Timeline for How to find the minimal path cost from left edge of a grid to the right edge using dynamic programming?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2017 at 17:49 | vote | accept | Yos | ||
| Dec 27, 2017 at 15:48 | answer | added | Marcelo Fornet | timeline score: 1 | |
| Dec 27, 2017 at 15:34 | comment | added | Marcelo Fornet | Actually the step 1 is not correct. The minimal path doesn't need to start in the cell with minimal cost. You need to run dynamic programming starting from all nodes on the leftmost edge at once. | |
| Dec 27, 2017 at 14:11 | history | edited | Yos | CC BY-SA 3.0 |
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| Dec 27, 2017 at 13:16 | comment | added | Yos | @Nehorai in this particular problem the index convention is different (which I admit is confusing). As stated in the OP i represents horizontal movement while j vertical | |
| Dec 27, 2017 at 13:14 | comment | added | Nehorai Elbaz | No, see here for example @Yos | |
| Dec 27, 2017 at 13:12 | comment | added | Yos | @Nehorai No $i$ represents columns | |
| Dec 27, 2017 at 13:10 | comment | added | Nehorai Elbaz | but $i$ is the rows indexes not columns so $(i,j)→(i+1,j)$ is down, no? | |
| Dec 27, 2017 at 13:06 | comment | added | Yos | @Nehorai yes, $i$ symbolizes horizontal movement so if $j$ doesn't change then we move to the right because $i$ increases | |
| Dec 27, 2017 at 12:28 | comment | added | Nehorai Elbaz | Are you sure that $(i,j)\to (i+1,j)$ is right? shouldn't be "down"? | |
| Dec 27, 2017 at 9:35 | history | asked | Yos | CC BY-SA 3.0 |