silentorgbusy

#!/bin/bash
# 2015.11.27

# Usage: silentorgbusy [file.org]
# Tells your average workload (hrs/days) to fulfill deadlines
# Format of the .org file: 
# ** <2014-06-05 Thu> Stuff			  :1:
# that is:
# [anything but "<" here] <date> Stuff :[number of hrs required]:
# number of hrs required can be any positive integer. It can also be 0, in which case "Stuff" is considered a full day programme that takes the given day out of those available for work (e.g., a conference or a family trip)
# Weekends are not taken out but I guess could be treated easily.


re='^[0-9]+$'
real='^[0-9]+([.][0-9]+)?$'
if [ $# == 1 ] ; then
 filenor="$1"
else
 echo "Usage: silentorgbusy [filename]"
 exit
fi

ffront=${filenor%.*}
file=$ffront.org

if [ ! -f $file ] ; then
 echo "Can't find the file $file."
 exit
fi

cat $file |

awk -F\: ' 
{
 if ((NF > 1) && ($(NF-1) + 0 == $(NF-1)) && ($NF == "")) {
  printf($1)
  for (i=2; i<NF; i+=1) printf(":"$i);
  print(":"$NF)
 };
}' |

sort |

awk -F\: ' 
{
 printf($1)
 for (i=2; i<=NF; i+=1) printf(":"$i);
 print(";"$(NF-1))
}' |

awk -F\< '
{
 year=substr($2,1,4)
 mo=substr($2,6,2)
 day=substr($2,9,2)
# print("year="year" month="mo" day="day";"$0)
# dayno=(year-2013)*365+mo*30+day
 if ((year % 4 == 0) && (year % 100 != 0)) leapy=1
 else leapy=0;
 if (mo == "01") mod=0;
 if (mo == "02") mod=31;
 if (mo == "03") mod=59+leapy;
 if (mo == "04") mod=90+leapy;
 if (mo == "05") mod=120+leapy;
 if (mo == "06") mod=151+leapy;
 if (mo == "07") mod=181+leapy;
 if (mo == "08") mod=212+leapy;
 if (mo == "09") mod=243+leapy;
 if (mo == "10") mod=273+leapy;
 if (mo == "11") mod=304+leapy;
 if (mo == "12") mod=334+leapy;
 dayno=mod+day
 for (i=2012; i<year; i+=1) {
  if ((i % 4 == 0) && (i % 100 != 0)) dayno=dayno+366
  else dayno=dayno+365
 }

 print(dayno";"$0)
}' |

awk -F\; -v tyear="$(date +"%Y")" -v tmo="$(date +"%m")" -v tday="$(date +"%d")" ' BEGIN {
# tyear=strftime("%Y")
# tmo=strftime("%m")
# tday=strftime("%d")
 #print(tyear", "tmo", "tday)
 #tdayno=(tyear-2013)*365+tmo*30+tday
 if ((tyear % 4 == 0) && (tyear % 100 != 0)) tleapy=1
 else tleapy=0;
 if (tmo == 1) tmod=0;
 if (tmo == 2) tmod=31;
 if (tmo == 3) tmod=59+tleapy;
 if (tmo == 4) tmod=90+tleapy;
 if (tmo == 5) tmod=120+tleapy;
 if (tmo == 6) tmod=151+tleapy;
 if (tmo == 7) tmod=181+tleapy;
 if (tmo == 8) tmod=212+tleapy;
 if (tmo == 9) tmod=243+tleapy;
 if (tmo == 10) tmod=273+tleapy;
 if (tmo == 11) tmod=304+tleapy;
 if (tmo == 12) tmod=334+tleapy;
 tdayno=tmod+tday
 for (i=2012; i<tyear; i+=1) {
  if ((i % 4 == 0) && (i % 100 != 0)) tdayno=tdayno+366
  else tdayno=tdayno+365
 }
# Put in 1 delay: we want to look at tomorrow.
 tdayno=tdayno+1
# print(tdayno)
 maxdays=-1
}
{
# Collect daily charge function p(days from today)

# The day from today distance
 dayft=$1-tdayno

# Set the max distance looking into future, set zeros for dummy days
 if (maxdays < dayft) {
  for (i=maxdays+1; i<=dayft; i+=1) {
   dummydays[i]=0;
  }
  maxdays=dayft;
 }

# Collect dummy days (those with 0)
 if ($NF == 0) dummydays[dayft]=1;

# Add the entry charge to the given day

 if ($NF+0==$NF) {
  if ((p[dayft]+0==p[dayft]) && (p[dayft]>0)) p[dayft]=p[dayft]+$NF;
  else p[dayft]=$NF;
 }
 #for (i=2; i<=NF-2; i+=1) printf($i";");
 #print($(NF-1))
}
END{
## For now just print p
#for (i=0; i<=maxdays; i+=1) {
# printf(p[i]", ")
#}
#print("")

p[-1]=0
# Now repack tasks from dummy days; p[-1] is the overflow if today is dummy and first dummy sequence has tasks.
 for (i=maxdays; i>=0; i-=1) {
  if (dummydays[i] == 1) {
   p[i-1]=p[i-1]+p[i]
   p[i]=0
  }
 }
## For now just print p
#for (i=-1; i<=maxdays; i+=1) {
# printf(p[i]", ")
#}
#print("")

# Now collapse dummy days in p[.] to get q[.]

 dtime=-1
 tmpmaxd=maxdays
 for (i=0; i<=tmpmaxd; i+=1) {
  if (dummydays[i] == 0) {
   dtime=dtime+1
   q[dtime]=p[i]
  }
  else {
   maxdays=maxdays-1
  }
 }

# Maybe some stuff is left in the first dummy sequence. Dont forget about it

 q[0]=q[0]+p[-1]

## For now just print q
#for (i=0; i<=maxdays; i+=1) {
# printf(q[i]", ")
#}
#print("")

###
#MORNING
###

# Compute the cumulative charge function F(days from today)
 for (i=0; i<=maxdays; i+=1) {
  F[i]=0
  for (j=0; j<=i; j+=1) F[i]=F[i]+q[j];
 }

## For now just print F
#for (i=0; i<=maxdays; i+=1) {
# printf(F[i]", ")
#}
#print("")

# Put concave bound on F. First a pointer to see where the critical points are.
# The origin is not necessarily critical: add imaginary point
 pointer[0]=-1
 F[-1]=0

# This will show where the max of the pointer is right now
 cvxday=-1

# This will show how many points we have in the cvx stuff
 cvxc=0

 while (cvxday<maxdays) {
# Max slope into the future
  maxslp=0
  cvxc=cvxc+1
# Look at all future days
  for (i=cvxday+1; i<=maxdays; i+=1) {
   currslp=(F[i]-F[cvxday])/(i-cvxday);
   if (currslp>=maxslp){
    pointer[cvxc]=i
    maxslp=currslp
   }
  }
  cvxday=pointer[cvxc]
 }

## Print the pointer
#for (i=0; i<=cvxc; i+=1) printf("no: %4.4f, day: %4.4f\n",i,pointer[i])

# Print the slopes
 for (i=0; i<=maxdays; i+=1) {
# Find the pointer values around on our left
  for (j=1; j<=cvxc; j+=1) {
   if ((pointer[j-1]<i) && (pointer[j]>=i)) poicurr=j
  }
  ch=(F[pointer[poicurr-1]]-F[pointer[poicurr]])/(pointer[poicurr-1]-pointer[poicurr])
  printf("%6.2f\n",ch)
 }
}'