I am new to Software testing and I am studying the basic techniques. I read the following problem:
Identify the Equivalence Classes for the following specification: The program accepts five to nine inputs which are 3 digit integers greater than 100.
I think that it doesn't matter how much inputs this program has, and the equivalence class is {99,100,101}. Am I right or not?
After the comments, I think the classes are:
1.(-00,99)
2.[100]
3.(101,999)
4.(1000,+00)
Inputs:
0-4 inputs
5-9 inputs
More than 9 inputs
Values:
0-100
101-999
Greater than 999
The program accepts when there are between 5 and 9 inputs and each input value is a 3-digit number between 101 and 999.
I suggest you use PICT for generating effective combinations to test.
Take a look at http://msdn.microsoft.com/en-us/magazine/ee819137.aspx
The tool can be downloaded from http://download.microsoft.com/download/f/5/5/f55484df-8494-48fa-8dbd-8c6f76cc014b/pict33.msi
You can look for similar tools at http://pairwise.org/tools.asp
equivalence classes for your problem are:
set of numbers that are not three digit and greater than hundred...
set of numbers that are less than one hundred
set of numbers that are greater than one hundred and less than 999
set consisting of the number 100
1 0<x<100 , value of x should contain 0-4
2 101<x<999 , value of x should contain 5-9
3 X>999 , value of x should be 0-9
Following should be the classes:
Inputs:
[0 - 4] invalid class
[5 - 9] valid class
[More than 9] invalid class
Values:
[Less than 99] invalid class
[100 to 999] valid class
[Greater than 1000] invalid class
Again a Decision table should be used to find out the valid combination of Inputs and Values.
Related
The following code gives a repeating sequence of specified numbers from 1 to 12, repeating it 12 times. However, the numbers generated have a different length (1,2,3...10,11,12). How can the formula be modified so that leading zeros are added in order to make all numbers the same length?
transpose(split(REPT(concat(JOIN(",",SEQUENCE(1,12)),","),ROUNDDOWN(ROWS(A1:A)/15)),",",true))
For example, in this example there are numbers with length = 2 and then the desired sequence would be 01,02...10,11,12. However if the sequence was up to 3 or more digits, for example:
...(",",SEQUENCE(1,150)),",")...
Then a desired sequence would be 001...010...150. A sequence going up to 4 would be 0001...1500 etc.
try:
=INDEX(TEXT(FLATTEN(MAKEARRAY(25, 12, LAMBDA(x, y, y))), "00"))
BASE formula has min_length argument. You can use it to set the LENgth of your sequences. It's also easy to create the sequence without TRANSPOSE/SPLIT/JOIN/SPLIT with just IF/FLATTEN. For eg, To create sequence of 12, 25 times,
=ARRAYFORMULA(FLATTEN(IF(SEQUENCE(25),BASE(SEQUENCE(1,12),10,LEN(12)))))
I was doing some ds and algo problems from faceprep and a question came:
Amy Farrah Fowler asks her students to give 3 examples for positive odd numbers. When the student specifies a correct answer, his/her score is incremented by 1. When the student specifies a positive even number, his/her score is decremented by 0.5. When the student specifies a negative number, he/she will not be given any more chances to correct his or her mistake and his/her score will be decremented by 1. So a student's turn comes to an end when he/she has correctly specified 3 positive odd numbers or when the student has specified a negative number. Some students didn't know the difference between odd numbers and even numbers and they made many mistakes and so it was difficult for her to maintain the scores. She asks for your help. Can you please help her by writing a program to calculate the score?
INPUT & OUTPUT FORMAT:
Input consists of a list of integers.
The output consists of a single line. The score needs to be corrected to 1 decimal place.
[For this exercise, don't worry about duplicate odd numbers. Even if the students specify the same odd number thrice, it is accepted].
input is given as:
1
2
3
4
5
Now How to take input and store it in a variable by single command?
till now I was doing but now number of inputs are different for each case study.
:
a=int(input())
b=int(input())
c=int(input())
I want to do this using a single command so that each input will store in a list.
Try this code snippet to store the input in the list "List_w".
List_w = []
try:
while(1):
List_w.append(int(input()))
except:
print(List_w)
if your input is separated by white space, you can use:
arr = list(map(int, input().split()))
print(arr) # input 1 2 3 4 # output [1,2,3,4]
if your input is based on given range use:
for _ in range(int(input())):
arr = list(map(int, input().split()))
print(arr)
if the inputs separated by new line you can use append method or one liner:
arr = [int(input()) for _ in range(int(input()))]
print(arr)
# input
3
4
5
6 # output [4, 5, 6]
know more about input here https://www.geeksforgeeks.org/python-get-a-list-as-input-from-user/
I want to check if a number is 50 or more using a regular expression. This in itself is no problem but the number field has another regex checking the format of the entered number.
The number will be in the continental format: 123.456,78 (a dot between groups of three digits and always a comma with 2 digits at the end)
Examples:
100.000,00
50.000,00
50,00
34,34
etc.
I want to capture numbers which are 50 or more. So from the four examples above the first three should be matched.
I've come up with this rather complicated one and am wondering if there is an easier way to do this.
^(\d{1,3}[.]|[5-9][0-9]|\d{3}|[.]\d{1,3})*[,]\d{2}$
EDIT
I want to match continental numbers here. The numbers have this format due to internal regulations and specify a price.
Example: 1000 EUR would be written as 1.000,00 EUR
50000 as 50.000,00 and so on.
It's a matter of taste, obviously, but using a negative lookahead gives a simple solution.
^(?!([1-4]?\d),)[1-9](\d{1,2})?(\.\d{3})*,\d{2}\b
In words: starting from a boundary ignore all numbers that start with 1 digit OR 2 digits (the first being a 1,2,3 or 4), followed by a comma.
Check on regex101.com
Try:
EDIT ^(.{3,}|[5-9]\d),\d{2}$
It checks if:
there 3 chars or more before the ,
there are 2 numbers before the , and the first is between 5 and 9
and then a , and 2 numbers
Donno if it answer your question as it'll return true for:
aa50,00
1sdf,54
But this assumes that your original string is a number in the format you expect (as it was not a requirement in your question).
EDIT 3
The regex below tests if the number is valid referring to the continental format and if it's equal or greater than 50. See tests here.
Regex: ^((([1-9]\d{0,2}\.)(\d{3}\.){0,}\d{3})|([1-9]\d{2})|([5-9]\d)),\d{2}$
Explanation (d is a number):
([1-9]\d{0,2}\.): either d., dd. or ddd. one time with the first d between 1 and 9.
(\d{3}\.){0,}: ddd. zero or x time
\d{3}: ddd 3 digit
These 3 parts combined match any numbers equals or greater than 1000 like: 1.000, 22.002 or 100.000.000.
([1-9]\d{2}): any number between 100 and 999.
([5-9]\d)): a number between 5 and 9 followed by a number. Matches anything between 50 and 99.
So it's either the one of the parts above or this one.
Then ,\d{2}$ matches the comma and the two last digits.
I have named all inner groups, for better understanding what part of number is matched by each group. After you understand how it works, change all ?P<..> to ?:.
This one is for any dec number in the continental format.
^(?P<common_int>(?P<int>(?P<int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<int_end>\.\d{3})*|0)(?!,)|(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})*,)|0,|,)(?=\d))(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_start>\d{3}\.)*(?P<frac_end>\d{1,3})))?$
test
This one is for the same with the limit number>=50
^(?P<common_int>(?P<int>(?P<int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<int_end>\.\d{3})+|(?P<int_short>[1-9]\d{2}|[5-9]\d))(?!,)|(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})+,)|(?P<dec_short_int>[1-9]\d{2}|[5-9]\d),)(?=\d))(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_start>\d{3}\.)*(?P<frac_end>\d{1,3})))?$
tests
If you always have the integer part under 999.999 and fractal part always 2 digits, it will be a bit more simple:
^(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})?,)|(?P<dec_short_int>[1-9]\d{2}|[5-9]\d),)(?=\d)(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_end>\d{1,2})))?$
test
If you can guarantee that the number is correctly formed -- that is, that the regex isn't expected to detect that 5,0.1 is invalid, then there are a limited number of passing cases:
ends with \d{3}
ends with [5-9]\d
contains \d{3},
contains [5-9]\d,
It's not actually necessary to do anything with \.
The easiest regex is to code for each of these individually:
(\d{3}$|[5-9]\d$|\d{3},|[5-9]\d)
You could make it more compact and efficient by merging some of the cases:
(\d{3}[$,]|[5-9]\d[$,])
If you need to also validate the format, you will need extra complexity. I would advise against attempting to do both in a single regex.
However unless you have a very good reason for having to do this with a regex, I recommend against it. Parse the string into an integer, and compare it with 50.
I'm trying to find an algorithm which "breaks the safe" by typing the keys 0-9. The code is 4 digits long. The safe will be open where it identifies the code as substring of the typing. meaning, if the code is "3456" so the next typing will open the safe: "123456". (It just means that the safe is not restarting every 4 keys input).
Is there an algorithm which every time it add one digit to the sequence, it creates new 4 digits number (new combinations of the last 4 digits of the sequence\string)?
thanks, km.
Editing (I post it years ago):
The question is how to make sure that every time I set an input (one digit) to the safe, I generate a new 4 digit code that was not generated before. For example, if the safe gets binary code with 3 digits long then this should be my input sequence:
0001011100
Because for every input I get a new code (3 digit long) that was not generated before:
000 -> 000
1 -> 001
0 -> 010
1 -> 101
1 -> 011
1 -> 111
0 -> 110
0 -> 100
I found a reduction to your problem:
Lets define directed graph G = (V,E) in the following way:
V = {all possible combinations of the code}.
E = {< u,v > | v can be obtained from u by adding 1 digit (at the end), and delete the first digit}.
|V| = 10^4.
Din and Dout of every vertex equal to 10 => |E| = 10^5.
You need to prove that there is Hamilton cycle in G - if you do, you can prove the existence of a solution.
EDIT1:
The algorithm:
Construct directed graph G as mentioned above.
Calculate Hamilton cycle - {v1,v2,..,vn-1,v1}.
Press every number in v1.
X <- v1.
while the safe isn't open:
5.1 X <- next vertex in the Hamilton path after X.
5.2 press the last digit in X.
We can see that because we use Hamilton cycle, we never repeat the same substring. (The last 4 presses).
EDIT2:
Of course Hamilton path is sufficient.
Here in summary is the problem I think you are trying to solve and some explanation on how i might approach solving it. http://www.cs.swan.ac.uk/~csharold/cpp/SPAEcpp.pdf
You have to do some finessing to make it fit into the chinese post man problem however...
Imagine solving this problem for the binary digits, three digits strings. Assume you have the first two digits, and ask your self what are my options to move to? (In regards to the next two digit string?)
You are left with a Directed Graph.
/-\
/ V
\- 00 ----> 01
^ / ^|
\/ ||
/\ ||
V \ |V
/-- 11 ---> 10
\ ^
\-/
Solve the Chinese Postman, you will have all combinations and will form one string
The question is now, is the Chinese postman solvable? There are algorithms which determine weather or not a DAG is solvable for the CPP, but i don't know if this particular graph is necessarily solvable based on the problem alone. That would be a good thing to determine. You do however know you could find out algorithmically weather it is solvable and if it is you could solve it using algorithms available in that paper (I think) and online.
Every vertex here has 2 incoming edges and 2 outgoing edges.
There are 4 (2^2) vertexes.
In the full sized problem there are 19683( 3 ^ 9 ) vertexs and every vertex has 512 ( 2 ^ 9 ) out going and incoming vertexes. There would be a total of
19683( 3 ^ 9 ) x 512 (2 ^ 9) = 10077696 edges in your graph.
Approach to solution:
1.) Create list of all 3 digit numbers 000 to 999.
2.) Create edges for all numbers such that last two digits of first number match first
two digits of next number.
ie 123 -> 238 (valid edge) 123 -> 128 (invalid edge)
3.) use Chinese Postman solving algorithmic approaches to discover if solvable and
solve
I would create an array of subsequences which needs to be updates upon any insertion of a new digit. So in your example, it will start as:
array = {1}
then
array = {1,2,12}
then
array = {1,2,12,3,13,23,123}
then
array = {1,2,12,3,13,23,123,4,14,24,124,34,134,234,1234}
and when you have a sequence that is already at the length=4 you don't need to continue the concatenation, just remove the 1st digit of the sequence and insert the new digit at the end, for example, use the last item 1234, when we add 5 it will become 2345 as follows:
array = {1,2,12,3,13,23,123,4,14,24,124,34,134,234,1234,5,15,25,125,35,135,235,1235,45,145,245,1245,345,1345,2345,2345}
I believe that this is not a very complicated way of going over all the sub-sequences of a given sequence.
I was reading an exercise of UVA, which I need to simulate a deterministic pushdown automaton, to see
if certain strings are accepted or not by PDA on a given entry in the following format:
The first line of input will be an integer C, which indicates the number of test cases. The first line of each test case contains five integers E, T, F, S and C, where E represents the number of states in the automaton, T the number of transitions, F represents the number of final states, S the initial state and C the number of test strings respectively. The next line will contain F integers, which represent the final states of the automaton. Then come T lines, each with 2 integers I and J and 3 strings, L, T and A, where I and J (0 ≤ I, J < E) represent the state of origin and destination of a transition state respectively. L represents the character read from the tape into the transition, T represents the symbol found at the top of the stack and A the action to perform with the top of the stack at the end of this transition (the character used to represent the bottom of the pile is always Z. to represent the end of the string, or unstack the action of not taking into account the top of the stack for the transition character is used <alt+156> £). The alphabet of the stack will be capital letters. For chain A, the symbols are stacked from right to left (in the same way that the program JFlap, ie, the new top of the stack will be the character that is to the left). Then come C lines, each with an input string. The input strings may contain lowercase letters and numbers (not necessarily present in any transition).
The output in the first line of each test case must display the following string "Case G:", where G represents the number of test case (starting at 1). Then C lines on which to print the word "OK" if the automaton accepts the string or "Reject" otherwise.
For example:
Input:
2
3 5 1 0 5
2
0 0 1 Z XZ
0 0 1 X XX
0 1 0 X X
1 1 1 X £
1 2 £ Z Z
111101111
110111
011111
1010101
11011
4 6 1 0 5
3
1 2 b A £
0 0 a Z AZ
0 1 a A AAA
1 0 a A AA
2 3 £ Z Z
2 2 b A £
aabbb
aaaabbbbbb
c1bbb
abbb
aaaaaabbbbbbbbb
this is the output:
Output:
Case 1:
Accepted
Rejected
Rejected
Rejected
Accepted
Case 2:
Accepted
Accepted
Rejected
Rejected
Accepted
I need some help, or any idea how I can simulate this PDA, I am not asking me a code that solves the problem because I want to make my own code (The idea is to learn right??), But I need some help (Some idea or pseudocode) to begin implementation.
You first need a data structure to keep transitions. You can use a vector with a transition struct that contains transition quintuples. But you can use fact that states are integer and create a vector which keeps at index 0, transitions from state 0; at index 1 transitions from state 1 like that. This way you can reduce searching time for finding correct transition.
You can easily use the stack in stl library for the stack. You also need search function it could chnage depending on your implementation if you use first method you can use a function which is like:
int findIndex(vector<quintuple> v)//which finds the index of correct transition otherwise returns -1
then use the return value to get newstate and newstack symbol.
Or you can use a for loop over the vector and bool flag which represents transition is found or not.
On second method you can use a function which takes references to new state and new stack symbol and set them if you find a appropriate transition.
For inputs you can use something like vector or vector depends on personal taste. You can implement your main method with for loops but if you want extra difficulties you can implement a recursive function. May it be easy.