Tuesday, February 2, 2010

TOM'S Golf Course

TOM's new golf course is a nine hole course. Yardages for the individual holes are as follows:

Hole 1 - 300 yds.
Hole 2 - 325 yds.
Hole 3 - 225 yds.
Hole 4 - 250 yds.
Hole 5 - 275 yds.
Hole 6 - 400 yds.
Hole 7 - 200 yds.
Hole 8 - 350 yds.
Hole 9 - 375 yds.

Karl, the course Pro, has received a new set of experimental clubs. Each club mysteriously hits the ball the same distance everytime (within a foot or two). The new clubs also never produce a slice or hook. The club distances are as follows:

1 Iron - 250 yds.
2 Iron - 225 yds.
3 Iron - 200 yds.
4 Iron - 175 yds.
5 Iron - 150 yds.
6 Iron - 125 yds.
7 Iron - 100 yds.
8 Iron - 75 yds.
9 Iron - 50 yds.
The new putter will sink any putt within a few feet of the hole.

Karl always uses a caddy to carry his bag. However, today, his ususal caddy, Knightmare, is asleep on his couch and did not arrive. Karl decides to play the course anyway and to avoid carrying the heavy bag, will play with just two irons and the putter.

If he choses the 8-iron, the 6-iron and the putter, he figures he can complete the round in 37 strokes (28 approach shots and 9 putts). If he overshoots a green, back shots are acceptable.

He really wishes Knightmare was here to help him decide which 2 clubs would give the best score.

If you were caddying for Karl, what 2 clubs (and the putter) would you recommend to beat a round of 37?

6 Comments:

Blogger Ross said...

What does "within a few feet of the hole" mean? I doubt it means as much as 25 yards...

You specified his clubs never slice or hook, but if he is allowed to AIM other than straight at the hole, he can cut out the putts altogether by doing trigonometry.

February 2, 2010 10:48 AM  
Blogger Zaux said...

Hi Ross ... the problem did not specify exactly how close to the pin a shot had to be for the putt to be automatic. I'm speculating that since the exact putt distance was not given, that a solution must be possible without that info.

February 2, 2010 10:57 AM  
Blogger Ross said...

All right, from recalculating, I find that the only way I can come up with 37 being the MINIMUM score for the 8-iron+6-iron is if I require every hole to add up exactly (e.g., 300 = 75+75+75+75, with zero left over) and then require a putt on every hole (i.e., you can't chip in with your 8-iron). I'll continue with this restriction then.

February 2, 2010 11:49 AM  
Blogger Ross said...

With that restriction, using a 6 iron and a 7 iron can result in a round of 36:

hole 300 = 100 + 100 + 100 + putt (4 strokes)
hole 325 = 125 + 100 + 100 + putt (4 strokes)
hole 225 = 125 + 100 + putt (3 strokes)
hole 250 = 125 + 125 + putt (3 strokes)
hole 275 = 125 + 125 + 125 - 100 + putt (5 strokes)
hole 400 = 125 + 125 + 125 + 125 - 100 + putt (6 strokes)
hole 200 = 100 + 100 + putt (3 strokes)
hole 350 = 125 + 125 + 100 + putt (4 strokes)
hole 375 = 125 + 125 + 125 + putt (4 strokes)
round = 36

(The negative terms indicate a stroke backwards from an overshoot)

I did an exhaustive search by writing a program. If the published solution is less than this, then I guess I still not have understood the conditions properly.

February 2, 2010 12:26 PM  
Blogger Zaux said...

hi Ross ...
that's good work ... but if you take another look at the problem, you can pick any 2 irons ... there is a better combo than 8-iron and
6-iron.

February 2, 2010 12:33 PM  
Blogger Zaux said...

Ross ... we had cross posts ... 36 is the best round ... good work, man

February 2, 2010 12:34 PM  

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