请教一个关于维序路由的问题

31 views
Skip to first unread message

da wang

unread,
Feb 19, 2020, 4:15:54 AM2/19/20
to Network-on-Chip Research in China
维序路由是公认的无死锁路由方法,有一个问题我一直不太明白,对于一个2x2网络,如果发包方向即存在环,比如(0,0)-->(0,1) (0,1)-->(1,1) (1,1)-->(1,0) (1,0)-->(0,0),这样还是会由于对于buffer的循环依赖而形成死锁,不知道为什么说维序路由无死锁呢?
谢谢

Wei Song

unread,
Feb 20, 2020, 8:05:54 AM2/20/20
to noc...@googlegroups.com
假设你的坐标是(x,y),然后是先x再y。那么,(0,1)->(1,1)和(1,1)->(1,0)之间不会存在依赖关系,同理(1,0)->(0,0)和(0,0)->(0,1)之间也不存在。


On 2020/2/19 17:15, da wang wrote:
维序路由是公认的无死锁路由方法,有一个问题我一直不太明白,对于一个2x2网络,如果发包方向即存在环,比如(0,0)-->(0,1) (0,1)-->(1,1) (1,1)-->(1,0) (1,0)-->(0,0),这样还是会由于对于buffer的循环依赖而形成死锁,不知道为什么说维序路由无死锁呢?
谢谢
--
--
You receive this email because you have subscribed the Google group “Network-on-Chip@CN”.
You can visit the group via https://groups.google.com/group/noc_cn
New discussion can be launched by an email to noc...@googlegroups.com
Group modulator: benjamin...@gmail.com

---
您收到此邮件是因为您订阅了Google网上论坛上的“Network-on-Chip Research in China”群组。
要退订此群组并停止接收此群组的电子邮件,请发送电子邮件到noc_cn+un...@googlegroups.com
要在网络上查看此讨论,请访问https://groups.google.com/d/msgid/noc_cn/8fd0b79d-4997-4df6-b1ce-4a7389b94fba%40googlegroups.com

-- 
Dr. Wei Song
Associate Professor
Institute of Information Engineering, CAS
C3 YiYuan 80 XingShiKou Road
Haidian Beijing 100195 China
Reply all
Reply to author
Forward
0 new messages