For a text adventure I'm writing, I need to establish an anagram X of a word Y such that letter Z of X is not letter Z of Y.

In other words, ABCDEE could go to EEDCBA or DCEEBA but not ABCEED, because ABCEED matches ABCDEE at slot 5.

# amak.py: this makes an anagram of a word with no identical letter slots.
# in other words, HEAT and HATE have the first letter identical, but EATH has no letter slots in common with HEAT.

import re
import sys
from collections import defaultdict

#option(s). There may be more later.
shift_1_on_no_repeat = False
try_rotating_first = False

# determine if we can still switch a pair. With 3 letters left, it is not possible. With 2, it should be.

def can_take_even(x):
    if x % 2 == 0: return x > 0
    else: return x > 3

# here is the explanation of the algorithm:
# 1. unless we have exactly 3 letters to place, we look for the 2 most frequent letters that have not been switched yet nd switch the earliest incidences of each
# 2. if there are 3 unique letters remaining, then we go a->b->c.
#   2a. Note that we can never have 2-1 left, because the previous would have to have 3-?-?. If we started with, say, 2-2-1, we would have 1-1-1 after. Similarly we can never have x-(summing less to x) unless we start with something unviable, because we'd have to have had x+1 and (something less than x+1) on the previous try. If we had x on the previous try, we would have deducted from it.
#   note having y>x/2 in x letters means we cannot have a unique anagram. That is because we would have x-y slots to move the y to, but x<2y so that doesn't work.

def find_nomatch_anagram(x):
    x = re.sub("[- '\.]", "", x.lower()) # allow for spaces, apostrophes, etc.
    old_string = list(x)
    new_string = ['-'] * len(x)
    f = defaultdict(list)
    letters_to_place = len(old_string)
    if not len(x):
        print("Blank string...")
        return ""
    for y in range(0, len(x)):
        if old_string[y] not in 'abcdefghijklmnopqrstuvwxyz':
            print("Nonalphabetical character in", x, 'slot', y, "--", old_string[y])
            return ""
    if shift_1_on_no_repeat and len(f) == len(old_string): return x[1:] + x[0] #abcde quickly sent to bcdea
    if try_rotating_first:
        for y in range(1, len(x)):
            retval = x[-y:] + x[:-y]
            print("Trying", retval)
            bad_matches = False
            for z in range(0, len(x)):
                bad_matches |= (retval[z] == old_string[z])
            if not bad_matches: return retval
    for q in f:
        if len(f[q]) > len(old_string) / 2:
            print(q, "appears too many times in", x, "to create an anagram with no letter slots in common.")
            return ""
    while can_take_even(letters_to_place):
        u = sorted(f, key=lambda x:len(f[x]), reverse=True)
        x1 = f[u[0]].pop(0)
        x2 = f[u[1]].pop(0)
        new_string[x1] = u[1]
        new_string[x2] = u[0]
        letters_to_place -= 2
    if letters_to_place == 3:
        u = sorted(f, key=lambda x:len(f[x]), reverse=True)
        new_string[f[u[0]][0]] = u[1]
        new_string[f[u[1]][0]] = u[2]
        new_string[f[u[2]][0]] = u[0]
    for y in range(0, len(x)):
        if old_string[y] == new_string[y]:
            print("Uh oh, failure at letter", y)
        if new_string[y] == '-':
            print("Uh oh, blank letter at", y)
    return ''.join(new_string)

def show_results(q, result_string = "has this anagram with no letters in common:"):
    temp = find_nomatch_anagram(q)
    if not temp: return
    print(q, result_string, temp)

if len(sys.argv) > 1:
    for q in sys.argv[1:]:
        if q == 's1': shift_1_on_no_repeat = True #this works for one option, but what if there are several?
        elif q == 'tr': try_rotating_first = True #this works for one option, but what if there are several?
    for q in sys.argv[1:]:
        if q != 's1' and q != 'tr': show_results(q, "<=>") # this feels like a real hack, again. I want to process meta commands before any results, though.
else: #these are just general test cases
    show_results("aabbb") #throw error
    show_results("try this")

What I have works. But I am wondering about a few things:

  1. is there any way I can write the command line better? I am taking two passes through it right now, but this seems inefficient. I want to be able to give the user the option of trying the obvious anagrams (shift everything 1/2/3/etc. letters over until you find one)
  2. While my algorithm seems to work provably, the code for it seems awkward. I plan (n/2) swaps where I match the 2 top remaining frequencies for unswapped letters, then take them, until I am at 3 or 0. Then I do a 3-way rotation for the final letters.

1 Answer 1


Things that stand out as particularly good:

  • Test cases, including tests that unsolvable inputs get handled correctly.
  • Comments, including comments that describe the high level algorithm.
  • The algorithm description makes it clear how it's avoiding painting itself into a corner.
  • Testing, as early as possible, for inputs that break your function rather than discovering part way through is fantastic.

Possible room for improvement:

  • Test cases should specify their expected output. Ideally, you'd have a testing rig that automatically checks they give a satisfying output.
  • Think about how you handle failures. In python, I'd suggest that exceptions are the way to go.
  • Where you have multiple approaches to solving the same problem, that is a particularly good sign that it's time to split them out of a mega-function.
  • Many of your variable names could do with elaboration, especially the single letter ones.
  • Run the code through a Pep-8 style linter and ensure it's laid out in standard pythonic form. For example single line condition and response is discouraged.
  • This algorithm produces an output that is determinisitic and depending on its place in your game may be a little bit boring and predictable. Having the rotate mode mixes it up a bit, but that actually makes for an even more obvious pattern. It may be better to see whether you can introduce some random element. The easist option would be to do a random shuffle, and then swap out any letters that violate the rule.
  • \$\begingroup\$ Thanks! I do tend to be lazy with variable names, and I am not good about using PEP8 editor. I like the randomization suggestion. I should maybe have mentioned that the first use would be for a test script, to make sure basic things work. But still, it would be nice to have a stress test of possible anagrams. I also agree the function can be broken up. I really appreciate the points you made and the quick response. I take 24 hours before accepting an answer but anything that beats yours will have to be quite good indeed. \$\endgroup\$
    – aschultz
    Commented Jun 29, 2019 at 12:38
  • 2
    \$\begingroup\$ Oh, one other thing. Maybe have a look at argparse for handling command line input (docs.python.org/3/howto/argparse.html) \$\endgroup\$
    – Josiah
    Commented Jun 29, 2019 at 12:44

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