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The solution is going to cover TASC eight from your final challenge, Task eight tells us to check every

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column for a straight exer castrato that still involves running through every character in a tiara,

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but in a different order.

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So I'm going to make a nested loop once again.

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This time, the outer loop is going to keep running as long as it is less than three and even though

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Baudot length equals three, I'm going to explain why I'm not using Baudot length and just a bit.

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And I'll keep the inner loop running as long as Jay is smaller than Baudot length.

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Wow, if you're confused, don't worry.

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I'll explain the science behind this madness in just a bit.

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So we're pretty used to indexing every element in the current row.

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But how are we going to pull off column indexing?

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Looking back at the article, I left you a very telling hint.

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Look at how the first index keeps changing while the second index is fixed.

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If you think in terms of a nested loop, I is always fixed.

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While Jay varies with the inner loop as the inner loop countered, Jay indexes every row, the eye counter

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is fixed.

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That one index.

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Accordingly, we're going to use J to index every row.

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And have I get an element in that row?

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And so now we're using AYT Index elements, so it doesn't make sense to have the condition based on

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the number of rows baudot length.

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But now we're using Jay to index every row, Jay is now the row index, and it makes sense that the

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inner loop keeps you running.

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As long as Jay is smaller than the number of rows, this loop is going to keep running until Jay indexes

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every single row.

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So from a functionality perspective, it wouldn't make a difference what you put because everything

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just happens to equal three.

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The number of rose is three.

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The number of columns, everything is three.

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So how is this going to behave, think about it.

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As the inner loop countered, Jay goes to every row, the outer loop counter indexes, the first value

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in each row.

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Then it indexes the second value in each row.

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And finally, it indexes the third value in each row.

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Ultimately, this allows the inner loop to run through every value in a column.

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And as we go through every value in a column, we're going to check if that value was in X.

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If so, add one to the count.

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Otherwise, subtract one.

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So after the inner loop, we just went through an entire column of characters and we're going to check

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if the column count resulted in a value of three.

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Or negative three.

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If so, we're going to break the function prematurely by returning the counts.

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Otherwise, we have to reset the count because you want your next inner loop to start counting again

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from zero.

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Kate, now the code is able to check every column for a straight story straight, so we're ready to

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run.

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I'll play X.

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Ex.

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Oh.

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And before I put my last X to declare victory, let me narrate what has happened so far.

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So right now I just put a No.

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Which print an updated board, obviously, check when he gets called.

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This chunk of our code is going to check every row for a straight Exaro Strato.

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With the first row was going to have a count of zero, the second row is going to have a count of zero,

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the third row is also going to have a count of zero.

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So overall, nothing happens here.

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And we move on to the next nested loop, this loop is going to check every column for a straight Cicero's,

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the first column is going to have a count of two, the second column negative to third column zero.

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So nothing happens during this nested loop.

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But if I take another turn now, X wins.

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In this case, our nested loop is going to count three axis along the first column.

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The return key word is going to break the function prematurely and return the counts.

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And since the return value is three X wins.

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All right, now we can check to see if our code works with straight O's.

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And it sure does.

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That is all for Tasket.