# My Weblog

## List with concatenation operation is monoid.

We will try to prove in Coq that list with concatenation ( ++ ) is monoid. Monoid is set S with binary operation $\bullet$ with following properties.
1). Closure.

$\forall a, b \in S,\hspace{1 mm} \exists c \in S \hspace{1 mm}such \hspace{1 mm}that\hspace{1 mm} c = a \bullet b.$

2). Associativity.

$\forall a, b, c \in S \hspace{1 mm} then \hspace{1 mm} a \bullet ( b \bullet c ) = ( a \bullet b) \bullet c$

3) Identity.

$\exists e \in S \hspace{1 mm} such \hspace{1 mm} that \hspace{1 mm} \forall a \in S,\hspace{1 mm} e \bullet a = a \bullet e = a$

Proof in Coq.

Require Import List.

Theorem Closure : forall ( X : Type ) ( l1 l2 : list X ) , exists ( l3 : list X ),
l1 ++ l2 = l3.
Proof.
intros X l1 l2. exists ( l1 ++ l2) .
reflexivity.
Qed.

Theorem Association : forall ( X : Type ) ( l1 l2 l3 : list X ) ,
l1 ++ ( l2 ++ l3 ) = ( l1 ++  l2 ) ++ l3.
Proof.
intros X l1 l2 l3.
simpl. rewrite -> app_assoc_reverse. reflexivity.
Qed.

Theorem Existence_identity_left : forall ( X : Type ) ( l : list X ), exists e,
e ++ l = l.
Proof.
intros X l. exists nil. simpl.
reflexivity.
Qed.

Theorem Existence_identify_right : forall ( X : Type ) ( l : list X ) , exists e,
l ++ e = l.
Proof.
intros X l. exists nil.
apply app_nil_r.
Qed.



We don’t need to prove the closure property because all the functions in Coq is total. See thec discussion . Source code on github[1] and other proofs from software foundation[2].
[1] https://github.com/mukeshtiwari/Coq/blob/master/Monoid.v
[2] https://github.com/mukeshtiwari/Coq/blob/master/Poly.v

September 10, 2014

## Proving Theorems in Coq

I recently started learning Coq from Software Foundations and I must say it’s highly addictive. Coq is theorem prover. When I started learning, the very first thing came to my mind is how is it possible for computer to prove theorem ?  In mathematics, a theorem is a statement that has been proven on the basis of previously established statements, such as other theorems—and generally accepted statements, such as axioms[1]. From Coq’Art, Page 3 “The Coq system is not just another programming language. It actually makes it possible to express assertion about the values being manipulated. These value may range over mathematical objects or over programs”. Great so let me write a program and prove some properties using Coq.  The definition of Boolean and couple of functions over boolean

Inductive bool : Type :=
| true : bool
| false : bool.

Definition negb ( b : bool ) : bool :=
match b with
| true => false
| false => true
end.

Definition andb ( x : bool ) ( y : bool ) : bool :=
match x with
| true => y
| false => false
end.

Definition orb ( x : bool ) ( y : bool ) : bool :=
match x with
| false => y
| true => true
end.

Definition xorb ( x : bool ) ( y : bool ) : bool :=
match x, y with
| true, true => false
| false, false => false
| _, _  => true
end.



Once we have some definitions, We can try to assert some properties about these functions. One assertion about XOR If output of xor is false it means both the inputs have same value. Lets write this theorem in Coq

Theorem xorb_equalleft : forall a b : bool, xorb a b = false -> a = b.


When Coq process this theorem, goal looks like this

1 subgoals, subgoal 1 (ID 97)

============================
forall a b : bool, xorb a b = false -> a = b



and proof of the theorem is

Proof.
intros a b H. destruct a. destruct b.
reflexivity. discriminate.
destruct b. discriminate. reflexivity.
Qed.


Once the proof is complete, we will get the message from Coq “No more subgoals. xorb_equalleft is defined”. Lets define one more assertion about AND function. If the output of andb function is true it means both inputs are true.

Theorem andbc_true : forall b c : bool, andb b c = true -> b = true /\ c = true.


Coq proof of this assertion is

Proof.
intros b c. destruct b.
Case "b = true".
simpl. intros H. split.
reflexivity.
rewrite <- H.
reflexivity.
Case "b = false".
simpl. intros H.
discriminate.
Qed.


We have proved both the theorems but what is destruct, reflexivity, split and discriminate ? These are tactics and Coq is interactive theorem prover so you have to guide it towards goal using tactics. Source code on github[2]. Famous Theorems proved in Coq[3]. Certified Dependent type programming[4]

August 22, 2014

## yatra.com left my family stranded in midnight

You want to book a room in hotel which is fully occupied? No worries. Yatra can book a room for you in even fully occupied hotel. I booked a room in Hotel President in Patna Reference No. 28051438884, Hotel confirmation number: YAT0001458518 for one day from 31-05-2014 to 01-06-2014. Train arrived five hours late in Patna, my mother and sister some how managed to reach Hotel President at 1AM. Now the biggest part of booking plan that there was no booking! The hotel staff were surprised how come Yatra was able to book a room in completely occupied hotel. The staff of Hotel President did not let my family check in because there was no confirmation of booking/payment from Yatra to Hotel. I tried to call yatra.com at 1AM but there was no response because they work from 7AM-12AM. I tried to convince hotel staff but he simply replied that there is no vacant room so we can’t accommodate. I was really worried because they were complete stranger in the city, had no place to stay and next day my sister was going to appear in entrance examination. Some how I managed to call my friend and arranged the place for their stay but the question remains, how is it possible for company like Yatra to behave like this. After looking at their page on facebook, you can see in the review section that there is no happy customer and they don’t allow you to write on their wall. I found couple of links on mouthshut and counsumercourt about their service and customer satisfaction.
I am planning to file a case against yatra.com because it’s not business but fraud. If it’s possible for you then please share this post and spread the word.

June 1, 2014

## Propositional Logic course ( Coursera )

Recently I took a logic course on Coursera. Although I am bit familiar with logic but this course was great. I got more clear grasp on Proposition logic, Proof tree, bit of Prolog and came to know about Quine–McCluskey algorithm ( in my TODO list of implementation ) while learning about circuit minimisation. In one of the assignments, we have to tell if the given formula is tautology, contradiction or contingent so I wrote some quick Haskell code to solve the assignment

import Text.Parsec.Prim
import Text.Parsec.Expr
import Text.Parsec.Char
import Text.Parsec.String ( Parser )
import Control.Applicative hiding ( ( <|> ) , many )
import qualified Data.Map as M
import Data.Maybe
import Data.List

data LExpr = Lit Char
| Not LExpr
| And LExpr LExpr
| Or LExpr LExpr
| Imp LExpr LExpr  -- ( P =>  Q )
| Red LExpr LExpr  -- ( P <=  Q )
| Eqi LExpr LExpr  -- ( P <=> Q )
deriving Show

exprCal :: Parser LExpr
exprCal = buildExpressionParser table atom

table = [  [  Prefix ( Not <$string "~" ) ] , [ Infix ( And <$ string  "&"  ) AssocLeft ]
, [  Infix  ( Or  <$string "|" ) AssocLeft ] , [ Infix ( Eqi <$ try ( string "<=>" ) ) AssocLeft
, Infix  ( Imp <$string "=>" ) AssocLeft , Infix ( Red <$  string "<="         ) AssocLeft
]
]

atom :: Parser LExpr
atom =  char '(' *>  exprCal   <* char ')'
<|> ( Lit <$> letter ) assignment :: LExpr -> [ M.Map Char Bool ] assignment expr = map ( M.fromList . zip vs ) ps where vs = variables expr ps = replicateM ( length vs ) [ True, False] variables :: LExpr -> [ Char ] variables expr = map head . group . sort . vars expr$ []  where
vars ( Lit c )    xs = c : xs
vars ( Not expr ) xs = vars expr xs
vars ( And exprf exprs ) xs = vars exprf xs ++ vars exprs xs
vars ( Or  exprf exprs ) xs = vars exprf xs ++ vars exprs xs
vars ( Imp exprf exprs ) xs = vars exprf xs ++ vars exprs xs
vars ( Red exprf exprs ) xs = vars exprf xs ++ vars exprs xs
vars ( Eqi exprf exprs ) xs = vars exprf xs ++ vars exprs xs

expEval :: LExpr -> M.Map Char  Bool -> Bool
expEval ( Lit v   )         mp =  fromMaybe False ( M.lookup v mp )
expEval ( Not expr  )       mp =  not . expEval  expr $mp expEval ( And exprf exprs ) mp = expEval exprf mp && expEval exprs mp expEval ( Or exprf exprs ) mp = expEval exprf mp || expEval exprs mp expEval ( Imp exprf exprs ) mp = ( not . expEval exprf$ mp  ) || expEval exprs mp
expEval ( Red exprf exprs ) mp =  expEval ( Imp exprs exprf  ) mp
expEval ( Eqi exprf exprs ) mp =  expEval exprf mp == expEval exprs mp

values :: LExpr -> [ Bool ]
values expr = map ( expEval expr ) ( assignment expr )

isTautology :: LExpr -> Bool
isTautology = and . values

isContradiction  = not . or . values

isContingent :: LExpr -> Bool
isContingent expr = not ( isTautology expr || isContradiction expr )

calculator :: String -> LExpr
calculator expr = case parse  exprCal ""  expr of
Left msg -> error "failed to parse"
Right val -> val

Mukeshs-MacBook-Pro:Compilers mukeshtiwari$ghci LogicPraser.hs GHCi, version 7.8.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer-gmp ... linking ... done. Loading package base ... linking ... done. [1 of 1] Compiling Main ( LogicPraser.hs, interpreted ) Ok, modules loaded: Main. *Main> calculator "p=>q|q=>p" Loading package transformers-0.3.0.0 ... linking ... done. Loading package array-0.5.0.0 ... linking ... done. Loading package deepseq-1.3.0.2 ... linking ... done. Loading package containers-0.5.5.1 ... linking ... done. Loading package bytestring-0.10.4.0 ... linking ... done. Loading package mtl-2.1.3.1 ... linking ... done. Loading package text-1.1.1.1 ... linking ... done. Loading package parsec-3.1.5 ... linking ... done. Imp (Imp (Lit 'p') (Or (Lit 'q') (Lit 'q'))) (Lit 'p') *Main> calculator "(p=>q)|(q=>p)" Or (Imp (Lit 'p') (Lit 'q')) (Imp (Lit 'q') (Lit 'p')) *Main> isTautology ( Or (Imp (Lit 'p') (Lit 'q')) (Imp (Lit 'q') (Lit 'p')) ) True *Main> calculator "(p=>q)&(q=>p)" And (Imp (Lit 'p') (Lit 'q')) (Imp (Lit 'q') (Lit 'p')) *Main> isTautology ( And (Imp (Lit 'p') (Lit 'q')) (Imp (Lit 'q') (Lit 'p')) ) False *Main> isCont ( And (Imp (Lit 'p') (Lit 'q')) (Imp (Lit 'q') (Lit 'p')) ) isContingent isContradiction *Main> isContingent ( And (Imp (Lit 'p') (Lit 'q')) (Imp (Lit 'q') (Lit 'p')) ) True *Main> isCont ( And (Imp (Lit 'p') (Lit 'q')) (Imp (Lit 'q') (Lit 'p')) ) isContingent isContradiction *Main> isContradiction ( And (Imp (Lit 'p') (Lit 'q')) (Imp (Lit 'q') (Lit 'p')) ) False  Proof tree for propositional logic was something which I learned from this course. Why proof tree method is great ? We will take a example to understand it more clearly. $p \Rightarrow ( q \vee r ) \newline s \Rightarrow ( \neg r) \newline \line(1,0){70}\newline ( p\wedge s ) \Rightarrow q$ A set of premises Δ logically entails a conclusion ϕ (written as Δ |= ϕ) if and only if every interpretation that satisfies the premises also satisfies the conclusion. Logical formulas above the line are premises and below is conclusion. $p \hspace{5 mm} q \hspace{5 mm} r \hspace{5 mm} s \hspace{5 mm} p \Rightarrow ( q \vee r ) \hspace{5 mm} s \Rightarrow ( \neg r) \hspace{5 mm}( p\wedge s ) \Rightarrow q$ $\begin{tabular}{ | l | l | l | l | p{2cm} | p{2cm} | p{2cm} | } \hline 0 & 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 1 \\ \hline \end{tabular}$ We can see from the truth table that all the interpretation which satisfy premises also satisfy conclusion. Looks great but lengthy. Proof tree rules $A \wedge B \hspace*{10mm} A \vee B \hspace*{10mm} A \Rightarrow B\hspace*{10mm} A \equiv B \hspace*{10mm} A \newline \hspace*{5mm}| \hspace*{15mm} / \hspace*{8mm} \backslash \hspace*{10mm} / \hspace*{8mm}\backslash \hspace*{10mm} / \hspace*{8mm} \backslash \hspace*{7mm} \neg A\newline \hspace*{4mm} A \hspace*{13mm} A \hspace*{8mm} B \hspace*{5mm} \neg A \hspace*{8mm}B \hspace*{6mm} A \hspace*{10mm} \neg A \hspace*{5mm} \times \newline \hspace*{4mm} B \hspace*{55mm} B\hspace*{10mm} \neg B\newline$ $\neg ( A \wedge B ) \hspace*{5mm} \neg ( A \vee B) \hspace*{5mm} \neg ( A \Rightarrow B) \hspace*{5mm} \neg ( A \equiv B) \hspace*{5mm} \neg \neg A \newline \hspace*{4mm} / \hspace*{8mm} \backslash \hspace*{15mm} | \hspace*{23mm} | \hspace*{15mm} /\hspace*{8mm} \backslash \hspace*{15mm} | \newline \hspace*{2mm} \neg A \hspace*{6mm} \neg B \hspace*{8mm} \neg A \hspace*{20mm} A \hspace*{13mm} A\hspace*{8mm} \neg A \hspace*{10mm} A \newline \hspace*{28mm} \neg B \hspace*{18mm} \neg B \hspace*{11mm} \neg B\hspace*{8mm} B \newline$ Theorem: Δ |= ϕ if and only if Δ ∪ {¬ϕ} is unsatisfiable. Following the theorem, our problem translates into proving that $p \Rightarrow ( q \vee r ) \newline s \Rightarrow ( \neg r) \newline \neg ( ( p\wedge s ) \Rightarrow q )$ is unsatisfiable. We can see that tree is closed and there is no interpretation which makes it satisfiable. Chose any branch and start moving up the tree and if it contains both A and ~A then we close the branch because it’s not possible to make A and ~A true simultaneously. You can see the course notes. . See Propositional logic library from hackage. Source code on github May 2, 2014 ## While Interpreter Now a days I am trying to learn about static program analysis and started reading the Principal of Program Analysis. In order to understand the concepts, the book introduces a small programming language While. I thought about writing interpreter for While to improve my Haskell skills. The very first task is in compilation/interpretation is breaking the source code into keywords, identifiers, operators, numbers and other tokens, known as lexical analysis. These tokens are passed to syntax analysis phase to combine the tokens into well formed expressions, statements and programs according to the grammar. The output of syntax analysis is abstract syntax tree which gives structural representation of of the input. Once we have abstract syntax tree, we can interpret or manipulate it for code generation. If you are interested in code generation then see Stephen Dielh blog. The grammar for while program is Grammar for expression a ::= x | n | - a | a opa a b ::= true | false | not b | b opb b | a opr a opa ::= + | - | * | / opb ::= and | or opr ::= > | < Grammar for statements S ::= x := a | skip | S1; S2 | ( S ) | if b then S1 else S2 | while b do S  The very first task is to define abstract syntax tree and Haskell algebraic data type makes this task almost trivial. module AST ( Opa (..), Opb (..), Opr (..), AExpr (.. ) , BExpr ( .. ), Stmt ( .. ) ) where data Opa = Add | Sub | Mul | Div deriving Show data Opb = And | Or deriving Show data Opr = Greater | Less deriving Show data AExpr = Var String | Num Integer | Neg AExpr | ABin Opa AExpr AExpr deriving Show data BExpr = Con Bool | Not BExpr | BBin Opb BExpr BExpr | AL Opr AExpr AExpr deriving Show data Stmt = List [ Stmt ] | Assing AExpr AExpr | If BExpr Stmt Stmt | While BExpr Stmt | Skip deriving Show  The next task is design the lexical analyzer. We will use Parsec for this purpose. module Lexer where import Text.Parsec import qualified Text.Parsec.Token as T import Text.Parsec.Language ( emptyDef ) import Text.Parsec.String ( Parser ) lexer :: T.TokenParser () lexer = T.makeTokenParser emptyDef { T.commentStart = "{-" , T.commentEnd = "-}" , T.reservedOpNames = [ "+", "-", "*", "/", ":=", ">", "<", "not", "and", "or" ] , T.reservedNames = [ "if", "then", "else", "while", "do", "skip", "true", "false" ] } identifier :: Parser String identifier = T.identifier lexer whiteSpace :: Parser () whiteSpace = T.whiteSpace lexer reserved :: String -> Parser () reserved = T.reserved lexer reservedOp :: String -> Parser () reservedOp = T.reservedOp lexer parens :: Parser a -> Parser a parens = T.parens lexer integer :: Parser Integer integer = T.integer lexer semi :: Parser String semi = T.semi lexer semiSep :: Parser a -> Parser [ a ] semiSep = T.semiSep lexer  Now lexical analyzer will assist the parser for creating the abstract syntax tree. A while program is list of statements and statement consists of expressions. We can build parse the expression by Parsec expression builder by giving the table of operators and associativity. module Parser ( whileParser ) where import Text.Parsec import Text.Parsec.Expr import Text.Parsec.String ( Parser ) import Control.Applicative hiding ( (<|>) ) import Lexer import AST aTable = [ [ Prefix ( Neg <$ reservedOp "-" )                ]
, [   Infix  ( ABin Mul <$reservedOp "*" ) AssocLeft , Infix ( ABin Div <$ reservedOp "/" ) AssocLeft ]
, [   Infix  ( ABin Add <$reservedOp "+" ) AssocLeft , Infix ( ABin Sub <$ reservedOp "-" ) AssocLeft ]
]

bTable = [  [  Prefix ( Not  <$reservedOp "not" ) ] , [ Infix ( BBin And <$ reservedOp "and" ) AssocLeft ]
, [  Infix  ( BBin Or  <$reservedOp "or" ) AssocLeft ] ] aExpression :: Parser AExpr aExpression = buildExpressionParser aTable aTerm where aTerm = parens aExpression <|> Var <$> identifier
<|> Num <$> integer bExpression :: Parser BExpr bExpression = buildExpressionParser bTable bTerm where bTerm = parens bExpression <|> ( Con True <$ reserved "true"  )
<|> (  Con False  <$reserved "false" ) <|> try ( AL Greater <$>  ( aExpression  <* reserved ">" )
<*>    aExpression )
<|> (  AL Less    <$> ( aExpression <* reserved "<" ) <*> aExpression ) whileParser :: Parser Stmt whileParser = whiteSpace *> stmtParser <* eof where stmtParser :: Parser Stmt stmtParser = parens stmtParser <|> List <$> sepBy stmtOne semi
stmtOne :: Parser Stmt
stmtOne =  ( Assing <$> ( Var <$> identifier )
<*> ( reserved ":=" *> aExpression ) )
<|> ( If <$> ( reserved "if" *> bExpression <* reserved "then" ) <*> stmtParser <*> ( reserved "else" *> stmtParser ) ) <|> ( While <$> ( reserved "while" *> bExpression <*  reserved "do" )
<*>   stmtParser )
<|> ( Skip <$reserved "nop" )  We have abstract syntax tree so we can interpret our program. You can think of a program as collection of commands which manipulates some memory location. module Interpreter ( evalProgram ) where import qualified Data.Map as M import AST type Store = M.Map String Integer evalA :: AExpr -> Store -> Integer evalA ( Var v ) s = M.findWithDefault 0 v s evalA ( Num n ) _ = n evalA ( Neg e ) s = - ( evalA e s ) evalA ( ABin Add e1 e2 ) s = evalA e1 s + evalA e2 s evalA ( ABin Sub e1 e2 ) s = evalA e1 s - evalA e2 s evalA ( ABin Mul e1 e2 ) s = evalA e1 s * evalA e2 s evalA ( ABin Div e1 e2 ) s = div ( evalA e1 s ) ( evalA e2 s ) evalB :: BExpr -> Store -> Bool evalB ( Con b ) _ = b evalB ( Not e ) s = not ( evalB e s ) evalB ( BBin And e1 e2 ) s = ( && ) ( evalB e1 s ) ( evalB e2 s ) evalB ( BBin Or e1 e2 ) s = ( || ) ( evalB e1 s ) ( evalB e2 s ) evalB ( AL Greater e1 e2 ) s = ( evalA e1 s ) > ( evalA e2 s ) evalB ( AL Less e1 e2 ) s = ( evalA e1 s ) < ( evalA e2 s ) interpret :: Stmt -> Store -> Store interpret ( Assing ( Var v ) expr ) s = M.insert v ( evalA expr s ) s interpret ( List [] ) s = s interpret ( List ( x : xs ) ) s = interpret ( List xs ) ( interpret x s ) interpret ( If e st1 st2 ) s | evalB e s = interpret st1 s | otherwise = interpret st2 s interpret ( While e st ) s | not t = s | otherwise = interpret ( While e st ) w where t = evalB e s w = interpret st s evalProgram :: Stmt -> Store evalProgram st = interpret st M.empty  Now every thing is complete so let’s add main and write some while program. module Main where import System.Environment import Text.Parsec import Parser import Interpreter main = do ( file : _ ) <- getArgs program <- readFile file case parse whileParser "" program of Left e -> print e >> fail "Parse Error" Right ast -> print ( evalProgram ast )  While program to compute the greatest common divisor Mukeshs-MacBook-Pro:whileinterpreter mukeshtiwari$ cat Gcd.while
a := 10 ;
b := 100  ;
while ( b > 0 ) do
(
t := b ;
b := a - ( a / b ) * b ;
a := t
)



Factorial program

Mukeshs-MacBook-Pro:whileinterpreter mukeshtiwari$cat Fact.while x := 10 ; y := x ; z := 1 ; while ( y > 1 ) do ( z := z * y ; y := y - 1 ); y := 0  Since our language doesn’t have IO instruction so we will have to see which variable store the result. In gcd program, the variable t stores the result and variable z stores the factorial of number. Mukeshs-MacBook-Pro:src mukeshtiwari$ ghc -fforce-recomp Main.hs
[1 of 5] Compiling AST              ( AST.hs, AST.o )
[2 of 5] Compiling Lexer            ( Lexer.hs, Lexer.o )
[3 of 5] Compiling Interpreter      ( Interpreter.hs, Interpreter.o )
[4 of 5] Compiling Parser           ( Parser.hs, Parser.o )
[5 of 5] Compiling Main             ( Main.hs, Main.o )
Mukeshs-MacBook-Pro:src mukeshtiwari$./Main ../Gcd.while fromList [("a",10),("b",0),("t",10)] Mukeshs-MacBook-Pro:src mukeshtiwari$ ./Main ../Fact.while
fromList [("x",10),("y",0),("z",3628800)]


I am not expert in either Haskell or compiler so if you have any comments then please let me know. The complete source code on github.

January 30, 2014

## 2013 in review

The WordPress.com stats helper monkeys prepared a 2013 annual report for this blog.

Here’s an excerpt:

The concert hall at the Sydney Opera House holds 2,700 people. This blog was viewed about 18,000 times in 2013. If it were a concert at Sydney Opera House, it would take about 7 sold-out performances for that many people to see it.

## We, after 67 years of Independence

India has been invaded by many emperors since its inception but if we start looking back from colonial era, it is mainly ruled by British from 1757 – 1947. British ruled only here because of our nature to fight with each other[Battle of Plassey]. We were declared as free country on 15 August 1947 and from that day onwards we stand as one big nation. Did we learn anything from our mistakes in the past we committed. Probably not! Out of these 67 years, Indian National Congress is ruling the country for 48 of the 60 years since independence in 1947. Now below are some facts about how we are doing as a nation or we should be proud of

[1]Scams
We are very good at this and those who raised their voice against corruption is either killed or punished[ Satyendra Dubey and Ashok Khemka]. You haven’t seen enough! Our Minister of Home affairs Sushil Kumar Shinde on coal scam “Earlier, the Bofors was a talking point. People forgot about it. Now it is coal. This too will be forgotten. Once hands are washed off coal, they again become clean.” The union minister later said he was just joking. Certainly, this is joke on us that we elected you.

[2] Poverty
Our 33% population earns less than US$1.25 per day while 81 % live on less than US$ 2.50 per day[World Bank]. Quoting from The Guardian “A pledge to eliminate poverty has figured prominently in the election campaigns of all of India’s political parties since the country gained independence in 1947. Yet the ruling Congress party – and particularly prime minister Manmohan Singh, labelled as ‘Mr Silent’ for failing to answer questions on the landmark food security bill – is accused of exploiting poverty rather than putting in place effective measures to tackle it”.

[3] Literacy
The literacy rate in India stand to 74.04% in 2011 and the level is well below the world average literacy rate of 84%, and of all nations, India currently has the largest illiterate population. There are many reasons for this but I have a strong believe that lack of strong will power among political leaders to educate their people is the major reason.

[4] Communal violence
It’s very common among India to have a fight on very petty things. We are still fighting with each other for something happened 500 years ago[ RamJamanbhoomi ] and this led to many unfortunate events including Mumbai Bomb blast, Mumbai Riots and emergence of many terrorist groups. One of our leading political party BJP always promises to construct a temple to Rama at the site. They are still not aware that we have already payed a heavy price and lost many beloved ones!

Now why I am writing this ? Today a political party named Aam Aadmi Party formally launched on 26 November 2012 by Arvind Kejriwal defeated the Indian Nation Congress in Delhi elections. This is certainly not a miracle but because of their dedication to clean the Indian politics. Delhi is ruled by Indian Nation Congress for 15 years and Delhi holds the top position in crime against women. In 2012, We had seen the worst crime in Delhi[2012 Delhi gang rape]. Delhi government and Central government tried to downplay the crime but it was the protest of common man which pushed the government to act fast. It was the anger of common man against the corrupt politicians and promise of Arvind Kejriwal to clean the politics which swept away the Indian National Congress from Delhi. I hope Arvind Kejriwal would fulfill his promise to clean the politics and I will remember this day ( 8 December 2013 ) as a day in history which changed the India and Indian Politics.

December 8, 2013

## Proving the correctness of code using SMT sovler.

SMT solvers are great tool in computer science. One use of these solvers to prove the correctness of code. I wrote Haskell code to compute the inverse of polynomial using extended euclidean algorithm in Galois Field GF ( 2n ). If we try to analyze the code of inversePolynomial then every thing is normal except passing the depth parameter. It is unnecessary because we know that recursion is going to terminate. Initially I wrote the code without depth and the program was not terminating so I wrote mail to Levent Erkok and according to him

Your code suffers from what’s known as the “symbolic termination”. See Section 7.4 of: http://goo.gl/0E7wkm for a discussion of this issue in the context of Cryptol; but the same comments apply here as well. Briefly, when you recurse in inversePoly, SBV does not know when to stop the recursion of both branches in the ite: Thus it keeps expanding both branches ad infinitum.

Luckily, symbolic termination is usually easy to deal with once you identify what the problem is. Typically, one needs to introduce a recursion depth counter, which would stop recursion when it reaches a pre-determined depth that is determined to be safe. In fact, SBV comes with the regular GCD example that talks about this issue in detail:

I’d recommend reading the source code of the above; it should clarify the problem. You’ll need to find out a recursion-depth that’ll work for your example (try running it on concrete values and coming up with a safe number); and then prove that the bound is actually correct as explained in the above. The recursion depth bound just needs to be an overapproximation: If the algorithm needs 20 iterations but you use 30 as your recursion depth, nothing will go wrong; it’ll just produce bigger code and thus be less efficient and maybe more complicated for the backend prover. Of course, if you put a depth of 18 when 20 is needed, then it will be wrong and you won’t be able to establish correctness.

My verification condition is inverse of inverse of polynomial is same as polynomial. Right now I have no idea about proving the upper bound on depth so I just took 20. If you have any suggestion/comments then please let me know.


import Data.SBV.Bridge.Z3
import Data.Maybe

inversePolynomial ::  [ Int ] -> [ Int ] ->  SWord16
inversePolynomial poly reducer =  inversePoly  reducer' rem' first' second' depth'  where
poly' =  polynomial poly :: SWord16
reducer' =  polynomial reducer :: SWord16
first' =  0 :: SWord16
second' = 1 :: SWord16
depth' = 20 :: SWord16
( quot', rem' ) = pDivMod poly' reducer'

inversePoly :: SWord16 -> SWord16 -> SWord16 -> SWord16  ->  SWord16 -> SWord16
inversePoly  reducer'' poly'' first'' second'' depth'' =
ite ( depth'' .== 0 ||| rem'' .== ( 0 :: SWord16 ) ) (  second'' )  ( inversePoly  poly'' rem'' second'' ret ( depth'' - 1 ) )  where
( quot'', rem'' ) = pDivMod reducer'' poly''
ret =  pAdd ( pMult ( quot'', second'', [] ) ) first''

isInversePolyCorrect :: SWord16 -> SWord16 -> SBool
isInversePolyCorrect poly reducer = inversePoly reducer ( inversePoly reducer rem first second depth ) first second depth .== rem where
( quot, rem ) = pDivMod  poly reducer
first = 0 :: SWord16
second = 1 :: SWord16
depth = 20 :: SWord16

*Main> showPoly ( inversePolynomial  [ 6,4,1,0] [ 8,4,3,1,0] )
"x^7 + x^6 + x^3 + x"
*Main> proveWith z3 { verbose=False }  $forAll ["x"]$ \x ->  pMod x ( 283 :: SWord16 ) ./=0 ==> isInversePolyCorrect x ( 283 :: SWord16 )
Q.E.D.
*Main> proveWith cvc4 { verbose=False }  $forAll ["x"]$ \x ->  pMod x ( 283 :: SWord16 ) ./=0 ==> isInversePolyCorrect x ( 283 :: SWord16 )
*** Exception: fd:10: hFlush: resource vanished (Broken pipe)
*Main> proveWith yices { verbose=False }  $forAll ["x"]$ \x ->  pMod x ( 283 :: SWord16 ) ./=0 ==> isInversePolyCorrect x ( 283 :: SWord16 )
Q.E.D.


September 21, 2013

## Counting Inversions

First week of algorithm course programming assignment is to compute the number of inversion count of given array. We have a given array A[0…n – 1] of n distinct positive integers and we have to count all the pairs ( i , j ) such that i < j and A[i] > A[j].
1. Brute Force
Brute force idea is very simple. Run two loops and keep counting the respective pairs for which i < j and A[i] > A[j]. The complexity of brute force is O( n2 ).

bruteForce :: [ Int ] -> Int
bruteForce [] = 0
bruteForce [_] = 0
bruteForce ( x : xs ) = ( length . filter ( x > ) $xs ) + bruteForce xs *Main> bruteForce [ 3 , 1 , 2] Loading package array-0.4.0.1 ... linking ... done. Loading package deepseq-1.3.0.1 ... linking ... done. Loading package bytestring-0.10.0.0 ... linking ... done. 2 *Main> bruteForce [ 2,3,8,6,1] 5  2. Divide and Conquer We can modify the merge sort to compute the inversion count. The idea is to compute the inversion count of during the merge phase. Inversion count of array C resulting from merging two sorted array A and B is sum of inversion count of A, inversion count of B and their cross inversion. The inversion count of single element is 0 ( base case ). The only trick is compute the cross inversion. What will be the inversion count of following array if we merge them ? A = [ 1 , 2 , 3 ] B = [ 4 , 5 , 6 ] Clearly 0 because all element of A is less than the first element of B. What is the inversion count of C = [ 4 , 5 , 6 ] and D = [ 1 , 2 , 3 ] ? First element of D ( 1 ) is less than first element of C ( 4 ) so we have total three pairs ( 4 , 1 ) ( 5 , 1 ) ( 6 , 1). Similarly for 2 and 3 ( total 9 inversion count ). It reveals that if we have any element y in D is less than certain element x in C then all the element of A after x will also be greater than y and we will have those many cross inversion counts. More clear explanation. Wrote a quick Haskell code and got the answer. import Data.List import Data.Maybe ( fromJust ) import qualified Data.ByteString.Lazy.Char8 as BS inversionCnt :: [ Int ] -> [ Int ] -> [ Int ] -> Int -> ( Int , [ Int ] ) inversionCnt [] ys ret cnt = ( cnt , reverse ret ++ ys ) inversionCnt xs [] ret cnt = ( cnt , reverse ret ++ xs ) inversionCnt u@( x : xs ) v@( y : ys ) ret cnt | x <= y = inversionCnt xs v ( x : ret ) cnt | otherwise = inversionCnt u ys ( y : ret ) ( cnt + length u ) merge :: ( Int , [ Int ] ) -> ( Int , [ Int ] ) -> ( Int , [ Int ] ) merge ( cnt_1 , xs ) ( cnt_2 , ys ) = ( cnt_1 + cnt_2 + cnt , ret ) where ( cnt , ret ) = inversionCnt xs ys [] 0 mergeSort :: [ Int ] -> ( Int , [ Int ] ) mergeSort [ x ] = ( 0 , [ x ] ) mergeSort xs = merge ( mergeSort xs' ) ( mergeSort ys' ) where ( xs' , ys' ) = splitAt ( div ( length xs ) 2 ) xs main = BS.interact$ ( \x -> BS.pack $show x ++ "\n" ) . fst . mergeSort . map ( fst . fromJust . BS.readInt :: BS.ByteString -> Int ) . BS.lines Mukeshs-MacBook-Pro:Puzzles mukeshtiwari$ ghc  -fforce-recomp -O2  Inversion.hs
[1 of 1] Compiling Main             ( Inversion.hs, Inversion.o )
Mukeshs-MacBook-Pro:Puzzles mukeshtiwari$time ./Inversion < IntegerArray.txt 2407905288 real 0m8.002s user 0m7.881s sys 0m0.038s  I tried to compare this solution with GeeksforGeeks‘s solution written in C and it was a complete surprise to me. Mukeshs-MacBook-Pro:Puzzles mukeshtiwari$ time ./a.out <  IntegerArray.txt
Number of inversions are 2407905288

real	0m0.039s
user	0m0.036s
sys	0m0.002s


Almost 200 times faster! My Haskell intuition told me that I am doing certainly some wrong. I started analyzing the code and got the point. In inversionCnt function, I am doing extra work for computing the length of list u ( O ( n ) ) so rather than doing this extra work, I can pass the length.

import Data.List
import Data.Maybe ( fromJust )
import qualified Data.ByteString.Lazy.Char8 as BS

inversionCnt :: [ Int ] -> [ Int ] -> [ Int ] -> Int -> Int -> ( Int , [ Int ] )
inversionCnt [] ys  ret _ cnt = ( cnt , reverse ret ++ ys )
inversionCnt xs []  ret _ cnt = ( cnt , reverse ret ++ xs )
inversionCnt u@( x : xs ) v@( y : ys )  ret n cnt
| x <= y = inversionCnt xs v  (  x : ret   )  ( pred n ) cnt
| otherwise = inversionCnt u ys  ( y : ret ) n ( cnt + n )

merge ::  ( Int , [ Int ] ) -> ( Int , [ Int ] ) -> ( Int , [ Int ] )
merge ( cnt_1 , xs ) ( cnt_2 , ys ) = ( cnt_1 + cnt_2 + cnt , ret ) where
( cnt , ret ) = inversionCnt xs ys [] ( length xs )  0

mergeSort ::  [ Int ] -> ( Int , [ Int ] )
mergeSort [ x ] = ( 0 , [ x ] )
mergeSort  xs   = merge ( mergeSort xs' ) ( mergeSort ys' ) where
( xs' , ys' ) = splitAt ( div ( length xs ) 2 ) xs

main =  BS.interact $( \x -> BS.pack$ show x ++ "\n" ) . fst
.  mergeSort . map ( fst . fromJust .  BS.readInt :: BS.ByteString -> Int  )
. BS.lines

Mukeshs-MacBook-Pro:Puzzles mukeshtiwari\$ time ./Inversion <  IntegerArray.txt
2407905288

real	0m0.539s
user	0m0.428s
sys	0m0.031s


Now 10 times slower and still lot of room for improvement but for now I am happy with this :). You can also try to solve INVCNT. Run this code on Ideone.

July 4, 2013

## Computing a^n using binary representation of natural number in Agda

While following the Agda tutorial, I wrote this code to compute an. Not very elegant example probably because I am not expert and still learning about dependent types. Maybe after getting some more experience and knowledge, I will try to prove the correctness and complexity of this algorithm. See the post of Twan van Laarhoven about proving correctness and complexity of merge sort in Agda.


module PowerFunction where

data ℕ⁺ : Set where
one : ℕ⁺
double : ℕ⁺ → ℕ⁺
double⁺¹ : ℕ⁺ → ℕ⁺

add : ℕ⁺ → ℕ⁺ → ℕ⁺
add one one = double one
add one ( double x ) = double⁺¹ x
add one ( double⁺¹ x ) = double ( add x one )
add ( double x ) one =  double⁺¹ x
add ( double x ) ( double y ) = double ( add x y )
add ( double x ) (  double⁺¹ y ) =  double⁺¹ (  add x y )
add (  double⁺¹ x ) one =   double (  add x one )
add (  double⁺¹ x ) (  double y ) =  double⁺¹ (  add x y  )
add (  double⁺¹ x ) (  double⁺¹ y ) = double ( add (  add x y ) one )

mult : ℕ⁺ → ℕ⁺ → ℕ⁺
mult x one = x
mult x ( double y ) = double ( mult x y )
mult x ( double⁺¹ y ) = add x ( double ( mult x y ) )

pow : ℕ⁺ → ℕ⁺ → ℕ⁺
pow one _ = one
pow x  one =  x
pow x ( double one ) = mult x x
pow x ( double y ) =  pow ( pow x y ) ( double one )
pow x ( double⁺¹ y ) =  mult x ( pow ( pow x y ) ( double one ) )



Computing 2 ^ 6
pow ( double one ) ( double ( double⁺¹ one ))
double (double (double (double (double (double one)))))
If you have suggestion then please let me know.

June 12, 2013