This week the #puzzle is: Can You Hack #Bowling? #minimum #maximum #permutations

If you knock down 100 pins over the course of a game [of bowling], you are guaranteed to have a score that’s at least 100. But what’s the minimum total number of pins you need to knock down such that you can attain a score of at least 100?

And for extra credit:

You and your opponent each play a single game in which you both knock down the same number of pins. However, your scores are quite different.

Your opponent remarks, “Given only the information that we knocked down the same number of pins in our two games, there’s no way the difference between our scores could have been any greater!”

What is this difference between your two scores?

Can You Hack Bowling?

Highlight to reveal (possibly incorrect) solution:

Some knowledge and some assumptions:

• A high score with few pins is obtained by getting a lot of strikes, thereby tripling the score value of a lot of the pins.
• The strikes have to come in series, with high scores from the very 1st frame.
• Therefore this is a valid way to get 100 points with few pins:

10	10	10	10	4/0
10+10+10=30	30+10+10+10=60	60+10+10+4=84	84+10+4+0=98	98+4+0=102

0/0	0/0	0/0	0/0	0/0
102+0+0=102	etc.

• If I only score 3 in the 5th frame, I won’t reach 100.
• So I have found the best possible score. 102 points with 44 pins.

And for extra credit:

Program

Again, some knowledge and some assumptions.

• First we look at the low scores.
• Given n pins, I can get a score of n points, and not lower. E.g. like this, n = 34:

0/10	0/10	0/10	0/4	0/0
0+10+0=10	10+0+10+0=20	20+0+10+0=30	30+0+4=34	34+0+0=34

0/0	0/0	0/0	0/0	0/0
etc.

• For n > 100, there’s a twist at the end (1 extra bowl), e.g. n = 109:

0/10	0/10	0/10	0/10	0/10
0+10+0=10	10+0+10+0=20	20+0+10+0=30	30+0+10+0=40	40+0+10+0=50

0/10	0/10	0/10	0/10	0/10	9
50+0+10+0=60	60+0+10+0=70	70+0+10+0=80	80+0+10+0=90	90+0+10+9=109

• And for n > 110, there’s are 2 small twists, at the end (2 extra bowls) and not quite at the end (9th frame, could be earlier though). n = 111:

0/10	0/10	0/10	0/10	0/10	0/10
0+10+0=10	10+0+10+0=20	20+0+10+0=30	30+0+10+0=40	40+0+10+0=50	50+0+10+0=60

0/10	0/10	0/9	10	10	2
60+0+10+0=70	70+0+10+0=80	80+0+9=89	89+10+10+2=111

• Going all the way to n = 119:

0/10	0/10	0/10	0/10	0/10	0/10
0+10+0=10	10+0+10+0=20	20+0+10+0=30	30+0+10+0=40	40+0+10+0=50	50+0+10+0=60

0/10	0/10	0/9	10	10	10
60+0+10+0=70	70+0+10+0=80	80+0+9=89	89+10+10+10=119

• This won’t work for n = 120, where the score is higher than 120.

0/10	0/10	0/10	0/10	0/10	0/10
0+10+0=10	10+0+10+0=20	20+0+10+0=30	30+0+10+0=40	40+0+10+0=50	50+0+10+0=60

0/10	0/10	0/10	10	10	10
60+0+10+0=70	70+0+10+0=80	80+0+10+10=100	100+10+10+10=130

• Let’s look at the high scores.
• Using the method at the top, I can get a score of 3n – 30 = 3(n – 10) for a lot of n, like n = 87. (The first 2 frames aren’t just tripled.)

10	10	10	10	10
10+10+10=30	30+10+10+10=60	60+10+10+10=90	… 120	… 150

10	10	10	7/0	0/0
… 180	180+10+10+7=207	207+10+7+0=224	224+7+0=231	231+0+0=231

• Also n > 100, a little lower than 3n – 30. If n = 100 + a + b, where a < 10 => b = 0. Now the score is 270 + 2a + b. Like n = 103:

10	10	10	10	10
10+10+10=30	30+10+10+10=60	60+10+10+10=90	… 120	… 150

10	10	10	10	10	3/0
… 180	180+10+10+10=210	210+10+10+10=240	240+10+10+3=263	263+10++3+0=276

• Or n = 111:

10	10	10	10	10	10
10+10+10=30	30+10+10+10=60	60+10+10+10=90	… 120	… 150	… 180

10	10	10	10	10	1
180+10+10+10=210	210+10+10+10=240	240+10+10+10=270	270+10++10+1=291

• Clearly we’re looking for a high number of pins, to get a big difference between the numbers.

Pins	Max	Min	Diff
100	270	100	170
101	272	101	171
102	274	102	172
…
110	290	110	180
111	291	111	180
…
119	299	119	180
120	300	130	170

• Given this information, there can be no higher difference between scores than 180.
• Given all this information I can actually go back to the fiddler and see if my assumptions hold up. In general the scores are like this, with f being the number of frames in a game (not counting extra frames because of spares/strikes at the end):

Pins	Max	Min	Diff
10f-m	30(f-1)-3m	10f-m	20f-30-2m
…
10f-1	30(f-1)-1*3	10f-1	20f-30-2*1
10f	30(f-1)	10f	20f-30
10f+1	30(f-1)+2*1	10f+1	20f-30+1
10f+2	30(f-1)+2*2	10f+2	20f-30+2
…
10f+10	30(f-1)+10*2	10f+10	20f-30+10
10f+10+1	30(f-1)+10*2+1	10f+10+1	20f-30+10
…
10f+10+9	30(f-1)+10*2+9	10f+10+9	20f-30+10
10f+10+10	30(f-1)+10*2+10	10f+10+10	20f-30

• If m = 56 in the 1st row, we get 30*9 – 3*56 = 102 as max. This is with 10*10-56 = 44 pins. Fiddler confirmed. (If m = 57, the max score is 99, and the higher m is, the lower the score is.)
• Note that I haven’t really talked about the lowest scores. Like, with 9 pins in all, you get a max and min score of 9.
• Now that my predictions only depend on f, I can make a toy example, with f being 2, 3 or 4, to see if my assumptions hold up. For this I make a program. Everything is confirmed.