Go lang: Return True specified Pct% of time

When writing test for state machines, I came across a need for a helper that would all call flow into a path some percentage% of times.

Eg:

func TestSelfDrivingCar_BasicRun(t *testing.T) {
    car := NewMockCar(t)
    car.Start()
    if (/* 50% probability that this will happen */) {
        car.SimulateLidarFailure()        
    } else if /* 25% probability that this will happen */ {
        car.SimulateVisionFailure()
    } else {
        /* 25% probability that this will happen*/
        car.SimulateGearBoxFailure()
    }
    car.VerifySafeTrajectory()
}

You can write a small Go helper like this:

package testutils

import (
  "math/rand"
  "time"
)

// Odds returns True to pct% number of Check() calls
type Odds interface {
  Check() bool
}

// odds returns `true` pct% of times
type odds struct {
  randgen *rand.Rand
  pct     int
}

// NewOdds returns a odds instance with the given percent vaule
func NewOdds(pct int) Odds {
  src := rand.NewSource(time.Now().UnixNano())
  return &odds{
    randgen: rand.New(src),
    pct:     pct,
  }
}

// Check returns True to pct% of Check() calls
func (o *odds) Check() bool {
  return o.randgen.Intn(100) < o.pct
}

Usage:

func TestSelfDrivingCar_BasicRun(t *testing.T) {
    odds50pct, odds25pct := NewOdds(50), NewOdds(25)
    car := NewMockCar(t)
    car.Start()
    if odds50pct.Check() {
        car.SimulateLidarFailure()        
    } else if odds25pct.Check() {
        car.SimulateVisionFailure()
    } else {
        /* 25% probability that this will happen*/
        car.SimulateGearBoxFailure()
    }
    car.VerifySafeTrajectory()
}

ps: No, I don’t work on self-driving cars – this is just a code sample ūüôā


		
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InFix to PostFix and PreFix conversion [C++]

To Convert from InFix to Postfix:

  • Print out constants from InFix directly onto the Postfix output
  • For operators, insert them into a stack
  • ¬† ¬† Before insertion, make sure to pop out (and print to Postfix output) all operator that have higher precedence than that of the operator currently being inserted.
  • After the entire Infix is processed, flush the stack (if it is not empty) onto the Postfix output

-In case of InFix to Prefix, we must start processing the string from the RIGHT most to LEFT most instead of the usual LEFT to RIGHT.

-This code can handle only numbers from 0-9 (single digit) and +-*/ operators.

//Evaluate the given expression
#include<iostream>
#include<stack>
#include<vector>
using namespace std;

//Only single digit numbers permitted

int GetPrecedence(char op)
{
 switch (op)
 {
 case '+':
 case '-':
 return 1;
 case '*':
 case '/':
 default:
 return 2;
 }
}
char* ConvertInorderToPreOrder(char* InOrderExp, int Size)
{
 stack<char> Operators;
 vector<char> PostOrderResult;
 for (int i = Size-1; i >=0; i--)
 if (InOrderExp[i] >= '0'&& InOrderExp[i] <= '9')
 PostOrderResult.push_back(InOrderExp[i]);
 else
 {
 while (!Operators.empty() && GetPrecedence(Operators.top()) >= GetPrecedence(InOrderExp[i]))
 {
 PostOrderResult.push_back(Operators.top());
 Operators.pop();
 }
 Operators.push(InOrderExp[i]);
 }
 while (!Operators.empty())
 {
 PostOrderResult.push_back(Operators.top());
 Operators.pop();
 }
 char *ResultString = (char*)malloc(sizeof(char)*PostOrderResult.size()+1);
 memset(ResultString, 0, sizeof(char)*(PostOrderResult.size() + 1));
 for (int i = 0; i < PostOrderResult.size(); i++)
 ResultString[i] = PostOrderResult[i];
 ResultString[PostOrderResult.size()] = '\0';
 _strrev(ResultString);
 return ResultString;
}
char* ConvertInorderToPostOrder(char* InOrderExp, int Size)
{
 stack<char> Operators;
 vector<char> PostOrderResult;
 for (int i = 0; i < Size; i++)
 if (InOrderExp[i] >= '0'&& InOrderExp[i] <= '9')
 PostOrderResult.push_back(InOrderExp[i]);
 else
 {
 while (!Operators.empty() && GetPrecedence(Operators.top()) >= GetPrecedence(InOrderExp[i]))
 {
 PostOrderResult.push_back(Operators.top());
 Operators.pop();
 }
 Operators.push(InOrderExp[i]);
 }
 while (!Operators.empty())
 {
 PostOrderResult.push_back(Operators.top());
 Operators.pop();
 }
 char *ResultString = (char*)malloc(sizeof(char)*PostOrderResult.size()+1);
 memset(ResultString, 0, sizeof(char)*(PostOrderResult.size()+1));
 for (int i = 0; i < PostOrderResult.size(); i++)
 ResultString[i] = PostOrderResult[i];
 ResultString[PostOrderResult.size()] = '\0';
 return ResultString;
}

int main()
{

char InExp[] = "2+3*7-5*2-4/2*5+1-4+2/2+1";
 cout << "\nInput: " << InExp;
 cout << "\n\n";
 cout << "\n\PostFix = " << ConvertInorderToPostOrder(InExp, strlen(InExp));
 cout << "\n\PreFix = " << ConvertInorderToPreOrder(InExp, strlen(InExp));
 cout << "\n\nEnd of code\n\n";
 system("pause");
 return 0;
}

Anagram Solver

I was coding out a simple string permuting function and I thought of writing out an AnagramSolver just for completion.

The Dictionary can be provided as a wordlist in the form of a text file with a word string per line. You can find several word lists here: http://wordlist.sourceforge.net/


//Sources
#include<iostream>
#include<string>
#include<fstream>
#include<map>

using namespace std;
class AnagramChecker
{
public:
map<string, bool> Dictionary;
map<string, bool> ResultList;

//Recursive string permuter
void RecurveStrPerm(string Buffer, string Test, int Cur)
{
 if (Cur >= Test.length())
 {
 if (Dictionary.count(Buffer) > 0)
 if (ResultList.count(Buffer) == 0)
 ResultList[Buffer] = true;
 return;
 }

for(int i = 0; i <= Buffer.length(); i++)
 {
 Buffer.insert(i, 1, Test[Cur]);
 RecurveStrPerm(Buffer, Test, Cur + 1);
 Buffer.erase(i, 1);
 }
}

//Build a table out of the strings
void BuildInMemDic()
{
 Dictionary.clear();
 ifstream DicReader;
 DicReader.open("WordList.txt");
 string CurrentWord= "";
 while (!DicReader.eof())
 {
 getline(DicReader, CurrentWord);
 for (int i = 0; i < CurrentWord.length(); i++)
 CurrentWord[i] = tolower(CurrentWord[i]);
 Dictionary[CurrentWord] = true;

 }
 DicReader.close();
}

//Get Result
void GetResult()
{
 cout << "\nAnagrams: \n";
 for (map<string, bool>::iterator ResultListPtr = ResultList.begin(); ResultListPtr != ResultList.end(); ResultListPtr++)
 cout << "\n" << ResultListPtr->first;

}

public:

AnagramChecker()
 {
 BuildInMemDic();
 }

void Find(string Test)
 {
 ResultList.clear();
 int cur = 0, n = Test.length();
 RecurveStrPerm("", Test, 0);
 GetResult();
 }

};

void main()
{
 string Test = "Slate";
 cout << "\nBuilding In memory Dictionary...";
 AnagramChecker AnaObj;
 cout << "\n\nInmemory dictionary built!...\n\n";

char ExitChoice = 'n';
 while (ExitChoice!='y')
 {
 cout << "\n\nEnter Anagram: ";
 cin >> Test;
 for (int i = 0; i < Test.length(); i++)
 Test[i] = tolower(Test[i]);

cout << "\n\nAnagrams for " << Test << ":\n\n";
 AnaObj.Find(Test);
 cout << "\n\nDo you want to continue: y /n :";
 cin >> ExitChoice;

 }
 cout << "\nEnd of code\n";
 system("pause");
 return;
}

The code is NOT optimized. It can be sped up with simple multi-threading, but I have let go of those for simplicity.

Some Binary Tree Tricks

This code can do the following:

  • Depth First Search of a Binary Tree
  • Breadth¬†First Search of a Binary Tree
  • Finding sum of numbers formed by a path from root to leaf, in a binary tree that can contain numbers from 0-9.
  • Mirror a binary tree

I have made use of STL – Queues and vectors here.

</pre>
//Source file

#include<iostream>
#include<queue>
#include<vector>
using namespace std;

class Node
{
public:
 int Value;
 Node *Left, *Right;

 Node(int ValueParam)
 {
 Value = ValueParam;
 Left = Right = NULL;
 }
};

class BTree
{
 private:
 Node* Root;

void _Insert(Node* Ptr, int ValueParam)
 {
 if (Ptr->Value == ValueParam) return;
 if (ValueParam < Ptr->Value)
 if (Ptr->Left == NULL) Ptr->Left = new Node(ValueParam);
 else _Insert(Ptr->Left,ValueParam);
 else if (ValueParam >Ptr->Value)
 if (Ptr->Right == NULL) Ptr->Right= new Node(ValueParam);
 else _Insert(Ptr->Right, ValueParam);
 }

void _DFS(Node* Ptr)
 {
 if (!Ptr) return;
 _DFS(Ptr->Left);
 cout << " - " << Ptr->Value;
 _DFS(Ptr->Right);
 }

//Mirroring a Binary Tree
 void _Mirror(Node* Ptr)
 {
 if (!Ptr) return;
 _Mirror(Ptr->Left);
 _Mirror(Ptr->Right);
 Node* Temp = Ptr->Left;
 Ptr->Left = Ptr->Right;
 Ptr->Right = Temp;
 }

public:
 BTree()
 {
 Root = NULL;
 }

void Insert(int ValueParam)
 {
 if (Root == NULL) Root = new Node(ValueParam);
 else _Insert(Root, ValueParam);
 }

void DFS()
 {
 cout << "\nDFS:\n";
 _DFS(Root);
 }
 //Breadth First Search
 queue<Node*> NodalQueue;
 void BFS()
 {
 cout << "\nBFS:\n";
 //Empty the queue
 while (!NodalQueue.empty())
 NodalQueue.pop();

if (!Root) return;
 else
 {
 cout << " - " << Root->Value;
 NodalQueue.push(Root);
 }

int CLev = 1, NLev = 0;

//Start working
 while (!NodalQueue.empty())
 {
 Node* Ptr = NodalQueue.front();
 NodalQueue.pop();
 CLev--;

&nbsp;

if (CLev <= 0)
 {
 cout << endl;
 CLev = NLev;
 NLev = 0;
 }

 if (Ptr->Left)
 {
 cout << " - " << Ptr->Left->Value;
 NodalQueue.push(Ptr->Left);
 NLev++;
 }
 if (Ptr->Right)
 {
 cout << " - " << Ptr->Right->Value;
 NodalQueue.push(Ptr->Right);
 NLev++;
 }

}
 }

//Get Numbers
 vector<vector<int>> NumberMap;
 void GetNumberMap()
 {
 cout << "\nNumberMap:\n";
 NumberMap.clear();
 queue<Node*> NumNodalQueue;
 if (!Root) return;
 else
 {
 NumNodalQueue.push(Root);
 vector<int> RootStage;
 RootStage.push_back(Root->Value);
 NumberMap.push_back(RootStage);
 }

//Start working
 int Stage = 0;
 int CLev = 1;
 int NLev = 0;
 while (!NumNodalQueue.empty())
 {
 Node* Ptr = NumNodalQueue.front();
 vector<int> NewStage;
 NumNodalQueue.pop();
 CLev--;
 if (CLev<=0)
 {
 Stage++;
 NumberMap.push_back(NewStage);
 CLev = NLev;
 NLev = 0;
 }
 if (Ptr->Left)
 {
 NumberMap[Stage].push_back(Ptr->Left->Value);
 NumNodalQueue.push(Ptr->Left);
 NLev++;
 }
 if (Ptr->Right)
 {
 NumberMap[Stage].push_back(Ptr->Right->Value);
 NumNodalQueue.push(Ptr->Right);
 NLev++;
 }
 }

//Numbermap is ready
 //Calcualte combinations from NumberMap
 NumberMap.pop_back();
 GetCombinations();
 }

void GetCombinations()
 {
 GlobalTot = 0;
 RecCombi();
 }

long int fnPow(int Base, int Power)
 {
 if (Power == 0) return 1;
 return Base*fnPow(Base, Power - 1);
 }

long int GlobalTot = 0;
 void RecCombi(vector<int>* Temp=NULL, int Stage=0)
 {
 if (Temp && Stage == NumberMap.size())
 {
 cout << "\n";
 for (int i = 0; i < Temp->size(); i++)
 {
 long int pval = fnPow(10, (Temp->size() - 1 - i));
 GlobalTot += (*Temp)[i]*(pval);
 cout << " " << (*Temp)[i];
 }
 cout << "\n";
 //Adding number to Global
 }
 if (Stage < NumberMap.size())
 {
 if (!Temp) Temp = new vector<int>();
 for (int i = 0; i < NumberMap[Stage].size(); i++)
 {
 Temp->push_back(NumberMap[Stage][i]);
 RecCombi(Temp, Stage + 1);
 Temp->pop_back();
 }
 }
 }

void Mirror()
 {
 _Mirror(Root);
 cout << "\nTree Mirrored\n";
 }

};

int main()
{
 BTree MyTree;

MyTree.Insert(6);
 MyTree.Insert(2);
 MyTree.Insert(10);
 MyTree.Insert(0);
 MyTree.Insert(4);
 MyTree.Insert(8);
 MyTree.Insert(15);
 MyTree.Insert(-1);
 MyTree.Insert(1);
 MyTree.Insert(3);
 MyTree.Insert(5);
 MyTree.Insert(7);
 MyTree.Insert(9);
 MyTree.Insert(12);
 MyTree.Insert(20);
 MyTree.DFS();
 MyTree.BFS();
 MyTree.Mirror();
 MyTree.DFS();
 MyTree.BFS();
 MyTree.GetNumberMap();

cout << "\nGloabl total =" << MyTree.GlobalTot << endl;
 cout << "\n\nEnd of code\n\n";
 system("pause");

return 0;
}

&nbsp;
<pre>

Please note: The third problem (finding sum of numbers) does not require a binary search tree which has been used here. A simple binary tree would do. But the code would be the same whether the tree is a BST or not.

Finding the square root (of a perfect square) using binary search c#

Well, the logic is pretty straightforward. We start a binary search from a range of 0 to NUM, where NUM is the number whose root we are looking for. Each time, we calculate the middle item of the range and see if it is the square root. If not, we search further either in the right side or the left side of the mid item depending on whether the mid^2 is lesser or greater than NUM.

double GetRoot(double Num,double High=0, double Low=0)
{
if(High&lt;Low || High-Low==1) return -1; //End case
if(High==Low &amp;&amp; Low==0) High=Num; //Start case
int Mid= Low+((High-Low)/2);
if(Mid*Mid==Num) return Mid;
else
{
if(Mid*Mid&gt;Num) return GetRoot(Num,Mid,Low);
else if(Mid*Mid&lt;Num) return GetRoot(Num,High,Mid);
}
}

If you have thoughts/ suggestions, post em on comments.