Consider the following search problem: next(A,B), next(A,C), next(B,D), next(B,E), next(B,F), next(C,D), next(D,E), next(F,E), next(E,G). H(A)=20, H(B)=12, H(C)=16, H(D)=(8), H(E)=6, H(F)=7, H(G)=0. Where next(X,Y) means that Y is a child of X and H(X) represent the heuristic value of X. Use hill climbing algorithm to find the path between A and G.

Show how the binary search algorithm discussed searches for 25 in the sorted listbelow: 4 6 17 25 32 39 41 43 45 49 Here I got  i=4 j=10 m=7 I know that 25<41....but I don't know what to do from here when it's like this. Is it the case where j is moved over to the left? Please help? I'm very confused and not sure if I'm understanding this correctly. Steps would be a great help here!

Design a bottom-up decrease-and-conquer algorithm for the following task. For any even n, mark n cells on an infinite sheet of graph paper so that each marked cell has an odd number of marked neighbors. Two cells are considered neighbors if they are next to each other either horizontally or vertically but not diagonally. The marked cells must form a contiguous region, that is a region in which there is a path between any pair of marked cells that goes through a sequence of marked neighbors. Below are examples of solutions for n=2 and n=4 respectively.

Let S be a set of n distinct positive integers, where n is odd.  The median of S is the (n+1)/2th smallest number of this set - i.e., the number in the middle of the set once the set is sorted. For example, if S = [70, 10, 20, 60, 30], then the sorted set is [10, 20, 30, 60, 70], from which we see that the median is 30. There is an obvious linearithmic algorithm to determine the median of S.  First you sort the set, which takes O(n log n) time.  Then you output the value of S[(n+1)/2}, which takes O(1) time.  But is this the best we can do? Here is a bold claim: there is an O(n) algorithm to determine the median of set S. Determine whether the above claim is TRUE or FALSE.