Line 216: | Line 216: | ||
suppose have n numbers array of n elements | suppose have n numbers array of n elements | ||
log n | log n | ||
+ | =========================================== | ||
+ | Wudi Zhou, April 19th, Prof. Lu's section | ||
+ | |||
+ | Hints for last assignment: | ||
+ | Steps: | ||
+ | 1,Integer partition | ||
+ | |||
+ | 2,Generate pieces by shifting | ||
+ | |||
+ | 3,Eliminate duplicates | ||
+ | |||
+ | use for loop to shift two rows: | ||
+ | |||
+ | ie for(shift = 1 - b; shift <= a - b; shift ++) | ||
+ | |||
+ | p = head; | ||
+ | while((p != NULL) && (checkDuplicate == 0)) | ||
+ | { | ||
+ | p = p -> next; | ||
+ | } | ||
+ | if(p == NULL) | ||
+ | { | ||
+ | insert(n); | ||
+ | } | ||
+ | |||
+ | For binary tree: | ||
+ | n step can be done by log(n) steps |
Revision as of 13:41, 19 April 2012
/*ECE 264 Lecture Wed Apr18 Peachanok Lertkajornkitti Professor Elmqvst (Section 1)
IPA2-5
need to use: ipa2-1 AND 2-3,2-4
Steps: 1.IPA2-1 2.IPA2-2 3.IPA2-4 4.Print list
- /
- include<stdio.h>
- include<stdlib.h>
- include<string.h>
- include"Blokus.h"
void print_partition(int size,int *partition) { printf("["); int i; for(i=0;i<size;i++) { printf(" %d",partition[i]); } printf("]\n");
Blokus *gen_partition(int budget,int pos,int *partition,int dim) { //Base case: no more budget if(budget==0) { print_partition(pos,partition); Blokus *list = gen_shifts(partition,pos,dim); return list; }
//Recursive case: have budget Blokus *head = NULL; int spending; int (spending = 1;spending<=budget;spending++) { partition(pos) = spending; Blokus *list = gen_partition(budget-spending,pos+1,partition,dim); head = BlokusL_append(head, list); //takes the head and add a new list to the end }
} Blokus *make_unique(Blokus *head) //takes in the head n make sure it's unique { //IPA2-4 return head; } Blokus *generate_pieces(int n) { int *partition = malloc(sizeof(int)*n); //Step 1+2 Blokus *head = gen_partition(n,0,partition,n); free(partition); //Step 3 head = make_unique(head); return head; }
int main(int argc,char *argv[]) { if(argc != 2) { return EXIT_FAILURE; } int n = strtol(argv[1],NULL,10);
Blokus *head = generate_pieces(n); BlokusL_printf(head); BlokusL_destroy(head);
return EXIT_SUCCESS; }
//Blokus.h
- ifndef BLOKUS_H
- define BLOKUS_H
typedef struct Blokus_t{ char *data; int dim; struct Blokus_t *next; }Blokus;
Blokus *Blokus_create(char *data,int dim); void Blokus_destroy(Blokus *blokus); void Blokus_print(Blokus *blokus);
Blokus BlokusL_insert(char *data,int dim,Blokus *head); void BlokusL_print(Blokus *head); void BlokusL_destroy(Blokus *head); Blokus *BlokusL_append(head, list);
- endif /*Blokus.h */
//Blokus.c
- include<stdio.h>
- include<stdlib.h>
- include<string.h>
- include"Blokus.h"
Blokus *Blokus_create(char *data,int dim) { Blokus *blokus = malloc(sizeof(Blokus)); blokus->data = malloc(sizeof(char) * dim * dim); memcpy(blokus->data,data,sizeof(char)*dim*dim); blokus->dim = dim; blokus->next = NULL;
return blokus; } void Blokus_destroy(Blokus *blokus) { free(blokus->data); free(blokus); }
void Blokus_print(Blokus *blokus); { int i; for(i=0;i<blokus->dim*blokus->dim;i++) { printf("%c",blokus->data[i]); } printf("\n");
}
Blokus BlokusL_insert(char *data,int dim,Blokus *head) { Blokus *blokus = Blokus_create(data,dim); blokus->next = head; return head; }
void BlokusL_print(Blokus *head)
{
while(head != NULL)
{
Blokus_piece(head);
head = head->next;
}
}
void BlokusL_destroy(Blokus *head)
{
while(head != NULL)
{
Blokus *tmp = head;
head = head->next;
Blokus_destroy(tmp);
}
}
Blokus *BlokusL_append(Blokus *head, Blokus *head2) //COULD COME OUT IN AN EXAM { if(head == NULL) return head2; Blokus *curr = head; while(curr->next != NULL) { curr = curr->next; }
curr->next = head2; return head; }
===================
Hanye Xu April 19th
Binary search tree v
/ \ left right
everything at left <v right >v integer partition generate piece by shift eliminate duplicates invalid piece 8 squares
how many ways can you shift: a squares & b squares b-1 distance to shift or a-1 distance to shift
for(shift = 1-b; shift<=a-1; shift++) { next two rows;
n= a new piece
p=head;
while('p' != NULL)&&checkDuplicate(P,n) == 0){
p=p->next;
}
if (p==NULL){insert(nth list);
}
binary tree
for any real number x you can find a real number y such that x = 2^y or x = -2^y
n steps to log n steps
suppose have n numbers array of n elements log n
===============================
Wudi Zhou, April 19th, Prof. Lu's section
Hints for last assignment: Steps: 1,Integer partition
2,Generate pieces by shifting
3,Eliminate duplicates
use for loop to shift two rows:
ie for(shift = 1 - b; shift <= a - b; shift ++)
p = head; while((p != NULL) && (checkDuplicate == 0)) { p = p -> next; } if(p == NULL) { insert(n); }
For binary tree: n step can be done by log(n) steps