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Biton while loop.

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So the while loop syntax in Python that while looping python is used to iterate over a block of code,

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as long as the test expression that that is a condition is true, when the condition is well, then

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it will come out from the white block.

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So we generally use the loop when we don't know the beforehand and the number of times to iterate.

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So whenever we know that how many times we are going to iterate, we are having the form.

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But if you don't know that, how many times we're going to iterate, then we shall go for this while

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looping.

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Because when the condition is through, the loop will get executed.

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When the condition is false, then it will come out of the loop construct.

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So here is the respective body.

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There's a while test and don't forget to get this on there and then body of white.

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So there is one indentation is there that will indicate that this these are the statements which will

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be under this while so intertwined.

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Look, the content will come this expression.

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If the expression is true, then the body of oil will get executed again.

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The test expression will be tested.

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And if the expression is false, then it will exit from the loop.

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So in this way, the respective python while loop will get executed.

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While loop with LTH syntax in Python, so let's think about that for the fort, we can have an optional

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S block with the while loop as well.

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The A's part is executed in the condition.

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In the while Loop evaluates to false when the connection will become false, then the start getting

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executed.

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When the condition is true, then the body will go on keeping their execution so the Y loop can be terminated

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with the break.

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A statement in such cases of ill spot is ignored.

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If there is some Bryk statement, Danneels part will get ignored.

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Hennesey while loops and spot runs if no break occurs and the condition is false.

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So whenever there is no public statement, whenever the condition is false, then only the part will

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get its control to execute its its respective code written under the else block.

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So let us go for one practical demonstration for the is the understanding of this while loop in python

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coding in this demonstration we are discussing the use of while loop in our python coding so that we

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have made Aisgill two three has been initialized with three while I is greater than Gittel Colon.

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So whenever you are giving colon then the next line will be coming with the indentation that wins.

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This line is within the while block printing the value of PI and also decreasing the value of PI by

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one hit.

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So let us go for the execution.

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You can find that three to one they are getting printed because when the value up I will become one.

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So one is that the condition is true.

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So print one is getting printed now.

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One is getting decreased by one, so it will become zero.

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So I will have the zero zero zero zero.

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The condition is false.

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So that is why it is coming out from the Y look.

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So let us go for a while else structure.

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So while I less than 10 print I and then I blast equal to one.

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That means when the value of I is not equal to ten below ten, that means when the eye is less than

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ten then the I will get printed and each and every time that I will be increased by one here and when

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this loop will get terminated then you are going to print that displayed successfully.

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So let us go for the execution.

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Say it for some just initializing decide with the values of five here.

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So now see the five is getting printed so five has got increased to six, six is less than ten, so

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six has got printed.

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Similarly seven, eight and nine has got printed.

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Nine has been increased by one.

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So it has become ten.

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So ten less than ten.

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The condition is also will be coming to the L spot and displayed successfully.

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The message is getting printed here.

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So in this we can find that how this respective while with LS structure is there, the blocks are getting

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printed accurately, are getting executed accordingly.

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Let us go for this pattern printing.

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These are the very common problems in our exams, so any will be taken from the keyboard as input.

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There is a number of players.

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If the number of player is equal to six, that means here will be having one, two, three, four,

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five, six or six.

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Number of players also can consider six is the number of here.

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So to print this particular pattern here will be having some blank characters and then some esoteric

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symbols will get printed.

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But if I write only the blank characters, it will be not easy for us to count that.

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How many blank characters are there?

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So in place of blank, we have written here got so in the coding also we can replace this door by blank

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if I want to get the final pattern here.

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So here you have made one dressing table.

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This is known as dressing table where we are trying to trace the logic of our code so far and the result

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is but the pattern to be printed.

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So I is a clear number or the height number.

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It is ranging from one to six.

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And here we are counting for each and every layer.

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We are counting the number of blanks and the number of starts in this particular dressing table.

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So here we are having five blank spaces and then one starts.

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We have written five and one here we are getting four blank spaces and then three starts.

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We have written this for entry here.

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We are getting no blank spaces, but here we are having eleven starts.

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So zero number of blank spaces and one number of starts or static symbols are going to get printed.

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So from here we are supposed to get the logic for blank logic, for the esoteric symbol.

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So what would be the logic for blank here?

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You can find that I is increasing at the blank is decreasing.

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So if you subtract this I from some fixed quantity, let it be in is equal to six.

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So if you subtract this I from the six then obviously when the I'll be increasing that difference value

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will be decreasing.

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So anybody who are finding some logic that is something is increasing and another one is decreasing,

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then obviously the value, the variable which is increasing should be subtracted from some fixed quantity

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so that when the variable will be increasing, that difference value will be decreasing.

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So that's why six minus one is five, six minus two is for six, minus three is six three and six,

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minus six is zero.

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So here the respective logic for the blank will be in minus eight.

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Next one is increasing by one study, is also increasing by two.

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So what we shall do if I is one start is a one I used to namaroff start will be three Isogawa five,

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number of studies nine.

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So we think that here the logic will be to start I minus one because doing two one two minus one is

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one, two into three is six, minus one is five to into six is twelve minus one is eleven here.

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So they hear the respective logic will be to start ie minus one.

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So we have got the logic now we are going to implement it here.

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So N is equal to in the input place and the number of players and then is equal to one report.

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And when that this value of EI is less than equal to end, that means I'll be ranging from one to six

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if the value up end is equal to six.

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So here we are, put that on everything in the while loop construct.

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So Jessica, we considering another control variable so will be equal to one.

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And while G is less than.

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So now what will happen for each and every player will be printing a certain number of blank characters

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and then certain number of asterisk symbols.

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Then you shall go for to continue to break.

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So that will be the skeleton of our program.

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So at first we're going to print some blank characters and then we're are going to bring some esoteric

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symbols.

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So how many blank characters we're having?

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We're having N minus a number of blank characters.

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So here we are putting this one as in minus one in minus eight.

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So basically one day less than equal to N minus I print here.

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We're printing this dot and end is equal to double double what means here we are just omitting the trailing

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new line characters so all these dots will get printed in the same line.

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So now again we are making this jiggle.

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The one and G is less than equal to.

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So here we are putting this start logic that is to start I minus one.

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So we are putting this logic here and then print start and continuity will be there.

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All the starts will get printed in the same row.

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So that's why it's difficult to double double code.

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So this is a print which will break the continuity.

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So we'll be printing certain number of dark, certain number of starts.

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So the next it'll be coming from the next line.

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So that's why to break the continuity here to uprooting and putting out of the print segment.

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So print function has been called here.

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So now let me go for the execution.

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We can find that if we give the value a six, I'm getting this pattern here.

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So I think so I should make this one to start at minus one.

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I should go for I increase.

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So that statement is missing.

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So let's go for this.

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I increase.

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So I is equal to eight plus one.

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So here this has to be increased by one.

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And also let us put the same in a while.

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So what we shall do is that.

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So Jay is equal to Jay plus one, so that statement is required, otherwise my program will fall in

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in an infinite loop.

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Jessica two plus one here.

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So let's go for the execution.

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So here you are, putting the value of an individual 060.

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You're getting this pattern.

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We're getting this.

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So now if you put a different value for in here, if we put the value of any element, we are getting

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the pattern like this, you can easily fill that.

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One, two, three, four, five, six, seven, eight, 11.

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So one, two, three, four, five, six, seven, eight, nine, 10, 11.

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Yes.

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So here we are getting this 11 number of layers and the pattern is getting printed accordingly.

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So after writing any code, you should test your code with the multiple different values for inputs.

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So let me check again.

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One, two, three, four, five, six, seven, eight, nine.

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Yes, we're having the nine layers, so each and every layer will be having this preceding dots and

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then start.

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So now in place of DOT, if you put a blank character, if you put a blank characters, and then the

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pattern will be looking like this.

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And the pattern will be looking like this in this way.

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OK, now let us go for another example, so I want to bring this pattern like this, so any eleven dollars

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a valuable and should be all here.

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So it is the middle point.

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I might be regretting it.

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So let us at first go for the pattern.

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So this is a pattern we are going to print.

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So in the precinct we have written items ranging from one to 11 because element has been given and it

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must be odd.

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Otherwise this pattern will not have the symmetry if we have the invisibility even number.

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So pattern will be will not be having the symmetry.

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So now we are counting the number of dark, the stars, you know, so similarly with whatever we did

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earlier.

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So just consider this rule arbitrarily so that having four dots and three starts, so have written that

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one has four dogs and three starts in this way, we filled up this DOT column and the star column.

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So now let's go for the logic derivation.

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You know that I is increasing throughout.

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It is ranging from one to 11.

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But here this is at first increasing, then it is decreasing.

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So I cannot expect that same formula will work for the world hubs.

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So that's why you are going for a separate formula, for this respective logic, for this dot printing

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in the first up and in the lower up separately.

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So how to do that?

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Let us go for the.

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Calculations here.

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So if it is one, then it is zero, if it is true, then it is one.

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If it is six, then it is five.

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So it would be a minus one.

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OK, now come to this point, if it is seven, it is four.

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If it is eight, it is three.

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If it is nine, it is two.

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So that means I should subtract this from the from some fixed quantity so that when the value up I will

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be increasing, the different value will be decreasing here.

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So here we are having this I n is equal to eleven so eleven minus seven for eleven minus eight.

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Three eleven minus ten.

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One eleven minus eleven is zero here.

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So here the logic will be in minus I.

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Next one is that for this particular individual relevant.

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My middle point is the middle, that is the sixth layer.

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So how to get six here.

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So that is very simple.

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Eleven plus one, twelve by two is equal to six.

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So that's why you are calculating another variable.

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So the variable is getting initialized with six when the value of N is eleven, M stands for the middle

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layer.

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So now let us go for the calculation, the logic derivation for this start.

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So this explosion's, whatever you are going to derive here can contain, I can contain N and also can

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contain M here.

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So let us go for this derivation.

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So here you can find that I is increasing by one and start is decreasing by two.

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So that means when the easier it must be subtracted from this fixed quantity so that different value

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will be decreasing when the value of I will be increasing.

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So let's go for the derivation.

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So what is the logic are finding.

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Just consider here.

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So six minus one there is a five five in the two 502.

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That is extent plus one that is eleven.

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So I'm going to give you two clues.

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Clue number one, when I is increasing, starting, decreasing, that means that I has to be subtracted

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from some fixed quantity so that when the value up I will be increasing the difference value will be

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decreasing.

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There is no point.

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Number one.

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Point number two here is that here you can find that I is increasing by one, but the start is decreasing

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by two.

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So that different value has to be multiplied by some by two so that that difference will get decreased

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the by the factor of two.

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So six minus one, five to ten plus one eleven six minus four is two into two four plus one five six

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minus six is zero two zero plus one one.

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So here the formula will be to start here, the formula will be to start in minus I plus one.

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Lost in this case, but having this one, there is this particular case.

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So if I use seven stories to three, I use eight stories five so we can see that it is increasing by

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one.

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It is also increasing, but increasing by two.

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So to multiplication factor will be there.

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So that is our clue.

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OK, now see.

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So seven minus six, one, two, two, one, two, plus one is three nine minus six, what is that,

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three in the two six plus one seven.

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So that's why here the logic will be to start to start.

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I might assume so to start at 8:00 a.m. So we are going to get here plus one.

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So we have got the logic now how to implement the same.

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So let me put this one here in the code I'm going to put here.

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I'm going to object in.

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So here you have got the logic now how to implement it, what we can do that after reading this valuable

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when we are calculating that middle muswell.

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So what does that mean?

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Plus one whole.

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OK, this entered the odd number of ballots, OK, next.

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If I get it, then Im not what is what is the meaning of bigger than perhaps a lower eye, greater than

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it means it is the lower half, so blank no blank will be equal to in the lower up in magnesite.

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We got this one from here.

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This one and four, they started having this formula.

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So that is our starter, so here's what I think is equal to this formula else.

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So we shall be going for this blank and blank is minus one, and the start will be equal to two minus

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I thought to start in minus eight, plus one.

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So if I get it, then there is a blank disaster or else there is a blank disaster.

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So in this way, I've got this logic.

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So now how many number of blanks will get printed, will get printed B number of and how many no upstarts

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will get printed its numbers.

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So in this way, I can easily implement my code.

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So let me go for the cross-check.

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If it is good, this code, I keep the value of it, then I'm getting this pattern.

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So here you see we are having this dressing table is that we are having and it's the respective implementation.

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You can easily find that how this code has got implemented.

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So if you give the value, say, 13, I'm getting the pattern like this.

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Let me count one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13.

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Yes, we are getting this our total number of layers here.

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Obviously, layer number will be starting from one to 13 in this fashion.

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So let me go for another execution if I make this one a seven.

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We're getting this better like this.

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OK, so if a user gives if user gives the event value for this end, then what will happen?

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The pattern will not have the respect of symmetry here.

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Now, I'm going to give you one example, one practice example here who just which has to be solved

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by you.

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So these are practice work.

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We're having a program to display a pattern using loops.

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So here the pattern is given here the pattern is given and here also prepared this dressing table you

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want and here you are supposed to derive the logic and then that respective logic has to be implemented

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using python coding.

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So I'm giving this particular assignment as a practice work for your better understanding and for for

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your better performance.

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So please solve this problem.

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I've made the problem partially done in the planning phase are now implemented using the Python code

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and similar type of code already written.

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So you are supposed to check that one out from this example.

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So in this particular demonstration, we have discussed how to use a while loop in Python code.

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Thanks for watching this video.