#### Transcript math

```Approximate Knapsack
Get as much value
as you can
into the knapsack
Approximate Knapsack
Ingredients:
•Instances: The volume V of the knapsack.
The volume and price of n objects
<<v1,p1>,<v2,p2>,… ,<vn,pn>>.
•Solutions: A set of objects that fit in the knapsack.
•i.e. iS vi  V
•Cost of Solution: The total value of objects in set.
•i.e. iS pi
•Goal: Get as much value as you can
into the knapsack.
Approximate Knapsack
•Dynamic Programming
Running time = ( V × n )
= ( 2#bits in V × n )
•Poly time if size of knapsack is small
•Exponential time if size is an arbitrary integer.
Approximate Knapsack
•Dynamic Programming
Running time = ( V × n )
= ( 2#bits in V × n )
•NP-Complete
•Approximate Algorithm
•In poly-time (n3/),
solution can be found
•that is perfect in iS vi  V
•(1+) as good as optimal wrt iS pi
•Eg,  = .001, Time is 1000n3
Approximate Knapsack
Subinstance:  V’[0..V], i[0..n],
knapsack(V’,i) = maximize iS pi
subject to S  {1..i} and iS vi  V
•Recurrence Relation
knapsack(V’,i)
= max( knapsack(V’,i-1),
knapsack(V’-vi,i-1)+pi )
No
Yes
Approximate Knapsack
 V’[0..V], i[0..n], knapsack(V’,i)
OptSol price
No
Yes
0 1 2
V’-vi
V’
V
0
1
2
same + pi
i-1
i
Our price? n
same
Take best
of best.
Approximate Knapsack
 V’[0..V], i[0..n], knapsack(V’,i)
OptSol price
0 1 2
0
1
2
i-1
i
Time = O(nV)
n
V’
V
Approximate Knapsack
Ingredients: (strange version)
•Instances: The price P wanted from the knapsack.
The volume and price of n objects
<<v1,p1>,<v2,p2>,… ,<vn,pn>>.
•Solutions: A set of objects with total value P.
•i.e. iS pi ≥ P
•Cost of Solution: The total volume of objects in set.
•i.e. iS vi
•Goal: Minimize the volume needed
to obtain this value P.
Approximate Knapsack
Subinstance:  P’[0..P], i[0..n],
knapsack’(P’,i) = minimize iS vi
subject to S  {1..i} and iS pi ≥ P
•Recurrence Relation
knapsack’(P’,i)
= min( knapsack’(P’,i-1),
knapsack’(P’-pi,i-1)+vi )
No
Yes
Approximate Knapsack
 V’[0..V], i[0..n], knapsack’(P’,i)
OptSol volume
No
Yes
0 1 2
P’-pi
P’
P
0
1
2
same + pi
i-1
i
Our volume?n
same
Take best
of best.
Approximate Knapsack
Original problem knapsack(V,n)
OptSol volume
0 1 2
P’
0
1
2
i
Time = O(nP)
n
P = i pi
Find largest price
not using more than V volume
P
Approximate Knapsack
•Dynamic Programming
Running time = ( V × n )
= ( 2#bits in V × n )
•Poly time if size of knapsack is small
•Exponential time if size is an arbitrary integer.
•Strange Dynamic Programming
Running time = ( P × n )
= ( 2#bits in P × n )
•Poly time if prices are small
•Exponential time if prices are arbitrary integers.
Approximate Knapsack
Approximation Algorithm:
•Given V, <<v1,p1>,<v2,p2>,… ,<vn,pn>>, & 
•Let k be some # of bits chosen later
•Let p’i = pi with low k bits zeroed
eg pi = 101101101011
p’i = 1011011000002 k=4
(Lost precision in prices, but that is ok)
Solve knapsack using the strange algorithm.
Approximate Knapsack
Original problem knapsack(V,n)
OptSol volume
0 1 2
0
1
2
i
P = i pi
P’
n
Find largest price
not using more than V volume
P
Approximate Knapsack
Original problem knapsack(V,n)
OptSol volume
0 1 2
P’
0
1
2
i
Time
= O(n P/2k) n
Only every 2k th column matters
P
Approximate Knapsack
Approximation Algorithm:
•Let Salg be the set of items selected by our alg.
•Let Sopt be the set of items in the optimal sol.
•Let Palg = pi be the price returned by our alg.
iSalg
•Let Popt = pi be the price returned by our alg.
iS
opt
•Need Palg ≥ Popt (1-)
Approximate Knapsack
Palg = pi
iSalg
because rounded down
≥ p’i
iSalg
because Salg is an optimal solution
≥ p’i for the <vi,p’i> problem
iSopt
k.
because
rounded
by
at
most
2
≥ (pi -2k)
iSopt
= Popt – n 2k
Approximate Knapsack
Palg ≥ Popt – n2k
k
n2
= 1Popt
Popt
Popt
2k
(need)
=1-

Popt
=
n
Time = O(n P/2k) = O( n2 P )
 Popt
Popt ≥ maxi pi ≥
P = i pi
Time = O( n3)

P
n
done
```