## Le Monde puzzle [#1000…1025]

Posted in Kids, R with tags , , , , , , on March 28, 2017 by xi'an

Le Monde mathematical puzzle launched a competition to celebrate its 1000th puzzle! A fairly long-term competition as it runs over the 25 coming puzzles (and hence weeks). Starting with puzzle #1001. Here is the 1000th puzzle, not part of the competition:

Alice & Bob spend five (identical) vouchers in five different shops, each time buying the maximum number of items to get close to the voucher value. In these five shops, they buy sofas at 421 euros each, beds at 347 euros each, kitchen appliances at 289 euros each, tables at 251 euros each and bikes at 211 euros each, respectively. Once the buying frenzy is over, they realise that within a single shop, they would have spent exactly four vouchers for the same products. What is the value of a voucher?

## Le Monde puzzle [#1001]

Posted in Kids, R with tags , , , , on March 27, 2017 by xi'an

After a long lag (due to my missing the free copies distributed at Paris-Dauphine!), here is a Sudoku-like Le Monde mathematical puzzle:

A grid of size (n,n) holds integer values such that any entry larger than 1 is the sum of one term in the same column and one term in the same row. What is the maximal possible value observed in such a grid when n=3,4?

This can be solved in R by a random exploration of such possible grids in a simulated annealing spirit:

mat=matrix(1,N,N)
goal=1

targ=function(mat){ #check constraints
d=0
for (i in (1:(N*N))[mat>1]){
r=(i-1)%%N+1;c=(i-1)%/%N+1
d=d+(min(abs(mat[i]-outer(mat[-r,c],mat[r,-c],"+")))>0)}
return(d)}

cur=0
for (t in 1:1e6){
i=sample(1:(N*N),1);prop=mat
prop[i]=sample(1:(2*goal),1)
d=targ(prop)
if (10*log(runif(1))/t<cur-d){
mat=prop;cur=d}
if ((d==0)&(max(prop)>goal)){
goal=max(prop);maxx=prop}}

returning a value of 8 for n=3 and 37 for n=4. However, the method is quite myopic and I tried instead a random filling of the grid, using each time the maximum possible sum for empty cells:

goal=1
for (v in 1:1e6){
mat=matrix(0,N,N)
#one 1 per row/col
for (i in 1:N) mat[i,sample(1:N,1)]=1
for (i in 1:N) if (max(mat[,i])==0) mat[sample(1:N,1),i]=1
while (min(mat)==0){
parm=sample(1:(N*N)) #random order
for (i in parm[mat[parm]==0]){
r=(i-1)%%N+1;c=(i-1)%/%N+1
if ((max(mat[-r,c])>0)&(max(mat[r,-c])>0)){
mat[i]=max(mat[-r,c])+max(mat[r,-c])
break()}}}
if (goal<max(mat)){
goal=max(mat);maxx=mat}}


which recovered a maximum of 8 for n=3, but reached 48 for n=4. And 211 for n=5, 647 for n=6… For instance, here is the solution for n=4:

[1,]    1    5   11   10
[2,]    2    4    1    5
[3,]   48    2   24    1
[4,]   24    1   22   11


## how large is 9!!!!!!!!!?

Posted in Statistics with tags , , , , , , , , , on March 17, 2017 by xi'an

This may sound like an absurd question [and in some sense it is!], but this came out of a recent mathematical riddle on The Riddler, asking for the largest number one could write with ten symbols. The difficulty with this riddle is the definition of a symbol, as the collection of available symbols is a very relative concept. For instance, if one takes  the symbols available on a basic pocket calculator, besides the 10 digits and the decimal point, there should be the four basic operations plus square root and square,which means that presumably 999999999² is the largest one can  on a cell phone, there are already many more operations, for instance my phone includes the factorial operator and hence 9!!!!!!!!! is a good guess. While moving to a computer the problem becomes somewhat meaningless, both because there are very few software that handle infinite precision computing and hence very large numbers are not achievable without additional coding, and because it very much depends on the environment and on which numbers and symbols are already coded in the local language. As illustrated by this X validated answer, this link from The Riddler, and the xkcd entry below. (The solution provided by The Riddler itself is not particularly relevant as it relies on a particular collection of symbols, which mean Rado’s number BB(9999!) is also a solution within the right referential.)

## a riddle at the end of its tether

Posted in R, Statistics with tags , , on February 24, 2017 by xi'an

A simply worded riddle this week on The Riddler, about four ropes having non-uniform and unknown burning rates, the only constraint being they all burn completely in one hour. With the help of a lighter (or even a single match), what are the possible units of time one can measure by burning them?

While I had worked out a range of possible times for each of the ropes to complete its combustion, I went for a simulation based confirmation. The starting point is that a new fire can only be started when a burning rope ends up burning. Which only happens for a finite number of possibilities. I found 24 of them, consisting of

> total*prec
[1]  0.000 0.5000 0.750 0.875 0.9375 1.000 1.125 1.1875 1.25 1.3125
[11] 1.375 1.4375 1.500 1.625 1.7500 1.875 2.000 2.1250 2.25 2.5000
[21] 2.750 3.0000 3.500 4.000

i.e., some combinations of 1, 2⁻¹, …, 2⁻⁴, with the comment that those times cannot all be obtained within a single arson experiment.

## A knapsack riddle [#2]?

Posted in Kids, R, Statistics with tags , , , on February 17, 2017 by xi'an

Still about this allocation riddle of the past week, and still with my confusion about the phrasing of the puzzle, when looking at a probabilistic interpretation of the game, rather than for a given adversary’s y, the problem turns out to search for the maximum of

$\mathbb{E}[L(x,Y)]=\sum_{i=1}^{10} i\{P(Y_ix_i)\}$

where the Y’s are Binomial B(100,p). Given those p’s, this function of x is available in closed form and can thus maximised by a simulated annealing procedure, coded as

utility=function(x,p){
ute=2*pbinom(x[1]-1,100,prob=p[1])+
dbinom(x[1],100,p[1])
for (i in 2:10)
ute=ute+2*i*pbinom(x[i]-1,100,prob=p[i])+
i*dbinom(x[i],100,p[i])
return(ute)}
#basic term in utility
baz=function(i,x,p){
return(i*dbinom(x[i],100,p[i])+
i*dbinom(x[i]-1,100,p[i]))}
#relies on a given or estimated p
x=rmultinom(n=1,siz=100,prob=p)
maxloz=loss=0
newloss=losref=utility(x,p)
#random search
T=1e3
Te=1e2
baza=rep(0,10)
t=1
while ((t<T)||(newloss>loss)){
loss=newloss
i=sample(1:10,1,prob=(10:1)*(x>0))
#moving all other counters by one
xp=x+1;xp[i]=x[i]
#corresponding utility change
for (j in 1:10) baza[j]=baz(j,xp,p)
proz=exp(log(t)*(baza-baza[i])/Te)
#soft annealing move
j=sample(1:10,1,prob=proz)
if (i!=j){ x[i]=x[i]-1;x[j]=x[j]+1}
newloss=loss+baza[j]-baza[i]
if (newloss>maxloz){
maxloz=newloss;argz=x}
t=t+1
if ((t>T-10)&(newloss<losref)){
t=1;loss=0
x=rmultinom(n=1,siz=100,prob=p)
newloss=losref=utility(x,p)}}


which seems to work, albeit not always returning the same utility. For instance,

> p=cy/sum(cy)
> utility(argz,p)
[1] 78.02
> utility(cy,p)
[1] 57.89


or

> p=sy/sum(sy)
> utility(argz,p)
[1] 82.04
> utility(sy,p)
[1] 57.78


Of course, this does not answer the question as intended and reworking the code to that purpose is not worth the time!

## Le Monde puzzle [#990]

Posted in Books, Kids, pictures, Statistics, Travel, University life with tags , , , , on January 12, 2017 by xi'an

To celebrate the new year (assuming it is worth celebrating!), Le Monde mathematical puzzle came up with the following:

Two sequences (x¹,x²,…) and (y¹,y²,…) are defined as follows: the current value of x is either the previous value or twice the previous value, while the current value of y is the sum of the values of x up to now. What is the minimum number of steps to reach 2016 or 2017?

By considering that all consecutive powers of 2 must appear at least one, the puzzles boils down to finding the minimal number of replications in the remainder of the year minus the sum of all powers of 2. Which itself boils down to deriving the binary decomposition of that remainder. Hence the basic R code (using intToBits):

deco=function(k=2016){
m=trunc(log2(k))
while (sum(2^(0:m))>k) m=m-1
if (sum(2^(0:m))==k){ return(rep(1,m+1))
}else{
res=k-sum(2^(0:m))
return(rep(1,m+1)+as.integer(intToBits(res))[1:(m+1)])


which produces

> sum(deco(2016))
[1] 16
> sum(deco(2017))
[1] 16
> sum(deco(1789))
[1] 18


## untangling riddle

Posted in Statistics with tags , , on December 23, 2016 by xi'an

This week riddle is rather anticlimactic [my rewording]:

N wires lead from the top of a tower down to the ground floor. But they have become hopelessly tangled. The wires on the top floor are all labelled, from 1 to N. The wires on the bottom, however, are not.

You need to figure out which ground-floor wire’s end corresponds to which top-floor wire’s end. On the ground floor, you can tie two wire ends together, and you can tie as many of these pairs as you like. You can then walk to the top floor and use a circuit detector to check whether any two of the wires at the top of the tower are connected or not.

What is the smallest number of trips to the top of the tower you need to take in order to correctly label all the wires on the bottom?

Indeed, first, if N=2, the wires cannot be identified, if N=3, they can be identified in 2 steps, and if N=4, they can also be identified in 2 steps (only pair two wires first, which gives their values and the values of the other two up to a permutation, then pair one of each group). Now, in general,  it seems that, by attaching the wires two by two, one can successively cut the number of wires by two at each trip. This is due to the identification of pairs: if wires a and b are connected in one step, identifying the other end of a will automatically determine the other end of b if one keeps track of all connected pairs at the top. Hence, if $N=2^k$, in (k-2) steps, we are down to 4 wires and it takes an extra 2 steps to identify those 4, hence all the other wires. Which leads to identification in N trips. If N is not a power of 2, things should go faster as “left over” wires in the binary division can be paired with set-aside wires and hence identify 3 wires on that trip, then further along the way… Which leads to a $2^{\lfloor \log_2(N) \rfloor}$ number of trips maybe (or even less). Continue reading