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 Base Denomination Efficiencies, for weights and currencies
DavidKennedy
Posted: Sep 20 2016, 07:43 PM


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I analysed bases with different systems for denominations of currency or weights. I chose to investigate denominations incrementing approximately logarithmically and are rounded to a near factor of the base.

If the base is a power of two, and the denominations are all the factors of the base, the system is efficient because there would be neither more nor fewer than are needed to make up any amount.

Bases with systems of denominations that are deficient require some of the denominations to be available more than once in order to be able to make up any amount.

Sufficient systems do not require more than one copy of any of their denominations. However, they may lack efficiency by having more denominations or kinds of value than are necessary to make up amounts.

It is better for a system to be sufficient than deficient.
By this analysis, decimal is deficient and requires carrying more weight items than an efficient system.
There is a dozenal system that is sufficient and is nearly efficient because its denominations sum to the base. The results show that dozenal requires fewer coins or weights to be carried than decimal to make up any amount up to a given value, because the number of amounts that can be made using the same number of weights is superior for the dozenal system.

BaseDecimal Pseudo-sexagesimal ≈ 10^(n/4) ≈ 1.78^nSexagesimal ≈ 60^(n/7) ≈ 1.79^nDozenal ≈ 12^(n/4) ≈ 1.86^nSexagesimal ≈ 60^ (n/6) ≈ 1.979^nLong Hundred ≈ 120^(n/7) ≈ 1.982^n= 2^nDecimal ≈ 10^(n/3) ≈ 2.15^nSexagesimal ≈ 60^(n/5) ≈ 2.27^n
EfficiencyExcessiveExcessiveSufficientSufficientSlightly sufficientEfficientDeficientVery deficient
n=011111111
122222222
22
233344455
5
36666881012
10 12
41010121515162030
20
52020243030325060
6303036606064100120
7606072120120128
8100120144
#999988910
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DavidKennedy
Posted: Sep 20 2016, 10:53 PM


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I omitted to show in the table that the base of the exponential sequence for the decimal pseudosexagesimal can be approximated by the fourth root of ten.

There is also a system based on the gross. Its exponential base is about the seventh root of the square of a dozen. This system is slightly deficient and up to its base requires one denomination to be duplicated, giving a total of nine weights, which is the same as for the equivalent amount in the dozenal system. However, the sum of the weights in the dozenal can reach a slightly higher value and hence can represent more amounts than the gross system. Nevertheless, the gross system is more efficient than decimal.

The value for the third power in the sixth root sexagesimal system could be either ten or six, affecting which denomination is required in double stock; just a little something extra to think about.
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DavidKennedy
Posted: Sep 21 2016, 07:41 AM


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For clarification, when I calculated the third power of the sixth root sexagesimal I originally opted for eight, but since this is not a factor of the base I changed it to ten, but neglected to duplicate a denomination and increase the number of weights. So I edited the post a few hours after posting, and correct it again this morning.
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wendy.krieger
Posted: Sep 21 2016, 08:36 AM


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The proper value to derive, is to suppose that enough weights to make all the values from 1 (inc) to b (exc) are taken, giving n, then quote b^(1/n), or the logrithm of that.

The closer to 2 the value is, the more desirable the base.

In base 120 (given as 'slightly sufficient), the weights run 1,2,4,8,15,30,60 but this is actually quite efficient, since 8 is very near 15/2, and little is lost using 8 as test for 1/2 of 15.

This test is actually to do a binary-like tree using the named weights, to get as close to the given weight as possible. In 12, for example, 1,1,3,6 suffice, since you use the sequence of 6/ 3, 31 311 to get to 5, and 6/ 3 31 311/ to get to 4.x (/ means too heavy).

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Shaun
Posted: Sep 21 2016, 08:49 AM


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In Britain we were quite happy with the 1-3-6 pattern for £sd, with 1d, 3d and 6d coins; for larger values the same pattern can be used - 1s, 3s, 6s.
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wendy.krieger
Posted: Sep 21 2016, 08:54 AM


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That's the same as what we had with money (1,3,6), but to make a set of weights out of this, you need two ones, since you can't make 5 (5d = 1+1+3).

The dozenal series 1,2,4,8,15,30,60 is fine for weights, since they make all numbers from 1 to 120, but i don't use them as money. Instead i use 1,2,5,10,30,60 for these. Dropping off the 1,2 and dividing by 10, gives the same series as the pennies in a shilling.
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DavidKennedy
Posted: Sep 21 2016, 09:13 AM


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wendy.krieger Posted on Sep 21 2016, 08:36 AM
QUOTE
The proper value to derive, is to suppose that enough weights to make all the values from 1 (inc) to b (exc) are taken, giving n, then quote b^(1/n), or the logrithm of that.

This is what is conveyed in the base headings of the table. The base is your b, while your n is the number of distinct denominations less than the base. This is a different use for n than in the table where it stands for the nth denomination.
QUOTE
The closer to 2 the value is, the more desirable the base.

The situation is complicated by the choice that the denomination should be a factor of the base. For example, with the sexagesimal base and six denominations the system becomes less efficient than the twelfty.

Shaun Posted on Sep 21 2016, 08:49 AM
QUOTE
In Britain we were quite happy with the 1-3-6 pattern for £sd

The dozenal system proposed here is to let the base twelve correspond to a shilling, dollar, or sixty european cent.
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wendy.krieger
Posted: Sep 21 2016, 12:49 PM


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The reason why ye should use logrithms, and exclude the fencepost error (ie including 1 and B ), is that for example, 12 does a better cover than 10, but not as good as 60 or 120,

10: 1, 1, 2, 5 ... 1.778
18: 1, 2, 3, 6, 9 1.782
120: 1 1 2 5 10 10 30 60: 1.8192
12: 1, 1, 3, 6 ... 1.861
28: 1 2 4 7 14 1.9473
60: 1 2 4 8 15 30: 1.978
120: 1 2 4 8 15 30 60: 1.981
496 1 2 4 8 16 31 62 124 248: 1.992
16: 1 2 4 8: 2.000

This particular test tells us how good we can get to binary scales. The better one gets to powers of 2, the more 'efficient' the weights are. On the other hand, it gives no really good indication on how versatile the base is otherwise.

For example, adding an extra unit to 120 (so it is 10*12) means that it takes a harder hit on divisors.

You can apply all sorts of tests like this, which make various numbers appear optimal, but the number in question is pretty useless as a number system.

For example, the numbers to 120 can be expressed as the sum of no more than four distinct divisors, and for every subsequent power, no more than three for each additional power. This is as good as 18, for example.

The optimum for n distinct divisors, is of course, (n+1)! But generally, factorial bases are not efficient means.
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DavidKennedy
Posted: Sep 21 2016, 01:32 PM


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wendy.krieger Posted on Sep 21 2016, 12:49 PM
QUOTE
12 does a better cover than 10, but not as good as 60

QUOTE
60: 1 2 4 8 15 30: 1.978

Here you have included 8 as a denomination although it is not a factor of the base. This sequence is just twelfty in disguise. Hence, for small amounts and without the suggestion of another sexagesimal sequence, twelve is better than sixty.
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DavidKennedy
Posted: Sep 23 2016, 11:50 PM


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Bases which are slightly sufficient can be constructed by taking two adjacent binary denominations (powers of two, 2^(n-1) and 2^n), subtracting unity from the higher, and multiplying them to obtain their lowest common multiple. This provides that all the denominations will be factors of the base. Bases of this form 2^(n-1).(2^n-1), including the current year, are a subset of the triangular numbers, which are the sum of all the natural numbers up to a given number. Where the denominations are to be all the factors of the base, other factors than the denominations must be avoided, which is achieved by a Mersenne prime, and this causes the base to be also a so-called perfect number.

When hearing or reading about perfect numbers in elementary number theory, this condition for perfection must have appeared to be rather unqualified, and an unsatisfactory explanation (by Philo of Alexandria according to Wikipedia) for the number of days of creation (six) or the days of a month (supposedly two dozen plus four, but this is now inaccurate). Nowadays we would see a number as being more perfect if it had more prime factors.
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DavidKennedy
Posted: Sep 24 2016, 04:41 PM


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Theorem: The denominations of a slightly sufficient base, so constructed, sum to the base.
Proof:
Preliminary lemma: Since an efficient base is a power of two, having a posteriori no other prime numbers in its prime factorisation, all of its factors are powers of two. Their sequence forms a geometric series summing to 2^(n+1)-1, which is just less than the base 2^(n+1).

Proposition 1: The denominations with a base should approximate a geometric sequence.
Argument: Deviations of the system from a smooth exponential sequence decrease the efficiency.

Proposition 2: Suppose that all denominations of a system should be factors of the base.
Argument: If the denominations were not factors of the base, then, given Proposition 1, there would be little reason for letting a system of denominations have one of them as the base instead of another.

Proposition 3: Given Proposition 2, the base should be the lowest common multiple of the denominations.
Argument: Given Proposition 2, the base cannot be less than the lowest common multiple. Given Proposition 1, increasing the base to a "denomination" larger than the lowest common multiple would not increase the efficiency. However, in actuality, because of the deviation of systems from perfectly exponential, the efficiency can be made to increase by increasing the base, but since this improvement in efficiency is only slight and manifests only for amounts larger than the base it does not justify increasing the base, particularly for currency coins which are not used for amounts much exceeding the maximum base practicable for humans.

For the constructed slightly sufficient systems, the lowest common multiple of the denominations is the multiplicative product of the highest power of two and the next denomination because they have no common factor, since the next denomination as an odd number is not divisible by two while the power of two is divisible by no other prime number than two.

The sum of the denominations up to yet excluding their slightly sufficient base is the sum conjointly of:
1. the geometric series formed by the denominations up to the lower denomination that is the highest power of two and
2. the geometric series formed of the higher next denomination and the multiples of it by a power of two.
The sum from the running variable k=0 to k=l of the first geometric series whose general term is 2^k is 2^(l+1)-1.
The sum of the second geometric series with initial term 2^(l+1)-1 and common ratio 2 is the sum from running variable j=0 to j=l-1 of the general term (2^(l+1)-1).2^j. This sum is (2^(l+1)-1)(2^l-1).
The conjoint sum of both geometric series is 2^(l+1)-1 + (2^(l+1)-1)(2^l-1) = 2^l.(2^(l+1)-1), which is of the constructed form for a slightly sufficient base.
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DavidKennedy
Posted: Sep 24 2016, 08:10 PM


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The argument to the first proposition requires refinement. It would be better to break the smoothness of an exponential function based on three.

The argument to the second proposition gives the impression of being there so that the third proposition would have something to stand on. If the denominators were not factors of the base, then they would not really belong to the base, would they?
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wendy.krieger
Posted: Sep 25 2016, 09:01 AM


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It is presented by way of argument, but is wrong in experiment. In any case, the logic is not how one approaches fracturing of bases for other ends.

Consider twelfty.

For a set of weights, one might go for 1,2,4,8,15,30,60. This is pretty much a binary set. One would need no more than six of the above to express any amount from 1 to 119 inclusive.

But this is not how one divides twelfty. In terms of money, it goes 1,2,5,10,30,60, the complete amount can be made by adding an extra 1 and 10. When this is done, the numbers add to 119, which means that with the exception of 2 mod 5, every amount has a unique spread of coins.

When one considers a different set of divisors, 1,2,3,4,5,10,15,20,40,60, one needs no more than four coins to make any amount, the excess over 120 is 40. This is seen because this set of coins represent the fractions of the 'factorial base'.

Money does not work by single coins, the aim is to minimalise the denominations, without having a very large number of coins.

Consider the step of 5. In the USA, and some other places, there is no coin between 1c and 5c. This is a fairly large jump, and one tends to accumulate 1c pieces more than in those countries where there is an intermediate 2c coin. On the other hand, there seems never a desire to insert a coin between 1d and 3d, the run of coins are 1d, 2d, 4d 1s or 1d 3d 6d 1s.

The run between 1s and 5s is to have varyingly 2s or 2s 6d or both. The latter represents 1/4 of 10s or 1/8 of the pound, while the florin (2s), comes only as a decimalisation attempt (ie 1/10 of the l.) In the USA, they eschew the 2 dime piece in favour of the 2½ dime, that is ¼$.

Base 18

Base 18 requires 5 coins, and the five divisors 1,2,3,6,9, are sufficient that any three makes an amount from 1 to 17, the total is all together 21.

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DavidKennedy
Posted: Sep 25 2016, 03:01 PM


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QUOTE (wendy.krieger @ Sep 25 2016, 09:01 AM)
wrong in experiment

Experience is not the same as experiment. Although I can imagine how a trial could be conducted, no prospective results have been presented. Examples from the real world cannot be assumed to be optimal.
QUOTE
Money does not work by single coins, the aim is to minimalise the denominations, without having a very large number of coins.

Your priorities are wrong and from the viewpoint of the mint rather than the purchaser. The goal primarily is to minimise the number of coins that have to be carried, and only secondarily the number of different denominations.
QUOTE
But this is not how one divides twelfty. In terms of money, it goes 1,2,5,10,30,60, the complete amount can be made by adding an extra 1 and 10.

Adding an extra 1 and ten increases the number of coins to more than the number of denominations in the more efficient sequence.
QUOTE
Base 18 requires 5 coins, and the five divisors 1,2,3,6,9, are sufficient that any three makes an amount from 1 to 17, the total is all together 21.

The denominations of base one and a half dozen can make up to the base by three coins, and produce an excessively inefficient system. Base twelve with five coins 1, 2, 3, 6, and twelve can make up to a dozen plus nine by three coins, and the total is two dozen.
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DavidKennedy
Posted: Sep 25 2016, 09:52 PM


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The questions of weights and currencies are different. Disregarding divisibility, the most efficient system of weights is without doubt binary. Dozenal is more efficient than decimal, and certainly much more efficient than one-and-a-half dozen, for weights, and benefits from the divisibility and sum of the denominations being the base.

Once something has been weighed, the same weights can be used to weigh again.
However, currency differs in that once something has been bought, the payer has fewer coins to spend. The difference is crucial, and means that the probability of a coin being used in the transaction should be taken into account.

For coins, dozenal and one-and-a-half dozen are practically equally efficient in terms of number of coins, except that dozenal can be used to pay for higher amounts with the same number of denominations, despite that the other base is larger, because in the system based on a dozen plus six, the denominations are more tightly packed and therefore smaller.

I discovered that the sexagesimal system and the base two dozen plus four are two of the more efficient coinage systems. Surprisingly, decimal is about as good as twelfty. However, dozenal benefits from having lower probabilities for the two lowest denominations, whereas in decimal, the second lowest denomination has a relatively high probability of being used in a transaction.
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DavidKennedy
Posted: Sep 26 2016, 11:41 AM


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For coins the optimal base is larger than binary.

m1n1f1g Posted: Aug 21 2013, 08:04 PM
QUOTE
It appears that the sequence of coins generated by this is F2n+1, where F is the Fibonacci sequence with F0 = 0.
QUOTE
With m1n1f1g's denominations, the ratio between consecutive denominations converges to phi^2 = 2.61803398874989 = 2;74EE6772802X5.

It's interesting that this is fairly close to e.

Examining powers of the square of the golden ratio, the third power is nearly a dozen plus six, such that the base twelve plus six set for coins should be {1, 3, 9, 18} with the 6 removed from the weights.

Searching for other roots of bases approximating the base of the natural logarithm, the fifth root of a gross gives the sequence of coins to near factors {1, 3, 8, 18, 48}, which appears to be considerably better than decimal. This sequence is given as an optimal by:

Dan Posted: Feb 8 2009, 07:15 AM
QUOTE
For 5 denominations: {1, 3, 8, 18, 48} averages 4.041666666666667 coins per transaction.

Divisible bases for optimal systems for coins could be constructed by varying between bases two and three, sometimes doubling or trebling, or by either of these combined with adding to a doubled or subtracting from a tripled denomination while hoping or designing that the resulting denomination will have small prime factors.

The dozenal set {1, 2, 3, 6} does not seem too bad compared to others for its range:

Dan Posted: Mar 21 2014, 03:40 AM
QUOTE
QUOTE (Pinbacker @ Mar 20 2014, 06:05 PM)

    {1, 2, 3, 6}
    {1, 2, 4, 6}

1;8 coins
QUOTE
For 4 denominations, {1, 2, 3, 7}, {1, 2, 3, 8}, {1, 2, 4, 7}, {1, 2, 4, 9}, {1, 2, 5, 8}, {1, 2, 5, 9}, and {1, 2, 6, 9} all average 1.7 coins per transaction.

Sometimes explaining why one system is better than another is not obvious, and the differences in optimality are too small to justify changing the base. The dozenal set {1, 2, 3, 6} appears to be just about as good for its range as decimal {1, 2, 5, 10}.
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wendy.krieger
Posted: Sep 26 2016, 11:47 AM


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QUOTE
Your priorities are wrong and from the viewpoint of the mint rather than the purchaser.


Coin denominations is not just a mint issue. Every shop keeper, merchant and banker would need separate slots in the draws for each kind of coin.

While it is true that a set of coins like 1,2,3,6 would serve to give 1 to 11 in no more than three coins, people tend to carry multiple sets of coins, and deal with multiple sets of transactions. This is why 1,3,6 is sufficient: it handles a step of 12 with just three coins.

Of course, unlike weights, the geometric mean step can be much larger than two, since one has little objection to handling three or four coins of the same kind. The internment camps over here during the world war, issued coins of denomination of 1 and 3 pence, and 1 and 5 shillings. You still need a collection of coins to handle this, but the mean power here is close to steps of 4.

In practice, the way change is given is that one 'counts up' to the tendered amount. That is, one rounds the pence to the shilling, and shillings to the pound. Thus coins that follow the arithmetic structure of the base are better than those that dont. For example, 1,2,5, 10,30,60 are easier to round a measure to the next ten, the next hundred, which 1,2,4,8,15,30,60 do not afford. You have to count to the nearest 15 here.

QUOTE

Experience is not the same as experiment.


One does not have to be a user of base 15, to evaluate the effectiveness of 1,1,3,5,5 as a set of weights or coins in that base. One can simply posit a set of coins and argue the various cases. This is not experience: it's experiment. It's taking individual cases and testing the theory.

QUOTE

... number of denominations in the more efficient sequence.


Efficiency is just a buzzword for making some parameter better. For example, something like 496 or 28 are rated as 'efficient' in this test, although in the main are largely unworkable bases. (I have used 28).

In any case, for a process of coins, one must understand the most common forms of arithmetic is to count upwards to the round numbers. So 28 is efficient in that given a number, you count to the multiple of 7, 28, 196, &c.

If the series of coins do not contain the measures of column, and its multiples, then it is less efficient. The quarter is less efficient than the 20c coin, although logrithmically, they divide the space of 5 from 10c to 50c into a step of 2 and 2½. Once one has counted to the 10c level, the next step is to count upwards to the $1 level, and this is easier if the steps are 1 and 2, rather than 1 and 2½.
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DavidKennedy
Posted: Sep 29 2016, 11:09 PM


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QUOTE (wendy.krieger @ Sep 26 2016, 11:47 AM)
Coin denominations is not just a mint issue.  Every shop keeper, merchant and banker would need separate slots in the draws for each kind of coin.

Drawers aren't people and don't have feelings to mind how many slots they have. If a currency is designed to have no more coins than are already being used, then I imagine that existing slots would do.

wendy.krieger Posted on Sep 25 2016, 09:01 AM
QUOTE
But this is not how one divides twelfty. In terms of money, it goes 1,2,5,10,30,60,

For twelfty currency, have you considered the sets {1, 2, 6, 20, 60, 120}, {1, 3, 8, 15, 40, 120}, or {1, 2, 5, 15, 40, 120}?

wendy.krieger Posted on Sep 26 2016, 11:47 AM
QUOTE
This is why 1,3,6 is sufficient: it handles a step of 12 with just three coins.

The one could be omitted instead of the two, if nothing can be bought for a penny. I originally investigated different systems of coins to decrease the probabilities of the lower coins being used because the coppers of the euro currency are being removed from circulation.

I remember objecting and asking at the time of the euro currency being introduced why such a small value was chosen for the smallest coin, since nothing is sold at that value. A primary school kid could work out that one couldn't buy a penny sweet with it. The only excuse I heard was like "A system was wanted in which the smallest value was capable of being represented." The smallest coin is only useful for purchases with all those senseless in-between values, which wouldn't even exist if the smallest value had not been so low in the first place. The values of the coppers are so low that they can often be seen dropped or discarded without care. They are very wasteful because the cost of producing these coins is more than they are worth. What a bunch of idiots it must have been who planned this system!

I reasoned that the smallest coins are used so frequently because of the ratios between the denominations in decimal being too widely spaced, and that this problem could be fixed by having them more tightly packed by changing the base to six with three denominations or twelve with four denominations less than the base, as {1, 2, 3, 6, twelve}. I determined that the probabilities of the lowest coins being needed to pay exactly for a random amount in this dozenal system are lower than for the larger coins. Thus, this proposal should decrease the quantity of small change being carried. People hate small change. Wouldn't we prefer to carry larger coins with which we could buy more than so many small coins being needed for everything? Other advantages are that this dozenal set allows more values to be paid in a greater number of equally efficient ways. The cost of this versatility is a slightly greater number of denominations. However, we could nevertheless maintain the same number of coins by eradicating the coppers and introducing paper notes for the values larger than the largest decimal coin. Now I have been thinking that the smallest dozenal coin could correspond to the ten cent of the euro currency.

Does anyone think there is anything really wrong with the {1, 3, 8, 18, 48, gross} twelve squared system for coins? In dozenal it looks less strange: {twelfth, quarter, two thirds, three halves, four, twelve}.
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DavidKennedy
Posted: Jun 4 2017, 04:06 PM


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A picture speaks a *duzand words.
user posted image
The plot assumes the Weighted Product Modified Efficiency Model at Base Subdivisions, an alternative way of thinking of bases, which applies to currency. I tested some other sets of denominations in different bases, but omit them from the graphs for clarity. If the model is correct, then we no longer have to answer to question of what the most efficient set of denominations is, we only have to answer objectively "What is the most efficient base for currency?"
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wendy.krieger
Posted: Jun 5 2017, 07:44 AM


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A double set of coins of [1, 3, 8, 19, 47]z, suffices to make every amount from 1 to 100, with no more than six coins. The base here is 2.75 (doz), which i imagine is fairly efficient. What happens is that most dozenal folk would not buy this set, but rather a set like [1, 3, 6, 10, 30, 60]z. This second set is also more efficient when time comes to 'drop off' the lowest coins. So any number of coins can be dropped off, as each is a divisor of the next. This second set is also how one percieves division of a gross: halves, quarters, uncia, half-uncia, quarter-uncia and unc-uncia.

One is more likely to recognise 53z as [3, 10, 10, 30] rather than [8, 47]. This is because one is essentially 'finding' 53 by a dozenal address, rather than as a sum of fibonacci numbers.

In twelfty, we have a similar set of 'efficient weights' [1, 3, 9, 27, 81], a single set of these serves to weigh every number from 1 to 100. But while it is possible to construct a given number, the set is not very well given for the binary division making that something like [1, 2, 4, 8, 15, 30, 60] is.

One might recall that a set of weights would live in a box near the weighel, and would be used in the binary discisions. The loss of efficiency in dividing 15 into 8 and 7, is more than amply made up in that these 7 weights make two digits, and the next seven (100, 200, &c), would make a similar hundred-place. So it's easier to figure out what 400, 100, 60, 15, 1 makes, against 432, 108, 32, 4 makes.

On the other coins, a fist full of cash from the bottom of ones purse or pocket, is not likely to produce all the neccessary denominations. In the dozenal fibonacci set, the absence of any 47c coins, would make our 53c busfare, say 19, 19, 19 or 19, 19, 8, 8, 3, 1, 1. The absence of a 30c would make it 10, 10, 10, 10, 10, 3.
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DavidKennedy
Posted: Jun 5 2017, 04:39 PM


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QUOTE (wendy.krieger @ Jun 5 2017, 07:44 AM)
A double set of coins of  [1, 3, 8, 19, 47]z, suffices to make every amount from 1 to 100, with no more than six coins.  The base here is 2.75 (doz), which i imagine is fairly efficient.  What happens is that most dozenal folk would not buy this set, but rather a set like [1, 3, 6, 10, 30, 60]z.

By the model, if the most efficient base is the square of the Golden Ratio, then the dozenal set {①, ③, ⑥, ①⓪} would not be as efficient as {①, ②, ③, ⑥, ①⓪}. The alternate Fibonacci terms are not all factors of the base. In the set {①, ③, ⑧, ①⑥, ④⓪, ①⓪⓪}, the denominations are factors of the base, but the base is twelve squared, not a dozen. I do also prefer that the denominations be whole multiples or unit fractions of each other, as they are in {①, ③, ①⓪}. But it is still possible to pay for all values if the lowest of the set {①, ②, ③, ⑥, ①⓪} be removed after time.
QUOTE (DavidKennedy @ Sep 29 2016, 11:09 PM)
wendy.krieger  Posted on Sep 26 2016, 11:47 AM
QUOTE
This is why 1,3,6 is sufficient: it handles a step of 12 with just three coins.

The one could be omitted instead of the two, if nothing can be bought for a penny.


QUOTE
In twelfty, we have a similar set of 'efficient weights' [1, 3, 9, 27, 81], a single set of these serves to weigh every number from 1 to 100.  But while it is possible to construct a given number, the set is not very well given for the binary division making that something like [1, 2, 4, 8, 15, 30, 60] is.

The ternary set of weights with one of each denomination could be used only in a kind of subtractive weighing, where some of the standard weights are mixed with the quantity to be weighed on the same pan of a lever balance. To tare a spring balance, additive weighing with a binary set would be much more convenient.

QUOTE
On the other coins, a fist full of cash from the bottom of ones purse or pocket, is not likely to produce all the neccessary denominations.

Similarly, for currency it is better to suppose that the purchaser rather than the seller should have the exact change needed. This causes me to doubt the most efficient base being ternary rather than binary. The dozenal set {①, ②, ③, ⑥, ①⓪} appears to beat binary at its own game. By providing more than one way to pay using the minimum number of coins, it also helps with the ticket fare problem.
QUOTE (DavidKennedy @ Sep 29 2016, 11:09 PM)
Other advantages are that this dozenal set allows more values to be paid in a greater number of equally efficient ways.


This post has been edited by DavidKennedy on Jun 5 2017, 04:48 PM
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