using System;
namespace SteadyStateTempCylinder
{
public class PotPoint : IComparable
{
private double x, y, u;
public double X
{
get
{
return x;
}
set
{
x = value;
}
}
public double Y
{
get
{
return y;
}
set
{
y = value;
}
}
public double U
{
get
{
return u;
}
set
{
u = value;
}
}
public PotPoint(double x, double y, double u)
{
this.x = x;
this.y = y;
this.u = u;
}
public int CompareTo(object obj)
{
if (obj == null)
return 1;
PotPoint pp = (PotPoint)obj;
if (u > pp.u)
return 1;
else if (u == pp.u)
return 0;
else
return -1;
}
}
}
The description of this alphabetic letter game is very facile. Given a word make as many other words as possible using the letters of the given initial word. But first before we enumerate the game solution, we need to refresh the reader’s memory of some elementary mathematics.
The binary number system also known as the base 2 number system is used by computers to perform arithmetic. The digits in the binary number system are 0 and 1. The numbers 0 to 15 in binary using only four binary digits are where ^ is the exponentiation operator (raising a number to a power) are:
0 0000
1 0001 2 ^ 0 = 1
2 0010 2 ^ 1 = 2
3 0011 2 ^ 1 + 2 ^ 0 = 2 + 1 = 3
4 0100 2 ^ 2 = 4
5 0101 2 ^ 2 + 2 ^ 0 = 4 + 1 = 5
6 0110 2 ^ 2 + 2 ^ 1 = 4 + 2 = 6
7 0111 2 ^ 2 + 2 ^ 1 + 2 ^ 0 = 4 + 2 + 1 = 7
8 1000 2 ^ 3 = 8
9 1001 2 ^ 3 + 2 ^ 0 = 8 + 1 = 9
10 1010 2 ^ 3 + 2 ^ 1 = 8 + 2 = 10
11 1011 2 ^ 3 + 2 ^ 1 + 2 ^ 0 = 8 + 2 + 1 = 11
12 1100 2 ^ 3 + 2 ^ 2 = 8 + 4 = 12
13 1101 2 ^ 3 + 2 ^ 2 + 2 ^ 0 = 8 + 4 + 1 = 13
14 1110 2 ^ 3 + 2 ^ 2 + 2 ^ 1 = 8 + 4 + 2 = 14
15 1111 2 ^ 3 + 2 ^ 2 + 2 ^ 1 + 2 ^ 0 = 8 + 4 + 2 + 1 = 15
An algorithm to convert a base 10 (decimal) number to base 2 (binary) number is given below:
Input n a base 10 number
Output b[0], b[1], b[2], … a finite binary string representing the decimal number
Integer i = 0
Do
Integer nmod2 = n mod 2
Integer ndiv2 = n / 2
b[i] = nmod2 + ‘0’
i = i + 1
n = ndiv2
While n > 0
The b[i] will be in reverse order. For example, convert 12 from decimal to using four binary digits:
12 mod 2 = 0
12 div 2 = 6
b[0] = ‘0’
i = 1
n = 6
6 mod 2 = 0
6 div 2 = 3
b[1] = ‘0’
i = 2
n = 3
3 mod 2 = 1
3 div 2 = 1
i = 3
b[2] = ‘1’
n = 1
1 mod 2 = 1
1 div 2 = 0
b[3] = ‘1’
n = 0
So, the reversed binary string of digits is “0011”. And after reversing the string we have 12 is represented by the binary digits “1100”.
Next, we need to define a power set and its binary representation. The index power set of 4 objects which has 2 ^ 4 = 16 entries is specified in the following table:
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
A permutation of the set of three indices is given by the following list:
012, 021, 102, 120, 201, 210
A permutation of n objects is a list of n! = n * (n – 1) * … * 2 * 1. A permutation of 4 objects has a list of 24 – 4-digit indices list since 4! = 4 * 3 * 2 * 1 = 24 has the table:
0123, 0132, 0213, 0231, 0312, 0321,
1023, 1032, 1203, 1230, 1320, 1302,
2013, 2031, 2103, 2130, 2301, 2310,
3012, 3021, 3102, 3120, 3201, 3210
Suppose our word is “lost” then we first find the power set:
1 0001 t
2 0010 s
3 0011 st ts
4 0100 o
5 0101 ot to
6 0110 os so
7 0111 ost ots sot sto tos tso
8 1000 l
9 1001 lt tl
10 1010 ls sl
11 1011 lst lts slt stl tsl tls
12 1100 lo ol
13 1101 lot lto olt otl tlo tol
14 1110 los lso slo sol osl ols
15 1111 lost lots slot etc.
Using a dictionary of 152,512 English words my program finds 16 hits for the letters of “lost”:
Dictionary Length: 152512
Word: lost
0 l
1 lo
2 lost
3 lot
4 lots
5 ls
6 o
7 s
8 slot
9 so
10 sol
11 sot
12 st
13 t
14 to
15 ts
Total letters and/or words 16
Next, we use “tear” as our word:
Dictionary Length: 152512
Word: tear
0 a
1 are
2 art
3 at
4 ate
5 e
6 ea
7 ear
8 eat
9 era
10 et
11 eta
12 r
13 rat
14 rate
15 re
16 rt
17 rte
18 t
19 tar
20 tare
21 tea
22 tear
23 tr
Total letters and/or words 24
Finally, we use the word “company”:
Dictionary Length: 152512
Word: company
0 a
1 ac
2 am
3 amp
4 an
5 any
6 c
7 ca
8 cam
9 camp
10 campy
11 can
12 canopy
13 cap
14 capo
15 capon
16 cay
17 cm
18 co
19 com
20 coma
21 comp
22 company
23 con
24 cony
25 cop
26 copay
27 copy
28 coy
29 cyan
30 m
31 ma
32 mac
33 man
34 many
35 map
36 may
37 mayo
38 mo
39 moan
40 mop
41 mp
42 my
43 myna
44 n
45 nap
46 nay
47 nm
48 no
49 o
50 om
51 on
52 op
53 p
54 pa
55 pan
56 pay
57 pm
58 pony
59 y
60 ya
61 yam
62 yap
63 yo
64 yon
Total letters and/or words 65
The C++ program's source code is given below:
// WordToWords.cpp : This file contains the 'main' function. Program execution begins and ends there.
//
#include "pch.h"
#include <algorithm>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<string> dictWords;
bool ReadDictionaryFile()
{
fstream newfile;
newfile.open("C:\\Users\\james\\source\\repos\\WordToWords\\Dictionary.txt", ios::in);
if (newfile.is_open()) {
int index = 0, length = 128;
char cstr[128];
while (newfile.getline(cstr, length)) {
string str;
str.clear();
for (int i = 0; i < (int)strlen(cstr); i++)
str.push_back(cstr[i]);
dictWords.push_back(str);
}
newfile.close();
sort(dictWords.begin(), dictWords.end());
return true;
}
else
return false;
}
string ConvertBase2(char cstr[], int n, int len)
{
int count = 0;
string str, rev;
do
{
int nMod2 = n % 2;
int nDiv2 = n / 2;
str.push_back(nMod2 + '0');
n = nDiv2;
} while (n > 0);
n = str.size();
for (int i = n; i < len; i++)
str.push_back('0');
n = str.size();
for (int i = n - 1; i >= 0; i--)
if (str[i] == '1')
rev.push_back(cstr[i]);
return rev;
}
vector<string> PowerSet(char cstr[], int len)
{
vector<int> index;
vector<string> match;
for (long ps = 0; ps <= pow(2, len); ps++)
{
string str = ConvertBase2(cstr, ps, len);
int psf = 1;
for (int i = 2; i <= len; i++)
psf *= i;
for (int i = 0; i < psf; i++)
{
if (binary_search(dictWords.begin(), dictWords.end(), str))
{
if (!binary_search(match.begin(), match.end(), str))
{
match.push_back(str);
sort(match.begin(), match.end());
}
}
next_permutation(str.begin(), str.end());
}
sort(match.begin(), match.end());
}
return match;
}
int main()
{
bool done = false;
char cstr[128];
int len;
string str;
vector<int> index;
vector<string> match;
if (!ReadDictionaryFile())
return -1;
cout << "Dictionary Length: " << dictWords.size() << endl << endl;
cout << "Word: ";
cin >> cstr;
cout << endl;
len = strlen(cstr);
if (len != 0)
{
vector<string> match = PowerSet(cstr, len);
for (int i = 0; i < match.size(); i++)
{
cout << setprecision(3) << setw(3) << i << "\t";
cout << match[i] << endl;
}
cout << endl;
cout << "Total letters and/or words " << match.size() << endl;
cout << endl;
}
}
Alpha is defined as 0.5 * n. In our case n = 1, 2, 3, 4.
The solutions for the four cases above are graphed using Microsoft Mathematics:
Elliptic Partial Differential Equation for n = 1
Elliptic Partial Differential Equation for n = 2
Elliptic Partial Differential Equation for n = 3
Elliptic Partial Differential Equation for n = 4
The Microsoft Mathematics Application Window is illustrated below:
Microsoft Mathematics Application
Finite Element Method Elliptic PDF Solver Main Form 1Finite Element Method Elliptic PDF Solver Main Form 2Finite Element Method Elliptic PDF Solver Main Form 3Finite Element Method Elliptic PDF Solver Equation Class 1Finite Element Method Elliptic PDF Solver Equation Class 2
On Sunday, August 28, 2016, 3:57:53 PM I created a C# program to fit a given curve to a polynomial of a specified degree and number of points. Here are ten experiments with a continuous function to be discretely fitted by a polynomial.
I created a C# application to test the preceding equations against numerical methods of calculating the trajectory of a baseball. The baseball has an initial velocity of 90 miles per hour and an angle of inclination of 20 degrees. The classical model certainly overestimates the trajectory.
Numerical Explorations 