pseudocode

Index of single bit in long integer (in C) [duplicate]

六眼飞鱼酱① 提交于 2021-01-27 13:32:52
问题 This question already has answers here : Bit twiddling: which bit is set? (15 answers) Closed 7 years ago . I am trying to find an optimal code to locate a single bit index in a long integer (64 bit). The long integer has exactly one set bit. (Using C language) Currently, I am just shifting the whole thing by one bit, then checking for zero. I have read about lookup tables, but it won't do for the whole 64 bits. I thought about checking each 8 bits for zero, if not use a lookup, but still I

Uniformly distributed bit sequence

喜欢而已 提交于 2020-03-25 16:07:45
问题 Suppose you have a regular number generator which is able to produce uniformly distributed random 32 bit numbers. And suppose you look for a way to generate a pseudo-random sequence of bits where ones (i.e., the set bits) appear in the sequence with predefined probability. A naive way of producing such sequence would be running number generator on per bit level but it's terribly inefficient for small probabilities like 0.01 or 1% of bits in the sequence most of the bits will be zero. On

Finding unoccupied bounding boxes in an image given a set of occupied boxes

拈花ヽ惹草 提交于 2020-03-22 09:03:40
问题 Following up on my previous post that handled this problem in 1-D, I'm trying to extend the solution into two dimensions. Let's say I have an MxN image with predefined bounding boxes of sizes (m1xn1, m2xn2, ... m_nxn_n). How would I go about efficiently finding boxes with a given size lxw. So far I have the following function from my previous question: def random_exclude(low: int, high: int, exclude: List[Tuple[int]]) -> int: N = -low+sum([(l-h) for l,h in exclude])+high r = np.random.randint

Wikipedia's pseudocode for Breadth-first search: How could n's parent be u if “n is adjacent to u”?

生来就可爱ヽ(ⅴ<●) 提交于 2020-02-08 02:23:41
问题 Studying the Breadth-first search algorithm, I encountered the following pseudo-code: 1 Breadth-First-Search(G, v): 2 3 for each node n in G: 4 n.distance = INFINITY 5 n.parent = NIL 6 7 create empty queue Q 8 9 v.distance = 0 10 Q.enqueue(v) 11 12 while Q is not empty: 13 14 u = Q.dequeue() 15 16 for each node n that is adjacent to u: 17 if n.distance == INFINITY: 18 n.distance = u.distance + 1 19 n.parent = u 20 Q.enqueue(n) My question is regarding line 19 (n.parent = u): How could n's

Trouble with threads in a function to exec another function after X time

别来无恙 提交于 2020-01-24 01:13:11
问题 I am trying to create this function in order to exec another function after X time: void execAfter(double time, void *(*func)(void *), t_params *params); I have made an Thread encapsulation and a Time encapsulation (objects Thread and Time). What I want to do in pseudo code: Call execAfter instantiate Thread call thread->create(*funcToExec, *params, timeToWait) [inside thread, leaving execAfter] instanciate Time object wait for X time exec funcToExec delete Time object [leaving thread, back

Merging multiple transformation matrices

只谈情不闲聊 提交于 2020-01-17 01:21:09
问题 I am trying to develop a gesture based transform function. I am just confused about merging many transformation matrices by their own pivot point. Actually I am trying to use some functions like this: matrix4x4 transform(matrix4x4 prevMatrix, vector3 pivot, vector3 deltaTranslation, float deltaRotation, float deltaScale){ 'magic is here.. } 'First call var temp = transform(identity, (.5,.5,0), (0.1,0.2), 0.1, 105deg); renderCanvas(temp); 'Second call temp = transform(temp, (0, 0), (-1, 2), -

Mips translating problems

眉间皱痕 提交于 2020-01-16 17:15:23
问题 I am having trouble translating this pseudo code to mips assembly i have added the registers i am using to help understand what is going on # if (n == 1) # return 1 # else if (n == 2) # return 6 # else # return 2*hex(n-1) - hex(n-2) + 4 # (hex) Registers: # $a0 - n, the argument # $v0 - the result the n-th Hexamorphic number # $t0 - holds hex(n-2) # $t1 - holds constants 1 # $t2 - holds constant 2 here is where i am at in my code i feel confident about hex: and elseif: but the else: is where

Mips translating problems

荒凉一梦 提交于 2020-01-16 17:14:19
问题 I am having trouble translating this pseudo code to mips assembly i have added the registers i am using to help understand what is going on # if (n == 1) # return 1 # else if (n == 2) # return 6 # else # return 2*hex(n-1) - hex(n-2) + 4 # (hex) Registers: # $a0 - n, the argument # $v0 - the result the n-th Hexamorphic number # $t0 - holds hex(n-2) # $t1 - holds constants 1 # $t2 - holds constant 2 here is where i am at in my code i feel confident about hex: and elseif: but the else: is where

Can you quickly tell me if this pseudocode makes sense or not?

北城余情 提交于 2020-01-11 14:45:07
问题 I believe my code is now foolproof. I will write up the pseudocode now. But I do have one question. Why does DRJava ask that I return something outside of my if statements? As you can see I wrote for ex: "return 1;" just because it asked. It will never return that value however. Can someone explain this to me? public class assignment1question2test { public static void main(String[] args) { int[] a = new int[50]; int l = 0; int r = a.length; for(int i=0; i<r; i++) { a[i] = 1; } a[0] = 10; for

AVL Tree: Finding the key with the smallest data values in keys between two values in O(logn) time

人走茶凉 提交于 2020-01-11 14:10:35
问题 So I'm given an AVL tree. And im trying to figure out at least the pseudocode to find the key with the minimum data values among all of the keys between two values k1 and k2. This is assuming that the field data stored in each node is an integer. I want to make sure that my pseudocode runs in O(logn) time. I know that I can do it by storing an extra field in the node structure..and showing how this field can be maintained during updates, but I don't know where to go from there. 回答1: Let's