hw1 xiweipan

hw1/xiweipan/Readme.md

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+# Task 9.6
+
+In `knapsack.jl`, `knapsack_dp` realizes an exact algorithom with the complexity $O(lV)=O(l^2 v_\text{max})$, where $V=lv_\text{max}$ is the upper bound of the maximum value.
+
+This doesn't mean $\mathbf{P}=\mathbf{NP}$. Because although the algorithm runs in polynomial time with respect to the numeric values of the inputs (namely, the number of items $l$ and the maximum item value $v_{\text{max}}$), it is not polynomial in the size of the input, which is measured in **bits**. And this algorithm is exponential in the bit size of the input.
+
+
+# Task 9.7
+
+In this task,we need to design a FPTAS version of the algorithm in Task 9.6 with approximation ratio $1-\epsilon$ and time complexity $\text{poly}(l, \frac{1}{\epsilon})$.
+
+The main idea is to scale down the item values by a factor of $K$, i.e. $v'_i = \lfloor v_i / K \rfloor$.
+
+However, in this process, $K v'_i \leq v_i < K (v'_i + 1)$, so for the whole problem, the maximum value loss is $lK$.
+
+Considering the scaled-down problem, we apply the exact dynamic programming algorithm from Task 9.6 to the instance with scaled values $v'_i$, and denote the result of this algorithm as $\text{APX}$. Then we have 
+
+$\text{APX} \ge \text{OPT} - lK$,
+
+where $\text{OPT}$ is the optimal value of the original problem.
+
+To ensure the approximation ratio is $1-\epsilon$, i.e.
+
+$\text{APX} \ge (1-\epsilon) \text{OPT}$,
+
+we can simply let 
+
+$\text{OPT}-l K \geq (1-\epsilon) \text{OPT}$,
+
+which is equivalent to
+
+$l K \leq \epsilon \text{OPT}$.
+
+Since the optimal solution must at least match the best item that fits into the knapsack, we have $\text{OPT} \ge v^* $, where $v^* = \max \{ v_i \mid w_i \le W \}$. We can simply assume for all $i$, $w_i \le W$ (otherwise, we can remove it from the problem).
+
+Therefore, we have the sufficient condition
+
+$l K \leq \epsilon v_\text{max}$.
+
+Thus, for any $\epsilon > 0$, by choosing $K = \frac{\epsilon v_\text{max}}{l}$, we can ensure the approximation ratio is $1-\epsilon$.
+
+For the time complexity, the dynamic programming algorithm is still $O(lV')=O(l^2 v'_\text{max}) = O(l^3 \frac{1}{\epsilon})$, which is polynomial of $l$ and $\frac{1}{\epsilon}$.
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hw1/xiweipan/knapsack.jl

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+function knapsack_dp(weights::Vector{Int}, values::Vector{Int}, W::Int)
+    l = length(weights)
+    vmax = maximum(values)  
+    V = l * vmax
+    INF = typemax(Int)  # mark the impossible case
+
+    # dp is a (l+1) x (V+1) matrix
+    # dp[i, j] is the minimum weight of the knapsack with value j-1 using the last i to l items
+    dp = fill(INF, l + 1, V + 1)
+    # i = l+1: no available item, only value 0 is possible with weight 0
+    dp[l + 1, 1] = 0
+    # j = 1: max value is 0
+    for i in 1:l
+        dp[i, 1] = 0
+    end
+
+    for i in l:-1:1
+        w, v = weights[i], values[i]
+        for val in 0:V
+            # without the i-th item
+            without = dp[i + 1, val + 1]
+            # with the i-th item
+            with = INF
+            if val >= v && dp[i + 1, val - v + 1] != INF
+                with = w + dp[i + 1, val - v + 1]
+            end
+            dp[i, val + 1] = min(without, with)
+        end
+    end
+
+    result = 0
+    for val in 0:V
+        if dp[1, val + 1] <= W
+            result = val
+        end
+    end
+
+    return result
+end
+
+# test
+weights = [2, 3, 4, 5]
+values = [3, 4, 5, 6]
+W = 5
+println(knapsack_dp(weights, values, W))