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Summary
| Description |
English: With the transfer event, D,T and L events must be assigned costs to define most parsimonious reconciliation.
Different costs give different most parsimonious solutions. In the top case, D is assigned a cost of 1, T of 3 and L of 1, and if we look at the number of events in the proposed scenario we find 1 duplication and 1 loss, so a cost of 2. If we compute that cost again with the bottom cost assignment we have a scenario cost of 3. With that new cost assignment the bottom scenario is more parsimonious as its cost is 2, while it had a cost of 3 with the top cost assignment. |
| Date | |
| Source | Own work |
| Author | Hugo Menet |
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| Date/Time | Thumbnail | Dimensions | User | Comment | |
|---|---|---|---|---|---|
| current | 12:34, 29 March 2022 | | 215 × 190 (22 KB) | Hmenet | Uploaded own work with UploadWizard |
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