# Author Archives: Techbuzz admin

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## Top Viva questions for Theory Of computation

#credit careerride.com

Theory of Computation questions and answers

(1) From the options given below, the pair having different expressive power is

(A) Deterministic Push Down Automata (DPDA) and Non-deterministic Push Down Automata (NPDA)

(B) Deterministic Finite Automata (DFA) and Non-deterministic Finite Automata(NFA)

(C) Single tape turning machine and multi tape turning machine.

(D) Deterministic single tape turning machine and Non-Deterministic single tape turning machine

View Answer / Hide Answer

ANSWER: Deterministic Push Down Automata (DPDA) and Non-deterministic Push Down Automata (NPDA)

(2) The problem that is undecidable –

(A) Finiteness problem for FSA’s

(B) Membership problem for CFG’s

(C) Equivalence problem for FSA’s

(D) Ambiguity problem for CFG’s

View Answer / Hide Answer

ANSWER: Ambiguity problem for CFG’s

(3) The language which is generated by the grammar S-> aSa I bSb I a I b over the alphabet {a, b} is the set of

(A) Strings that begin and end with the same symbol

(B) All odd and even length palindromes

(C) All odd length palindromes

(D) All even length palindromes

View Answer / Hide Answer

ANSWER: All odd length palindromes

(4) Two persons X and Y have been asked to show that a certain problem p is NP-complete. X shows a polynomial time reduction from the 3-SAT problem to p and Y shows a polynomial time reduction from p to 3-SAT. From these reduction it can be inferred that

(A) π is NP-complete

(B) π is NP-hard but not NP-complete

(C) π is in NP but not NP-complete

(D) π is neither NP-hard nor in NP

View Answer / Hide Answer

ANSWER: π is NP-complete

(5) Out of the three problems S, Q and R, S is an NP-complete problem and Q and R are the two other problems not known to be in NP. Which one of the following statements is true if Q is polynomial time reducible to S and S is the polynomial time reducible to R?

(A) Q is NP-complete

(B) R is NP-complete

(C) Q is NP-hard

(D) R is NP-hard

View Answer / Hide Answer

ANSWER: R is NP-complete

(6) From the options given below the statement, which is not necessarily true if X1 is the recursive language and X2 and X3 are the languages that is recursively enumerable but not recursive is

(A) X2 ∩ X1 is recursively enumerable

(B) X2 ∪ X1 is recursively enumerable

(C) X2 – X1 is recursively enumerable

(D) X1 – X3 is recursively enumerable

View Answer / Hide Answer

ANSWER: X1 – X3 is recursively enumerable

(7) For the language {ap I P is a prime}, the statement which hold true is

(A) It is not regular but context free

(B) It is regular but not context free

(C) It is neither regular nor context free, but accepted by a turing machine

(D) It is not accepted by turing machine

View Answer / Hide Answer

ANSWER: It is neither regular nor context free, but accepted by a turing machine

(8) The statement that holds true is

(A) Infinite union of finite sets is regular

(B) The union of two non-regular set is not regular

(C) Every finite subset of a non-regular set is regular

(D) Every subset of a regular set is regular

View Answer / Hide Answer

ANSWER: Every finite subset of a non-regular set is regular

(9) The language described by the regular expression (0+1)*0(0+1)*0(0+1)* over the alphabet {0 1} is the set of

(A) All strings containing at least two 1’s

(B) All strings containing at least two 0’s

(C) All strings that begin and end with either 0’s or 1’s

(D) All strings containing the substring 00

View Answer / Hide Answer

ANSWER: All strings containing at least two 0’s

(10) 3-SAT and 2-SAT problems are

(A) NP-complete and in P respectively

(B) Undecidable and NP-complete

(C) Both NP-complete

(D) Both in P

View Answer / Hide Answer

ANSWER: NP-complete and in P respectively

(11) Which one of the following statement is true?

(A) The intersection of two context free languages is context free

(B) A context free language can always be accepted by a deterministic push down automaton

(C) The union of two context free languages is context free

(D) The complement of a context free language is context free.

View Answer / Hide Answer

ANSWER: The union of two context free languages is context free

(12) Let n be the positive integer constant and L be the language with alphabet {a}. To recognize L the minimum number of states required in a DFA will be

(A) 2k + 1

(B) k + 1

(C) 2n + 1

(D) n + 1

View Answer / Hide Answer

ANSWER: n + 1

(13) Consider a stack, which is limited to 10 items. The language accepted by a push- down automaton in such stack is best described as

(A) Regular

(B) Deterministic context free

(C) Context free

(D) Recursive

View Answer / Hide Answer

ANSWER: Regular

(14) Which one of the following statement is true if L denotes the language generated by the grammar S->0S0/00?

(A) L is not context free

(B) L is regular but not 0+

(C) L = 0+

(D) L is context free but not regular

View Answer / Hide Answer

ANSWER: L is regular but not 0+

(15) Consider the regular expression 0 * (10 *) which is similar to the same set as

(A) 0 + (0 + 10) *

(B) (0 +1) * 10 (0 + 1) *

(C) (1 * 0) * 1*

(D) None of the above

View Answer / Hide Answer

ANSWER: None of the above

(16) W is any string whose length is n in {0, 1}* and L is the set of all sub-strings of W. The minimum number of states in a non-deterministic finite automaton that accepts L is

(A) n

(B) 2n

(C) n + 1

(D) n – 1

View Answer / Hide Answer

ANSWER: n + 1

(17) The DFA shown below accepts the set of all strings over {0, 1} that

(A) End with 00

(B) End with 0

(C) Begin either with 0 or 1

(D) Contain the substring 00

View Answer / Hide Answer

ANSWER: End with 00

(18) We have two statements S1 and S2 whose definition are as follows:

S1 – {02n In ≥ I} is a regular language.

S2 – {0m 1n 0 1m+n Im=1 and n≥1I is a regular language.

Which one of the following statements is correct?

(A) Both S1 and S2 are correct

(B) Only S2 is correct

(C) Only S1 is correct

(D) Neither S1 nor S2 is correct

View Answer / Hide Answer

ANSWER: Only S1 is correct

(19) Consider a string s over (0+1)*. The number of 0’s in s is denoted by no(s) and the number of 1’s in s is denoted by n1(s). The language that is not regular is

(A) L = {s ∈ (0+1)* I for every prefix s’ of s, I no(s’)-n1(s’) I ≤ 2}

(B) L = {s ∈ (0+1)* I no(s) mod 7 = n1(s) mod 5 = 0}

(C) L = {s ∈ (0+1)* I no(s) is a 3 digit prime}

(D) L = {s ∈ (0+1)* I no(s)-n1(s) I ≤ 4

View Answer / Hide Answer

ANSWER: L = {s ∈ (0+1)* I no(s)-n1(s) I ≤ 4

(20) Which one of the following statement is false?

(A) In Chomsky Normal Form the derivative trees of strings generated by a context-free grammar are always binary trees

(B) If W is the string of a terminals and Y is a non-terminal, the language generated by a context free grammar, all of whose productions are of the form x->W or X->WY is always regular

(C) By using suitable transformation all ε-productions can be removed from any context-free grammar.

(D) Every left recursive grammar can be converted to a right recursive grammar and vice-versa

View Answer / Hide Answer

ANSWER: If W is the string of a terminals and Y is a non-terminal, the language generated by a context free grammar, all of whose productions are of the form x->W or X->WY is

always regular

(21) State table of an FSM is given below. There are two states A And B, one input and one output.

Let the initial state be A = 0 and B = 0. To take the machine to the state A = 0 and B = 1 with output = 1 the minimum length of input string required will be

(A) 2

(B) 7

(C) 4

(D) 3

View Answer / Hide Answer

ANSWER: 3

For questions 22 and 23 refer to the data given below:

The figure shown below is a finite state automaton

(22) Which one of the following is true for this automaton?

(A) b*ab*ab*ab*

(B) b*a(a+b)*

(C) b*ab*ab*

(D) (a+b)*

View Answer / Hide Answer

ANSWER: b*a(a+b)*

(23) For the above FSA the equivalent minimum state automaton has the following number of states

(A) 1

(B) 2

(C) 3

(D) 4

View Answer / Hide Answer

ANSWER: 2

(24) Out of the three decision problems P1, P2 and P3, P1 is decidable and P2 is undecidable. The statement that holds true is

(A) P3 is decidable if P3 is reducible to compliment of P2

(B) P3 is decidable if P1 is reducible to P3

(C) P3 is undecidable if P1 is reducible to P3

(D) P3 is undecidable if P2 is reducible to P3

View Answer / Hide Answer

ANSWER: P3 is undecidable if P2 is reducible to P3

(25) G = {S->SS, S->ab, S->ba, S->?} is the context free grammar whose statements are given below:

a. G is ambiguous

b. G produces all strings with equal number of a’s and b’s.

c. Deterministic PDA accepts G

Which of the following statement is true about G?

(A) a, b, c all are true

(B) Only and b are true

(C) Only b and c are true

(D) Only a is true

View Answer / Hide Answer

ANSWER: a, b, c all are true

(26) The minimum number of states in any DFA accepting the regular language L = (111+11111)* is

(A) 5

(B) 7

(C) 9

(D) 11

View Answer / Hide Answer

ANSWER: 9

(27) Consider the language L = {W I W ∈ {0, 1}*, where 0’s and 1’s in W are divisible by 3 and 5 respectively. The minimum state deterministic finite automaton accepting the language L has

(A) 20 states

(B) 5 states

(C) 10 states

(D) 15 states

View Answer / Hide Answer

ANSWER: 15 states

(28) We have an undirected graph G(V, E) with two problems given below:

α – Does G have an independent set of size IVI – 4?

β – Does G have an independent set of size 5?

The statement that holds true is

(A) α is NP-complete and β is in P

(B) α is in P and β is NP-complete

(C) Both α and β are NP-complete

(D) Both α and β are in P

View Answer / Hide Answer

ANSWER: α is in P and β is NP-complete

(29) Figure shows deterministic finite state automaton M. Let the set of seven bit binary strings whose 1st, 4th and the last bits are 1 is denoted by S. How many strings in S is accepted by M?

(A) 1

(B) 9

(C) 3

(D) 5

View Answer / Hide Answer

ANSWER: 5

(30) Which one of the following statement is true for a regular language L over {a} whose minimal finite state automation has two states?

(A) L must be either {an I n is odd} or {an I n is even}

(B) L must be {an I n is odd}

(C) L must be {an I n is even}

(D) L must be {an I n = 0}

View Answer / Hide Answer

ANSWER: L must be either {an I n is odd} or {an I n is even}

(31) Consider a graph G = (V, E) where I V I is divisible by 3. The problem of finding a Hamiltonian cycle in a graph is denoted by SHAM3 and the problem of determining if a Hamiltonian cycle exits in such graph is denoted by DHAM3. The option, which holds true, is

(A) Only DHAM3 is NP-hard

(B) Only SHAM3 is NP-hard

(C) Both SHAM3 and DHAM3 are NP-hard

(D) Neither SHAM3 nor DHAM3 is NP-hard

View Answer / Hide Answer

ANSWER: Both SHAM3 and DHAM3 are NP-hard

(32) Which one of the following is true for the language {am bn c m+n I m, n≥1}?

(A) It is context-free but not regular

(B) It is regular

(C) It is type-0 but not context-sensitive

(D) It is context-sensitive but not context-free

View Answer / Hide Answer

ANSWER: It is regular

(33) We have decision problems P1 and P2 as described below:

P1: Does a given finite state machine accept a given string?

P2: Does a given context-free grammar generate an infinite number of strings?

The statement that holds true for P1 and P2 is

(A) Only P2 is decidable

(B) Only P1 is decidable

(C) Neither P1 nor P2 are decidable

(D) Both P1 and P2 are decidable

View Answer / Hide Answer

ANSWER: Both P1 and P2 are decidable

(34) Problem X is given below:

We have a turing machine M over the input alphabet ∑, any state q of M and a word W ∈ ∑*, does the computation of M on W visit the state q? The statement, which holds true

about X, is

(A) X is undecidable but partially decidable

(B) X is decidable

(C) X is not a decision problem

(D) X is undecidable and not even partially decidable.

View Answer / Hide Answer

ANSWER: X is undecidable but partially decidable

(35) The state diagram describes the finite state machine. A is the starting state and an arc label is x/y where x stands for 1 bit input and y stands for 2 bit output

(A) Whenever the input sequence is 10 it outputs 00

(B) Whenever the input sequence is 11 it outputs 01

(C) It outputs the sum of the present and the previous bits of the input

(D) None of the above

View Answer / Hide Answer

ANSWER: It outputs the sum of the present and the previous bits of the input

(36) Which one of the following statement is true for the C language?

(A) It is a regular language

(B) It is context-sensitive language

(C) It is context-free language

(D) It is parsable fully only by a turing machine

View Answer / Hide Answer

ANSWER: It is context-free language

(37) How many states are present in the smallest finite automaton which accepts the language {x I length of x is divisible by 3}?

(A) 5

(B) 4

(C) 3

(D) 2

View Answer / Hide Answer

ANSWER: 4

(38) The last two symbols of L which is the set of all binary strings are same. In the minimum state deterministic finite state automaton, which is accepting L _____, states are present

(A) 4

(B) 6

(C) 3

(D) 5

View Answer / Hide Answer

ANSWER: 5

(39) The true regular expression is

(A) (r*s*)* = (r+s)*

(B) (r+s)* = r* + s*

(C) r*s* = r* + s*

(D) r(*) = r*

View Answer / Hide Answer

ANSWER: (r*s*)* = (r+s)*

(40) Let n be the length of a character string. How many substrings (of all lengths inclusive) can be formed from n?

(A) n(n-1)/2

(B) n²

(C) (n (n+1)/2) + 1

(D) n

View Answer / Hide Answer

ANSWER: (n (n+1)/2) + 1

(41) The set which is not countable if we have ∑ = {a, b}, is

(A) Set of all languages over ∑ accepted by turing machine

(B) Set of all regular languages over ∑

(C) Set of all strings over ∑

(D) Set of all languages over ∑

View Answer / Hide Answer

ANSWER: Set of all languages over ∑

(42) How many states are present in the minimum state finite automaton that recognizes the language represented by the regular expression (0+1)(0+1)…..N times?

(A) n+1

(B) n+2

(C) n

(D) 2n

View Answer / Hide Answer

ANSWER: n+2

(43) Consider the state table of a finite state machine that has input x and a single output z. The shortest input sequence to reach the final state C if the initial state is unknown is

(A) 10

(B) 01

(C) 101

(D) 110

View Answer / Hide Answer

ANSWER: 10

(44) The set that can be recognized by a deterministic finite state automaton is

(A) The set {1, 101, 11011, 1110111, …….}

(B) The set of binary string in which the number of 0’s is same as the number of 1’s

(C) 1, 2, 4, 8……2n ….. written in binary

(D) 1, 2, 4, 8……2n ….. written in unary

View Answer / Hide Answer

ANSWER: 1, 2, 4, 8……2n ….. written in binary

(45) Consider the four regular expressions given below;

a. (00)*( ε+0)

b. (00)*

c. 0*

d. 0(00)*

The equivalent regular expression out of the four is

(A) b and c

(B) c and d

(C) a and b

(D) a and c

View Answer / Hide Answer

ANSWER: a and c

(46) L1 = Φ and L2 = {a} are the two languages. Out of the following four options the one that represents L1L2* U L1* is

(A) Φ

(B) a*

(C) {ε}

(D) {ε, a}

View Answer / Hide Answer

ANSWER: {ε}

(47) We have the language L = {ab, aa, baa} and the four strings given below:

I) abaabaaabaa

II) aaaabaaaa

III) baaaaabaaaab

IV) baaaaabaa

The strings present in L* are

(A) I, II and IV

(B) I, II and III

(C) II, III and IV

(D) I, III and IV

View Answer / Hide Answer

ANSWER: I, II and IV

(48) Which one of the following is true regarding FOTRAN?

(A) It is a context free language

(B) It is a context sensitive language

(C) It is a regular language

(D) None of the above

View Answer / Hide Answer

ANSWER: It is a context sensitive language

(49) The feature that cannot be captured by context free grammar is

(A) Recursive procedure Syntax

(B) Syntax of if-then-else statement

(C) Arbitrary length of variable names

(D) Variable declared before its use

View Answer / Hide Answer

ANSWER: Variable declared before its use

(50) Which one of the following is applicable for context free languages?

(A) These are closed under union, Kleene closure

(B) These are closed under complement, Kleene closure

(C) These are closed under union, intersection

(D) These are closed under intersection, complement

View Answer / Hide Answer

ANSWER: These are closed under union, Kleene closure

(51) S -> a α b I b a c I ab

S -> α S I b

S -> α bb I ab

S -> bdb I b

The grammar described above is

(A) Context free

(B) Context sensitive

(C) Regular

(D) LR(k)

View Answer / Hide Answer

ANSWER: Context sensitive

(52) Match the following

a. Regular expression I. Syntax analysis

b. Pushdown automata II Code generation

c. Dataflow analysis III Lexical analysis

d. Register allocation IV Code optimization

(A) a – III, b – IV, c – I, d – II

(B) a – IV, b – III, c – I, d – II

(C) a – III, b – I, c – IV, d – II

(D) a – II, b – III, c – IV, d – I

View Answer / Hide Answer

ANSWER: a – III, b – I, c – IV, d – II

(53) Which of the following statement is false for a turing machine?

(A) There exists an equivalent deterministic turing machine for every non-deterministic turing machine

(B) Turing decidable languages are closed under intersection and complementation

(C) Turing recognizable languages are closed under union and intersection

(D) Turing recognizable languages are closed under union and complementation

View Answer / Hide Answer

ANSWER: Turing recognizable languages are closed under union and complementation

(54) The problem, which is not NP-hard, is

(A) Finding bi-connected problem of a graph

(B) The graph colouring problem

(C) Hamiltonian circuit problem

(D) The 0/1 Knapsack problem

View Answer / Hide Answer

(55) If P≠NP the statement which holds true is

(A) NP-hard = NP

(B) NP-complete ∩ P = Φ

(C) P=NP-complete

(D) NP-complete=NP

View Answer / Hide Answer

ANSWER: NP-complete ∩ P = Φ

## Earn mobile recharge online 100% working trick

**by dhruv J mk**

Hey friends , today i am going to share a trick to earn free mobile recharge from your pc. (checked by me)

You need to have the following softwares ……..

1> Bluestacks (search in google for Bluestacks thin installer )

2> VMware workstation 8.0 ( recommended )

3> WIndows 7 iso (to be installed in VMware workstation )

*HOW TO EARN*

Assume that you have installed Blustacks.

Now open Bluestacks and sign in too google play services with your gmail.

**HOW TO GET RECHARGE : Method 1**

Now in Google play search for ‘toi’ (times of india app) . Download it and install it. After installation just open TOI app for the 1st time . Now you will notice a pop-up will open which will ask you to login with facebook or google+ . Just select login with Facebook (Note : You must have a fb account , haha … fb is common to all now a days )

After logged in you will see a pop-up from TOI giving you a voucher code . Just copy the voucher code to clipboard for a while . Now again go to Google Play and download an app called “Paytm” (search without quote).

After paytm installation open it . Here you can see a form asking your mobile no and operator . Fill up the form and in ‘Enter amount’ section type 50 and click proceed.

Below you can see enter promotional code option . Just paste the voucher code you have copied earlier . And click apply. Now click proceed to Pay Rs 50.

You’r done …………. you will get Rs 50 moble recharge.

**NOTE :*** You can not use same facebook account to get another Rs. 50 . You have to create another fb account with separate mobile number than you can again earn free Rs. 50 recharge.** *

http://bdv.bidvertiser.com/BidVertiser.dbm?pid=685241&bid=1712757

**HOW TO GET RECHARGE : Method 2**

Open Google Play and search for an app “Pocketmoney”. This app will give you almost 100 Rs recharge. Open pocket money and you will see some list applications (Amazon, Flipkart, Snapdeal, Gaana etc ) . Just download those applications one by one . Note that after downloading you must keep your app open for 1 minutes , otherwise you will not get money. ( Poketmoney snap is shown below ) .

** Trick **after getting the money you can uninstall those app (amazon ,flipkart etc ) and retry installing them after 1 day . You will again get money .

After you have got the money you can proceed to recharge.

http://bdv.bidvertiser.com/BidVertiser.dbm?pid=685241&bid=1712757

**HOW TO GET RECHARGE : Method 3**

Open Google play and search for “mCent ” . Download it & install it. This app is same as poket money which allows to download apps and for successful download they will give you some money. (Snapshot is given below)

**Note : **Before downloading an app you must uninstall the app if it is exist already in your installed apps. Otherwise you will not get paid.

**HOW TO GET RECHARGE : Method 4**

Open Google play and search for “FreeBuster ” . Download it & install it. This app is same as poket money which allows to download apps and for successful download they will give you some money. (Snapshot is given below)

**Note : **Before downloading an app you must uninstall the app if it is exist already in your installed apps. Otherwise you will not get paid.

**HOW TO GET RECHARGE : Method 5**

Open Google play and search for “Ladoo” . Download it and install it. Now open ladoo and provide your no. to verification. If verification is successful then you will see app list available for download . Download and open those apps you will get money.

In Ladoo an option is available to refer a friend to install Ladoo . Invite a friend by using facebook or imo application if your friend installs Ladoo you will get money.

*****IMPORTANT ***NOW ALL of the above are one time payment for a single pc or system . If you try to uninstall bluestack and install it again from the same PC or system in order to be paid for the second time you will not get money. ( You trey it if you are lucky you may get but chance is 50 %)**

**S0 how you will get benefited again and again from the same system or pc ???????**

TRICK : **Download VMware Workstation 8 . Vmware is a tool or software using which you can virtually install or create some PCs inside your own pc. That is you can install mulitiple no. of operating systems (Win7,win8,Linux etc) inside the vmware and access them from your original windows operating system.**

So now install vmware workstation and use and windows7.iso file to install windows virtually. This will not override your existing windows.

**After windows installation install BlueStacks in your virtual OS . And repeat the methods (1-5) above and thus you can again earn money or mobile recharge.**

**After some days when you feel that you will not getting money from your virtual Os ; just delete it and again install a new OS inside your Vmware . Thus you will earn recharge in an unlimited way.**

ENJOY …………………

*(By Dhrubajyoti Malakar : Checked by me) *

## How to watch live match online bypassing the preview in starsports.com

This trick is for those sports lover who want to watch live match in starsports official site. But the starsports.com only allows

to watch the match for 4 minute free preview and after that the match stops and you can not able to watch it again.

(adsbygoogle = window.adsbygoogle || []).push({});

Trick : You have to do only one thing. Open starsports.com and before the free preview of a live match has gone you just change your pc date to previous day. For eg : If you are watching it in 10/12/2014 change it to 9/12/2014. Then the free preview counter will increase to 14 minute and never go below 13 minute and you can enjoy the full match.

## Europe’s oldest footprints uncovered on English coast …Date: February 7, 2014

The earliest human footprints outside of Africa have been uncovered, on the English coast, by a team of scientists led by Queen Mary University of London, the British Museum and the Natural History Museum.

Up to five people left the series of footprints in mud on the bank of an ancient river estuary over 800,000 years ago at Happisburgh in northeast Norfolk.

Dr Simon Lewis from Queen Mary’s School of Geography has been helping to piece together the geological puzzle surrounding the discovery — made in May 2013 — which is evidence of the first known humans in northern Europe.

Dr Lewis’s research into the geology of the site has provided vital information on the sediments in which the prints were found. “My role is to work out the sequence of deposits at the site and how they were laid down. This means I can provide a geological context for the archaeological evidence of human occupation at the site.”

The importance of the Happisburgh footprints is highlighted by the rarity of footprints surviving elsewhere. Only those at Laetoli in Tanzania at about 3.5 million years and at Ileret and Koobi Fora in Kenya at about 1.5 million years are older.

A lecturer in physical geography, and co-director of the Happisburgh project (http://www.ahobproject.org/), Dr Lewis added that the chance of encountering footprints such as this was extremely rare; they survived environmental change and the passage of time.

Timing was also crucial as “their location was revealed just at a moment when researchers were there to see it” during a geophysical survey. “Just two weeks later the tide would have eroded the footprints away.”

“At first we weren’t sure what we were seeing,” explains Dr Nick Ashton of the British Museum “but as we removed any remaining beach sand and sponged off the seawater, it was clear that the hollows resembled prints, and that we needed to record the surface as quickly as possible.”

Over the next two weeks researchers used photogrammetry, a technique that can stitch together digital photographs to create a permanent record and 3D images of the surface. It was the analysis of these images that confirmed that the elongated hollows were indeed ancient human footprints.

In some cases the heel, arch and even toes could be identified, equating to modern shoes of up to UK size 8. While it is not possible to tell what the makers of these footprints were doing at the time, analysis has suggested that the prints were made from a mix of adults and children.

Their discovery offers researchers an insight into the migration of pre-historic people hundreds of thousands of years ago when Britain was linked by land to continental Europe.

At this time, deer, bison, mammoth, hippo and rhino grazed the river valley at Happisburgh. The land provided a rich array of resources for the early humans with edible plant tubers, seaweed and shellfish nearby, while the grazing herds would have provided meat through hunting or scavenging.

During the past 10 years the sediments at Happisburgh have revealed a series of sites with stone tools and fossil bones; this discovery is from the same deposits.

The findings are published in the science journal *PLOS ONE*.

The work at Happisburgh forms part of a new major exhibition at the Natural History Museum Britain: One Million Years of the Human Story opening on February 13.

**Story Source:**

The above story is based on materials provided by **Queen Mary University of London**. *Note: Materials may be edited for content and length.*

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