recursively), but each with size (n/68)x(n/68), then when two 68x68 matrices can be multiplied by 132464 scalar multiplications, we'll have the recurrence T(n) = 132464 T(n/68). Similarly, the other two ways will have their recurrence: T(n) = 143640 T(n/70) and T(n) = 155424 T(n/72), respectively. The solutions are (respectively) (with big O sign omitted) n^log(68, 132464) = n^2.795128..., n^log(70, 143640) = n^2.795122..., and n^log(72, 155424) = n^2.795147.... So, the 2nd method is the fastest (in theory), and all of the three Victor Y. Pan's algorithms are better than Strassen's. Notes: (1) I used log(b, v) to mean "log v with base b". (2) The solution to the recurrence T(n) = a T(n/c) is n^(log(c, a)) in our case, which is covered by the "Master Theorem". See Theorem 3.1 on http://chern.ie.nthu.edu.tw/alg2003/Suppl_3_rec-sol.PDF or http://en.wikipedia.org/wiki/Master_theorem. 圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_35.gif
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