# 3 and 5 Litre Jug Puzzle

You may have seen this one in Die Hard: With a Vengeance... This question is based on the famous 3 and 5 Litre Jug Puzzle, but with a slightly different slant.

Golf up some code that when given an integer between 1 and 100 will provide you with the quickest instructions to measure out into a tank, the corresponding number of litres of water from a fountain, using a 3 litre jug and a 5 litre jug.

There are no gradations on either of the jugs; the fountain is abundant in supply of water, and the tank is assumed to be emptied out upon the start of each execution of the code.

You cannot access water from the tank once it goes into the tank.

The format of execution is as follows:

Input:

4 for example.

Output

Output each numbered step, as shown, followed by a tally of the volumes of the 5L jug, the 3L jug and the tank. Tally format also shown below. The number of steps must also be outputted at the end of the steps.

1) Fill 5L jug

5L: 5, 3L: 0, T: 0

2) Pour from 5L jug into 3L jug

5L: 2, 3L: 3, T: 0

3) Empty 3L jug

5L: 2, 3L: 0, T: 0

4) Pour from 5L jug into 3L jug

5L: 0, 3L: 2, T: 0

5) Fill 5L jug

5L: 5, 3L: 2, T: 0

6) Pour from 5L jug into 3L jug

5L: 4, 3L: 3, T: 0

7) Pour from 5L jug into tank

5L: 0, 3L: 3, T: 4

Volume measured out in 7 turns

Example 2

Input: 8

Output:

1) Fill 5L jug

5L: 5, 3L: 0, T: 0

2) Pour from 5L jug into tank

5L: 0, 3L: 0, T: 5

3) Fill 3L jug

5L: 0, 3L: 3, T: 5

4) Pour from 3L jug into tank

5L: 0, 3L: 0, T: 8

Volume measured out in 4 turns

Conventions

1. Fill xL jug - fills the associated jug to the top from the fountain
2. Empty xL jug - empties the contents of the associated jug into the fountain
3. Pour from xL jug into yL jug - Pours the contents of the xL jug into the yL jug
4. Pour from xL jug into tank - Pours the contents of the xL jug into the tank

Shortest code wins.

• possible duplicate of Water-Bucket problem – Howard Feb 13 '14 at 5:20
• @Howard, the old question is ill-specified (has no winning criteria) and was abandoned, so I think that this one is better and should not be closed. – Victor Stafusa Feb 13 '14 at 10:01
• Call me crazy, but isnt the optimal solution going to be 1. Add as many 5L as possible, 2. Add 3L if needed, 3. Add an already solved 2L or 1L portion as required? – user8777 Feb 13 '14 at 22:11
• @LegoStormtroopr When it all boils down, true. But I'm expecting it to be code-golfed accordingly. – Eliseo D'Annunzio Feb 13 '14 at 22:13
• @LegoStormtroopr I thought that too, but aren't 6 and 9 counterexamples? – Paul Prestidge Feb 14 '14 at 1:47

## Ruby, 407376365331324 323

This is getting kind of hard to read...

x=y=n=d=0
g=gets.to_i
"#{[43435,102,t=45,t,12,t,12,t,t][g+~d]||12}".chars{|c|n+=1
puts [eval(["x-=t=[3-y,x].min;y+=t"+t=";'Pour from 5L jug into 3L jug'","x=5;'Fill 5L jug'","d+=x;x=0"+t.sub(/3.+/,"tank'")][c.ord%3].tr t='35xy',c<?3?t:'53yx'),"5L: #{x}, 3L: #{y}, T: #{d}"]}while g>d
$><<"Volume measured out in #{n} turns"  Takes input on STDIN. Example run for N=10: Fill 5L jug 5L: 5, 3L: 0, T: 0 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 5 Fill 5L jug 5L: 5, 3L: 0, T: 5 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 10 Volume measured out in 4 turns  • "This is getting kind of hard to read..." - Dude, ain't that the point of code golf...? ;) – Eliseo D'Annunzio Feb 14 '14 at 4:22 • @WallyWest Nope. We have an [obfuscation] tag for those! My humble opinion would be that of two codegolf solutions of the same length, the most readable one would be the best. – Mr Lister Feb 14 '14 at 14:29 • @MrLister Fair enough, but sometimes obfuscation is the only way to achieve the desired shrinkage... – Eliseo D'Annunzio Jun 11 '14 at 23:37 # T-SQL 2012: 1410 1302 Another quixotic attempt at a question in SQL, but this one offered an enjoyable opportunity to play with some of the new window function options in version 2012. In addition, it exploits recursive CTEs, which may be nothing impressive in most programming languages, but recursion in SQL is like switching from horse and buggy to a Ferrari. The engine at the heart of this is in lines 5-12, which uses a recursive CTE and a window function to build a table of most of the numbers needed to solve the problem. Note in particular the test for 3, 4, 6, or 9, which ensures an optimal approach to the solution by 3s from those numbers onward. (Technically, it's a tie for 4 between the 3-1 approach and the 2-2, but doing it this way golfed me a lot of characters.) Then it's a simple matter to join to a lookup table of the optimal steps for different chunks of the problem and use another window function to properly number the steps. If you don't have MS SQL lying around, play with it on SQLFiddle. DECLARE @i INT=42,@l VARCHAR(9)='L jug ',@k VARCHAR(99)='into tank 5L: 0, 3L: 0, T: ',@o VARCHAR(99)=' 5L: 5, 3L: 0, T: ',@n CHAR(1)=' ',@5 VARCHAR(99)=') Pour from 5',@3 VARCHAR(99)=') Pour from 3' ;WITH t AS (SELECT @i i,(@i-@i%5)%5 j UNION ALL SELECT i-5,(i-i%5)%5+5 FROM t WHERE i>=5 AND i NOT IN(6,9) UNION ALL SELECT i-3,3FROM t WHERE i in(3,4,6,9) UNION ALL SELECT i-i,i FROM t WHERE i<3 AND i>0) SELECT t.i,t.j,v.s,ROW_NUMBER()OVER(PARTITION BY t.j ORDER BY t.i DESC)x,SUM(t.j)OVER(ORDER BY t.i DESC ROWS UNBOUNDED PRECEDING)y INTO #q FROM(VALUES(1,5),(2,3),(3,2),(5,2))v(i,s) JOIN t ON t.j = v.i SELECT z.b FROM(SELECT ROW_NUMBER()OVER(ORDER BY q.i DESC,w.s)a,CAST(ROW_NUMBER()OVER(ORDER BY q.i DESC,w.s)AS VARCHAR)+w.v+CAST(y-CASE WHEN q.s!=w.s THEN q.j ELSE 0 END AS VARCHAR)b FROM(VALUES(5,1,') Fill 5'+@l+@o),(5,2,@5+@l+@k),(3,1,') Fill 3'+@l+@n+'5L: 0, 3L: 3, T: '),(3,2,@3+@l+@k),(2,1,') Fill 5'+@l+@o),(2,2,@5+@l+' into 3'+@l+@n+'5L: 2, 3L: 3, T: '),(2,3,@5+@l+@k),(1,1,') Fill 3'+@l+@n+'5L: 0, 3L: 3, T: '),(1,2,@3+@l+'into 5'+@l+@n+'5L: 3, 3L: 0, T: '),(1,3,') Fill 3'+@l+@n+'5L: 3, 3L: 3, T: '),(1,4,@3+@l+'into 5'+@l+@n+'5L: 5, 3L: 1, T: '),(1,5,@3+@l+'into tank'+@o))w(i,s,v)JOIN #q q ON w.i=q.j UNION SELECT 99,'Volume measured out in '+CAST(COUNT(*)AS VARCHAR)+' turns' FROM #q)z  Results for the input 42: 1) Fill 5L jug 5L: 5, 3L: 0, T: 0 2) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 5 3) Fill 5L jug 5L: 5, 3L: 0, T: 5 4) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 10 5) Fill 5L jug 5L: 5, 3L: 0, T: 10 6) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 15 7) Fill 5L jug 5L: 5, 3L: 0, T: 15 8) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 20 9) Fill 5L jug 5L: 5, 3L: 0, T: 20 10) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 25 11) Fill 5L jug 5L: 5, 3L: 0, T: 25 12) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 30 13) Fill 5L jug 5L: 5, 3L: 0, T: 30 14) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 35 15) Fill 5L jug 5L: 5, 3L: 0, T: 35 16) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 40 17) Fill 5L jug 5L: 5, 3L: 0, T: 40 18) Pour from 5L jug into 3L jug 5L: 2, 3L: 3, T: 40 19) Pour from 5L jug into tank 5L: 0, 3L: 0, T: 42 Volume measured out in 9 turns  ### Edit: Golfed out a decent score improvement by • eliminating an unnecessary +5 in the first row of the CTE, and the WHERE clause it necessitated • in-lining the VALUES tables, saving costly DECLARE statements • remembering to convert Windows double-byte CRLFs to Unix style this time. • +1 for bravery, dude... Very impressive and thanks for the MS SQL fiddle link! – Eliseo D'Annunzio Feb 20 '14 at 5:55 • Haha, thanks man! I actually believed this one might be winnable when I started and had the recursive core query. But even with extensive string golfing, the verbosity of adding all the necessary text doomed my final solution. :) – Jonathan Van Matre Feb 20 '14 at 11:50 • If I could do a "most-creative" bounty, you'd get it... – Eliseo D'Annunzio Feb 20 '14 at 12:32 • +1 for making me LOL. T-SQL is certainly an odd club to carry in your code-golf bag. – Comintern Feb 23 '14 at 1:20 # Javascript: 481 First attempt at golfing, advice appreciated n=["3L jug","5L jug","tank"];l=[0,0,0];t=[3,5,0];h=0;c=console;function e(d){l[d]=t[d];c.log(++h+") Fill "+n[d]);k()}function m(d,g){s=l[d];f=l[g];b=s+f>t[g];l[g]=b?t[g]:f+s;l[d]=b?s-(t[g]-f):0;c.log(++h+") Pour from "+n[d]+" into "+n[g]);k()}function k(){c.log("5L: "+l[1]+", 3L: "+l[0]+", T: "+l[2])}a=prompt();for(t[2]=a;4<a;)e(1),m(1,2),a-=5;2<a&&(e(0),m(0,2),a-=3);1<a&&(e(1),m(1,0),m(1,2),a=0);0<a&&(e(0),m(0,1),e(0),m(0,1),m(0,2));c.log("Volume measured out in "+h+" turns")  It messes up with some numbers because it doesn't check if it's better to pour 3 or 5, example: 9 gives 9 turns instead of 6, I might fix it later Paste it in console From 553 to 481 thanks to @WallyWest • You could try: n=["3L jug","5L jug","tank"];l=[0,0,0];t=[3,5,0];h=0;c=console;function e(d){l[d]=t[d];c.log(++h+") Fill "+n[d]);k()}function m(d,g){s=l[d];f=l[g];b=s+f>t[g];l[g]=b?t[g]:f+s;l[d]=b?s-(t[g]-f):0;c.log(++h+") Pour from "+n[d]+" into "+n[g]);k()}function k(){c.log("5L: "+l[1]+", 3L: "+l[0]+", T: "+l[2])}a=prompt();for(t[2]=a;4<a;)e(1),m(1,2),a-=5;2<a&&(e(0),m(0,2),a-=3);1<a&&(e(1),m(1,0),m(1,2),a=0);0<a&&(e(0),m(0,1),e(0),m(0,1),m(0,2));c.log("Volume measured out in "+h+" turns") for 481 characters... – Eliseo D'Annunzio Jun 11 '14 at 23:42 • @WallyWest thanks, didn't think about using logical operators instead of ifs – Sam Jun 12 '14 at 8:15 ## Java, 610 class X{int n,c=0,t=0;public void static main(String[]a){n=Integer.parseInt(a[0]);String s,b,f,k,m,u;b="5L";s="3L";k="tank";u="Fill %s jug\n5L: %d, 3L: %d, T: %d";m="\nPour from %s jug into %s\n5L: %d, 3L: %d, T: %d";f=u+m;for(;n>4;)z(f,2,5,b,5,0,t,b,k,0,0,t+=5);while(n!=0){if(n==1)z(f+f+m,5,1,s,0,3,t,s,b,3,0,t,s,3,3,t,s,b,5,1,t,s,k,5,0,t+1);if(n==3)z(f,2,3,s,0,3,t,s,k,0,0,t+3);z(f+m,3,2,b,5,0,t,b,s,2,3,t,b,k,0,3,t+=2);if(n==2)z("Empty 3L jug\n5L: 0, 3L: 0,T: %d",1,0,t)}z("Volume measured out in %d turns",0,0,c)}void z(String s,int o,int w,Object...a){c+=o;n-=w;System.out.println(String.format(s,a))}}  I took the solution of Sumedh and golfed it. I wanted to put it in the comments but my reputation isn't enough :( . It's a 40% less, I think it should at least be shared. Still far from first though... Here is ungolfed:  class X{ int n,c=0,t=0; public void static main(String[] a){ n=Integer.parseInt(a[0]); String s,b,f,k,m,u; b="5L"; s="3L"; k="tank"; u="Fill %s jug\n5L: %d, 3L: %d, T: %d"; m="\nPour from %s jug into %s\n5L: %d, 3L: %d, T: %d"; f=u+m; for(;n>4;)z(f,2,5,b,5,0,t,b,k,0,0,t+=5); while(n!=0) { if(n==1)z(f+f+m,5,1,s,0,3,t,s,b,3,0,t,s,3,3,t,s,b,5,1,t,s,k,5,0,t+1); if(n==3)z(f,2,3,s,0,3,t,s,k,0,0,t+3); z(f+m,3,2,b,5,0,t,b,s,2,3,t,b,k,0,3,t+=2); if(n==2)z("Empty 3L jug\n5L: 0, 3L: 0,T: %d",1,0,t); } z("Volume measured out in %d turns",0,0,c); } void z(String s,int o, int w,Object... a){ c+=o; n-=w; System.out.println(String.format(s,a)); } }  NB: it works only on the first run. Rerun it and the result will be wrong (due to a global variable). The following version is safe, but we lose 2 char, going from 610 to 612:  class X{ int n,c,t; public void static main(String[] a){ n=Integer.parseInt(a[0]); String s,b,f,k,m,u; t=c=0; b="5L"; s="3L"; k="tank"; u="Fill %s jug\n5L: %d, 3L: %d, T: %d"; m="\nPour from %s jug into %s\n5L: %d, 3L: %d, T: %d"; f=u+m; for(;n>4;)z(f,2,5,b,5,0,t,b,k,0,0,t+=5); while(n!=0) { if(n==1)z(f+f+m,5,1,s,0,3,t,s,b,3,0,t,s,3,3,t,s,b,5,1,t,s,k,5,0,t+1); if(n==3)z(f,2,3,s,0,3,t,s,k,0,0,t+3); z(f+m,3,2,b,5,0,t,b,s,2,3,t,b,k,0,3,t+=2); if(n==2)z("Empty 3L jug\n5L: 0, 3L: 0,T: %d",1,0,t); } z("Volume measured out in %d turns",0,0,c); } void z(String s,int o, int w,Object... a){ c+=o; n-=w; System.out.println(String.format(s,a)); } }  Sample output for N=69: Fill 5L jug 5L: 5, 3L: 0, T: 0 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 5 Fill 5L jug 5L: 5, 3L: 0, T: 5 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 10 Fill 5L jug 5L: 5, 3L: 0, T: 10 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 15 Fill 5L jug 5L: 5, 3L: 0, T: 15 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 20 Fill 5L jug 5L: 5, 3L: 0, T: 20 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 25 Fill 5L jug 5L: 5, 3L: 0, T: 25 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 30 Fill 5L jug 5L: 5, 3L: 0, T: 30 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 35 Fill 5L jug 5L: 5, 3L: 0, T: 35 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 40 Fill 5L jug 5L: 5, 3L: 0, T: 40 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 45 Fill 5L jug 5L: 5, 3L: 0, T: 45 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 50 Fill 5L jug 5L: 5, 3L: 0, T: 50 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 55 Fill 5L jug 5L: 5, 3L: 0, T: 55 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 60 Fill 5L jug 5L: 5, 3L: 0, T: 60 Pour from 5L jug into tank 5L: 0, 3L: 0, T: 65 Fill 5L jug 5L: 5, 3L: 0, T: 65 Pour from 5L jug into 3L 5L: 2, 3L: 3, T: 65 Pour from 5L jug into tank 5L: 0, 3L: 3, T: 67 Empty 3L jug 5L: 0, 3L: 0,T: 67 Fill 5L jug 5L: 5, 3L: 0, T: 67 Pour from 5L jug into 3L 5L: 2, 3L: 3, T: 67 Pour from 5L jug into tank 5L: 0, 3L: 3, T: 69 Volume measured out in 33 turns  # Java: 984 Here's the code class X{public static void main(String[] s){int n=Integer.parseInt(s[0]);int t=0;int c=0;while(n>4){n-=5;System.out.println("Fill 5L jug\n5L: 5, 3L: 0, T: "+t+"\nPour from 5L jug into tank\n5L: 0, 3L: 0, T: "+(t+5));t+=5;c+=2;}while(n!=0){switch(n){case 1:System.out.println("Fill 3L jug\n5L: 0, 3L: 3, T: "+t+"\nPour from 3L jug into 5L jug\n5L: 3, 3L: 0, T: "+t+"\nFill 3L jug\n5L: 3, 3L: 3, T: "+t+"\nPour from 3L jug into 5L jug\n5L: 5, 3L: 1, T: "+t+"\nPour from 3L jug into tank\n5L: 5, 3L: 0, T: "+(t+1));n=0;c+=5;break;case 3:System.out.println("Fill 3L jug\n5L: 0, 3L: 3, T: "+t+"\nPour from 3L jug into tank\n5L: 0, 3L: 0, T: "+(t+3));n=0;c+=2;break;default:System.out.println("Fill 5L jug\n5L: 5, 3L: 0, T: "+t+"\nPour from 5L jug into 3L jug\n5L: 2, 3L: 3, T: "+t+"\nPour from 5L jug into tank\n5L: 0, 3L: 0, T: "+(t+2));n-=2;c+=3;t+=2;if(n==2){System.out.println("Empty 3L jug\n5L: 0, 3L: 0,T: "+t);c++;}break;}}System.out.println("Volume measured out in "+c+" turns");}}  Input is from command line. for example: java X 4 • I cant comment anywhere else, so I'm commenting here. @Lego Stormtroopr, there's an alternate optimum solution, where for 4L remaining, you can do same as that for 2L(3 steps) then empty 3L jug, & then repeat for remaining 2L, thus making it complete in 7 steps....which is same for your solution where 4L is divided into: 3L into tank(2 steps) & a 5 step method for remaining 1L. – Sumedh Feb 14 '14 at 11:00 • @Chron,does your code work for values of N, where N%5 is 1 or 4 ? I dont understand ruby, thats why I couldnt test it myself... – Sumedh Feb 14 '14 at 11:03 • it should, for example have a look here at N=11: ideone.com/3ZDuOS You can hit edit in the top-left and change STDIN to other values if you want to check. – Paul Prestidge Feb 14 '14 at 21:34 • Wow, you've got more optimum solution than mine....how do you decide when to stop 5L & use 3L instead? I mean, if input is 81,then you get upto 75L using 5L, & then use 3L. if ip is 89, then 5L is used upto 80L, & remaining are 3L. – Sumedh Feb 15 '14 at 12:41 • Save some chars: main(String[]s), int n=Integer.parseInt(s[0]),t=0,c=0;, java.io.PrintStream q=System.out;. Also, it may be possible to write the first while as a one or two characters shorter for. Further, your Strings are repetitive, you might try to store repetitive parts in variables or create functions that build them using only one prefab String. – Victor Stafusa Feb 16 '14 at 14:00 ## Python 2.7 - 437 Not the shortest code, but I think this is the most optimal way of solving this. As I stated in the comments, the most optimal way to calculate this: 1. Take as many chunks of 5L as possible - divmod(amount,5). This will give you one of 4,3,2,1 as the remainder. 2. Take 3 (if possible) from the remainder. 3. Which leaves either 1 or 2 as the remainder. Use the optimal solution for either which can be known ahead of time as: 1. 1L, 5 steps : 3L -> 5L, 3L -> 5L, leaving 1L in the 3L, 3L (holding 1L) -> tank 2. 2L, 3 steps : 5L -> 3L, leaves 2L in the 5L, 5L (holding 2L) -> tank The code: j,T="%dL jug","tank" A="\n5L: %d, 3L: %d, T: %d" F,P="Fill "+j+A,"Pour from "+j+" into %s"+A f,r=divmod(input(),5) o,t=f*5,[] for i in range(f):o+=[F%(5,5,0,5*i),P%(5,T,0,0,5*i+5)] if r>2:o+=[F%(3,0,3,t),P%(3,T,0,0,t+3)];r-=3;t+=3 if r==2:o+=[F%(5,5,0,t),P%(5,j%3,2,3,t),P%(5,T,0,3,t+2)] if r==1:o+=[F%(3,0,3,t),P%(3,j%5,3,0,t),F%(3,3,3,t),P%(3,j%5,5,1,t),P%(3,T,5,0,t+1)] print"\n".join(o),'\n',"Volume measured out in %d turns"%len(o)  And an output for 4L in 7 steps: Fill 3L jug 5L: 0, 3L: 3, T: 0 Pour from 3L jug into tank 5L: 0, 3L: 0, T: 3 Fill 3L jug 5L: 0, 3L: 3, T: 3 Pour from 3L jug into 5L jug 5L: 3, 3L: 0, T: 3 Fill 3L jug 5L: 3, 3L: 3, T: 3 Pour from 3L jug into 5L jug 5L: 5, 3L: 1, T: 3 Pour from 3L jug into tank 5L: 5, 3L: 0, T: 4 Volume measured out in 7 turns  • You are assigning an int to o and then trying to add a list. I think you meant to assign o,t=[],f*5 on line 5. – psion5mx Feb 14 '14 at 13:31 • Lose those for, range, and if statements and you can get it down to 399 on one line: j,T="%dL jug","tank";A="\n5L: %d, 3L: %d, T: %d";F,P="Fill "+j+A,"Pour from "+j+" into %s"+A;f,r=divmod(input(),5);t,o=f*5,[];o=[F%(5,5,0,5*i),P%(5,T,0,0,5*i+5)]*f+[F%(3,0,3,t),P%(3,T,0,0,t+3)]*(r>2)+[F%(5,5,0,t),P%(5,j%3,2,3,t),P%(5,T,0,3,t+2)]*(r==2)+[F%(3,0,3,t),P%(3,j%5,3,0,t),F%(3,3,3,t),P%(3,j%5,5,1,t),P%(3,T,5,0,t+1)]*(r in[1,4]);print"\n".join(o),"\nVolume measured out in %d turns"%len(o) – psion5mx Feb 14 '14 at 13:53 • Impressive manipulation... @psion5mx I didn't think this kind of programming was possible on Python? No range, recursion or if statements? – Eliseo D'Annunzio Feb 16 '14 at 22:57 • The power of lists. Multiplying a list by an integer replaces the 'loops'. Multiplying by a boolean replaces the 'ifs'. – psion5mx Feb 17 '14 at 10:40 • Also - I managed to get it down to a single space (outside of quotes), but could eliminate all spaces at the cost of a character by replacing (r in[1,4]) with (r%5in[1,4]) in this case. – psion5mx Feb 17 '14 at 10:44 # Smalltalk (Smalltalk/X), 568 560 516 input in n:  T:=j:=J:=c:=0.m:={'Pour from'.' into'.' 3L jug'.' 5L jug'.[j:=j+3.'Fill'].[J:=J+5.'Fill'].[t:=j.j:=0.''].[t:=J.J:=0.''].[r:=j min:5-J.j:=j-r.J:=J+r.''].[r:=J min:3-j.J:=J-r.j:=j+r.''].[T:=T+t.' into tank'].[c:=c+1.'\5L: %1 3L: %2 T: %3\'bindWith:J with:j with:T].['Volume measured out in %1 turns'bindWith:c]}.[n>=0]whileTrue:[s:=n.n:=0.(s caseOf:{0->[n:=-1.'<'].1->'42;02813;42;02813;062:;'.2->'53;03912;073:;'.3->'42;062:;'.4->[n:=1.'42;062:;']}otherwise:[n:=s-5.'53;073:;'])do:[:c|(m at:c-$/)value withCRs print]]


boy this is definitely the most obfuscated program I've ever written...

Edit: Some other Smalltalks may not allow autodeclared workspace variables and you'll have to prepend declarations. Also bindWith: may be different (expandWith:'<p>').

sample output for n=17:

Fill 5L jug
5L: 5, 3L: 0, T: 0
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
Fill 5L jug
5L: 5, 3L: 0, T: 5
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 10
Fill 5L jug
5L: 5, 3L: 0, T: 10
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 15
Fill 5L jug
5L: 5, 3L: 0, T: 15
Pour from 5L jug into 3L jug
5L: 2, 3L: 3, T: 15
Pour from 5L jug into tank
5L: 0, 3L: 3, T: 17
Volume measured out in 9 turns


# C, 567 609

#define r printf
#define l r("5L: %d, 3L: %d, T: %d\n", a, b, T);
#define j(x,y,z,w) r("%d) "#x" %dL jug\n", i++, y),z=w,l
#define e j(Empty,3,b,0)
#define f j(Fill,5,a,5)
#define g j(Fill,3,b,3)
#define o(x,y,z,w) r("%d) Pour from %dL jug into "x"\n", i++, y,z),w;l
#define t(x,y) T+=x,o("tank",y,0,x=0)
#define p(x) o("%dL jug",5,3,(a-=x,b+=x))
int N,T,i;q(x){int a=0,b=0;switch(x){case 5:f t(a,5) break;case 3:g t(b,3) break;case 1:case 2:case 4:f if(x-2){e p(2)f p(1)if(x-4){e p(3)}}t(a,5)}N-=x;}main(){T=0,i=1,scanf("%d",&N);while(N>5)q((N-6)&&(N-9)?5:3);q(N);r("Volume measured out in %d turns",i-1);}


previous invalid version:

#define r printf
#define l r("5L: %d, 3L: %d, T: %d\n", a, b, T);
#define j(x,y,z,w) r("%d) "#x" %dL jug\n", i++, y),z=w,l
#define e j(Empty,3,b,0)
#define f j(Fill,5,a,5)
#define g j(Fill,3,b,3)
#define o(x,y,z,w) r("%d) Pour from %dL jug into "x"\n", i++, y,z),w;l
#define t o("tank",5,0,a=0)
#define p(x) o("%dL jug",5,3,(a-=x,b+=x))
int N,T,i;q(x){int a=0,b=0;switch(x){case 5:f t break;case 3:g t break;case 1:case 2:case 4:f if(x-2){e p(2)f p(1)if(x-4){e p(3)}}t}N-=x;}main(){T=0,i=1,scanf("%d",&N);while(N>5)q(5);q(N);r("Volume measured out in %d turns",i-1);}


and here is the code degolfed:

#define r printf
#define l r("5L: %d, 3L: %d, T: %d\n", a, b, T);
#define j(x,y,z,w) r("%d) "#x" %dL jug\n", i++, y),z=w,l
#define e j(Empty,3,b,0)
#define f j(Fill,5,a,5)
#define g j(Fill,3,b,3)
#define o(x,y,z,w) r("%d) Pour from %dL jug into "x"\n", i++, y,z),w;l
#define t o("tank",5,0,a=0)
#define p(x) o("%dL jug",5,3,(a-=x,b+=x))
int N,T,i;
q(x)
{
int a=0,b=0;
switch(x)
{
case 5:
f
t
break;
case 3:
g
t
break;
case 1:
case 2:
case 4:
f
if(x-2)
{
e
p(2)
f
p(1)
if(x-4)
{
e
p(3)
}
}
t
}
N-=x;
}
main()
{
T=0,i=1,scanf("%d",&N);
while(N&gt;
5)q(5);
q(N);
r("Volume measured out in %d turns",i-1);
}

• This doesn't give the optimal solution for 9 (8 turns, should be 6 - fill and empty the 3L 3 times). – Comintern Feb 23 '14 at 1:24
• Also doesn't work at all for an input of 1. – Comintern Feb 23 '14 at 7:07
• Doesn't work for 1? Pity... But admirable effort... :) – Eliseo D'Annunzio Feb 23 '14 at 10:23
• yep, there are some bugs and the solution is not allways optimal. but it makes 1 L in 8 steps... – V-X Feb 23 '14 at 11:18
• Couple golf tips - You can save 6 bytes by replacing int N,T,i; with N,T,i,a,b; and int a=0,b=0; with a=b=0;. You also get 3 bytes by adding the ( to your printf definition. I think the biggest gain would be reducing the switch statement to a nested ternary though - the case and break statements really add up. – Comintern Feb 23 '14 at 17:38

# C (480 465 bytes)

#define P printf(
#define O(x) P"%d) Pour from %dL jug into "x"\n"
#define S P"5L: %d, 3L: %d, T: %d\n",F,H,g);}
F,H,s,g,x;l(j){P"%d) Fill %dL jug\n",++s,j);St(j,o,m){O("%dL jug"),++s,j,(j^5)?5:3);Se(j,i){O("tank"),++s,j);Smain(){scanf("%d",&x);while(x>4){x-=5;l(F=5);g+=5;e(5,F=0);}while(x>2){x-=3;l(H=3);g+=3;e(3,H=0);}(x^2)?(x^1)?0:(l(H=3),t(3,H=0,F=3),l(H=3),t(3,H=1,F=5),g++,e(3,H=0)):(l(F=5),t(5,F=2,H=3),g+=2,e(5,F=0));P"Volume measured out in %d turns",s);}


Optimal version (adds 10 bytes)

#define P printf(
#define O(x) P"%d) Pour from %dL jug into "x"\n"
#define S P"5L: %d, 3L: %d, T: %d\n",F,H,g);}
F,H,s,g,x;l(j){P"%d) Fill %dL jug\n",++s,j);St(j,o,m){O("%dL jug"),++s,j,(j^5)?5:3);Se(j,i){O("tank"),++s,j);Smain(){scanf("%d",&x);while(x>4&&x^6&&x^9){x-=5;l(F=5);g+=5;e(5,F=0);}while(x>2){x-=3;l(H=3);g+=3;e(3,H=0);}(x^2)?(x^1)?0:(l(H=3),t(3,H=0,F=3),l(H=3),t(3,H=1,F=5),g++,e(3,H=0)):(l(F=5),t(5,F=2,H=3),g+=2,e(5,F=0));P"Volume measured out in %d turns",s);}


Likely more golfing to be done here - the output functions were killing me. This should give the optimal solution (least number of steps). Similar to other code here, it fills and empties 5L jugs until it gets below 5 and then switches to 3L jugs. It tests for 2 special cases (6 and 9) and if it finds them switches to 3L jugs. The instructions for getting 1L and 2L are hard coded.

More readable version:

#define P printf(
#define O(x) P"%d) Pour from %dL jug into "x"\n"
#define S P"5L: %d, 3L: %d, T: %d\n",F,H,g);}
F,H,s,g,x;
l(j)
{
P"%d) Fill %dL jug\n",++s,j);S

t(j,o,m)
{
O("%dL jug"),++s,j,(j^5)?5:3);S

e(j,i)
{
O("tank"),++s,j);S

main()
{
scanf("%d",&x);
//while(x>4&&x^6&&x^9)     <--optimal version
while(x>4)
{
x-=5;l(F=5);g+=5;e(5,F=0);
}
while(x>2)
{
x-=3;l(H=3);g+=3;e(3,H=0);
}
(x^2)?
(x^1)?
0
:
(l(H=3),t(3,H=0,F=3),l(H=3),t(3,H=1,F=5),g++,e(3,H=0))
:(l(F=5),t(5,F=2,H=3),g+=2,e(5,F=0));
P"Volume measured out in %d turns",s);
}


Edits:

• Removed 10 bytes that gave the optimum performance for the scored version based on the OP's clarification.
• Shave 5 bytes by converting function to definition.

Test output for n = 11 (optimal version):

11
1) Fill 5L jug
5L: 5, 3L: 0, T: 0
2) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
3) Fill 3L jug
5L: 0, 3L: 3, T: 5
4) Pour from 3L jug into tank
5L: 0, 3L: 0, T: 8
5) Fill 3L jug
5L: 0, 3L: 3, T: 8
6) Pour from 3L jug into tank
5L: 0, 3L: 0, T: 11
Volume measured out in 6 turns

• Why don't you count 11 as a special case? What do you do for 14? One five, three threes (8) is going to beat two fives and making up the other four litres (which can't be done in four turns). – Bill Woodger Feb 23 '14 at 7:47
• 11 and 14 both include the special cases. After you subtract the first 5L, they leave 6 and 9 respectively and these are handled by the special cases. 9 is the largest number that can be done in less steps using only a 3L jug. Input of 14 gives the 8 step solution, the output for 11 is above. – Comintern Feb 23 '14 at 17:13

## T-SQL(2012): 794689 580

Inspired by @Jonathan-Van-Matre 's T-SQL answer in combination with @Lego-Stormtroopr's algorithm. I wanted to do this because I enjoyed the 99 Bottles of Beer challenge so much.

I tried to keep window (OVER) functions at a minimum in preference of math/bool functions.

SQLFiddle is here.

WITH n AS(SELECT 11 n UNION ALL SELECT n-IIF(n>4,5,3)FROM n WHERE n>2)SELECT n, a,LEN(a)L,i=IDENTITY(INT,1,1),'L jug'j INTO #t FROM n JOIN(VALUES(3303),(33900),(5550),(55900),(2550),(259323),(25903),(1303),(139530),(1333),(139551),(13950))x(a)ON RIGHT(LEFT(12335,n),1)=LEFT(a,1)ORDER BY n DESC SELECT LTRIM(i)+') '+REPLACE(IIF(L=4,'Fill ','Pour ')+RIGHT(a/100,L-3),9,j+' into ')+IIF(L=5,'tank',j)  +'
5L: '+LTRIM((A%100)/10)+', 3L: '+LTRIM(A%10)+', T: '+LTRIM(SUM(IIF(L=5,LEFT(a,1),0))OVER(ORDER BY i))FROM #t UNION SELECT 'Volume measured out in ' +LTRIM(MAX(i))+' turns'FROM #t


Input: 11

1) Fill 5L jug
5L: 5, 3L: 0, T: 0
2) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
3) Fill 5L jug
5L: 5, 3L: 0, T: 5
4) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 10
5) Fill 3L jug
5L: 0, 3L: 3, T: 10
6) Pour from 3L jug into
5L jug 5L: 0, 3L: 3, T: 10
7) Fill 3L jug
5L: 3, 3L: 3, T: 10
8) Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 10
9) Pour from 3L jug into tank
5L: 5, 3L: 0, T: 11
Volume measured out in 9 turns


Human-readable:

WITH n AS(
SELECT 11 n
UNION ALL
SELECT n-IIF(n>4,5,3)
FROM n
WHERE n>2
)
SELECT n, a,LEN(a) L, i = IDENTITY(INT,1,1), 'L jug'j
INTO #t
FROM n
JOIN(VALUES
(3303),(33900),
(5550),(55900),
(2550),(259323),(25903),
(1303),(139530),(1333),(139551),(13950)
)x(a)
ON RIGHT(LEFT(12335,n),1) = LEFT(a,1)
ORDER BY n DESC

SELECT LTRIM(i)+') '
+ REPLACE(IIF(L=4,'Fill ','Pour ')
+ RIGHT(a/100,L-3),9,j+' into ')+IIF(L=5,'tank',j)
+'
5L: ' + LTRIM((A%100)/10) + ', 3L: ' + LTRIM(A%10) + ', T: '
+ LTRIM(SUM(IIF(L=5,LEFT(a,1),0))OVER(ORDER BY i)) FROM #t
UNION ALL
SELECT 'Volume measured out in ' +LTRIM(MAX(i))+' turns'FROM #t
DROP TABLE #t

• Do you have some sample output? – Bill Woodger Jun 16 '14 at 19:38
• @BillWoodger added output for input = 8 – comfortablydrei Jun 16 '14 at 19:49
• Thanks. 8 is quite easy. One to 11 gives the code a good stretch :-) I'm upvoting, so don't come back and tell me it doesn't work. – Bill Woodger Jun 16 '14 at 20:19
• @Bill Thanks. Changed to input = 11 – comfortablydrei Jun 16 '14 at 20:38
• @comfortablydrei Amazing stuff using T-SQL... Taking a page from Jonathon's book... – Eliseo D'Annunzio Jun 19 '14 at 23:30

# Python 3 (417 chars)

P=print
D=divmod
N=['3L jug','5L jug','tank',0]
M=999
R=[0,0,0,M]
F=[3,5,M,M]
def o(a,b):k=a==3;P(['Pour from %s into %s','Empty %s','Fill %s'][k*2+(b==3)]%[(N[a],N[b]),(N[b])][k]);d=min(R[a],F[b]-R[b]);R[a]-=d;R[b]+=d;P('5L:',R[1],'3L:',R[0],'T:',R[2]);N[3]+=1
k,r=D(int(input()),5)
for i in'0'*k:o(3,1);o(1,2)
for x in['','c1c12','d46','c2','d434d46'][r]:o(*D(int(x,16),4))
P('Volume measured out in',N[3],'turns')


## Explained

Note that we have 4 objects, namely, the 3L jug, the 5L jug, the tank, and the foutain. The only operations we can do is to move water from object a to object b. This is what function o(a, b) do in my code, it move water and print it and keep counting.

## Tricks

• N=['3L jug','5L jug','tank',0]. Here I need the last element to avoid IndexError. Also, it can be used as the global counting variable, without the expansive global keyword. For example, N[3] += 1

• Since 0 <= a < 4, 0 <= b < 4 in function o(a, b), we can encode (a, b) into a hex digit using (a << 2) | b, and decode it using divmod(x, 4). With this trick, all 5 solutions(reminder=0, 1, 2, 3, 4), can be encoded into array ['','c1c12','d46','c2','d434d46'], which is a little bit shorter than it's original form:

A=[ (), ((3,0),(0,1),(3,0),(0,1),(0,2)), ((3,1),(1,0),(1,2)), ((3,0),(0,2)), ((3,1),(1,0),(0,3),(1,0),(3,1),(1,0),(1,2)) ]

## Sample Output(n = 17)

17
Fill 5L jug
5L: 5 3L: 0 T: 0
Pour from 5L jug into tank
5L: 0 3L: 0 T: 5
Fill 5L jug
5L: 5 3L: 0 T: 5
Pour from 5L jug into tank
5L: 0 3L: 0 T: 10
Fill 5L jug
5L: 5 3L: 0 T: 10
Pour from 5L jug into tank
5L: 0 3L: 0 T: 15
Fill 5L jug
5L: 5 3L: 0 T: 15
Pour from 5L jug into 3L jug
5L: 2 3L: 3 T: 15
Pour from 5L jug into tank
5L: 0 3L: 3 T: 17
Volume measured out in 9 turns


COBOL (IBM Enterprise COBOL) 192 lines of 72 characters

This is a Proof of Concept for the Question, and the start of one for Golf-COBOL :-)

The question asks for the quickest. So, implement parallelism. Even one person can readily fill one 3L jug and one 5L jug at the same time.

Simply divide the input by eight, also leaving the remainder. Do some quick 5L/3L fills to the number of times eight fits exactly, then deal with the one through seven litres remaining.

The most interesting of the remainder is for four litres. Doing it as one litre plus three litres pushes a lot less water around, only 18 litres vs 23 for the other possibilities.

The Code (working)

   ID DIVISION
PROGRAM-ID
DATA DIVISION
WORKING-STORAGE SECTION
1.
88 g1 VALUE ' '.
2  PIC X
88 H VALUE 'F'.
88 I VALUE 'E'.
88 J VALUE 'T'.
2 PIC X
88 K VALUE 'F'.
88 L VALUE 'E'.
88 M VALUE 'T'.
1 R
2 A1 PIC 999
2 B PIC 99
2 C PIC 9
1 E
2 e2 PIC X(120) VALUE "  ) Fill both jugs"
2 e3 PIC X(120)
88 O VALUE "5L: 0, 3L: 0, T: 000".
2 e4 PIC X(120) VALUE "  ) Empty both jugs"
2 e5 PIC X(120)
2 e1 occurs 32 depending on p pic x(240)
2 e6 pic x(99)
1 F PIC 999 VALUE 0
1 P PIC 99 VALUE 0
1 P1 PIC 99
PROCEDURE DIVISION
ACCEPT A1
DIVIDE A1 BY 8 GIVING B REMAINDER C
set o to true
move e3 to e5
move 5 to e3(5:1)
move 3 to e3(12:1)
PERFORM D1 B TIMES
EVALUATE C
WHEN 1
MOVE ZERO TO R
SET K TO TRUE
PERFORM N
SET M TO TRUE
PERFORM N
SET K TO TRUE
PERFORM N
SET M TO TRUE
PERFORM N
SET L TO TRUE
PERFORM N
WHEN 2
MOVE ZERO TO R
SET H TO TRUE
PERFORM N
SET J TO TRUE
PERFORM N
SET I TO TRUE
PERFORM N
WHEN 3
MOVE ZERO TO R
SET K TO TRUE
PERFORM N
SET L TO TRUE
PERFORM N
WHEN 4
MOVE ZERO TO R
SET K TO TRUE
PERFORM N
SET M TO TRUE
PERFORM N
SET K TO TRUE
PERFORM N
SET M TO TRUE
PERFORM N
SET L TO TRUE
PERFORM N
SET K TO TRUE
PERFORM N
SET L TO TRUE
PERFORM N
WHEN 5
MOVE ZERO TO R
SET H TO TRUE
PERFORM N
SET I TO TRUE
PERFORM N
WHEN 6
MOVE ZERO TO R
SET K TO TRUE
PERFORM N
SET L TO TRUE
PERFORM N
SET K TO TRUE
PERFORM N
SET L TO TRUE
PERFORM N
WHEN 7
MOVE ZERO TO R
SET H TO TRUE
PERFORM N
SET I TO TRUE
PERFORM N
SET H TO TRUE
PERFORM N
SET J TO TRUE
PERFORM N
SET I TO TRUE
PERFORM N
END-EVALUATE
string "Volume measured out in " delimited size P " turns"
delimited size into e6
if  e6(24:1) = 0
move e6(25:) to e6 (24:)
end-if
move p to p1
perform d2 p times
DISPLAY E(481:)
GOBACK
D1
ADD 1 TO P
MOVE P TO E(1:2)
move e2 to e1(p)
move e3 to e1(p)(121:)
ADD 1 TO P
MOVE P TO E(241:2)
ADD 8 TO F
MOVE F TO E(378:3)
move e4 to e1(p)
move e5 to e1(p)(121:)
MOVE F TO E(138:3)
N
ADD 1 TO P
SET O TO TRUE
EVALUATE TRUE
WHEN K

MOVE 3 TO B
string p delimited size ") Fill 3L jug" delimited by size
into e1(p)
WHEN M
COMPUTE C = C + B
IF  C > 5
COMPUTE B = C - 5
MOVE 5 TO C
ELSE
MOVE 0 TO B
END-IF
string  P delimited size ") Pour from 3L jug into 5L jug"
delimited size into e1(p)
WHEN L
ADD B TO F
MOVE 0 TO B
string  P delimited size ") Empty 3L jug into tank"
delimited size into e1(p)
END-EVALUATE
EVALUATE TRUE
WHEN H
MOVE 5 TO C
string  P delimited size ") Fill 5L jug"
delimited size into e1(p)
WHEN J
COMPUTE B = C + B
IF  B > 3
COMPUTE C = B - 3
MOVE 3 TO B
ELSE
MOVE 0 TO C
END-IF
string  P delimited size ") Pour from 5L jug into 3L jug"
delimited size into e1(p)
WHEN I
ADD C TO F
MOVE 0 TO C
string  P delimited size ") Empty 5L jug into tank"
delimited size into e1(p)
END-EVALUATE
string  "5L: " delimited size
C delimited size ", 3L: " delimited size B(2:)
", T: " delimited size F delimited size
into e1(p)(121:)
SET g1 TO TRUE
d2
perform d3 2 times
if  e1(p1)(1:1) = 0
move e1(p1)(2:) to e1(p1)(1:120)
end-if
subtract 1 from p1
d3
if  e1(p1)(138:1) = 0
move e1(p1)(139:) to e1(p1)(138:)
end-if


This gets an absolute shed-load of diagnostic messages for code starting in the wrong place and shortage of required full-stops.

None of the diagnostics indicate any impact on the object code. So, despite it being a busted RC=8 I know the object will be OK, so linked it and ran it.

Here are the outputs for one to eight litres. After that, all results can be intuited. 17 and 100 are included as examples of the parallelism.

There is still much that can be done to squeeze the program down in characters, the correct output was the important thing first. Counting the characters when they are on fixed-length lines is another thing entirely.

Sample output:

1) Fill 3L jug
5L: 0, 3L: 3, T: 0
2) Pour from 3L jug into 5L jug
5L: 3, 3L: 0, T: 0
3) Fill 3L jug
5L: 3, 3L: 3, T: 0
4) Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 0
5) Empty 3L jug into tank
5L: 5, 3L: 0, T: 1
Volume measured out in 5 turns

1) Fill 5L jug
5L: 5, 3L: 0, T: 0
2) Pour from 5L jug into 3L jug
5L: 2, 3L: 3, T: 0
3) Empty 5L jug into tank
5L: 0, 3L: 3, T: 2
Volume measured out in 3 turns

1) Fill 3L jug
5L: 0, 3L: 3, T: 0
2) Empty 3L jug into tank
5L: 0, 3L: 0, T: 3
Volume measured out in 2 turns

1) Fill 3L jug
5L: 0, 3L: 3, T: 0
2) Pour from 3L jug into 5L jug
5L: 3, 3L: 0, T: 0
3) Fill 3L jug
5L: 3, 3L: 3, T: 0
4) Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 0
5) Empty 3L jug into tank
5L: 5, 3L: 0, T: 1
6) Fill 3L jug
5L: 5, 3L: 3, T: 1
7) Empty 3L jug into tank
5L: 5, 3L: 0, T: 4
Volume measured out in 7 turns

1) Fill 5L jug
5L: 5, 3L: 0, T: 0
2) Empty 5L jug into tank
5L: 0, 3L: 0, T: 5
Volume measured out in 2 turns

1) Fill 3L jug
5L: 0, 3L: 3, T: 0
2) Empty 3L jug into tank
5L: 0, 3L: 0, T: 3
3) Fill 3L jug
5L: 0, 3L: 3, T: 3
4) Empty 3L jug into tank
5L: 0, 3L: 0, T: 6
Volume measured out in 4 turns

1) Fill 5L jug
5L: 5, 3L: 0, T: 0
2) Empty 5L jug into tank
5L: 0, 3L: 0, T: 5
3) Fill 5L jug
5L: 5, 3L: 0, T: 5
4) Pour from 5L jug into 3L jug
5L: 2, 3L: 3, T: 5
5) Empty 5L jug into tank
5L: 0, 3L: 3, T: 7
Volume measured out in 5 turns

1) Fill both jugs
5L: 5, 3L: 3, T: 0
2) Empty both jugs
5L: 0, 3L: 0, T: 8
Volume measured out in 2 turns

1) Fill both jugs
5L: 5, 3L: 3, T: 0
2) Empty both jugs
5L: 0, 3L: 0, T: 8
3) Fill both jugs
5L: 5, 3L: 3, T: 8
4) Empty both jugs
5L: 0, 3L: 0, T: 16
5) Fill 3L jug
5L: 0, 3L: 3, T: 16
6) Pour from 3L jug into 5L jug
5L: 3, 3L: 0, T: 16
7) Fill 3L jug
5L: 3, 3L: 3, T: 16
8) Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 16
9) Empty 3L jug into tank
5L: 5, 3L: 0, T: 17
Volume measured out in 9 turns

1) Fill both jugs
5L: 5, 3L: 3, T: 0
2) Empty both jugs
5L: 0, 3L: 0, T: 8
3) Fill both jugs
5L: 5, 3L: 3, T: 8
4) Empty both jugs
5L: 0, 3L: 0, T: 16
5) Fill both jugs
5L: 5, 3L: 3, T: 16
6) Empty both jugs
5L: 0, 3L: 0, T: 24
7) Fill both jugs
5L: 5, 3L: 3, T: 24
8) Empty both jugs
5L: 0, 3L: 0, T: 32
9) Fill both jugs
5L: 5, 3L: 3, T: 32
10) Empty both jugs
5L: 0, 3L: 0, T: 40
11) Fill both jugs
5L: 5, 3L: 3, T: 40
12) Empty both jugs
5L: 0, 3L: 0, T: 48
13) Fill both jugs
5L: 5, 3L: 3, T: 48
14) Empty both jugs
5L: 0, 3L: 0, T: 56
15) Fill both jugs
5L: 5, 3L: 3, T: 56
16) Empty both jugs
5L: 0, 3L: 0, T: 64
17) Fill both jugs
5L: 5, 3L: 3, T: 64
18) Empty both jugs
5L: 0, 3L: 0, T: 72
19) Fill both jugs
5L: 5, 3L: 3, T: 72
20) Empty both jugs
5L: 0, 3L: 0, T: 80
21) Fill both jugs
5L: 5, 3L: 3, T: 80
22) Empty both jugs
5L: 0, 3L: 0, T: 88
23) Fill both jugs
5L: 5, 3L: 3, T: 88
24) Empty both jugs
5L: 0, 3L: 0, T: 96
25) Fill 3L jug
5L: 0, 3L: 3, T: 96
26) Pour from 3L jug into 5L jug
5L: 3, 3L: 0, T: 96
27) Fill 3L jug
5L: 3, 3L: 3, T: 96
28) Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 96
29) Empty 3L jug into tank
5L: 5, 3L: 0, T: 97
30) Fill 3L jug
5L: 5, 3L: 3, T: 97
31) Empty 3L jug into tank
5L: 5, 3L: 0, T: 100
Volume measured out in 31 turns

• Admirable approach, @BillWoodger; I didn't however set a "fill both jugs" command within the available instructions... wonfderful work, and props for a) using COBOL, b) going with the quickest method route. – Eliseo D'Annunzio Feb 23 '14 at 10:20
• @WallyWest Thanks. If I think the business spec can be improved upon, I always do so :-). I didn't need the "empty into fountain" either, so two new and two not used - double failure! – Bill Woodger Feb 23 '14 at 21:26