A drawer does not have to be a single color. It can be a whole column pattern, a repeated row-pair signature, a line-sum value, or a checkerboard color count.
A signature records exactly the feature that must repeat. For two equal columns, the signature is the complete vertical color pattern. For a one-color rectangle, the signature is a pair of rows together with their repeated color.
Read column codes from top to bottom; column numbers run left to right. Click a cell, or focus it and press Enter or Space, to switch red/yellow. The complete code—not one cell—is the drawer label.
A whole column is one object, read in a fixed order.
Each of three positions has two choices: 2×2×2=8 patterns. RRY and RYR are different drawers because the row positions differ.
Every column has three colored cells. Red or yellow in each position gives only eight possible vertical patterns, but the board has ten columns.
Count possible signatures before counting columns.
Ten column objects enter eight possible-pattern drawers. At least two columns have exactly the same three entries.
Nine columns force two identical complete patterns. This is the worked example guided practice.
Five columns force two identical complete patterns. This is Exercise 4.
After listing every possible pattern once, the next must repeat.
Three binary positions give 8 patterns and a duplicate at 9 columns. Two positions give 4 and a duplicate at 5; four positions give 16 and a duplicate at 17.
With four cells and three colors, how many columns are sufficient to guarantee three identical patterns?
For three copies, allow two of every pattern before adding one.
Four positions with three colors give 3⁴=81 patterns. Two of each make 162 without a triple. Column 163 forces three identical patterns.
The theorem guarantees a match, but a solver must still read the grid accurately. Click cells to change the coloring, or generate a new board.
Compare all three rows in your selected columns.
Choose distinct columns with the same complete signature, then check both theory answers. A matching top cell alone is insufficient. After checking your attempt, the separate worked review identifies a matching pair in the current board without changing your selection.
Added tightness check: the six nonconstant column patterns can avoid a one-color rectangle; adding any seventh binary column forces one.
Within each column, two of its three cells share a color.
There are three possible row pairs and two colors: six signatures. Assign one such signature per column. Seven columns force a repeat, giving four same-color rectangle corners. The six-column escape example shows why six need not force it.
Fill a 5×5 grid with 1, 2, or 3. There are five row sums, five column sums, and the two corner-to-corner diagonal sums: twelve line-sum objects.
Click a cell to cycle 1 → 2 → 3. Try to make all twelve line sums different.
Distinct sums now: 0 of 12
Count the lines and the possible sums separately.
Five rows, five columns, and two diagonals give 12 line sums. Each sum is an integer from 5 through 15: eleven possibilities. Two sums must match.
Each domino covers one square of each color.
The removed opposite corners have the same color. The remaining counts are 32 and 30, a difference of 2. Dominoes cover equal color counts, so they cannot cover this board.
A column has four positions, each colored red or yellow. How many columns guarantee at least three identical complete patterns?
This fresh problem has its own checkpoint. Your written explanation is saved for comparison and is not automatically graded.
There are 2⁴=16 signatures. Two of each allow 32 columns without a triple. The 33rd forces three of one pattern.
Revisit the mission connected to each question.
Answers:8;10>8;4;81;6;7;12;the remaining color counts are unequal. Whole-column signatures use every row in order. A rectangle signature needs only a row pair and their matching color.
Correct all five answers and complete all ten missions to earn the certificate.
This certifies that
can use whole patterns, signatures, line sums, and coloring invariants as pigeonhole drawers.
Solve each item without using the answer shown in an earlier example.
Answers:32;33;27;11;2. Five binary positions give2⁵=32patterns, so33columns force a duplicate. Three positions with3colors give27patterns. Line sums5..15give11values. Removing two same-color corners leaves counts32and30.