YsZ HW2 Attempts

hw1/YushengZhao/p9_16.jl

@@ -0,0 +1,2 @@
+# min(b') s.t A' x' + b' = b x'>= 0 and b' >= 0 
+# trivial solution is x' = 0 and b' = b

hw1/YushengZhao/p9_6.jl

@@ -0,0 +1,2 @@
+# standard DP problem
+# needs $log(v_max)$ bits to describe $v_max$. Exponential time 

hw1/YushengZhao/p9_7.jl

@@ -0,0 +1,7 @@
+# floor division of K rounds some weights to 0, such weights maybe added in correctly to solution
+# very roughly
+# ϵ ∝ Σ_i floor(v_i / K)
+# ϵ ∝ floor(v_i / K) * l
+# K ∝ l/ϵ
+# but also need to supress v_max to be small, let 
+# K = v_max * l / ϵ

hw2/YushengZhao/9_26.jl

@@ -0,0 +1,7 @@
+# according to hint, when formula is satisfied, we have 1/2 - y_0/3 <= y_i <= 1/2 +y_0
+# and to minimize f(x) need to maximize (y_i-1/2)^2, which is y_i at its upper bound and gives (y_i - 1/2)^2 = y_0^2, that gives lower bound of f(x) = -1/2y_0^2
+# on the other hand if the formula is not satisfied, the bound of at least one y_i is 1/2 - y_0/3 <= y_i <= 1/2 +y_0, and the min f(x) is  num unsat * y_0^2 - y_0^2/9 + sum_(sat clause) (y_0^2 - y_0^2) - y_0^2/2 + num_unsat * y_0^ / 2*n*9 which is at least 7/18 y_0^2 
+
+# using hint, if formula is satisfied, y^* cannot be local minimum because can slide over y_0 axis a little bit, epsilon, and get lower. this is in the neighborhood of y^* because y_i' 's difference from y_i = 1/2 is bounded by the epsilon. And f(x)'s min is -y_0^2/2 when formula is satisfied. This min is smaller than f(y^*) = 0. 
+
+# so x_i means is y_i >= 1/2 - y_0/3 ?

hw2/YushengZhao/9_27.jl

@@ -0,0 +1 @@
+# x0 + sum ki vi where Avi = 0 and Ax0 = b , use https://en.wikipedia.org/wiki/Smith_normal_form to solve