Seximal: A better way to count [video]
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I'd be interested to know if others count by 1, 2, 3, or something else. (Another trick I learned working retail is to count out X coins, stack them vertically, and then make more stacks of the same height. Much faster than counting them all out individually!)
That's roughly how you count change too.
Ex: if I were to count 21 items my mental count would be 2,3 [5] 3,2 [10] 2,2,1 [15] 2,3 [20] and 1. [21]
Hope that makes sense!
You can't make off by 1 or 2 errors because then you'd end up with a count like 23 or 27. You're counting by 5... the last digit needs to be 0 or 5! And you're not going to make an off by 10 error.
You can visually pick out 2s and 3s to make your 5 before you add them in.
On your question, I feel less confident in my count, the more things I have to count. Since there will always be fewer groups of 5 than groups of 1, I always feel more confident counting in groups of 5 than of 1.
EDIT:
It's like divide and conquer. You divide the work by first making sure each group has only 5 items by glance counting (what kjeetgill means with 2 + 3), and then you count the groups, and add what didn't fit in any group.
Having the work divided like this makes it easier to verify that it was all done correctly. You'll end up feeling more confident about some groupings than others (because of the way the items are positioned) and you'll just want to recheck the ones you're not so confident about and readjust accordingly. Once you're confident in the groupings, you can recount the groups more easily than having to recount each individual thing.
For most sums, you'll end up adding less than 5 or 10 things at each step, which is easy.
For bills and coins, I count by object, not denomination. Then multiply total by denomination.
For counting large numbers of bills, I like creating stacks of 20. Then I count the number of stacks. After that I make stacks of 100 bills, and wrap them. In this way, I am never 75 bills deep, lose count and have to restart.
I know I have to count each stack of twenty twice, and ideally not three times, so I count 21 bills out, then if its 21 the second time, I take one off.
For piles of coins(and socks, et al), remove the most common/visible/largest type. For example, I would remove $1 coins into their own pile/bag, not bothering to count yet. Then as they become less obvious, I switch to $2 coins, then $0.25, then $0.05(since they are larger than $0.10 coins).
I had an electronic coin counter, but away from the office, that's how I sorted coins. If I wanted to count on site, it was a lot easier to count homogeneous bags. I could also weigh for a rough estimate.
I'm not sure I understand this. Is it just so you don't have to hunt for an extra bill somewhere else? If the second count doesn't match the first count you need to count a third time anyways, right? It sounds like you're implying using a 21 stack initially reduces recounts.
By the end of that, I could count up to about 18 donuts just by looking.
> The accuracy, speed, and confidence with which observers make judgments of the number of items are critically dependent on the number of elements to be enumerated. Judgments made for displays composed of around one to four items are rapid,[2] accurate[3] and confident.[4] However, as the number of items to be enumerated increases beyond this amount, judgments are made with decreasing accuracy and confidence.[1] In addition, response times rise in a dramatic fashion, with an extra 250–350 ms added for each additional item within the display beyond about four.[5]
I can't spot the references I thought I'd see here but I recall there being work around training this and seeing people improve up to around 7-8.
I also count coins by 5s if I have a bunch of them flat on a table - it's faster that way.
You better check the book, because it has been a long time since I read it, but he explained that most people counts 3 by 3, sometimes 2 by 2, and exceptionally 4 by 4. Only very exceptionally people can count 5 by 5. What most of us do when counting 5 elements, is to divide them in a chunk of 3 and one of 2.
In base 12, it is convenient to count 4 groups of 3. This is definitely easier and more efficient than decimal counting.
I do a similar thing when counting "manipulable" objects of the same size --- in semi-binary. I make a stack of height n, then another, and stack them on top of each other to create one of 2n, before making another of 2n. If there aren't enough, I use the next-lower power of 2, and so on. Then I add them all together at the end and it's usually faster for me to go from binary -> hex -> decimal than do the additions in decimal. Memorably, I once skipped the last step while tired:
coworker: how many?
me: E6.
coworker: huh?
(several more seconds later)
me: ...oh, I mean 230.Of course, counting items with fixed dimensions like coins or bottles, you group them appropriately in piles of 5 or 10 or 12 or 16 depending on the "natural" layout, and multiply from there.
I count in 1,2,3,4 or 5 blocks depending on the situation. For example if I am supposed to have 13 buttons and I want to verify it. I visually or physically separate 5, then another 5 and then check the remaining buttons. If it is 3, then I have verified it.
Anything above 5 gets harder and harder to distinguish. I could easily and naturally spot 1, 2, 3, 4 or 5 items. But if it gets higher than that, it gets harder for me to distinguish between a group of 7 or 8 or 9 items.
So 10 in base 12 can be exactly divided by 2, 3, 4, 6, 8, and 9.
Compare this to base 10 where 10 can only be exactly divided by 2, 4, 5, and 8.
10/4 = 2.5
vs.
10/3 = 3.33333...
10 / 4 = 2.5
2.5 can be represented exactly in decimal. As opposed to:
10 / 3 = 3.33333333333...
which cannot.
In duodecimal (10=dozen, A=ten): 10 / 3 = 4 A / 3 = 3.4
10 / 3 = 4
A / 3 = 3.4
10 / 3 = 4
A / 3 = 3.4They’re pretty nice if you use base 60, in which case you can figure out how to almost always approximate any division problem by something you already have in your reciprocal table.
I would consider a duodecimal (base-12) metric system to be ideal, personally.
The popularity of pentatonic music also suggests divisibility isn't unmissable.
Why do the divisors of 12 matter in music? I'm not aware of anything using the divisors of 12.
12 is mainly important in music because the (3/2)^12 is a power of 2. We divide up the octave into fractions of frequencies, not multiple os steps (a.k.a. fractions of number-of-steps-per-octave). The harmonious way of stepping through the 12 notes is in steps of size [2,2,1,2,2,2,1], which is importantly not by divisors of 12. (Stepping strictly in any arithmetic progression hits at least one low-harmony tone and misses high-harmony tones.).
Stepping in the sequence [2,2,1,2,2,2,1] leads to frequency ratios in these low-complexity (==harmonious) fractions: [1 9/8 5/4 4/3 3/2 5/3 15/8 2]
Now I'm plagued by off-by-one errors in my mental math, particularly when it comes to intervals. If you ask me how many days there are between now and Christmas, I will be uncertain about the answer unless I individually count them all off on my fingers.
Well, between day 1 and day 3 there is 1 day. By finite induction between day n>3 and day 1, there are n-2 days, and analogously between days n>k>1 there are n-k-1 days.
| H | e | l | l | o |
Index 0 1 2 3 4 5
Length |<--------5-------->|
This makes it easy to visualize that 1-based indexing has indices centered on the memory cell, so calculating a span requires reducing the subtractor by 1."well, if it spanned to row 19, there would be two elements, so N elements end at row 17+N, so 33-17 is 16, so there are 16 elements."
That's painstaking and I should fix it, lol.
Subtract the start and stop. If the start and stop are both excluded, subtract 1. If one of them is included, make no change. If both are included, add 1.
Excluded start and stops happen when you have external boundaries. Like, your last day on the old project was the 10th, and your first day on the new one is the 15th. You had no project for 15-10-1=4 days.
One included happens when you're measuring between numbers. If you start working at 2 and finish at 8, you worked 8-2=6 hours.
Double inclusion counts when you have internal boundaries, or are counting how many things are "touched". If you work on something from the 11th to the 15th, you worked on it 15-11+1=5 days. (Also, your example.)
Stop position - start position + (number of included positions - 2) = total
I was just taught counting using it, it was a quick bonus math class one day.
It is super useful for counting things off without losing place. Even being able to go up to 10 on just one hand is useful.
We just need a good naming scheme for counting in hex :)
Do you know of another unrelated language that works similarly?
I think base 1 is a little more fundamental than base 2, no?
1 dec == 1 unit (1x1^0)
2 dec == 11 unit (1x1^0 + 1x1^1)
3 dec == 111 unit (1x1^0 + 1x1^1 + 1x1^2)
etc. Just like the other bases, with "1" instead of "2" or "10".Well, maybe not "just like", the digit I have to use to represent each power of 1 is 1 and not 0, but I think that's a small concession.
In your system, 1 = 10 = 100 = 1000, and 11 == 101 = 110, and 101 = 1+1 = 11. It's useless as a notation for arithmetic resembling like what bases are used for.
In base 1, the only valid digit is 0, so the only valid number is unit.
Same for Italian.
German and English are even more closely related: they are both West Germanic languages.
As a native English speaker, I cannot see any similar patterns between German in English when I happen across random conversations on a site like Reddit in German. But looking at romance languages, I can see more similarities.
Maybe it's because of my childhood environment in Texas, where most English speakers learn some basic Spanish by interacting with other people.
I would love to see a passage of German that can be intuitively reasoned about by an English speaker with no experience with German solely based on context and perhaps root words.
In all honesty I'd probably get confused though. :(
There was an episode of Barney Miller where one of the cops was translating for a German woman. At some point she says "Das ist mein bebe." Bebe is pronounced very similar to baby and means the same thing.
I can't find a video of just the scene. It is Season 5, Episode 5 The Baby Broker.
The Daily Motion has the episode, but it keeps glitching on me. I haven't been able to get to the scene in question.
I also used to have German language resources that built on words that were readily understood by English speakers.
I = ich
have = habe
hear = hören
before = bevor [in some meanings to be fair]
and = und
[Your first half sentence is a good example :)]
is = ist
can = können [and in general the whole can/können, shall/sollen, will/wollen connection - though the meanings have drifted a bit apart]
when = wenn
learn = lernen
see = seeen
word = wort
More generally, my experience is that literal (i.e., word-for-word) translations from German to English sound weird, but often comprehensible, while literal translations from French usually end up as gibberish.
"Just so we're on the same page here: you're wrong"!
We should really be seeing a great education dividend from the availability of material these days.
>Are reposts ok?
>If a story has had significant attention in the last year or so, we kill reposts as duplicates. If not, a small number of reposts is ok.
On topic: I agree with Aardwolf, I prefer Hexadecimal but having a better symbols for A-F would improve it.
I assumed (incorrectly) that 2 links I saw via google would have made this 3 at least)
Regarding your search for a non-video link in the grandchild comment, you are in luck. Read about seximal here [2] and make sure to click through the hamburger menu, that's where the meat of the website is at. Unfortunately the website and the video content is not fully overlapping so you kinda have to watch and read both for the fullest understanding of seximal.
[0] https://news.ycombinator.com/item?id=10285392