Only numerical results are demonstrated, obtained by brute force enumerations. ]]>

My reasoning for the technique to work for any arbitrary number of mints is that the ratios produced are all unique.

Each mint creates a range of ascending, unique ratios, and no two ranges can overlap.

I am confident that the ranges will never overlap.

However, the ranges aren’t strictly ascending.

This is evident even from the results I posted, starting at just 4 mints.

This raises the possibility that, given enough mints, a ratio may be duplicated within a single range.

I have verified my general construction for up to 25 mints and no duplication has appeared.

But infinity is very large and I must therefore retract my “proof”.

Things were dropped out when I had less than and greater than symbols in the text. ]]>

I’ll remove all less than and greater than signs and post again.

f(2^n-1) is less than f(2^(n-2))

(2^(n+3)-4n-8)/(3^(n+1)-2n-3) is less than (2^n-2)/(3^(n-1)-1)

(2^(n+3)-4n-8)*(3^(n-1)-1) is less than (2^n-2)*(3^(n+1)-2n-3)

O(2^(n+3)*3^(n-1)) is less than O((2^n)*3^(n+1))

divide both sides by 2^n*3^(n-1)

8 is less than 9

The first assertion holds all n larger than 1.

f(2^(n-1)) is greater than f(2^(n+2)-1)

(2^(n+1)-2)/(3^n-1) is greater than (2^(n+5)-4n-16)/(3^(n+3)-2n-7)

(2^(n+1)-2)*(3^(n+3)-2n-7) is greater than (2^(n+5)-4n-16)*(3^n-1)

O(2^(n+1)*3^(n+3)) is greater than O(2^(n+5)*3^n)

divide both sides by 2^(n+1)*3^n

27 is greater than 16

The second assertion holds for all n larger than 0.

]]>f(2^n-1) < f(2^(n-2))

(2^(n+3)-4n-8)/(3^(n+1)-2n-3) < (2^n-2)/(3^(n-1)-1)

(2^(n+3)-4n-8)*(3^(n-1)-1) < (2^n-2)*(3^(n+1)-2n-3)

O(2^(n+3)*3^(n-1)) < O((2^n)*3^(n+1))

divide both sides by 2^n*3^(n-1)

8 f(2^(n+2)-1)

(2^(n+1)-2)/(3^n-1) > (2^(n+5)-4n-16)/(3^(n+3)-2n-7)

(2^(n+1)-2)*(3^(n+3)-2n-7) > (2^(n+5)-4n-16)*(3^n-1)

O(2^(n+1)*3^(n+3)) > O(2^(n+5)*3^n)

divide both sides by 2^(n+1)*3^n

27 > 16

The second assertion holds for all n larger than 0.

]]>Let me try a more verbose tactic.

We need to show that f(2^n-1) is less than f(2^(n-2))

AND

f(2^(n-1)) is greater than f(2^(n+2)-1).

We need to show that f(2^n-1) f(2^(n+2)-1) for all n.

If that is the case, the ranges never overlap and all values of f(x) are unique.

Pr = 1, 3, 7, 15, … 2^r-1

Qr = 1, 4, 13, 40, … (3^r-1)/2

Sp(n) = sum(r=1..n, Pr) = 2^(n+1)-n-2

Sq(n) = sum(r=1..n, Qr) = (3^(n+1)-2n-3)/4

Define a function which contains sums of Pr terms in all combinations

fp(1) = P1

fp(2) = P2

fp(3) = P2+P1

fp(4) = P3

fp(5) = P3+P1

fp(6) = P3+P2

fp(7) = P3+P2+P1

fp(8) = P4

…

fp(2^(a-1)) = Pa

fp(2^(a-1)+2^(b-1)) = Pa+Pb

fp(2^(a-1)+2^(b-1)+2^(c-1)+…) = Pa+Pb+Pc+…

fp(2^(a-1)+x) = Pa+fp(x)

For all x != y, fp(x) != fp(y).

All fp values are unique as a consequence of Pr > Sp(r-1).

Construct fq(n) similarly as the sum of Qn terms.

For all x != y, fq(x) != fq(y).

All fq values are unique as a consequence of Qr > Sq(r-1).

f(n)=fp(n)/fq(n)

I will show that for all x != y, f(x) != f(y).

This proves the conditions for a general construction for ApSimon’s mints.

Each new mint r creates 2^n new combinations given by f(2^(n-1)) to f(2^n-1).

Each range has an ascending set of unique values. We just need to show that the

values in each range never overlap.

—– 1 mint —–

f(1) = 1

—– 2 mints —–

f(2) = 0.75

f(3) = 0.8

—– 3 mints —–

f(4) = 0.538462

f(5) = 0.571429

f(6) = 0.588235

f(7) = 0.611111

—– 4 mints —–

f(8) = 0.375

f(9) = 0.390244

f(10) = 0.409091

f(11) = 0.422222

f(12) = 0.415094

f(13) = 0.425926

f(14) = 0.438596

f(15) = 0.448276

—– 5 mints —–

f(16) = 0.256198

f(17) = 0.262295

f(18) = 0.272

f(19) = 0.277778

f(20) = 0.283582

f(21) = 0.288889

f(22) = 0.297101

f(23) = 0.302158

f(24) = 0.285714

f(25) = 0.290123

f(26) = 0.29697

f(27) = 0.301205

f(28) = 0.304598

f(29) = 0.308571

f(30) = 0.314607

f(31) = 0.318436

—– 6 mints —–

f(32) = 0.173077

f(33) = 0.175342

f(34) = 0.179348

f(35) = 0.181572

f(36) = 0.185676

f(37) = 0.187831

f(38) = 0.191601

f(39) = 0.193717

f(40) = 0.193069

f(41) = 0.195062

f(42) = 0.198529

f(43) = 0.200489

f(44) = 0.203837

f(45) = 0.205742

f(46) = 0.209026

f(47) = 0.2109

f(48) = 0.193814

f(49) = 0.195473

f(50) = 0.198364

f(51) = 0.2

f(52) = 0.202811

f(53) = 0.204409

f(54) = 0.207171

f(55) = 0.208748

f(56) = 0.207619

f(57) = 0.209125

f(58) = 0.21172

f(59) = 0.213208

f(60) = 0.215613

f(61) = 0.217069

f(62) = 0.219557

f(63) = 0.220994

We need to show that f(2^n-1) f(2^(n+2)-1) for all n.

If that is the case, the ranges never overlap and all values of f(x) are unique.

f(2^(n-1)) = fp(2^(n-1))/fq(2^(n-1)) = Pn/Qn = (2^n-1)/((3^n-1)/2) = (2^(n+1)-2)/(3^n-1)

f(2^(n-2)) = (2^n-2)/(3^(n-1)-1)

fp(2^n-1) = fp(2^(n-1)+2^(n-2)+2^(n-2)+2^(n-1)+…1) = Pn + Pn-1 + Pn-2 + … P1 = Sp(n) = 2^(n+1)-n-2

fq(2^n-1) = Sq(n) = (3^(n+1)-2n-3)/4

f(2^n-1) = fp(2^n-1)/fq(2^n-1) = (2^(n+1)-n-2)/((3^(n+1)-2n-3)/4) = (2^(n+3)-4n-8)/(3^(n+1)-2n-3)

f(2^(n+2)-1) = (2^(n+5)-4n-16)/(3^(n+3)-2n-7)

f(2^n-1) < f(2^(n-2))

(2^(n+3)-4n-8)/(3^(n+1)-2n-3) < (2^n-2)/(3^(n-1)-1)

(2^(n+3)-4n-8)*(3^(n-1)-1) < (2^n-2)*(3^(n+1)-2n-3)

O(2^(n+3)*3^(n-1)) < O((2^n)*3^(n+1))

divide both sides by 2^n*3^(n-1)

8 f(2^(n+2)-1)

(2^(n+1)-2)/(3^n-1) > (2^(n+5)-4n-16)/(3^(n+3)-2n-7)

(2^(n+1)-2)*(3^(n+3)-2n-7) > (2^(n+5)-4n-16)*(3^n-1)

O(2^(n+1)*3^(n+3)) > O(2^(n+5)*3^n)

divide both sides by 2^(n+1)*3^n

27 > 16

The second assertion holds for all n larger than 0.

The general construction therefore holds for any number of mints and we have reduced the upper bound for n mints to (3^(n+1)-2n-3)/4.

]]>P=(4,6,6,7,3,13,15,3) Q=(4,0,1,6,12,12,1,27)

total=4+6+6+7+12+13+15+27=90

9 mints seems out of range for an exhaustive search for now.

Here is the best I’ve found so far.

P=(1,2,4,12,5,4,20,39,43) Q=(0,1,3,3,25,33,34,18,27)

total=1+2+4+12+25+33+34+39+43=193

If r is the number of mints, and r > 1, the number of total coins must be >= ceiling(2^(r/2)).

This is easier to talk about for a discrete case. Let’s take 8 mints as an example.

There are 256 possible outcomes that we must identify with just two weighings, so we must have at least 256 distinct ratios of weight values. For instance, the first weighing might have 20 possible results and the second weighing might only have 13. 20 * 13 > 256. That will work if there aren’t many that reduce to similar ratios. The ideal case, however, would be when the distinct results from both weighing are almost equal. Since the square root of 256 is 16, it is sufficient to have just 16 distinct weight outcomes from each weighing (assuming we can find two sets that have no common ratios). So how many coins do we need to guarantee 16 distinct outcomes?

Let t be the total number of coins.

One outcome is that they are all genuine. weight = t * W(1).

One coin could be fake. weight = (t – 1) * W(1) + W(1 + e).

Two coins could be fake. weight = (t – 2) * W(1) + 2 * W(1 + e).

etc.

We must have at least 15 coins that could be fake to generate 16 distinct outcomes. And this is our minimal answer, 15 total coins.

And now for the second weighing. If we use all the same coins, the ratios will not be unique. If we leave one or more out, we cannot achieve 16 distinct outcomes. Therefore we must add at least 1 more coin to our overall set of coins that we can include in the second weighing. So we must have at least 16 coins, the square root of 2^8.

And in general, that is ceiling(2^(r/2)).

That is more than r, but still very low considering actual results that have been posted. The minimum number of coins to test 6 mints with this formula is 8, but the best case uses 29 coins! ]]>