臨床統計もおもしろいですよ、その1 [無断転載禁止]©2ch.net
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内科認定医受験の最低限の知識、
製薬会社の示してくる臨床データ、
論文の考察、
論文を書くときの正当性、
というのが、臨床統計の今までの目的の大きい部分でしたが、
AI=機械学習の基本も、結局は統計学と確率に支配されます。
そういう雑多な話をするスレです。
library(rjags)
N=50
z=40
FP=0.01
shape1=1
shape2=1
dataList=list(N=N,z=z,FP=FP,shape1=shape1,shape2=shape2)
modelstring <- paste0("
model{
theta=TP*x+FP*(1-x)
z ~ dbinom(theta,N)
TP ~ dbeta(shape1,shape2)
x ~ dbeta(shape1,shape2)
}"
)
writeLines( modelstring , con="TEMPmodel.txt" )
jagsModel = jags.model( file="TEMPmodel.txt" , data=dataList, quiet=TRUE)
update(jagsModel)
codaSamples = coda.samples( jagsModel ,
variable=c("TP","x","theta"), n.iter=100000 )
js=as.matrix(codaSamples)
head(js)
BEST::plotPost(js[,'TP'],xlab='sensitivity')
BEST::plotPost(js[,'x'],xlab='prevalence')
BEST::plotPost(js[,'theta'],xlab='positive result',showMode = TRUE) 経理課の須藤は着服をやめろ!
勤務実態もないのに、グループ病院内から管理手当て(10万円)をもらうな!!!
意図的な給与操作、どうにかしろ! seqN <- function(N=100,K=5){
a=numeric(N)
for(i in 1:K) a[i]=2^(i-1)
for(i in K:(N-1)){
a[i+1]=0
for(j in 0:(K-1)){
a[i+1]=a[i+1]+a[i-j] # recursion formula
}
}
P0=numeric(N)
for(i in 1:N) P0[i]=a[i]/2^i # P0(n)=a(n)/2^n
P0
MP=matrix(rep(NA,N*K),ncol=K)
colnames(MP)=paste0('P',0:(K-1))
MP[,1]=P0
head(MP);tail(MP)
MP[1,2]=1/2
for(i in (K-2):K) MP[1,i]=0
for(k in 2:K){
for(i in 1:(N-1)) MP[i+1,k]=1/2*MP[i,k-1]
} # Pk(n+1)=1/2*P(k-1)(n)
ret=1-apply(MP,1,sum)
ret[N]
}
seqN(100,5)
seqN(1000,10) ## 表の出る確率がpであるとき、N回コインを投げて K回以上表が連続する確率
seqNp <- function(N=100,K=5,p=0.5){
q=1-p
a=numeric(N) # a(n)=P0(n)/p^n , P0(n)=a(n)*p^n
for(i in 1:K) a[i]=q/p^i # P0(i)=q
for(i in K:(N-1)){ # recursive formula
a[i+1]=0
for(j in 0:(K-1)){
a[i+1]=(a[i+1]+a[i-j])
}
a[i+1]=q/p*a[i+1]
}
P0=numeric(N)
for(i in 1:N) P0[i]=a[i]*p^i # P0(n)=a(n)*p^n
P0
MP=matrix(rep(NA,N*K),ncol=K)
colnames(MP)=paste0('P',0:(K-1))
MP[,'P0']=P0
head(MP);tail(MP)
MP[1,'P1']=p
for(i in (K-2):K) MP[1,i]=0
for(k in 2:K){
for(i in 1:(N-1)) MP[i+1,k]=p*MP[i,k-1]
} # Pk(n+1)=p*P(k-1)(n)
ret=1-apply(MP,1,sum)
ret[N]
} p:表
q=1-p
# Pk(n) (k=0,1,2,3,4)を途中、5連続して表が出ていなくて
# 最後のk回は連続して表が出ている確率とする。
#
P0(1)=q
P1(1)=p
P2(1)=P3(1)=P4(1)=0
P(k+1)(n+1)=p*Pk(n)
P0(n+1)=q*{P0(n)+P1(n)+P2(n)+P3(n)+P4(n)}
=q*{P0(n)+p*P0(n-1)+p^2*P0(n-2)+p^3*P0(n-3)+p^4*P0(n-4)}
P0(n)=a(n)*p^n
# a(n+1)p^(n+1)=q*p^n{a(n)+a(n-1)+a(n-2)+a(n-3)+a(n-4)}
# a(n+1)=q/p1*(a(n)+a(n-1)+a(n-2)+a(n-3)+a(n-4))
a(n)=P0(n)/p^n >>532
統計くらいできるのが国立卒の普通の臨床医。
おい、ド底辺
統計処理からはおまえは
都外のド底辺シリツ医大卒と推測されたが、あってるか? ## 表の出る確率がpであるとき、N回コインを投げて K回以上表が連続する確率に一般化してみた。
seqNp <- function(N=100,K=5,p=0.5){
q=1-p
a=numeric(N) # a(n)=P0(n)/p^n , P0(n)=a(n)*p^n
for(i in 1:K) a[i]=q/p^i # P0(i)=q
for(i in K:(N-1)){ # recursive formula
a[i+1]=0
for(j in 0:(K-1)){
a[i+1]=(a[i+1]+a[i-j])
}
a[i+1]=q/p*a[i+1]
}
P0=numeric(N)
for(i in 1:N) P0[i]=a[i]*p^i # P0(n)=a(n)*p^n
MP=matrix(rep(NA,N*K),ncol=K)
colnames(MP)=paste0('P',0:(K-1))
MP[,'P0']=P0
head(MP);tail(MP)
MP[1,'P1']=p
for(i in (K-2):K) MP[1,i]=0
for(k in 2:K){
for(i in 1:(N-1)) MP[i+1,k]=p*MP[i,k-1]
} # Pk(n+1)=p*P(k-1)(n)
ret=1-apply(MP,1,sum)
ret[N]
} # pdfからcdfの逆関数を作ってHDIを表示させて逆関数を返す
pdf2hdi <- function(pdf,xMIN=0,xMAX=1,cred=0.95,Print=TRUE){
nxx=1001
xx=seq(xMIN,xMAX,length=nxx)
xx=xx[-nxx]
xx=xx[-1]
xmin=xx[1]
xmax=xx[nxx-2]
AUC=integrate(pdf,xmin,xmax)$value
PDF=function(x)pdf(x)/AUC
cdf <- function(x) integrate(PDF,xmin,x)$value
ICDF <- function(x) uniroot(function(y) cdf(y)-x,c(xmin,xmax))$root
hdi=HDInterval::hdi(ICDF,credMass=cred)
print(c(hdi[1],hdi[2]),digits=5)
if(Print){
par(mfrow=c(3,1))
plot(xx,sapply(xx,PDF),main='pdf',type='h',xlab='x',ylab='Density',col='lightgreen')
legend('top',bty='n',legend=paste('HDI:',round(hdi,3)))
plot(xx,sapply(xx,cdf),main='cdf',type='h',xlab='x',ylab='Probability',col='lightblue')
pp=seq(0,1,length=nxx)
pp=pp[-nxx]
pp=pp[-1]
plot(pp,sapply(pp,ICDF),type='l',xlab='p',ylab='x',main='ICDF')
par(mfrow=c(1,1))
}
invisible(ICDF)
} library(rjags)
modelstring <- paste0("
model{
theta=TP*x+FP*(1-x)
z ~ dbinom(theta,N)
TP ~ dbeta(shape1,shape2)
x ~ dbeta(shape1,shape2)
}"
)
writeLines( modelstring , con="TEMPmodel.txt" )
N=100 ; FP=0.01 ; shape1=1 ; shape2=1
guess.TP <- function(z){
dataList=list(N=N,z=z,FP=FP,shape1=shape1,shape2=shape2)
jagsModel = jags.model( file="TEMPmodel.txt" , data=dataList, quiet=TRUE)
update(jagsModel)
codaSamples = coda.samples(jagsModel,variable=c("TP","x"), n.iter=10000)
js=as.matrix(codaSamples)
m.TP=mean(js[,'TP'])
ci.TP=HPDinterval(as.mcmc(js[,'TP']))
m.x=mean(js[,'x'])
ci.x=HPDinterval(as.mcmc(js[,'x']))
c(m.TP=m.TP,ci.TP=ci.TP,m.x=m.x,ci.x=ci.x)
}
zz=1:20*5
re=sapply(zz,guess.TP)
head(re[,1:4])
re=as.matrix(re)
plot(zz,re['m.TP',],bty='l',ylim=c(0,1),type='n',las=1,
xlab='n : positives out of 100',ylab='sensitivity')
segments(zz,re[2,],zz,re[3,],col=8,lwd=3)
points(zz,re['m.TP',],pch=16) guess.TP2 <- function(z,FP){
dataList=list(N=N,z=z,FP=FP,shape1=shape1,shape2=shape2)
jagsModel = jags.model( file="TEMPmodel.txt" , data=dataList, quiet=TRUE)
update(jagsModel)
codaSamples = coda.samples(jagsModel,variable=c("TP"), n.iter=10000,thin=5)
js=as.matrix(codaSamples)
mean(js[,'TP'])
}
vG=Vectorize(guess.TP2)
n=1:20*5
FP=seq(0,0.25,length=20)
TP=outer(n,FP,vG) # wait several minutes
contour(n,FP,TP, col='navy',
xlab='n : positives out of 100',ylab='FP : 1-specificity',bty='l',nlevels=64)
points(50,0.10,pch='+',col='red',cex=1.5) n=5 # prime number
nn=1:(n-1)
tasu <- function(x,y) (x+y)%%n
hiku <- function(x,y) (x-y)%%n # row - col
kake <- function(x,y) (x*y)%%n
g=function(x) nn[which(x==1)]
.M=outer(nn,nn,kake)
G=apply(.M,2,g)
gyaku <- function(x) nn[which(G==(x%%n))]
waru <- function(x,y) (x*gyaku(y))%%n # row / col
waru(3,2)
xx=yy=c(0,nn)
names(xx)=paste0('x',c(0,nn))
names(yy)=paste0('y',c(0,nn))
outer(xx,yy,tasu) # x + y
outer(xx,yy,hiku) # x - y
outer(xx,yy,kake) # x * y
X=Y=nn
outer(X,Y,waru) # WRONG!!
outer(X,Y,Vectorize(waru))
a=expand.grid(X,Y)
b=matrix(mapply(waru,a[,1],a[,2]),ncol=length(X))
rownames(b)=paste0('x',nn)
colnames(b)=paste0('y',nn)
b # x / y rule3 <- function(n,confidence.level=0.95){
p=1/n
q=1-p # q^n.sample > 1-confidence.level
n.sample = log(1-confidence.level)/log(q)
return(n.sample)
}
n.sample=log(0.05)/log(0.999) shuffle <- function(Cards){
n=length(Cards)
n1=ceiling(n/2)
n2=n-n1
C1=Cards[1:n1]
C2=Cards[(n1+1):n]
ret=NULL
for(i in 1:n1){
ret=c(ret,C1[i],C2[i])
}
ret[is.na(ret)==F]
}
x=as.character(c('A',2:10,'J','Q','K'))
cat(x,'\n') ; cat(shuffle(x))
Shuffles <- function(x){
tmp=shuffle(x)
i=1
while(!identical(x,tmp)){
tmp=shuffle(tmp)
i=i+1
}
return(i)
}
f =function(x)Shuffles(1:x)
nn=1:53
y=sapply(nn,f)
plot(nn,y,pch=16,bty='l',xlab='cards',ylab='shuffles')
cbind(nn,y) http://000013.blogspot.com/2010/12/99.html
inversion <- function(x){ # 転倒数
n=length(x)
ret=numeric(n)
for(i in 1:(n-1)){
ret[i] = sum(x[i] > x[(i+1):n])
}
sum(ret)
}
x=c(4, 3, 5, 2, 1)
inversion(x)
is.even= function(x) !inversion(x)%%2
is.even(x)
prisoner99 <- function(n=100){
indx=sample(1:n,1) # defective number
X=sample((1:n)[-indx]) ; is.even(X)
Y=numeric(n-1)
for (i in 1:(n-1)){
x1=X[-i]
x2=(1:n)[!(1:n) %in% x1] # 囚人iが見えない番号
tmp=X
tmp[i]=x2[1] ; tmp[n]=x2[2]
Y[i]=ifelse(is.even(tmp), x2[1],x2[2]) # 偶順列になるように選択
}
all(X==Y)
}
mean(replicate(1e3,prisoner99())) # http://000013.blogspot.com/2010/12/99.html
inversion <- function(x){
n=length(x)
ret=numeric(n)
for(i in 1:(n-1)){
ret[i] = sum(x[i] > x[(i+1):n])
}
sum(ret) # inversion number
}
is.even= function(x) !inversion(x)%%2 # is inverion number even?
prisoner99 <- function(n=100){
indx=sample(1:n,1) # defective number
X=sample((1:n)[-indx])
Y=numeric(n-1)
for (i in 1:(n-1)){ # select as even permutation
x1=X[-i]
x2=(1:n)[!(1:n) %in% x1] # two numbers unseen for i-th prisoner
tmp=X
tmp[i]=x2[1] ; tmp[n]=x2[2]
Y[i]=ifelse(is.even(tmp), x2[1],x2[2])
}
all(X==Y)
}
mean(replicate(1e3,prisoner99())) inversion <- function(x){ #inversion number
n=length(x)
ret=numeric(n)
for(i in 1:(n-1)){
ret[i] = sum(x[i] > x[(i+1):n])
}
sum(ret)
}
is.even <- function(x) !inversion(x)%%2 # even inversion?
even.perm <- function(n=100){
indx=sample(1:n,1) # defective number
X=sample((1:n)[-indx]) # row of 99 prisoner numbers
is.even(X)
}
mean(replicate(1e3,even.perm())) # probability of even permutation #ある大学の学生数は500以上1000人以下であることはわかっている。
#無作為に2人を抽出して調べたところ
#二人とも女子学生である確率は1/2であった。
#この大学の女子学生数と男子学生数は何人か?
girlsboys <- function(g,b) g*(g-1)/(g+b)/(g+b-1)==1/2
gr=expand.grid(1:1000,1:1000)
(re=gr[which(mapply(girlsboys,gr[,1],gr[,2])),])
girlsboys(re[nrow(re),1],re[nrow(re),2]) # ある大学の学生数は500以上1000人以下であることはわかっている。
# 無作為に2人を抽出して調べたところ
# 二人とも女子学生である確率は1/2であった。
# この大学の女子学生数と男子学生数は何人か?
girlsboys <- function(g,b) g*(g-1)/(g+b)/(g+b-1)==1/2
gr=expand.grid(1:1000,1:1000)
(re=gr[which(mapply(girlsboys,gr[,1],gr[,2])),])
# 検証
Vectorize(girlsboys)(re[,1],re[,2]) N=120
r=10
D=c(rep(1,r),rep(0,N-r))
hiseiki <- function(m){
found=0
for(i in 1:(N-r+m)){
found=found+sample(D,1)
if(found==m) break
}
return(i)
}
re=replicate(1e4,hiseiki(3))
mean(re)
sd(re)
BEST::plotPost(re,breaks=30) cereals <- function(n=5){
coupons=NULL
while(!all((1:n) %in% coupons)){
coupons=append(sample(1:n,1),coupons)
}
return(length(coupons))
}
re=replicate(100,mean(replicate(1e3,cereals(5))))
> mean(re)
[1] 11.43503 p=4:1
cereals <- function(p){
n=length(p)
coupons=NULL
while(!all((1:n) %in% coupons)){
coupons=append(sample(1:n,1,p=p),coupons)
}
return(length(coupons))
}
mean(replicate(1e3,cereals(p)))
re=replicate(100,mean(replicate(1e3,cereals(p))))
mean(re) blood.type <- function(p,need){
n=length(p)
ABO=NULL
enough <- function(x){
pool=numeric(n)
for(i in 1:n) pool[i]=sum(ABO==i)
all(pool >= need)
}
while(!enough(ABO)){
ABO=append(sample(1:n,1,p=p),ABO)
}
return(length(ABO))
}
p=4:1
need=c(10,10,5,2)
re=replicate(1e4,blood.type(p,need))
BEST::plotPost(re) BT <- function (a,b,c,d) 1/a + 1/b + 1/c + 1/d - 1/(a+b) - 1/(a+c) - 1/(b+c) - 1/(a+d) - 1/(b+d) - 1/(c+d) + 1/(a+b+c) + 1/(d+a+b) + 1/(c+d+a) + 1/(b+c+d) - 1/(a+b+c+d)
a=4
b=3
c=2
d=1
s =a+b+c+d
a=a/s
b=b/s
c=c/s
d=d/s
BT(a,b,c,d) even.tally <- function(a=3 , b=2){
idx=combn(1:(a+b),a)
n=ncol(idx)
mat=matrix(0,nrow=n,ncol=a+b)
for(i in 1:n) mat[i,idx[,i]]=1
tally <- function (x) any(cumsum(x)==cumsum(1-x))
mean(apply (mat,1,tally))
}
even.tally()
even.tally(5,10) a=750 ; b=250
v=c(rep(1,a),rep(0,b))
f <- function(v){
x=sample(v)
any(cumsum(x)==cumsum(1-x))
}
mean(replicate(1e5,f(v))) date=1:366
p=c(97/400,rep(1,365))
same.birth <- function(n,lwr=2,upr=1e6){
x=sample(date,n,replace=TRUE,prob=p)
di=max(table(x)
lwr<=di & di<=upr
}
birth <- function(n,lwr=2,upr=1e6,k=1e4){
mean(replicate(k,same.birth(n,lwr,upr)))
}
#
birth(100, 3)
vrb=Vectorize(birth)
x=1:50
y=vrb(x)
plot(x,y,pch=19)
abline(h=0.5,lty=3,col=4)
min(x[whic(y > 0.5)]) date=1:366
p=c(97/400,rep(1,365))
same.birth <- function(n,lwr=2,upr=1e6){
x=sample(date,n,replace=TRUE,prob=p)
di=max(table(x))
lwr<=di & di<=upr
}
birth <- function(n,lwr=2,upr=1e6,k=1e4){
mean(replicate(k,same.birth(n,lwr,upr)))
}
#
birth(100, 3)
vrb=Vectorize(birth)
x=1:50
y=vrb(x)
plot(x,y,pch=19)
abline(h=0.5,lty=3,col=4)
min(x[whic(y > 0.5)]) date=1:366
p=c(97/400,rep(1,365))
same.birth <- function(n,lwr=2,upr=1e6){
x=sample(date,n,replace=TRUE,prob=p)
di=max(table(x)
lwr<=di & di<=upr
}
birth <- function(n,lwr=2,upr=1e6,k=1e4){
mean(replicate(k,same.birth(n,lwr,upr)))
}
#
birth(100, 3)
vrb=Vectorize(birth)
x=1:50
y=vrb(x)
plot(x,y,pch=19)
abline(h=0.5,lty=3,col=4)
min(x[whic(y > 0.5)]) インフルエンザの迅速キットは特異度は高いが感度は検査時期によって左右される。
ある診断キットが開発されたとする。
このキットは特異度は99%と良好であったが、
感度については確かな情報がない。
事前確率分布として一様分布を仮定する。
50人を無作為抽出してこの診断キットで診断したところ40人が陽性であった。
この母集団の有病率の期待値と95%信用区間はいくらか?
またこの診断キットの感度の期待値と95%信用区間はいくらか
暇つぶしにこれをMCMCを使わずに解く方法を考えていた。
偽陽性率FP=0.01として
陽性確率p=TP*x+(1-x)*FP
尤度が50C40*p^40*(1-p)^10
TPは一様分布なので積分消去して
確率密度関数に比例する関数を作ってarea under the curveで割って確率密度関数化したのち積分して累積密度関数をつくる。この累積密度関数の逆関数を作って95%区間が最短になる区間を計算すれば信頼区間が算出できる。
この結果がstanでのシミュレーションの結果と一致すればよし。 # # choose(n,r) == gamma(n+1) / (gamma(r+1) * gamma(n-r+1))
same.birthday <- function(n) 1-choose(365+97/400,n)*factorial(n)/(365+97/400)^n
plot(x,y,bty='l',xlab='subjects',ylab='probability')
curve(same.birthday(x),add = TRUE)
abline(h=0.5,col=8)
same.birthday(22:23) library(rjags)
N=50
z=40
FP=0.01
shape1=1
shape2=1
dataList=list(N=N,z=z,FP=FP,shape1=shape1,shape2=shape2)
modelstring <- paste0("
model{
theta=TP*x+FP*(1-x)
z ~ dbinom(theta,N)
TP ~ dbeta(shape1,shape2)
x ~ dbeta(shape1,shape2)
}"
)
writeLines( modelstring , con="TEMPmodel.txt" )
jagsModel = jags.model( file="TEMPmodel.txt" , data=dataList, quiet=TRUE)
update(jagsModel)
codaSamples = coda.samples( jagsModel ,
variable=c("TP","x","theta"), n.iter=100000 )
js=as.matrix(codaSamples)
head(js)
BEST::plotPost(js[,'TP'],xlab='sensitivity')
BEST::plotPost(js[,'x'],xlab='prevalence')
BEST::plotPost(js[,'theta'],xlab='positive result',showMode = TRUE) N=50
z=40
FP=0.01
shape1=1
shape2=1
data = list(N=N,z=z,FP=FP,shape1=shape1,shape2=shape2)
stanString=paste0('
data{
int N;
int z;
real FP;
real shape1;
real shape2;
}
parameters{
real<lower=0,upper=1> TP;
real<lower=0,upper=1> x; //prevalence
}
transformed parameters{
real<lower=0,upper=1> theta;
theta=TP*x+FP*(1-x);
}
model{
z ~ binomial(N,theta);
TP ~ beta(shape1,shape2); // T[0.5,];
x ~ beta(shape1,shape2);
}
') # model=stan_model(model_code = stanString)
# saveRDS(model,'quick_kit.rds')
model=readRDS('quick_kit.rds')
fit=sampling(model,data=data,iter=10000)
print(fit,digits=3,probs=c(.025,.50,.975))
stan_trace(fit)
# stan_diag(fit)
stan_ac(fit)
stan_dens(fit,separate_chains = TRUE)
stan_hist(fit,fill='skyblue',bins=15,pars=c('x','TP'))
ms=rstan::extract(fit)
BEST::plotPost(ms$TP,showMode = TRUE,xlab='TP')
BEST::plotPost(ms$x,showMode = FALSE,xlab='prevalence',col=sample(colours(),1)) escape.cliff <- function(p=2/3,k=1000){
pos=1
while(pos<k){
if(pos==0) return(FALSE)
pos=pos+sample(c(1,-1),1,prob=c(p,1-p))
}
return(TRUE)
}
mean(replicate(1e2,escape.cliff())) escape.cliff <- function(pos=1,p=2/3,k=1000){
while(pos<k){
if(pos==0) return(FALSE)
pos=pos+sample(c(1,-1),1,prob=c(p,1-p))
}
return(TRUE)
}
mean(replicate(1e3,escape.cliff(10,0.5,100)))
#
totter <- function(p=2/3,pos=1,pos0=NULL,k=1e3){
if(is.null(pos0)) pos0=pos
for(i in 1:k) {
if(pos==0) return(FALSE)
pos=pos+sample(c(1,-1),1,prob=c(p,1-p))
}
return(pos>pos0)
}
mean(replicate(1e3,totter(0.5,10,5 ))) exam <- function(p=0.5,hit=0,money=5,k=5){
while(hit<k){
money=money-1
if(money==0) return(FALSE)
shoot=sample(c(1,0),1,prob=c(p,1-p))
if(shoot){
money=money+1
hit=hit +1
}
}
return(TRUE)
}
mean(replicate(1e5, exam())) ryunen <- function(p=0.6 ,money=10,grade=0,ryu=0){
while(grade<6){
test=rbinom(1,1,p)
if(test) grade=grade+1
else ryu=ryu+1
if(ryu==2){
money=money-5
ryu=ryu-1
}
if(money<=0) return(FALSE)
}
return(grade==6)
}
mean(replicate(1e5,ryunen())) escape.cliff <- function(pos=1,p=2/3,k=1000){
while(pos<k){
if(pos==0) return(FALSE)
pos=pos+sample(c(1,-1),1,prob=c(p,1-p))
}
return(TRUE)
}
mean(replicate(1e5,escape.cliff(1,2/3,3)))
4/7 gmbl <- function(money=20,p=18/38){
while(0<money & money <40){
money=money+sample(c(1,-1),1,p=c(p,1-p))
}
return(money==40)
}
mean(replicate(1e3, gmbl())) x=c(rep(1,4),rep(0,48))
mean(replicate(1e5,which(sample(x)==1)[1]))
y=numeric(48)
for(i in 1:48) y[i]=i*choose(48,i-1)*factorial(i-1)*4*factorial(52-i)/factorial(52)
sum(y) (a) A railroad numbers its locomotives in order, 1, 2, . . . , N. One day you see a locomotive and its number is 60.
Guess how many locomotives the company has.
(b) You have looked at 5 locomotives and the largest number observed is 60.
Again guess how many locomotives the company has. n=5
m=60
N=n:100
loco <- function(x){
max(sample(1:x,n))
}
vloco=Vectorize(loco)
loco.sim <- function() N[which(vloco(N)==m)]
locomotives=unlist(replicate(1e4,loco.sim()))
BEST::plotPost(locomotives,breaks=30)
HDInterval::hdi(locomotives) n=5
library(gtools)
y=permutations(n,n,1:n)
f=function(x) sum(x==1:n)
sum(apply(y,1,f))/length(y)
1/n f = function(n){
sum(sample(1:n)==1:n)/n
}
g=function(n)mean(replicate(1e4,f(n)))
vg=Vectorize(g)
x=1:100
plot(x,vg(x))
curve(1/x,add=T) gr=expand.grid(1:9,1:9,1:9)
u=10^(2:0)
f=function(a,b,c,x=1776) sum(c(a,b,c)*u+c(b,c,a)*u+c(c,a,b)*u)-x
# when A>B>C
idx=which(mapply(f,gr[,1],gr[,2],gr[,3])==0 & gr[,1]>gr[,2] & gr[,2]>gr[,3])
gr[idx,]
# otherwise
idx2=which(mapply(f,gr[,1],gr[,2],gr[,3])==0)
length (idx2)
gr[idx2,] n=5
m=60
N=n:100
loco <- function(x){
max(sample(1:x,n))
}
vloco=Vectorize(loco)
loco.sim <- function() N[which(vloco(N)==m)]
locomotives=unlist(replicate(1e4,loco.sim()))
BEST::plotPost(locomotives,breaks=30)
HDInterval::hdi(locomotives)
n=60:100
pmf=choose(59,4)/choose(n,5) #Pr(max=60|n)
pdf=pmf/sum (pmf)
sum( n*pdf) #E(n)
lines(n,pdf) own <- function(n){
mean(replicate(1e5,sum(sample(1:n)-1:n==0)))
}
own(100) r1 <- function(x){ # rotate by one bead
n=length(x)
y=numeric(n)
y[1]=x[n]
for(i in 1:(n-1)){
y[i+1]=x[i]
}
return(y)
}
rn <- function(x){ # every rotation
n=length(x)
mat=matrix(rep(NA,n^2),ncol=n)
mat[,1]=x
for(i in 1:(n-1)){
mat[,i+1]=r1(mat[,i])
}
return(t(mat))
}
same <- function(x,y){
if(sum(x)!=sum(y)) return(FALSE)
f=function(a,b=y){ # is equal to y
all(a==b)
}
mat=rbind(rn(x),rn(rev(x))) # with symmetric conversion
any(apply(mat,1,f))
} dec2bin <- function(num, digit=0){ # decimal to 0,1 vector
if(num <= 0 && digit <= 0){
return(NULL)
}else{
return(append(Recall(num%/%2,digit-1), num%%2))
}
}
vd2b=Vectorize(dec2bin)
#
bracelett <- function(n){
mat=t(vd2b(0:(2^n-1),n)) # make all permutation of beads
# head(mat)
ret=list() # list of the same bracelett
for(i in 1:2^n){
ret[[i]]=which(apply(mat,1,function(z)same(z,mat[i,])))
}
# head(ret) ; table(unlist(ret))
del=NULL
for(i in 1:2^n) del=append(del,ret[[i]][-1])
2^n - length(unique(del))
} > library(gtools)
> swap <- function(n){
+ perm=permutations(n,n,1:n)
+ mean(apply(perm,1,function(x)sum(x==1:n)))
+ }
> swap(9)
[1] 1
> swap.sim <- function(n,k=1e5){
+ mean(replicate(k,sum(sample(1:n)==1:n)))
+ }
> swap.sim(9)
[1] 1.0004 swap2 <- function(n){ # k*nCk*(1/n)^k*(1-1/n)^(n-k) k=0:n
re=numeric(n)
for(i in 0:n){
p=1/n
re[i]=i*choose(n,i)*p^i*(1-p)^(n-i)
}
sum(re)
}
swap2(7)
#
swap3 <- function(n){
re=numeric(n)
for(i in 0:n){
p=1/n
re[i]=i*dbinom(i,n,p)
}
sum(re)
}
swap3(7) f012 <- function(n){
args=list()
length(args)=n-1
args[[1]]=args[[n-1]]=1:2
for(i in 2:(n-2)){
args[[i]]=0:2
}
gr=do.call(expand.grid,args)
gr=as.matrix(gr)
ret=numeric()
for(i in 1:nrow(gr)){
ret[i]=all(diff(gr[i,])!=0)
}
sum(ret)
} /* SEND+MORE=MONEYの覆面暗算をC言語で解いてみる。*/
#include<stdio.h> /* SEND+MORE=MONEY */
int compare_int(const void *a, const void *b)
{return *(int*)a - *(int*)b;}
int i,j;
int unique(int num[8]){
qsort(num,8,sizeof(int),compare_int);
for(i=0;i<8;i++){
for(j=0;j<i;j++){
if(num[j]==num[j+1]){
return 0;
}}}
return 1;
}
main(){
int S,E,N,D,M,O,R,Y;
for(S = 1; S < 10; S++){
for(E = 0; E < 10; E++){
for(N = 0; N < 10; N++){
for(D = 0; D < 10; D++){
for(M = 1; M < 10; M++){
for(O = 0; O < 10; O++){
for(R = 0; R < 10; R++){
for(Y = 0; Y < 10; Y++){
if(S*1000+E*100+N*10+D+M*1000+O*100+R*10+E==M*10000+O*1000+N*100+E*10+Y){
int num[]={S,E,N,D,M,O,R,Y};
if(unique(num)==1){
printf("%d%d%d%d+%d%d%d%d=%d%d%d%d%d\n", S,E,N,D,M,O,R,E,M,O,N,E,Y);
}}}}}}}}}}}
C:\MinGW>gcc sendmore.c -o money
C:\MinGW>money
9567+1085=10652 #include<stdio.h>
int compare_int(const void *a, const void *b){
return *(int*)a - *(int*)b;}
int unique(int num[]){
int i,j,n=10;
qsort(num,n,sizeof(int),compare_int);
for(i=0;i<n;i++){
for(j=0;j<i;j++){
if(num[j]==num[j+1]){
return 0;
}}}
return 1;}
main(){ /* ド底辺+私立医=裏口馬鹿 */
int n=1,A,B,C,D,E,F,G,H,I,J;
for(A = 1; A < 10; A++){
for(B = 0; B < 10; B++){
for(C = 0; C < 10; C++){
for(D = 1; D < 10; D++){
for(E = 0; E < 10; E++){
for(F = 0; F < 10; F++){
for(G = 1; G < 10; G++){
for(H = 0; H < 10; H++){
for(I = 0; I < 10; I++){
for(J = 0; J < 10; J++){
if(A*100+B*10+C +D*100+E*10+F==G*1000+H*100+I*10+J){
int num[]={A,B,C,D,E,F,G,H,I,J};
if(unique(num)==1){
printf("%2d: %d%d%d + %d%d%d = %d%d%d%d\n", n,A,B,C,D,E,F,G,H,I,J);
n++;
}}}}} }}} }}} }} # 2つの整数があります。
# それらをたしてできた数は、
# 十の位と一の位の数字が等しい2けたの整数になり、
# それらをかけてできた数は、
# 百の位、十の位、一の位が等しい3けたの整数になりました。
# このような2つの整数の組をあるだけ答えなさい。
gr=expand.grid(1:99,1:99)
f <- function(x,y){ all(x>=y,c((x+y)%%10==(x+y)%/%10,(x*y)%/%100==((x*y)%/%10)%%10,(x*y)%/%100==(x*y)%%10))}
gr[which(mapply(f,gr[,1],gr[,2])==TRUE),] # 実行速度遅すぎて実用性なし
# Consider a function which, for a given whole number n,
# returns the number of ones required when writing out all numbers between 0 and n.
# For example, f(13)=6. Notice that f(1)=1.
# What is the next largest number n such that f(n)=n.
f0 <− function(n){
y=as.character(n)
nc=nchar(y)
z=NULL
for(i in 1:nc) z[i] <− substr(y,i,i)==1
return(sum(z))
}
f <− function(n){
re=numeric()
re[1]=1
for(i in 2:n){
re[i]=re[i−1]+f0(i)
}
return(re[n])
}
g <− function(n) n − f(n)
i=199979
while(g(i)!=0){
i<−i+1
}
i 6!=(5+1)5!=5*5!+5!
=5*5!+(4+1)*4!
=5*5!+4*4!+4!
=5*5!+4*4!+(3+1)*3!
=5*5!+4*4!+3*3!+3!
=5*5!+4*4!+3*3!+(2+1)*2!
=5*5!+4*4!+3*3!+2*2!+2!
=5*5!+4*4!+3*3!+2*2!+1*1!+1
factorial(n+1)= sum(factoral(n:1)*(n:1))+1 B,C,D,E,Fが0〜9の数字(同じ数字であってもよい)で
6!*B+5!*C+4!*D+3!*E+2!*F+1!*G=5555
が成立するときB+C+D+E+F+Gの最小となるB〜Gの組み合わせを求めよ。
n=5555
f <- function(B,C,D,E,F,G) sum(factorial(6:1)*c(B,C,D,E,F,G))-n
max=n%/%factorial(6)
gr=expand.grid(0:max,0:6,0:5,0:4,0:3,0:2)
ret=mapply(f, gr[,1],gr[,2],gr[,3],gr[,4],gr[,5],gr[,6])
bg=apply(gr[which(ret==0),],1,sum)
(min.sum=bg[which.min(bg)])
gr[names(min.sum),]
n%/%factorial(6)
n%%factorial(6)%/%factorial(5)
n%%factorial(6)%%factorial(5)%/%factorial(4)
n%%factorial(6)%%factorial(5)%%factorial(4)%/%factorial(3)
n%%factorial(6)%%factorial(5)%%factorial(4)%%factorial(3)%/%factorial(2)
n%%factorial(6)%%factorial(5)%%factorial(4)%%factorial(3)%%factorial(2)%/%factorial(1) #include<stdio.h>
#include<string.h>
int compare_int(const void *a, const void *b){
return *(int*)a - *(int*)b;
}
main( int argc, char *argv[] ){
char N = '0'+ atoi(argv[1]);
int a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t;
for(a = 0; a < 10; a++){
for(b = 0; b < 10; b++){
for(c = 0; c < 10; c++){
for(d = 0; d < 10; d++){
for(e = 0; e < 10; e++){
for(f = 0; f < 10; f++){
for(g = 0; g < 10; g++){
for(h = 0; h < 10; h++){
for(i = 0; i < 10; i++){
for(j = 0; j < 10; j++){
for(k = 0; k < 10; k++){
for(l = 0; l < 10; l++){ for(m = 0; m < 10; m++){
for(n = 0; n < 10; n++){
for(o = 0; o < 10; o++){
for(p = 0; p < 10; p++){
for(q = 0; q < 10; q++){
for(r = 0; r < 10; r++){
for(s = 0; s < 10; s++){
for(t = 0; t < 10; t++){
int num[]={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t};
qsort(num,20,sizeof(int),compare_int);
char NUM[20];
int i;
for(i=0;i<20;i++){
NUM[i]='0'+ num[i];
}
char N4[4]={N,N,N,N};
char N5[5]={N,N,N,N,N};
if(strstr(NUM,N4)!=NULL & strstr(NUM,N5)==NULL &
(100*a+10*b+c)*f == 100*g+10*h+i &
(100*a+10*b+c)*e == 100*j+10*k+l &
(100*a+10*b+c)*d == 100*m+10*n+o &
(100*m+10*n+o)*100 + (100*j+10*k+l)*10 + 100*g+10*h+i == p*10000+q*1000+r*100+s*10+t
){
printf("abc def ghi jkl mno pqrst = %d%d%d %d%d%d %d%d%d %d%d%d %d%d%d %d%d%d%d%d\n",
a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t);
}
}}}}} }}}}} }}}}} }}}}}
} options(scipen = 10)
fibon <- function(N){
f=numeric(N)
f[1]=1
f[2]=1
if(N>2){for(i in 3:N){
f[i]=f[i-2]+f[i-1]
}}
return(f[N])
}
Vectorize(fibon)(1:15)
fibon(50)
fibon(100)# wrong! c([],L,L).
c([X|L1],L2,[X|L3]) :- c(L1,L2,L3). ?- L= [1,2,3,4,5,6,7,8,9],c([_,_,_],X,L),c(Y,[_,_,_],X).
L = [1,2,3,4,5,6,7,8,9],
X = [4,5,6,7,8,9],
Y = [4,5,6] ;
No last([Last],Last).
last([_|Rest],Last) :- last(Rest,Last).
% last([1,2,3,4,5],Last). 日本人の血液型はA,O,B,ABの比率が4:3:2:1であるという。
それぞれの血液型の人を最低でも各々4、3、2、1人集めるためには必要な人数の期待値はいくらか? >>597
blood.type <- function(p,need){
n=length(p)
ABO=NULL
enough <- function(x){
pool=numeric(n)
for(i in 1:n) pool[i]=sum(ABO==i)
all(pool >= need)
}
while(!enough(ABO)){
ABO=append(sample(1:n,1,p=p),ABO)
}
return(length(ABO))
}
p=4:1
need=c(4,3,2,1)
re=replicate(1e5,blood.type(p,need))
mean(re) blood.samples <- function(p=c(4,3,2,1),need=c(1,1,1,1),k=1e5){
mean(replicate(k,blood.type(p,need)))
}
blood.samples()
blood.samples(need=4:1,k=1e4) subset([],[]).
subset([First|Rest],[First|Sub1]) :− subset(Rest,Sub1).
subset([_|Rest],Sub2) :− subset(Rest,Sub2).
?− subset([ド底辺,特殊,シリツ医大,イカサマ入試,裏口,馬鹿],_恥), writeln(_恥),fail. #https://i.imgur.com/Y81vbNp.jpg
N=5
library(gtools)
perm=permutations(n=2,r=N,v=c(1,5), repeats.allowed = TRUE)
move=t(apply(perm,1,cumsum))
p0=0
P=(p0+move)%%6
q0=2
Q=(q0+move)%%6
is.fe <- function(x,y){ # is first encounter after N sec?
all(x[-N]!=y[-N]) & x[N]==y[N]
}
re=NULL
for(i in 1:nrow(P)){
for(j in 1:nrow(Q)){
re=append(re,is.fe(P[i,],Q[j,]))
}
}
(Ans=mean(re)) ; 3^(N-1)/(2^N)^2
cat(Ans*(2^N)^2,'/',(2^N)^2) is.pe <- function(x,y,M) { # encounter M or more than M times within N sec.
sum(x==y) >= M
}
M=2
re=NULL
for(i in 1:nrow(P)){
for(j in 1:nrow(Q)){
re=append(re,is.pe(P[i,],Q[j,],M))
}
}
(Ans=mean(re))
cat(Ans*(2^N)^2,'/',(2^N)^2) hanoi <- function(n,from='A',via='B',to='C'){
if(n >= 1){
Recall(n-1,from,to,via)
cat('move',n,'from',from,' to ',to,'\n')
Recall(n-1,via,from,to)
}
}
hanoi(4) 掛け算を再帰関数で定義する。
VAT <- function(n,m=1.08){
if(n==0) 0
else Recall(n-1) + m
}
VAT(250) 総和を再帰関数で書く
sup <- function(v){
if (length(v)==0) 0
else v[1] + Recall(v[-1])
}
sup(1:10) # Ackermann
> A <- function(m,n){
+ if(m==0) return(n+1)
+ if(n==0) return(Recall(m-1,1))
+ else return(Recall(m-1,Recall(m,n-1)))
+ }
>
> A(2,4)
[1] 11
> A(3,4)
[1] 125
> A(4,1)
Error: evaluation nested too deeply: infinite recursion / options(expressions=)?
> 累積和
# cumsum with for-loop
cumsumL <- function(v){
n=length(v)
re=numeric(n)
re[1]=v[1]
for(i in 1:(n-1)) re[i+1]=re[i]+v[i+1]
re
}
# cumsum with recursive call
cumsumR <- function(v,res=NULL,i=1){
res[1]=v[1]
if(i==length(v)) return(res)
else{
res[i+1] = res[i] + v[i+1]
Recall(v,res,i+1)
}
}
} # 10進法をN進法でdigit桁表示する
dec2n <- function(num, N = 2, digit = 0){ # decimal to 0,1,..,n-1 vector
if(num==0 & digit==0) return(0)
if(num <= 0 & digit <= 0) return()
else{
return(append(Recall(num%/%N, N ,digit-1), num%%N))
}
}
> dec2n(0)
[1] 0
> dec2n(11,digit=5)
[1] 0 0 1 0 1 1
> dec2n(9,N=5,digit=3)
[1] 0 0 1 4
> dec2n(1000,N=16)
[1] 3 14 8 >>610 (degugged)
dec2n <- function(num, N = 2, digit = 1){ # decimal to 0,1,..,n-1 vector
if(num <= 0 & digit <= 0) return()
else{
return(append(Recall(num%/%N, N ,digit-1), num%%N))
}
}
dec2n(0)
dec2n(11,digit=5)
dec2n(9,N=5,digit=3)
dec2n(1000,N=16)
#
dec2hex <- function(x){ # decimal to hexa
hex=c(0:9,letters[1:6])
n=length(x)
re=numeric(n)
for(i in 1:n){
if(x[i]==0) re[i]='0'
else re[i]=hex[x[i]+1]
}
cat(re,'\n')
}
dec2hex(c(1,0,15))
dec2hex(dec2n(1000,16))
dec2sexa <- function(x) dec2hex(dec2n(x,16)) draft
X <- function(n,red=10,white=90){
rw=red+white
total=martix(0,nrow=n,ncol=2^n)
p=martix(0,nrow=n,ncol=2^n)
total[1,1:2]=c(rw-1,rw+1)
p[1,1:2]=c(white/rw,red/rw)
if(n > 1){
for(i in 1:n){
li=2^i
total[i+1,1:(2*li)]=c(total[i,1:li]-1, total[i,1:li]+1)
p[i+1,1:2*li)]=c(p[i,1:li]*(total[i,1:li]-red)/total[i,1:li], p[i,1:li]*(red)/total[i,1:li])
}
}
return(sum(p[n,]*total[n,]))
} rm(list=ls())
X = function(n,r=10,w=90){ #n:試行数 r: 赤玉数 w:白玉数
rw=r+w # 試行前総玉数
J=rw+n # 総玉数の上限
# s[i,j] i回試行後に総数がj個である確率の行列
s=matrix(0,nrow=n,ncol=J)
s[1,rw-1]=w/rw ; s[1,rw+1]=r/rw # 1回試行後
if(n > 1){
for(i in 2:n){
for(j in r:J){ # jはr未満にはならない
# if(j==1) s[i,j] = s[i-1,j+1]*(j+1-r)/(j+1)
if(j==J) s[i,j] = s[i-1,j-1] * r/(j-1)
else s[i,j] = s[i-1,j-1] * r/(j-1) + s[i-1,j+1]*(j+1-r)/(j+1)
}
}
}
total=sum((r:J)*s[n,r:J])# n回試行後総数の期待値
white=total-r
return(c(total=total,white=white))
}
> vX=Vectorize(X)
> vX(1000:1010)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
total 20.5 20.5 20.5 20.5 20.5 20.5 20.5 20.5 20.5 20.5 20.5
white 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 # シオマネキ
# http://i.imgur.com/uBTnujq.png
H=(sqrt(5+2*sqrt(5)))/2
A=0.5+0i
B=A+r
shiomaneki <- function(x,y){
xy=x+y*1i
(abs(xy-A)+abs(xy-B)+x)*2 + 1
}
x=seq(0,Re(B),length=100)
y=seq(0,H,length=100)
z=outer(x,y,shiomaneki)
contour(x,y,z,nlevels=20,lwd=2,bty='l')
x11() ;persp(x,y,z, theta=35, lty=3,col="lightblue",xlab='x',ylab='y',zlab='z',
ticktype='detailed',shade=0.75,phi=30,ltheta=-10,border=TRUE)
rgl::persp3d(x,y,z,col='blue')
sio <- function(xy) shiomaneki(xy[1],xy[2])
optim(c(0.4,0.4),sio,method='Nelder-Mead') # カブトガニ
# https://i.imgur.com/VhC8cug.png
H=(sqrt(5+2*sqrt(5)))/2
A=0.5+0i
B=A+r
C=0+H*1i
abs(C-B)
kabutogani <- function(y1,y2){
K=0+y1*1i
G=0.5+y2*1i
abs(K-C) + 2*abs(K-G) + 2*abs(G-A) +2*abs(G-B)
}
y1=seq(0,H,length=100)
y2=seq(0,H,length=100)
z=outer(y1,y2,kabutogani)
contour(y1,y2,z,nlevel=100,lwd=2,xlim=c(0.5,1.5),ylim=c(0,1.2),bty='l')
kab <- function(y1y2) kabutogani(y1y2[1],y1y2[2])
optim(c(1,0.5),kab, method = 'Nelder-Mead') DOP <- function(n,print=FALSE){ # diagonal length of regular plygon of size length 1
q=2*pi/n
r=cos(q)+1i*sin(q)
p=numeric(n+1)
for(i in 1:(n+1)) p[i]=r^(i-1)
D=NULL
if(n>3){
for(j in 3:(2+ceiling(n-3)/2)){
D= append(D,abs(p[1]-p[j]))
}
}
if(print){
plot(p,type='l',bty='l',axes=FALSE,ann=FALSE,lwd=2)
for(i in 3:(n-1)) segments(1,0,Re(p[i]),Im(p[i]),col='gray')
}
return(D)
}
DOP(17,p=T) DOP <- function(n,print=FALSE){ # diagonal length of regular plygon of size length 1
q=2*pi/n
r=cos(q)+1i*sin(q)
p=numeric(n+1)
for(i in 1:(n+1)) p[i]=r^(i-1)
D=NULL
if(n>3){
for(j in 3:(2+ceiling(n-3)/2)){
D= append(D,abs(p[1]-p[j])/abs(p[1]-p[2]))
}
}
if(print){
plot(p,type='l',bty='l',axes=FALSE,ann=FALSE,lwd=2)
for(i in 3:(n-1)) segments(1,0,Re(p[i]),Im(p[i]),col='gray')
}
return(D)
}
DOP(17,p=T) > (sin(2*pi/7)/((1 - cos(2*pi/7))^2 + sin(2*pi/7)^2) - (cos(4*2*pi/7)*sin(2*pi/7))/((1 - cos(2*pi/7))^2 + sin(2*pi/7)^2) - sin(4*2*pi/7)/((1 - cos(2*pi/7))^2 + sin(2*pi/7)^2) + (cos(2*pi/7)* sin(4*2*pi/7))/((1 - cos(2*pi/7))^2 + sin(2*pi/7)^2))
[1] 2.190643 rm(list=ls())
graphics.off()
ngon <- function(n,digit=TRUE,axis=FALSE,cex=1){
r=exp(2*pi/n*1i)
p=complex(n)
for(i in 1:(n+1)) p[i]= (1-r^i)/(1-r)
plot(p,bty='l',type='l',axes=axis, ann=FALSE,lwd=1,asp=1)
points(1/(1-r),pch='.')
if(digit) text(Re(p),Im(p),1:n,cex=cex)
if(axis){axis(1) ; axis(2)}
invisible(p)
} ngon <- function(n,digit=TRUE,axis=FALSE,cex=1,...){
r=exp(2*pi/n*1i)
p=complex(n)
for(i in 1:(n+1)) p[i]= (1-r^i)/(1-r)
plot(p,bty='l',type='l',axes=axis, ann=FALSE,asp=1,...)
points(1/(1-r),pch='.')
if(digit) text(Re(p),Im(p),paste('p',1:n),cex=cex)
if(axis){axis(1) ; axis(2)}
invisible(p)
}
seg <- function(a,b,...) segments(Re(a),Im(a),Re(b),Im(b),col=2,lwd=2,...)
pt <- function(x,y,...) text(Re(x),Im(x), y,...)
kabutogani3 <- function(xl,yl,xc,yc,xr,yr){
L=xl+yl*1i
C=xc+yc*1i
R=xr+yr*1i
abs(p[3]-C)+abs(C-R)+abs(C-L)+abs(R-p[2])+abs(L-p[4])+abs(R-p[1])+abs(L-p[5])
}
kabu3 <- function(par){
kabutogani3(par[1],par[2],par[3],par[4],par[5],par[6])
}
p=ngon(5,axis=T,col='lightblue',lwd=2)
opt=optim(runif(6),kabu3,method='Nelder-Mead')
kabu3(opt$par)
(par=opt$par)
L=par[1]+par[2]*1i
C=par[3]+par[4]*1i
R=par[5]+par[6]*1i
pt(C,'C') ; pt(L,'L') ; pt(R,'R')
seg(p[1],R);seg(p[2],R);seg(p[3],C);seg(p[4],L);seg(p[5],L);seg(L,C);seg(R,C) ngon <- function(n,digit=TRUE,axis=FALSE,cex=1,...){
r=exp(2*pi/n*1i)
p=complex(n)
for(i in 1:(n+1)) p[i]= (1-r^i)/(1-r)
plot(p,bty='l',type='l',axes=axis, ann=FALSE,asp=1,...)
points(1/(1-r),pch='.')
if(digit) text(Re(p),Im(p),paste('p',1:n),cex=cex)
if(axis){axis(1) ; axis(2)}
invisible(p)
}
seg <- function(a,b,...) segments(Re(a),Im(a),Re(b),Im(b),col=2,lwd=2,...)
pt <- function(x,y,...) text(Re(x),Im(x), y,...)
kabutogani3 <- function(xl,yl,xc,yc,xr,yr){
L=xl+yl*1i
C=xc+yc*1i
R=xr+yr*1i
abs(p[3]-C)+abs(C-R)+abs(C-L)+abs(R-p[2])+abs(L-p[4])+abs(R-p[1])+abs(L-p[5])
}
kabu3 <- function(par){
kabutogani3(par[1],par[2],par[3],par[4],par[5],par[6])
}
p=ngon(5,axis=T,col='lightblue',lwd=2)
opt=optim(runif(6),kabu3,method='CG')
kabu3(opt$par)
(par=opt$par)
L=par[1]+par[2]*1i
C=par[3]+par[4]*1i
R=par[5]+par[6]*1i
pt(C,'C') ; pt(L,'L') ; pt(R,'R')
seg(p[1],R);seg(p[2],R);seg(p[3],C);seg(p[4],L);seg(p[5],L);seg(L,C);seg(R,C) # how many ways of allocating 5 rooms to 6 people without vacancy?
# allocated to 1 room (4 vacant)
a1=choose(5,1)*1^6 ; a1
# allocated to 2 rooms (3 vacant)
a2=choose(5,2)*(2^6-2) ; a2
# allocated to 3 rooms (2 vacant)
a3=choose(5,3)*( 3^6-choose(3,2)*(2^6-2)-3 ) ; a3
# allocated to 4 rooms (1 vacant)
a4=choose(5,4)*( 4^6 - choose(4,3)*(3^6-choose(3,2)*(2^6-2)-3) - choose(4,2)*(2^6-2)-4 ) ; a4
5^6 - a1 - a2 - a3 - a4 dec2n <- function(num, N = 2, digit = 1){ # decimal to 0,1,..,n-1 vector
if(num <= 0 & digit <= 0) return()
else{
return(append(dec2n(num%/%N, N ,digit-1), num%%N))
}
}
room.allocation <- function(n,r){ # allocate n people to r rooms without vacancy
max=r^n
counter=0
x=0
while(x < max){
if(length(unique(dec2n(x,r,n))) == r) counter = counter+1
x= x + 1
}
return(counter)
} kagamimochi = function(b, h){
r = (b^2 + h^2)/(2*h)
V = (2/3*r^3 - r^2*(r-h) + (r-h)^3/3)*pi
return(c(radius=r,Volume=V))
} kagamimochi = function(b, h){
r = (b^2 + h^2)/(2*h)
if(b > h) V = (2/3*r^3 - r^2*(r-h) + (r-h)^3/3)*pi
else V = (r^2*(h-r) - 1/3*(h-r)^3 + 2*r^3/3)*pi
return(c(radius=r,Volume=V))
} draft
# how many ways of allocating 6 rooms to n people without vacancy?
# allocated to 1 room
a1=choose(6,1)*1^n
# allocated to 2 rooms
a2=choose(6,2)*(2^n-2)
# allocated to 3 rooms
a3=choose(6,3)*( 3^n-choose(3,2)*(2^n-2)-3 )
# allocated to 4 rooms
a4=choose(6,4)*( 4^n - choose(4,3)*(3^n-choose(3,2)*(2^n-2)-3) - choose(4,2)*(2^n-2)-4 )
# allocated to 5 rooms
a5=choose(6,5)*( 5^n - choose(5,4)*(4^n-choose(4,3)*(3^n-
choose(3,2)*(2^n-2)-3) - choose(4,2)*(2^n-2)-4)-choose (5,3)*(3^n-choose(3,2)*(2^n-2)-3) - choose (5,2)*(2^n -2) -5
6^n - a1 - a2 - a3 - a4 - a5 # how many ways of allocating 6 rooms to n people without vacancy?
library(Rmpfr)
six_rooms <- function(x){
if(x[1]<10) n=x else n=mpfr(x,100)
# allocated to 1 room
a1=choose(6,1)*1^n
# allocated to 2 rooms
a2=choose(6,2)*(2^n-2)
# allocated to 3 rooms
a3=choose(6,3)*( 3^n-choose(3,2)*(2^n-2)-3 )
# allocated to 4 rooms
a4=choose(6,4)*( 4^n - choose(4,3)*(3^n-choose(3,2)*(2^n-2)-3) - choose(4,2)*(2^n-2)-4 )
# allocated to 5 rooms
a5=choose(6,5)*(5^n-choose(5,4)*(4^n-choose(4,3)*(3^n-choose(3,2)*(2^n-2)-3)
-choose(4,2)*(2^n-2)-4)-choose(5,3)*(3^n-choose(3,2)*(2^n-2)-3)-choose(5,2)*(2^n -2) -5)
6^n - a1 - a2 - a3 - a4 - a5
} allocate.rooms <- function(m,n){ # m:rooms n:people
if(m==n) return(factorial(m))
else if(m==1) return(1)
else m*Recall(m,n-1) + m*Recall(m-1,n-1)
}
#include<stdio.h>
long factorial(long n) {
long re = 1;
long k;
for(k=1;k <=n;k++) {re *= k;}
return re;
}
long rooms(int m, int n){
if(m==n) { return factorial(m);}
else if(m==1){ return 1;}
else{
return m * rooms(m,n-1) + m * rooms(m-1,n-1);
}
}
void main( int argc, char *argv[] ){
int m,n;
long ways;
m=atoi(argv[1]);
n=atoi(argv[2]);
ways=rooms(m,n);
printf("%d\n",ways);
} 楕円体 x^2/a^2 + y^2/b^2 + z^2/c^2 = 1
高さ h のロケットおっぱいの体積
RocketPi <= function(a,b,c,h) 2/3*pi*a*b*c - 1/3*pi*b*c*(3*a^2-(a-h)^2)*(a-h)/a 楕円体 x^2/a^2 + y^2/b^2 + z^2/c^2 = 1
RocketPi <= function(a,β,γ,h) # a:楕円体の長軸長, β,γ:ロケットおっぱいの楕円底面の軸長,h:ロケットおっぱいの高さ
b=(a-h)*β/sqrt(a^2-(a-h)^2)
c=(a-h)*γ/sqrt(a^2-(a-h)^2)
2/3*pi*a*b*c - 1/3*pi*b*c*(3*a^2-(a-h)^2)*(a-h)/a
} # x^2/a^2 + y^2/b^2 + z^2/c^2 = 1
(1+sqrt(5))/2
f <- function(a=10,b=10,c=10){
xy2z <- function(x,y) c*sqrt(a^2*b^2-a^2*y^2-b^2*x^2)/(a*b)
x=seq(-a,a,le=50)
y=seq(-r*a,r*a,le=50)
z=outer(x,y,xy2z)
contour(x,y,z)
persp(x,y,z,
theta=35, lty=3,col="pink",xlab='x',ylab='y',zlab='z',ylim=c(-2*a,2*a),
ticktype='detailed',shade=0.75,phi=30,ltheta=-10,border=TRUE)
rgl::persp3d(x,y,z,col='pink')
}
f()
f(10,20,25) seg <- function(a,b,...){
segments(Re(a),Im(a),Re(b),Im(b),...)}
pt <- function(x,y=NULL,...){
text(Re(x),Im(x), ifelse(is.null(y),'+',y), ...)}
# solve α^2/a^2 + (b+β)^2/b^2=1 for a
α=3
β=-1
b=1
# α^2*b^2/(-β*(2*b+β))
(a = α*b/sqrt(-β*(2*b+β)))
(q=b+β)
f= function(x,y) x^2/a^2 + (y-q)^2/b^2 # = 1
x=seq(-5,5,length=100)
y=seq(-5,5,length=100)
z=outer(x,y,f)
contour(x,y,z,level=1,bty='l',xlim=c(-10,10),ylim=c(-2,10),
drawlabels=FALSE,col=sample(colours(),1),axes=FALSE,lwd=2)
pt(α,'●') ; pt(-α,'●') ; pt(β*1i,'●')
seg(-10,10,lty=3) ; seg(-4i,10i,lty=3)
pt(0+q*1i,'.')
eclipse <- function(b) {
# α^2*b^2/(-β*(2*b+β))
(a = α*b/sqrt(-β*(2*b+β)))
(q=b+β)
f= function(x,y) x^2/a^2 + (y-q)^2/b^2 # = 1
x=seq(-10,10,length=100)
y=seq(-10,10,length=100)
z=outer(x,y,f)
contour(x,y,z,level=1,drawlabels=FALSE,add=TRUE,col=sample(colours(),1),lwd=2)
pt(0+q*1i,'.')
}
for(i in 2:10) eclipse(i) ■ このスレッドは過去ログ倉庫に格納されています