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Please note that Graded Discussion Board (GDB) of CS502 will be started from July
16, 2013 and will be closed on July 17, 2013.

Topic for discussion:

As both Dynamic Programming and Greedy Strategy are used for solving Optimization
problems and influenced by optimal-sub property; it might possible that one can
attempt to find Greedy solution, if in fact Dynamic Programming is the right
approach or vice versa. With this background, Knapsack problem (either 0/1 or
Fractional) is a classic optimization problem to investigate with.

You are required to support or contradict the given below statement with proper
justification.

“Dynamic Programming yields optimal solution for 0/1 Knapsack problem while Greedy
approach does not. In addition, Greedy approach yields optimal solution for
Fractional Knapsack problem”.

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Replies to This Discussion

assignment tu assigmment , GDB subhanAllah...............!

Sir ko lgta ha puri book main say sirf Knapsack, activity selection and greedy algo say piar ha baki graphs aur NP etc say shadeed nafrat ha. chlain g lets see wot happend next. Allah ka naam lay kr start krain Discussion. 

B FOR FAZAL sis plz ??? bcz ladies first??

Game starts now..........

<>/p>

 B FOR FAZAL hor ki................. :(

given statement is true.....

Dear, sir the choice to choren aur GDB k topic per discuss kren to zyada acha ho. mera khyal hai statement true hai q k hum ye parh bhi chuke hen.

this GDB is not the case of yes or no, rather we have to support the statement with proper reasons  or  we have to provide the contradict of the the given statement with proper reasons

i am going to support the statement that would be easier than to provide contradict with reasoning. As  supporting material can be deducted from the handouts instead of searching goooooooooogle.

kindly give us more detail about the solution. In handouts please mention page n0s also ??

sir ko yahi topic mila..........

We can also observe that the greedy algorithm is not optimal for the 0-1 knapsack problem. Consider

the example shown in the Figure 7.9. If you were to sort the items by

i , then you would first take the

items of weight 5, then 20, and then (since the item of weight 40 does not fit) you would settle for the

item of weight 30, for a total value of $30 + $100 + $90 = $220. On the other hand, if you had been less

greedy, and ignored the item of weight 5, then you could take the items of weights 20 and 40 for a total

value of $100+$160 = $260. This is shown in Figure 7.10.

Reference: Hand-out page 110 

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