You could also see the original math problem turning up 6.5 as a solution, not a problem.

If you think of it as not 1 stitch being increased among 6.5, but 2 stitches increased among 13, you could do this:

x x x x x x x x x x x x O x O x x x x x x x x x x x x O x O...

By grouping your increases (or decreases) into groups of two, you allow the math to come out and I think it would look pretty nice, too.

I think this idea can scale, but you will need to figure out a few things:

1) How big do I want to make my groups?

2) How many stitches will go between groups?

If you want to keep 1-stitch divisions between increases (or decreases), you need to take the number of increases (or decreases) in your group (let's call this g) and subtract one. You can see the reason for this by holding out one of your hands. If you have five fingers, you will have 4 spaces in-between your fingers.

Revisiting the questions:

Multiply your increase factor (your original # / the # you want to increase by) by whole, prime numbers until your result equals a whole number (which I will call n). A good list of primes is here. The number you have to multiply by is the number of increases or decreases in your group (g). N is the number of stitches you have, total, in a given set of stitches for your increases or decreases. In the first example, above, n = 13 and the prime needed is 2. That makes g = 2.

To find out what you really need to do to make it work, you have one more step. To find out how many stitches should go in-between your groups of increases or decreases, take n - (g - 1). In the first example, that would be 13 - (2 - 1) = 12.

This seems complex and wordy, although it's really not, so I might make a worksheet for it. What do you think?

## Saturday, September 27, 2008

### Yaay! Comments!

Comments make this little blogger very happy.

Response to one: Sarah-Marie's Klein Bottle is here. She even has a klein bottle as a hat, and a photo of someone wearing it. Neat!

To sum up from my last post: what we were ultimately dealing with was a remainder. You don't have to break out in hives to deal with having last heard (or read) that word in the third grade. Just divide, say, 13 by 4 to find 3.25. Now, find that 3 * 4 = 12, so 13 = (3 * 4) + 1. Your remainder is 1. If you were increasing (or decreasing) by 4 stitches across 13, you would add or subtract stitches every 3 stitches and have on "extra" stitch to deal with however your heart desires. Numbers can be bigger or smaller, but they can all be tackled in this general way:

x = original number of stitches

y = number of stitches that need to be increased or decreased

r = remainder = "extra" stitches that need to be dealt with

x / y = either whole number or number in the form of a.bbbb. There could be many b's or not.

No b's? That means you have a whole number. On your way. Your math work is done. Simply increase or decrease every a number of stitches. Remember that, if you are decreasing, you will have to take into account the fact that the decreases use some of the a stitches. For example, if you need to decrease by 4 over 12 stitches, your a = 3. You will actually work 1 stitch then ssk or k2t four times, rather than work 3 and then k2t, because both a ssk and a k2t require two stitches to be worked.

If fraction, use x - (y * a) = r to find remainder.

Response to one: Sarah-Marie's Klein Bottle is here. She even has a klein bottle as a hat, and a photo of someone wearing it. Neat!

To sum up from my last post: what we were ultimately dealing with was a remainder. You don't have to break out in hives to deal with having last heard (or read) that word in the third grade. Just divide, say, 13 by 4 to find 3.25. Now, find that 3 * 4 = 12, so 13 = (3 * 4) + 1. Your remainder is 1. If you were increasing (or decreasing) by 4 stitches across 13, you would add or subtract stitches every 3 stitches and have on "extra" stitch to deal with however your heart desires. Numbers can be bigger or smaller, but they can all be tackled in this general way:

x = original number of stitches

y = number of stitches that need to be increased or decreased

r = remainder = "extra" stitches that need to be dealt with

x / y = either whole number or number in the form of a.bbbb. There could be many b's or not.

No b's? That means you have a whole number. On your way. Your math work is done. Simply increase or decrease every a number of stitches. Remember that, if you are decreasing, you will have to take into account the fact that the decreases use some of the a stitches. For example, if you need to decrease by 4 over 12 stitches, your a = 3. You will actually work 1 stitch then ssk or k2t four times, rather than work 3 and then k2t, because both a ssk and a k2t require two stitches to be worked.

If fraction, use x - (y * a) = r to find remainder.

## Thursday, September 25, 2008

### Sticky Problem and a Super-Math-Rich Post

I just moderated a comment from a listener that said she had to put 41 yarn-overs into 267 stitches evenly. She fudged it. I'm going to say she would have to - it averages to one yarn-over about every 6.5 stitches. This is nearly impossible, really. If I were working some other increase, say, a lifted increase, I would say, just do an *increase, work 6 stitches, increase, work 7 stitches* repeat, and it should even out. I think that would drive me nuts when making eyelets, since my switching between 6 and 7 stitches apart would be obvious, to me, at least.

A possible solution: factor your numbers out and look for common factors. The major issue of this is that 41 is a prime number, which means that we would have to add or subtract somewhere to make this work.

267 is not a prime number, so we can take it apart and see what we have. Unfortunately, we don't have far to go.

267 = 89 * 3

89 is a prime number. Luckily, 3 * 14 = 42. So, we could round 41 to 42 by deciding that one more increase in the mix is better than the crazy-making idea above.

We can mentally, and with stitch markers, divide our work by three - so we're facing putting 14 eyelets into 89 stitches. 14 * 6 = 84, which leaves us with 5 extra stitches for each "wedge" of 89, if we increase every six stitches.

Speaking of which, to increase every six stitches *really* evenly, we would do this:

(x = stitch, o = increase)

(stitch marker) xxx o xxxxxx o xxxxxx o xxxxxx o xxxxxx o xxx...

I'm not going to do it 14 times, but you get the idea.

What do we do about those extra 5 stitches? We could put them in-between wedges, so that at some point, our work will look like this:

o xxxxxx o xxxxxx o xxxxxxxx(stitch marker)xxx o xxxxxx o xxxxxx

Which is to say, that, around our stitch markers, we would have 5 + 6 = 11 stitches as spacers. You could, and I would, decide that this is a design feature and only have to decide where they should lie - it doesn't even have to be 100% evenly, but I would probably put two of them under the arms of a sweater and the third in the middle of the back or front.

Design Feature Solution #2:

Break the set of 42 increases into 3 sets of 14 stitches.

267 / 14 = just over 19.

So, the first set of eyelets will be an increase every 19 stitches.

Total stitches = 281

281 / 14 = just over 20.

The second set could be every 20 stitches.

Total stitches = 295 stitches

295 / 14 = just over 21

The second set could be every 21 stitches.

Total stitches = 309 stitches

So, by adding one total stitch, and I'm not going to try to draw it for you, you could have a perfectly neat little leaning line of yarn-overs, in either 3 rows or 6, depending on how you do it. You could even stagger them by as much as you like, by making your first increase sooner than you would, and letting the stitches fall where they will.

I recorded my show on grafting, but have not knit, grafted, or photographed it, so it is still in the works, so to speak. I may talk about the above math in my next show, because I don't think a lot of you read this blog. Am I wrong? Leave me a comment to prove it. :)

A possible solution: factor your numbers out and look for common factors. The major issue of this is that 41 is a prime number, which means that we would have to add or subtract somewhere to make this work.

267 is not a prime number, so we can take it apart and see what we have. Unfortunately, we don't have far to go.

267 = 89 * 3

89 is a prime number. Luckily, 3 * 14 = 42. So, we could round 41 to 42 by deciding that one more increase in the mix is better than the crazy-making idea above.

We can mentally, and with stitch markers, divide our work by three - so we're facing putting 14 eyelets into 89 stitches. 14 * 6 = 84, which leaves us with 5 extra stitches for each "wedge" of 89, if we increase every six stitches.

Speaking of which, to increase every six stitches *really* evenly, we would do this:

(x = stitch, o = increase)

(stitch marker) xxx o xxxxxx o xxxxxx o xxxxxx o xxxxxx o xxx...

I'm not going to do it 14 times, but you get the idea.

What do we do about those extra 5 stitches? We could put them in-between wedges, so that at some point, our work will look like this:

o xxxxxx o xxxxxx o xxxxxxxx(stitch marker)xxx o xxxxxx o xxxxxx

Which is to say, that, around our stitch markers, we would have 5 + 6 = 11 stitches as spacers. You could, and I would, decide that this is a design feature and only have to decide where they should lie - it doesn't even have to be 100% evenly, but I would probably put two of them under the arms of a sweater and the third in the middle of the back or front.

Design Feature Solution #2:

Break the set of 42 increases into 3 sets of 14 stitches.

267 / 14 = just over 19.

So, the first set of eyelets will be an increase every 19 stitches.

Total stitches = 281

281 / 14 = just over 20.

The second set could be every 20 stitches.

Total stitches = 295 stitches

295 / 14 = just over 21

The second set could be every 21 stitches.

Total stitches = 309 stitches

So, by adding one total stitch, and I'm not going to try to draw it for you, you could have a perfectly neat little leaning line of yarn-overs, in either 3 rows or 6, depending on how you do it. You could even stagger them by as much as you like, by making your first increase sooner than you would, and letting the stitches fall where they will.

I recorded my show on grafting, but have not knit, grafted, or photographed it, so it is still in the works, so to speak. I may talk about the above math in my next show, because I don't think a lot of you read this blog. Am I wrong? Leave me a comment to prove it. :)

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