CSE431/531 Homework 3 Solutions

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1 CSE431/531 Homework 3 Solutions Due: Oct 19, of 11

2 1 Problem 1 input : A convex polygon P with vertices [v 1, v 2,..., v n ] output: A minimum weight triangulation of P A[1 : n, 1 : n] ( 1) ; /* n n array, initialized to 1 */ B[1 : n, 1 : n] ( 1); Cost(1,n); PrintTriangulation(1,n); Function PrintTriangulation(i,j) if j i 3 then k B[i, j]; if k = n then Print((v i+1, v j )); Print((v i, v k )); PrintTriangulation(i,k); PrintTriangulation(k,j); /* Cost(i,j) returns the cost of the minimum weight triangulation of the vertices between v i and v j. A is used to memoize Cost() and B is used to store the diagonals used */ Function Cost(i,j) if j i 2 then return 0 ; /* Already triangulated */ if A[i, j] 1 then return A[i, j] ; /* Return memoized value */ C[1 : n] ; /* To handle edges originating from v i */ for k [i + 2, j 1] do C[k] Distance(v i, v j ) + Cost(i, k) + Cost(k, j); /* If there is no edge from v i there must be one from v i+1 to v j */ C[n] Distance(v i+1, v j ) + Cost(i + 1, j); A[i, j] min k B[i, j] argmin k (C[k]); (C[k]); 2 of 11

3 2 Problem D Case input : A string a = [a 1, a 2,..., a n ] and a string b = [b 1, b 2,..., b m ] output: A shortest common superstring of a and b A[0 : m, 0 : n] 0; /* A[i,j] = Cost of superstring(a[i:],b[j:]) */ for j [0, n] do A[m, j] n j; for i [0, m] do A[i, n] m i; for i [m 1, m 2,..., 0] do for j [n 1, n 2,..., 0] do if a i = b j then A[i, j] 1 + A[i + 1, j + 1]; A[i, j] 1 + min (A[i, j + 1], A[i + 1, j]); i 0; j 0; while i < m or j < n do if i = m or A[i, j] = 1 + A[i, j + 1] then Print(b j ); if j = n or A[i, j] = 1 + A[i + 1, j] then 3 of 11

4 2.2 3D Case input : Strings a = [a 1...a m ], b = [b 1...b n ], c = [c 1...c p ] output: A shortest common superstring of a, b and c A[0 : m, 0 : n, 0 : p] 0; /* A[i,j,k] = Cost of superstring(a[i:],b[j:],c[k:]) */ Initialize(); Compute(); PrintResult(); Function Initialize for i [0, m] do for j [0, n] do for k [0, p] do if i = m then A[i, j, k] 2Dsuperstring(b[j :], c[k :]); if j = n then A[i, j, k] 2Dsuperstring(a[i :], c[k :]); if k = p then A[i, j, k] 2Dsuperstring(a[i :], b[j :]); 4 of 11

5 Function Compute for i [m 1, m 2,..., 0] do for j [n 1, n 2,..., 0] do for k [p 1, p 2,..., 0] do cases[1 : 7] ; if a i = b j = c k then cases[1] 1 + A[i + 1, j + 1, k + 1]; if a i = b j then cases[2] 1 + A[i + 1, j + 1, k]; if b j = c k then cases[3] 1 + A[i, j + 1, k + 1]; if a i = c k then cases[4] 1 + A[i + 1, j, k + 1]; cases[5] 1 + A[i + 1, j, k]; cases[6] 1 + A[i, j + 1, k]; cases[7] 1 + A[i, j, k + 1]; A[i, j, k] 1 + min (cases[1 : 7]); 5 of 11

6 Function PrintResult i 0; j 0; k 0; while i < m or j < n or k < p do if i m and j n and k p and A[i, j, k] = 1 + A[i + 1, j + 1, k + 1] then k k + 1; if i m and j n and A[i, j, k] = 1 + A[i + 1, j + 1, k] then if i m and k p and A[i, j, k] = 1 + A[i + 1, j, k + 1] then k k + 1; if j n and k p and A[i, j, k] = 1 + A[i, j + 1, k + 1] then Print(b j ); k k + 1; if i m and A[i, j, k] = 1 + A[i + 1, j, k] then if j n and A[i, j, k] = 1 + A[i, j + 1, k] then Print(b j ); if k p and A[i, j, k] = 1 + A[i, j, k + 1] then Print(c k ); k k + 1; 6 of 11

7 3 Problem Part a Let cost(n) be the cost computing T (n) directly. We have: n 1 cost(n) = cost(i) + cost(i 1) i=1 n 1 2cost(i 1) = i=1 i=1 n 2 2 = 2 n 2 = ( 2) n (1) 3.2 Part b To solve the problem in O(n 2 ) time memoize the value T (i). Then cost(n) = 3.3 Part c n 2n = O(n 2 ) (2) i=1 input : Integer n output: T (n) A[0 : n] 2; for i [2, n] do A[i] A[i 1] + A[i 1] A[i 2]; return A[n] 7 of 11

8 4 Problem 4 input : Binary tree T with vertices v 1..v n and root v 1, Weight function w, Number of cuts k output: Minimum weight forest /* T Cost[i, j] = The cost of the optimal subtree at v i with j cuts */ T Cost[1 : n, 1 : k] 0; /* CCCost[i, j] = The cost of the connected component containing v i of the optimal subtree at v i with j cuts */ CCCost[1 : n, 1 : k] 0; /* S[i, j] = The set of edges to remove for the subtree v i with j cuts */ S[1 : n, 1 : k] ; for i [max(dist(v, v 1 ), 0] do for v V where v is distance i from v 1 do for j [0 : k] do Cost(v,j); return S [1,k]; 8 of 11

9 Function Cost(v,k) if v is a leaf or k = 0 then S[v, k] ; CCost[v, k] w(v); T Cost[v, k] w(v); return w (v); if left-child(v) is null then a max (w(v), T Cost[right-child(v), k 1]); b max (w(v) + CCost[right-child(v), k], T Cost[right-child(v), k]); if a<b then S[v, k] S[right-child(v), k 1] {right-edge(v)}; CCost[v, k] w(v); T Cost[v, k] a; S[v, k] S[right-child(v), k]; CCost[v, k] w(v) + CCost[right-child(v), k]; T Cost[v, k] b; if right-child(v) is null then a max (w(v), T Cost[left-child(v), k 1]); b max (w(v) + CCost[left-child(v), k], T Cost[left-child(v), k]); if a<b then S[v, k] S[left-child(v), k 1] {left-edge(v)}; CCost[v, k] w(v); T Cost[v, k] a; S[v, k] S[left-child(v), k]; CCost[v, k] w(v) + CCost[left-child(v), k]; T Cost[v, k] b; InteriorNode(v,k); 9 of 11

10 Function InteriorNode(v,k) /* Cut left edge */ a[0 : k 1] 0; for i [0 : k 1] do a[i] max( w (v)+ccost[right-child(v),k-i-1] TCost[left-child(v),i] TCost[right-child(v),k-i-1]); /* Cut right edge */ b[0 : k 1] 0; for i [0 : k 1] do b[i] max( w (v)+ccost[left-child(v),k-i-1] TCost[right-child(v),i] TCost[left-child(v),k-i-1]); /* Cut neither edge */ c[0 : k] 0; for i [0 : k] do c[i] max( w (v)+ccost[right-child(v),k-i]+ccost[left-child(v),i] TCost[left-child(v),i] TCost[right-child(v),k-i]); /* Cut both edges */ d[0 : k 2] 0; for i [0 : k 2] do d[i] max( w (v) TCost[left-child(v),i] TCost[right-child(v),k-i-2]); if min (a) = min (a, b, c, d) then i argmin (a); S[v, k] S[left-child(v), i] S[right-child(v), k i 1] {left-edge(v)}; CCost[v, k] w(v) + CCost[right-child(v), k i 1]; T Cost[v, k] min (a); if min (b) = min (a, b, c, d) then i argmin (b); S[v, k] S[right-child(v), i] S[left-child(v), k i 1] {right-edge(v)}; CCost[v, k] w(v) + CCost[left-child(v), k i 1]; T Cost[v, k] min (b); if min (c) = min (a, b, c, d) then i argmin (c); S[v, k] S[left-child(v), i] S[right-child(v), k i]; CCost[v, k] w(v) + CCost[right-child(v), k i] + CCost[left-child(v), i]; T Cost[v, k] min (c); if min (d) = min (a, b, c, d) then i argmin (d); S[v, k] S[left-child(v), i] S[right-child(v), k i 2] {left-edge(v), right-edge(v)}; CCost[v, k] w(v); T Cost[v, k] min (d); 10 of 11

11 5 Problem 5 input : Strings A = a 1...a m and B = b 1...b n output: Minimum cost conversion from A to B M[0 : m, 0 : n] 0 /* M[i, j] = the cost of converting a i...a m to b j...b n */; for i [m 1, m 2,..., 0] do M[i, n] M[i + 1, n] + 3; for j [n 1, n 2,..., 0] do M[m, j] M[m, j + 1] + 5; for i [m 1, m 2,..., 0] do for j [n 1, n 2,..., 0] do if a i = a j then M[i, j] M[i + 1, j + 1]; M[i, j] min (3 + M[i + 1, j], 5 + M[i, j + 1], 7 + M[i + 1, j + 1]); i 0; j 0; while i < m or j < n do if j < n and A[i, j] = 5 + M[i, j + 1] then Print(Insert b j ); if i < m and A[i, j] = 3 + M[i + 1, j] then Print(Delete a i ); if A[i, j] = M[i + 1, j + 1] then Print(Leave a i unchanged); i j + 1; Print(Replace a i with b j ); i j + 1; 11 of 11

Dynamic programming II - The Recursion Strikes Back

Chapter 5 Dynamic programming II - The Recursion Strikes Back By Sariel Har-Peled, December 17, 2012 1 Version: 0.4 No, mademoiselle, I don t capture elephants. I content myself with living among them.