Two quick quizzes for your Sunday pleasure:
1. Who bought the first HP 35 when it was first released in 1972?
2. Sometimes you will see a calculator display read:
123456789
as proof that the keybord is working. But there is more than one way to get the calculator to display this. What keystroke sequence (on the HP35 or HP35s) provides the FEWEST number of keys used. Obviously there are two easy answers:
 (1 key) Press Sigma+ 123 Million times, or
 (9 keys) Press the numbers in 19
is there another way??
Quote:
1. Who bought the first HP 35 when it was first released in 1972?
Hint: Check HP Key Notes, Volume 6 Number 1 (February 1982) Page 14 in the lefthand column for the answer.
Jake Schwartz
I'm not sure of the sequence that requires the fewest keystrokes, I've not thought up anything shorter than keying in the digits.
What might be more interesting would be a practical method (i.e. at most a couple of hundred keystrokes) that uses a minimum of *different* keys to get the number 123456789 on the display. I.e. what minimal functionality is required in order to get this display.
Off the top of my head, I can think of a simple algorithm that requires four different functions / five different keys (need that shift). To get the '1' digit do something like:
CLx
orange shift 10^x
ENTER
ENTER
+
ENTER
+
ENTER
+
+
orange shift 10^x
Repeat similar sequences for the other digits and add them together.
I think I could replace the shift 10^x sequence with a CLx, COS, +, ENTER, y^x series but this doesn't reduce the number of different keys which have to function and would increase the length significantly.
I'm sure somebody here can do better...
 Pauli
On the HP35, the best I was able to come up with was this 10key sequence:
10
ENTER
x^{y}
91

81
/
Simpling typing 123456789 is still better...
Gerson.
You mean y^{x} instead of x^{y} I hope :)
This sequence is one operation shorter but the same number of keystrokes:
10
10^{x}
91

81
/
 Pauli
Quote:
You mean y^{x} instead of x^{y} I hope :)
No, on the classic HP35, that's the way it is!
Unfortunately it also lacks 10^{x}. Anyway, your sequence is great on the newer HP35S.
Regards,
Gerson.
Oops, somewhere between between reading the question and attempting to solve it, I slipped from a 35 to a 35s :( My earlier algorithm is also wrong for a 35.
 Pauli
Quote:
I'm not sure of the sequence that requires the fewest keystrokes..
Not necessarily looking for the fewest keystrokes, but the fewest number of working keys. E.g.
81
+
91
+
would be 4 different keys: +,1,8 and 9.
huh?
That sequence doesn't seem like it would work properly...
Does it?
However, here is a sequence that only requires three working keys and 83 keystrokes:
1
ENTER + 1 +
ENTER + 1 +
ENTER +
ENTER + 1 +
ENTER +
ENTER + 1 +
ENTER + 1 +
ENTER +
ENTER + 1 +
ENTER + 1 +
ENTER + 1 +
ENTER + 1 +
ENTER +
ENTER +
ENTER + 1 +
ENTER + 1 +
ENTER +
ENTER + 1 +
ENTER +
ENTER +
ENTER +
ENTER + 1 +
ENTER +
ENTER + 1 +
ENTER +
ENTER + 1 +
You can get rid of the single numeric (1) keystrokes and use CLx, COS and extra ENTERs appropriately.
To understand what I did, take the binary expansion of 123456789 and read the bits left to right.
 Pauli
How about:
987654321
9

8
/
I started with a number that is the reverse of 123456789.
Also,
1111111111
10

9
/
Namir
Edited: 7 Oct 2007, 9:05 p.m.
11111.11111
ENTER
*
1.1

17 keystrokes but five keys. You're the winner!
Gerson.
Another almost solution at 7 commands:
5
10^{x}
9
/
x^{2}
1

To avoid the "10^{x}", use: "EEX 5 ENTER" instead of the first two lines. Likewise, replace the "x^{2}" with "ENTER *"
Eight keystrokes on my 15c, no idea on the 35. Looks okay in FIX 0 mode.
 Pauli
Change the final '1' to '1.1' to remove the dependency on the display setting at the cost of two additional keystrokes. This puts us more than typing the number in directly :(
 Pauli
What about this slight modification:
9
1/x
EEX
5
*
ENTER
*
1.1

11 key strokes, 9 different keys.
 Pauli
Edited: 7 Oct 2007, 9:35 p.m.
I just key in .0081 1/x Sam
You get:
123.45679
Notice that there is an 8 missing ...... so it's back to the drawing board my friend!!
Namir
PS: I saw this same mistake on another web site. Hmmmmmmmm .. did you copy your answer from that site?
Edited: 8 Oct 2007, 11:57 a.m.
Quote:
What keystroke sequence (on the HP35 or HP35s) provides the FEWEST number of keys used. Obviously there are two easy answers:
1. (1 key) Press Sigma+ 123 Million times, or
2. (9 keys) Press the numbers in 19
is there another way??
3 keys, but lots of keystrokes:
9
ENTER
ENTER
ENTER
+ (13717421 times)
This has the added "advantage" of allowing one to truthfully declare: "Tested basic math operation and numeric keyboard."
How about:
3 ENTER x 3803 x 3607 x
Thirteen keystrokes, seven different keys, only five working digit keys.
Stefan
Or the slightly shorter:
9 ENTER 3803 x 3607 x
Twelve keystrokes, but eight different keys of which six are digit keys.
Enter the number as the hexadecimal:
75BCD15h
And the convert it to the decimal. Still not less than 9 keys.
And another method. Store 1E9 in register A and then repeat:
RCL A
ENTER
<RS> RAND
*
+
You are generating random numbers between 1E9 and 2E9. A lucky random number will give you 123456789. The process uses 7 keys.
These keystrokes test how touch the keyboard is!!
I don't think the Classic HP35 has a RAND key.
Finally got one that is less than nine keystrokes on a 15c. Not applicable to a 35 though.
A total of 7 commands, 8 keystrokes and 7 working keys required:
9
ENTER
10^{x}
delta%
9
0
/
There are plenty of other ways to produce the 90 if we're in degrees mode (e.g. 0 COS^{1}; 1 SIN^{1} or ENTER TAN^{1}).
 Pauli
We are having fun with this problem. Not adhering 100% to the rules is part of the fun.
If we want to get serious, this challenge is downright meaningless!! But like I said, we are having fun.
Namir
Edited: 8 Oct 2007, 4:50 p.m.
You solution also works with the HP41C/CV/CX. Good show man!!
Namir
Wow! Excellent submissions!!
 Jake definitely got the award for the first part. He found the source of the trivia. HP Keynotes reports that a Professor R.J. Donnelly of the University of Oregon purchased the first production HP35, which as of 1982 is not in use, rather kept in a Bank box for safe keeping.
 The part 2 award goes to Pauli for both the number of solutions and for the solution with the fewest KEYS needed to create the display:
123456789
Namir's submission was the closest to my original solution requiring 4 keys, but Pauli managed to find a solution with only 3 keys!!! EXCELLENT!
My original 4key solution could be done a few different ways, but the general formula, starting from a cleared stack (this can be done without pressing any buttons) is:
1111111111


1

1

1

1

1

1

1

1

1

1

9
/
Four keys: 1,9,[] and [/]. This could also be done with (444444444440)/6/6 using 4 keys:4,6,[],and [/].
And of course if we're after a display of 1234567890 we can use this sequence which is one keystroke shorter but again not good for a 35 (6 commands, 7 keystrokes, 6 working keys on the 15c):
9
ENTER
10^{x}
delta%
9
/
If we're happy with 12345.6789 displayed FIX 4 there is this one again for the 15c (6 commands, 10 keystrokes, 9 working keys):
9
ENTER
TANH
CHS
delta%
x^{2}
 Pauli