Question
In: Psychology
Problem 1: Discover the rule where a sequence of three number operates on. Problem 2 (code...
Problem 1: Discover the rule where a sequence of three number operates on.
Problem 2 (code braking problem): Discover the rule of when the box opens.
Assignment: Give a discussion of the two problems and how it was solved. What made each problem difficult and how it relates to the Gestalt psychology discussed in class
Solutions
Expert Solution
According to gestalt psychology, a person has the tendency to look at things completely in context such that it would make a structure complete. This means that people learn to look at things as a whole.
Problem 1: Discover the rule where a sequence of three number operates on.
This is in relevance to the rule of closure and the rule of similarity. People tend to look at things keeping what is similar, and hence when people look at a sequence of three numbers, they tend to identify the numbers as a whole and identify the sequence. Also, the law of similarity tends to make people identify numbers based on the similarity of sequence.
Problem 2: discover the rule when the box opens.
When the box is closed, it is seen as a complete figure, completed by fitting in. Missing sequences and completing the figure. This is also shown by the law of closure. Closure means that the sequence is completed by the human perception. When the box opens, as the law of closure, it is intended to be closed, in order to fill in the gap. An open box is perceived as an incomplete design.
Hayet Jeridi answered 2 years agoRelated Solutions
Problem 1: Discover the rule where a sequence of three number operates on. Problem 2 (code...
Problem 1: Discover the rule where a sequence of three number
operates on.
Problem 2 (code braking problem): Discover the rule of when the
box opens.
Assignment: Give a discussion of the two problems and how it was
solved. What made each problem difficult and how it relates to the
Gestalt psychology discussed in class
The sequence 2, 4, 1, 3, 5 has three inversions (2,1), (4,1), (4,3). Using C++ Code...
The sequence 2, 4, 1, 3, 5 has three inversions (2,1), (4,1),
(4,3). Using C++ Code an O(nlog(n)) algorithm to count the number
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below: (Use the file below that to test)
countInv.cpp
-------------------------------------------------------------------------------------------------------------------------------
#include <vector>
#include <algorithm>
using namespace std;
int mergeInv(vector<int>& nums, vector<int>&
left, vector<int>& right) {
// Your code here
}
int countInv(vector<int>&nums) {
// Your code here
}
CountInv_test.cpp
-------------------------------------------------------------------------------------------------------------------------
#include <iostream>
#include <vector>
using namespace...
The sequence 2, 4, 1, 3, 5 has three inversions (2,1), (4,1), (4,3). Using C++ Code...
The sequence 2, 4, 1, 3, 5 has three inversions (2,1), (4,1),
(4,3). Using C++ Code an O(nlog(n)) algorithm to count the number
of inversions. Use the Merge sort Algorithm, and use the template
below: (Use the file below that to test)
countInv.cpp
-------------------------------------------------------------------------------------------------------------------------------
#include <vector>
#include <algorithm>
using namespace std;
int mergeInv(vector<int>& nums, vector<int>&
left, vector<int>& right) {
// Your code here
}
int countInv(vector<int>&nums) {
// Your code here
}
CountInv_test.cpp
-------------------------------------------------------------------------------------------------------------------------
#include <iostream>
#include <vector>
using namespace...
Where is the infinite loop in the code? Input for the code 1 2.5 2 40200000...
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Input for the code 1 2.5 2 40200000 1 0 2 80400000 2 ffffffff
3
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int count = 31;
void toBinary(int num, int n){
for(int i = 1; i < n; i++){
if((int)(num/pow(2,(n-i))) > 0){
num = num - pow(2,(n-i));
printf("1");
count--;
}else{
printf("0");
count--;
}
}
}
char checkSign (int sign)
{
char new_sign;
if (sign == 0)
{
new_sign = '+';
return new_sign;
}...
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MIPS code to calculate The 1- minimum 2- maximum 3- mid value of
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Problem 1:
Problem 2:
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C) Determine whether the sequence converges or diverges. State
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