20210915, 02:55  #12 
Romulan Interpreter
"name field"
Jun 2011
Thailand
3^{4}×11^{2} Posts 
That's a nice puzzle, actually. Pity I missed it but the RL^{(TM)} kept me busy...

20210916, 21:23  #13 
Jan 2017
2^{3}×3×5 Posts 
I got 40 for the base problem and 144 for bonus.

20210917, 17:26  #14 
Jan 2017
2^{3}×3×5 Posts 
By the way the 175 limit for the bonus question seems very lax. Can it actually be said to be harder than the main question? Is there any natural way to solve it which would fail to also get 175 for the bonus one?

20210922, 10:03  #15 
"Kebbaj Reda"
May 2018
Casablanca, Morocco
1011110_{2} Posts 
The list of solvers is published. Congrats Dieter!
The remark : "Remember you can use hyphens in both strings". by "dg211" was decisive for me. Thanks "dg211"! That allowed me to leave the framework that I was fix it by obstinacy! I think in my thought I would find it hard to think about two strings at the same time: The "Villain and Heroes" lines. So I persisted to find a solution for the sake of simplicity on a single string "Heroes" , Until I couldn't take anymore less. Since we are blocked at 54. A less than someone found better. The lesson I learned is "sometimes you have to go out of the box to find a solution". Thank you, 
20211004, 10:03  #16 
Oct 2017
1111001_{2} Posts 
The published bonus solution agrees with my submitted solution (174).
I would like to know the teams for the better solutions (<150). 
20211004, 11:24  #17 
Jan 2017
2^{3}·3·5 Posts 
["0381", "0645", "0896", "0923", "5279"] gives 144 damage for the bonus question.
I found this by writing an exact solver for the best possible damage value for given teams, and then starting from random teams and changing one number at a time until the teams could no longer be improved. It seems that one number at a time is enough in the sense that trying simultaneous multipleplace changes just makes the search slower  it's more efficient to do singleplace changes and start from a new random position if the search gets stuck. My implementation searched for the overall best change and then did that, instead of picking the first one that showed any improvement at all. I'm not sure whether that would matter for how likely it is to converge to a good solution. The 175 limit seems very easy in the sense that almost all random starting points seem to lead to better solutions than that with singledigit changes (though not literally all). I think finding the 144 one is something around 1 in 1000 to 10000 random starting positions with onedigit changes. My implementation finds that within a couple of minutes (exact time of course varies since it's a random chance to find it at each attempt). I wrote the "best damage with given teams" part in C, and other search logic in Python. I left a search running for a longer time, and it seems that if there is a solution better than 144, it's vastly harder to find with this kind of search. There are no alternative 144 solutions either (modulo order of teams and placement of '0' values within a team). Best solutions found: 144: 1 solution ('0381', '0645', '0896', '0923', '5279') 145: 9 solutions ('0284', '0379', '0645', '0821', '0896') ('0379', '0384', '0645', '0821', '0896') ('0279', '0384', '0645', '0821', '4896') ('0284', '0415', '0816', '0896', '0938') ('0284', '0415', '0816', '0897', '0938') ('0381', '0645', '0896', '0923', '5269') ('0381', '0645', '0896', '0923', '6279') ('0381', '0645', '0897', '0923', '5279') ('0381', '0645', '0923', '5279', '8196') 146: 29 solutions ('0279', '0284', '0645', '0821', '0896') ('0384', '0479', '0645', '0821', '0896') ('0284', '0379', '0645', '0821', '0897') ('0264', '0347', '0528', '0938', '0964') ('0254', '0347', '0528', '0938', '0964') ('0379', '0384', '0645', '0821', '0897') ('0279', '0384', '0645', '0821', '0896') ('0283', '0435', '0938', '2896', '8645') ('0435', '0916', '0938', '4382', '8649') ('0279', '0384', '0654', '0821', '4896') ('0283', '0514', '0896', '0938', '8645') ('0379', '0384', '0645', '0821', '8196') ('0284', '0514', '0896', '0937', '8645') ('0284', '0514', '0896', '0938', '8645') ('0279', '0384', '0645', '0821', '5896') ('0379', '0384', '0645', '0821', '4896') ('0384', '0479', '0821', '0896', '6415') ('0284', '0415', '0816', '0897', '9318') ('0415', '0694', '0816', '0897', '2843') ('0379', '0384', '0645', '0821', '8197') ('0381', '0645', '0823', '0896', '5279') ('0381', '0645', '0897', '0923', '6279') ('0284', '0379', '0645', '0821', '8196') ('0381', '0645', '0897', '0923', '5269') ('0284', '0379', '0645', '0821', '8197') ('0284', '0415', '0816', '0896', '9318') ('0381', '0645', '0923', '5279', '8197') ('0381', '0645', '0923', '5269', '8196') ('0381', '0645', '0923', '6279', '8196') One thing to note is that most solutions have '0' in most teams. This helps make the teams effectively 3 in length (you can always pair the 0 with a skip). Shorter teams are more likely to match random patterns. 
20211004, 12:57  #18  
Jan 2017
1111000_{2} Posts 
Quote:
4197 5284 with 1+1+1+3=6 damage. But using team 4809 instead of 5284, you get: 4197 4809 with 0+1+1+0+2=4 damage. So you get 160 damage for those teams, and they can be further improved with onedigit changes: ["4809", "5284", "1065", "1382", "3469"]: 160 ["4809", "6284", "1065", "1382", "3469"]: 157 ["4809", "6284", "1065", "0382", "3469"]: 155 Improving further would require a twodigit change: ["4809", "6284", "1065", "0382", "0459"]: 154 Improving from that requires changing 4 digits: ["4709", "5280", "1065", "0382", "0759"]: 152 ["4709", "5280", "9061", "0382", "0154"]: 151 Further improvements would require 5 or more changes, I didn't check for those. 

20211004, 22:56  #19 
Sep 2017
76_{16} Posts 
It would be great if people who solved it, can share their code.

20211005, 11:43  #20 
Jan 2017
2^{3}·3·5 Posts 
Well here is a slightly cleaned up version of the code I used. By default it uses a C routine to calculate optimal damage for given teams. You can disable that, and the pure Python code is still easily enough to solve the problem and give something like a 150 solution for the bonus problem, but waiting for the 144 solution might take quite a while. With the C version properly compiled, you should find the 144 solution within a couple of minutes.
Code:
#!/usr/bin/python3 import random import sys # Calculate best possible damage value for given teams def calc(seq, teams): seq = seq + [0] states = [(i, j) for i in range(len(teams)) for j in range(1, len(teams[i]))] + [None] this = dict.fromkeys(states, 1e99) this[None] = 0 for s in seq: nxt = dict.fromkeys(states, 1e99) for state in states: cost = this[state] nxt[state] = min(nxt[state], cost + s) if state is None: for teami2, team in enumerate(teams): cost2 = cost for j, tval in enumerate(team): state2 = (teami2, j+1) if j + 1 < len(team) else None nxt[state2] = min(nxt[state2], cost2 + abs(s  tval)) cost2 += tval break teami, pos = state team = teams[teami] tval = team[pos] state2 = (teami, pos+1) if pos + 1 < len(team) else None nxt[state2] = min(nxt[state2], cost + abs(s  tval)) this[state2] = min(this[state2], cost + tval) this = nxt return this[None] orig_calc = calc if 1: # Import C version for much faster calc() implementation. # It only works for 5 teams of 4. import ctypes clib = ctypes.CDLL("./ibm_ponderthis_202109.so") from itertools import chain def calc(seq, teams): res = clib.calc(bytes(list(chain(*teams))), bytes(seq + [0]), len(seq)+1) # assert res == orig_calc(seq, teams) return res # Print the matching sequences required for puzzle answer. I didn't # implement the actual format required by the puzzle, this just splits # the sequence into parts matched by a single team, printing each matching # part together with the team (or '' for skip outside team). # # To avoid cluttering the calc() code with keeping track of how exactly # the best solution is formed, this code instead indirectly reconstructs # a possible best match. def out(seq, teams): print('start') res = calc(seq, teams) while seq: for i in range(1, len(seq)+1): res1 = calc(seq[:i], teams) res2 = calc(seq[i:], teams) if res1 + res2 == res: print(seq[:i]) if sum(seq[:i]) == res1: print('') else: for team in teams: if orig_calc(seq[:i], [team]) == res1: print(team) break else: assert 0 res = res2 seq = seq[i:] print('diff:', res1) break print('end') def search(seq, teams, offset, maxdepth, bestval, bestteams=None): for a in range(offset, 20): t, i = divmod(a, 4) orig = teams[t] for k in range(10): if k in orig: continue teams[t] = orig.copy() teams[t][i] = k r = calc(seq, teams) if r < bestval: print(r, teams) bestval = r bestteams = [x.copy() for x in teams] sys.stdout.flush() if maxdepth > 1: bestval, bestteams = search(seq, teams, a + 1, maxdepth  1, bestval, bestteams) teams[t] = orig return bestval, bestteams def canon(l): ll = ['0'*y.count(0)+''.join(str(x) for x in y if x > 0) for y in l] ll.sort() return tuple(ll) seq = [int(x) for x in "314159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196"] bestval = allbest = 1e99 teams = [random.sample(range(10), 4) for _ in range(5)] c = None while 1: bestval, bestteams = search(seq, teams, 0, 1, bestval) print('iter done') if bestteams is None: c = canon(teams) allbest = min(allbest, bestval) print('canonical form:', c, bestval, '/', allbest) if bestval <= 144: break bestteams = [random.sample(range(10), 4) for _ in range(5)] bestval = 1e99 teams = bestteams out(seq, teams) Here's the C code. You need to compile this into a shared library in the same directory for the Python code to load. On Linux "gcc O3 shared thiscode.c o ibm_ponderthis_202109.so" should work. Code:
#include <assert.h> #include <stdlib.h> #include <string.h> #define MIN(x, y) (((x) < (y)) ? (x) : (y)) int calc(unsigned char *teams, unsigned char *seq, int len) { assert(seq[len  1] == 0); unsigned short best[3*5 + 2]; memset(best, 2, sizeof(best)); best[0] = 0; for (int seqp = 0; seqp < len; seqp++) { int s = seq[seqp]; int besti = 2; for (int teami = 0; teami < 5; teami++) { int cost1 = best[besti]; best[besti] += s; int tval = teams[4*teami + 1]; besti++; int cost2 = MIN(best[besti], cost1 + tval); best[besti] = MIN(cost1 + abs(s  tval), cost2 + s); tval = teams[4*teami + 2]; besti++; int cost3 = MIN(best[besti], cost2 + tval); best[besti] = MIN(cost2 + abs(s  tval), cost3 + s); tval = teams[4*teami + 3]; besti++; best[0] = MIN(best[0], cost3 + tval); best[1] = MIN(best[1], cost3 + abs(s  tval)); } besti = 2; for (int teami = 0; teami < 5; teami++) { int cost = best[0]; for (int j = 0; j < 3; j++) { int tval = teams[teami*4 + j]; best[besti] = MIN(best[besti], cost + abs(s  tval)); besti++; cost += tval; } int tval = teams[teami*4 + 3]; best[1] = MIN(best[1], cost + abs(s  tval)); } best[0] = MIN(best[1], best[0] + s); best[1] = 1; } return best[0]; } 
20211006, 00:52  #21  
Sep 2017
2×59 Posts 
Quote:


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